INTRODUCTORYSTATISTICS
4th Edition

By OpenStax College
Adapted by Riyanti Boyd & Natalia Casper
Special Acknowledgement to Mark Beintema & Gladys PomaThis page is purposely left blank.

Introductory Statistics
Susan Dean and Barbara Illowsky (Published 2013 by OpenStax College)
Adapted by: College of Lake County Faculty: Riyanti Boyd and Natalia Casper
Revised July 2021
This project was funded by a grant from the College of Lake County Foundation.
Adapted College of Lake County text can be found at:
http://dept.clcillinois.edu/mth/oer/IntroductoryStatistics.pdf
Original text materials for Introductory Statistics by Dean and Illowski 2013
available at: http://cnx.org/content/col11562/latest/.
Original Publication Under the following license:
This work is licensed under theCreative Commons Attribution-Noncommercial-Share Alike 3.0 unported license.
To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/
or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco,California, 94105, USA.

i

Table of Contents (4th Edition)

Chapter 1
Section 1.1
Section 1.2
Section 1.3
Chapter 2
Section 2.1
Section 2.2
Section 2.3
Chapter 3
Section 3.1
Section 3.2
Section 3.3
Chapter 4
Section 4.1
Section 4.2
Section 4.3
Section 4.4
Section 4.5
Chapter 5
Section 5.1
Section 5.2
Section 5.3
Section 5.4
Section 5.5
Chapter 6
Section 6.1
Section 6.2
Section 6.3

Sampling and Data
Basic Definitions
Data and SamplingExperimental Design and Ethics
Exercises
Solutions
Displaying Data
Frequency Distributions
Graphs for Qualitative Data
Graphs for Quantitative Data
Exercises
Solutions
Numerical Descriptors
Measures of Central Tendency
Measures of Variation
Measures of Relative Position
Exercises
Solutions
Probability Topics
Terminology
Independent and Mutually Exclusive Events
Two Basic Rules of Probability
Contingency Tables
Counting and Probability
Exercises
Solutions
Discrete Random Variables
Probability Distribution Function (PDF) for a Discrete
Random Variable
Mean, Expected Value, and Standard Deviation
Binomial Distribution
Geometric Distribution
Poisson Distribution
Exercises
SolutionsContinuous Random Variables
Continuous Probability Distributions
Uniform Distribution
Exponential Distribution
Exercises
Solutions

ii

1
2
7
19
29
38
41
43
49
55
71
78
83
84
94
102
119
134
137
138
145
153
164
169
180
202
205
206
209
216
226
231
241
260
265
266
270
281
294
307

Table of Contents (4th Edition)

Chapter 7
Normal Distribution
Section 7.1 Standard Normal Distribution
Section 7.2 Using Normal Distribution
Exercises
Solutions
Chapter 8
The Central Limit Theorem
Section 8.1 The Central Limit Theorem for Sample Means and Sample
Proportions
Section 8.2 Using the Central Limit Theorem
Section 8.3 The Central Limit Theorem for Sums
Exercises
Solutions
Chapter 9
Confidence Intervals
Section 9.1 Introduction
Section 9.2 Confidence Interval for Single Population Mean when σ
known
Section 9.3 Confidence Interval for Single Population Mean when σunknown
Section 9.4 Confidence Interval for Single Population Proportion
Section 9.5 Confidence Interval for Standard Deviation, σ (Optional)
Exercises
Solutions
Chapter 10
Hypothesis Testing with One Sample
Section 10.1 Elements of a Hypothesis Test
Section 10.2 The Z-test for a Single Population Mean using p-value
Section 10.3 The T-test for a Single Population Mean using p-value
Section 10.4 Hypothesis Test for a Single Population Proportion using
p-value
Section 10.5 Critical Values and the Critical Value Method
Section 10.6 Type I and Type II Errors
Exercises
Solutions
Chapter 11
Hypothesis Testing with Two Samples
Section 11.1 Two Population Means with Known σ1 and σ2
Section 11.2 Two Population Means when σ1 and σ2 are Unknown
Section 11.3 Comparing Two Population Proportions
Section 11.4 Matched Pairs Test
Section 11.5 Confidence Intervals from Two Samples
Exercises
Solutions

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313
315
322
341
352
355
358
368
371
378
389
391
392
396
406
413
420
429
449
453
456
463
467
471
476
484
491
508
513
515
521
531
539
546
555
571

Table of Contents (4th Edition)

Chapter 12
Section 12.1
Section 12.2
Section 12.3
Section 12.4
Section 12.5
Chapter 13
Section 13.1
Section 13.2
Section 13.3
Section 13.4
Section 13.5
Section 13.6
Section 13.7
Chapter 14
Section 14.1
Section 14.2
Appendix

The Chi-Square Distribution Continued
Introduction to Chi-square Distribution
Confidence Intervals for Variance/Standard Deviation
Test of a Single Variance
Goodness of Fit Test
Test of Independence and Test for Homogeneity
Exercises
SolutionsLinear Regression and Correlation
Linear Equations
Scatter Plots
The Regression Equation
Testing the Significance of the Correlation Coefficient
Prediction
Outliers
Summary of Concepts using TI-83/84 and Excel
Exercises
Solutions
F Distribution and One-Way ANOVA
Test of Two Variances/Standard Deviations
One-way ANOVA
Exercises
Solutions

575
576
580
583
587
596
607
625
629
630
633
638
649
658
660
666
676
691
695
696
703
712
724

Standard Normal Table
T-distribution Tables
Chi-Square Tables
r critical value Table

A1
B1
C1
D1

This Photo by Unknown Author is licensed under CC BY-

iv

1 | SAMPLING AND DATA

Figure 1.1 Photo by Markus Spiske on Unsplash

Introduction
Chapter Objectives: By the end of this chapter, the student should be able to• Understand basic statistical terminology• Recognize and differentiate between key terms.
• Apply various types of sampling methods to data collection.
You are probably asking yourself the question, "When and where will I use statistics?" If you
read any newspaper, watch television, or use the Internet, you will see statisticalinformation. There are statistics about crime, sports, education, politics, and real estate.
Typically, when you read a newspaper article or watch a television news program, you are
given sample information. With this information, you may make a decision about thecorrectness of a statement, claim, or "fact." Statistical methods can help you make "besteducated guess."
Since you will undoubtedly be given statistical information at some point in your life, youneed to know some techniques for analyzing the information thoughtfully. Think aboutbuying a house or managing a budget. Think about your chosen profession. Think about theCovid-19 pandemic. Think about the political elections. As well, the fields of economics,
business, psychology, education, biology, law, computer science, police science, and earlychildhood development require at least one course in statistics.
Included in this chapter are some basic ideas and terms used in Statistics. We will learn how
data are gathered and how to distinguish "good" data from "bad."1

1.1 | Basic Definitions

The science of Statistics deals with the collection, analysis, interpretation, and presentationof data. We see and use data in our everyday lives. There are two basic branches of
Statistics: Descriptive and Inferential.
••

Descriptive Statistics involves organizing and summarizing data. Two generalways to summarize data are with graphs and numerical values, such as an average.
Inferential Statistics. Involved learning formal methods for drawing conclusions
from data, which we will learn later in the text. Statistical inference uses probability
to determine how confident we can be that our conclusions are correct.

Effective interpretation of data (inference) is based on good procedures for producing data
and thoughtful examination of the data. We will encounter what may seem to be too manymathematical formulas for interpreting data. The goal of statistics is not to perform
numerous calculations using the formulas, but to gain an understanding of your data. The
calculations can be done using a calculator or a computer; the understanding must come fromyou. If you can thoroughly grasp the basics of statistics, you can be more confident in thedecisions you make in life and in your chosen career.

Key Terms
In Statistics, we generally want to study a population. You can think of a population as acollection of persons, things, or objects under study. For logistical reasons, it is usually notpossible to gain access to all of the information from the entire population. So when want
to study a population, we usually select a sample. The idea of sampling is to select a
portion (or subset) of the larger population and study that portion (the sample) to gain
information about the population. Data are the result of sampling from a population.

https://openclipart.org/detail/216524/many-and-few

Because it takes a lot of time and resources to examine an entire population, sampling is avery practical technique. If you wished to compute the overall grade point average at yourschool, it would make sense to select a sample of students who attend the school. The datacollected from the sample would be the students' grade point averages. In presidential2elections, samples of between 1,000 and 2,000 prospective voters are used for opinion polls.
The opinion poll is supposed to represent the views of the people in the entire country.
Manufacturers of canned carbonated drinks take samples to determine if a 16 ounce
can contains 16 ounces of carbonated drink.
From the sample data, we can calculate a statistic. A statistic is a number that represents
a property of the sample. For example, if we consider one math class to be a sample of the
population of all math classes, then the average number of points earned by students in that
one math class at the end of the term is an example of a statistic. The statistic is an estimateof a population parameter. A parameter is a number that represents a property of the
population. Since we considered all math classes to be the population, then the average
number of points earned per student over all the math classes is an example of a parameter.

One of the main concerns in the field of statistics is how accurately a statistic estimates aparameter. The accuracy really depends on how well the sample represents the
population. The sample must contain the characteristics of the population in order to bea representative sample. We are interested in both the sample statistic and the population
parameter in inferential statistics. In a later chapter, we will use the sample statistic to test
the validity of the established population parameter.
A variable, denoted by capital letters such as X and Y, is a characteristic of interest for each
person or thing in a population. Variables may be numerical or categorical.

3

•
•

Numerical variables take on numerical values with equal units such as weight inpounds and time in hours.
Categorical variables place the person or thing into a category.

If we let X equal the number of points earned by one math student at the end of a term,then X is a numerical variable. If we let Y be a person's party affiliation, then Y is a
categorical variable, and some possible values of Y would be Republican, Democrat, andIndependent. Y is a categorical variable. We could do some math with values of X (calculate
the average number of points earned, for example), but it makes no sense to do math withvalues of Y (calculating an average party affiliation makes no sense).
The actual values of a variable are called data (a single value is a datum); these values
may be numbers or words.
Example 1.1
Determine what the key terms refer to in the following study: We want to know theaverage amount of money first-year college students spend at ABC College on school supplies
that do not include books. We randomly survey 100 first-year students at the college. Threeof those students spent $150, $200, and $225, respectively.
Solution 1.1:
The population is all first year students attending ABC College this term.
The sample is the 100 first year students surveyed at the college (although this sample
may not represent the entire population).
The parameter is the average amount of money spent (excluding books) by first year
college students at ABC College this term. This average would be represented by µ.
The statistic is the average amount of money spent (excluding books) by first year college
students in the sample. This average would be represented by 𝑥𝑥̅ .
The variable is the amount of money spent (excluding books) by one first year student.
Let X = the amount of money spent (excluding books) by one first year student attending
ABC College.

The data would be the actual dollar amounts spent by the first-year students. Examples ofthe data would be $150, $200, and $225.

1.1 Determine what the key terms refer to in the following study. We want to know theaverage amount of money spent on school uniforms each year by families at Knoll Academy.
We randomly survey 100 families with children in the school. Three of the families spent$65, $75, and $95, respectively.

4

Example 1.2
Determine what the key terms refer to in the following study.
A study was conducted at a local college to analyze the average cumulative GPA’s ofstudents who graduated last year. Fill in the letter of the phrase that best describes each of
the items below.
1. ___ Population 2. ___ Statistic 3.___ Parameter
4. ___ Sample 5.
Variable 6.
Dataa) the cumulative GPA of a student who graduated from the college last yearb) the average cumulative GPA of students surveyed who graduated from the college lastyearc) 3.65, 2.80, 1.50, 3.90d) a group of students who graduated from the college last year, randomly selectede) all students who graduated from the college last yearf) the average cumulative GPA of all students in the study who graduated from thecollege last year
Solution 1.2: 1. e; 2. b; 3. f; 4. d; 5. a; 6. cExample 1.3
Determine what the key terms refer to in the following study.
As part of a study designed to test the safety of automobiles, the National Transportation
Safety Board collected and reviewed data about the effects of an automobile crash on test
dummies. Cars with dummies in the front seats were crashed into a wall at a speed of 35
miles per hour. We want to know the proportion of dummies in the driver’s seat that would
have had head injuries if they had been actual drivers. We start with a simple random
sample of 75 cars.
Solution 1.3
The population consists of all cars containing dummies in the front seat.
The sample is the set of 75 randomly selected cars.
The parameter is the proportion of driver dummies (if they had been real people) who
would have suffered head injuries in the population.
The statistic is proportion of driver dummies (if they had been real people) who would
have suffered head injuries measured in the sample.
The variable X = whether or not the driver dummies (if they had been real people) havesuffered head injuries.
The possible data values would be either: yes, had head injury, or no, did not.
5

e of a Discrete Random Variable
Example 1.4
Determine what the key terms refer to in the following study.
An insurance company would like to determine the proportion of all medical doctors who
have been involved in one or more malpractice lawsuits. The company selects 500 doctors at
random from a professional directory and determines the number in the sample who havebeen involved in a malpractice lawsuit.
Solution to 1.4:
The population is all medical doctors listed in the professional directory.
The parameter is the proportion of medical doctors who have been involved in one or moremalpractice suits in the population.
The sample is the 500 doctors selected at random from the professional directory.
The statistic is the proportion of medical doctors who have been involved in one or moremalpractice suits in the sample.

6

1.2 | Data and Sampling
Data can come from a population or from a sample. Small letters like x or y generally areused to represent data values. Most data can be put into the two major categories ofqualitative or quantitative.
Qualitative data are the result of categorizing or describing attributes of a population. Haircolor, blood type, ethnic group, type of car a person drives, and the street a person lives onare examples of qualitative data. Qualitative data are generally described by words or letters.
For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood
type might be AB+, O-, or B+.
Researchers often prefer to use quantitative data over qualitative data because it lends itself
more easily to mathematical analysis. For example, it does not make sense to find an averagehair color or blood type.
Quantitative data are always numbers. Quantitative data are the resultof counting or measuring attributes of a population. Amount of money, pulse rate, weight,
number of people living in your town, and number of students who take statistics areexamples of quantitative data. Quantitative data may be either discrete or continuous.
•

Quantitative discrete data is data that are the result of counting. These data takeon only certain numerical values. If you count the number of phone calls you receivefor each day of the week, you might get values such as zero, one, two, or three.

•

Quantitative continuous data is all data that are the result of measuring assumingthat we can measure accurately. Measuring angles in radians might result in such
numbers as π, π/3, 5π/6, etc. If you and your friends carry backpacks with books inthem to school, the numbers of books in the backpacks would be discrete data and theweights of the backpacks would be continuous data.

The data are the number of books students carry in their backpacks. You sample five
students. Two students carry three books, one student carries four books, one student carries
two books, and one student carries one book. The numbers of books (three, four, two, and one)are the quantitative discrete data.

1.5 The data are the number of machines in a gym. You sample five gyms. One gym has
12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22
machines, and the other gym has 20 machines. What type data is this?
7
1.6 The data are the areas of lawns in square feet. You sample five houses. The areas ofthe lawns are 144 sq. ft, 190 sq. ft, 180 sq. ft, and 210 sq. ft. What type of data is this?

7

Example 1.7
You go to the supermarket and purchase three cans of soup (19 ounces) tomato bisque,14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts andpeanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), andtwo desserts (16 ounces Cherry Garcia ice cream and 32 ounces chocolate chip cookies).
Identify data sets that are quantitative discrete, quantitative continuous, and qualitative.
Solution 1.7: Answers will vary, but one possibility is:•••

The three cans of soup, two packages of nuts, four kinds of vegetables and twodesserts are quantitative discrete data because you count them.
The weights of the soups (19 ounces, 14.1 ounces, 19 ounces) and weights ofdesserts are quantitative continuous data
because we measure weights as precisely as possible.
Types of soups, nuts, vegetables and desserts are qualitative data because they
are categorical.

Example 1.8
The data are the colors of backpacks. Again, you sample the same five students. One
student has a red backpack, two students have black backpacks, one student has a green
backpack, and one student has a gray backpack. The colors red, black, black, green, andgray are qualitative data.
NOTE: You may collect data as numbers and report it categorically. For example,
exam scores for students are recorded throughout the term. At the end of the term, letter
grades are reported as A, B, C, D, or F.
Example 1.9
Work collaboratively to determine the data is quantitative or qualitative. For quantitative
data, indicate whether they are continuous or discrete.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.

the number of pairs of shoes you own
the type of car you drive
where you go on vacation
the distance it is from your home to the nearest grocery store
the number of classes you take per school year
the tuition for your classes
the type of calculator you use
movie ratings (PG, R, …)
political party preferences
weights of sumo wrestlers
amount of money (in dollars) won playing poker
number of correct answers on a quiz8m. peoples’ attitudes toward the government
n. IQ scores (This may cause some discussion.)
Solution to 1.9:
Quantitative Discrete (a, e, l); Quantitative Continuous (d, f, j, k, n); Qualitative (b, c, g,h, i, m)

1.9 Determine the correct data type (quantitative or qualitative) for the number of cars in aparking lot. If quantitative indicate whether it is continuous or discrete.

Example 1.10
A statistics professor collects information about the classification of her students as
freshmen, sophomores, juniors, or seniors. Figure 1.1 summarizes the data she collected.
What type of data does this graph show?

Figure 1.1
Solution 1.10 The chart shows the students in each year, which is qualitative data.

Sampling
Gathering information about an entire population often costs too much or is virtually
impossible. Instead, we usually use data from a sample of the population.

9

A sample should have the same characteristics as the population it isrepresenting.
Most statisticians use various methods of random sampling in an attempt to achieve thisgoal. This section will describe a few of the most common sampling methods. There areseveral different methods of random sampling. In each form of random sampling, eachmember of a population initially has an equal chance of being selected for the sample. Eachmethod has pros and cons.
The easiest method to describe is called a simple random sample. Any groupof n individuals is equally likely to be chosen by any other group of n individuals if the simple
random sampling technique is used. In other words, each sample of the same size has anequal chance of being selected.
For example, suppose Lisa wants to form a four-person study group (herself and three otherpeople) from her pre-calculus class, which has 31 members not including Lisa. To choose a
simple random sample of size three from the other members of her class, Lisa could put all
31 names in a hat, shake the hat, close her eyes, and pick out three names. A more
technological way is for Lisa to first list the last names of the members of her class together
with a two-digit number, as in Table 1.2:
ID
00
01
02
03
04
05
06
07
08
09
10

NAME
ID NAMEID
Anselmo
11 King
21
Bautista
12 Legeny
22
Bayani13 Lundquist 23
Cheng
14 Macierz
24
Cuarismo
15 Motogawa 25
Cunningham 16 Okimoto 26
Fontecha
17 Pate;27
Hong
28
18 Price
Hoobler
19 Quizon
29
Jiao
20 Reyes30
Khan
Table 1.2

NAME
RoqueroRothRowellSalangsang
Slade
Statcher
Tallai
Tran
Wai
Wood

Lisa can then use a table of random numbers (found in many statistics books
and mathematical handbooks), a calculator, or a computer to generate random numbers. For
this example, suppose Lisa chooses to generate random numbers from a calculator. The
numbers generated are as follows:
0.94360; 0.99832; 0.14669; 0.51470; 0.40581; 0.73381; 0.04399
Lisa reads two-digit groups until she has chosen three class members (that is, she reads
0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one classmember. If she needed to, Lisa could have generated more random numbers. The random
numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However thethird random number, 0.14669, contains 14 (the fourth random number also contains 14),10

the fifth random number contains 05, and the seventh random number contains 04. The
two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04
corresponds to Cuarismo. Besides herself, Lisa’s group will consistof Marcierz, Cuningham, and Cuarismo.

Using the TI-83, 83+, 84, 84+ Calculator
To generate random numbers:• Press MATH.
• Arrow over to PRB.
• Press 5: randInt( : Enter 0, 30.
• Press ENTER for the first random number.
• Press ENTER two more times for the other 2 random numbers.
If there is a repeat press ENTER again.
Note: We can provide third input get a specified number of random values. E.g. randInt(0,30, 3) will generate 3 random numbers, which would yield the answer {29, 28, 4}.
Besides simple random sampling, there are other forms of sampling that involve random
chance. Other well-known random sampling methods are the stratified samples, cluster
samples, and systematic samples.
To choose a stratified sample, divide the population into groups called “strata” and thentake a proportionate number from each stratum. For example, you could stratify (group)
your college population by department and then choose a proportionate simple random
sample from each stratum (each department) to get a stratified random sample. To choose a
simple random sample from each department, number each member of the first department,
number each member of the second department, and do the same for the remaining
departments. Then use simple random sampling to choose proportionate numbers from thefirst department and do the same for each of the remaining departments. Those numbers
picked from the first department, picked from the second department, and so on represent
the members who make up the stratified sample.
To choose a cluster sample, divide the population into clusters (groups) and thenrandomly select some of the clusters. All the members from these clusters are in the cluster
sample. For example, if you randomly sample four departments from your college population,
the four departments make up the cluster sample. Divide your college faculty by department.
The departments are the clusters. Number each department, and then choose four different
numbers using simple random sampling. All members of the four departments withthose numbers are the cluster sample.
To choose a systematic sample, randomly select a starting point and take every nth pieceof data from a listing of the population. For example, suppose you have to do a phone survey.
Your phone book contains 20,000 residence listings. You must choose 400 names for the
sample. Number the population 1 to 20,000 and then use a simple random sample to pick a
number that represents the first name in the sample. Then choose every fiftieth namethereafter until you have a total of 400 names (you might have to go back to the beginning ofyour phone list). Systematic sampling is frequently chosen because it is a simple method.

11

A type of sampling that is non-random is convenience sampling. Conveniencesampling involves using results that are readily available. For example, a computer
software store conducts a marketing study by interviewing potential customers who happen
to be in the store browsing through the available software. The results of conveniencesampling may be very good in some cases and highly biased (favor certain outcomes) inothers.
Sampling data should be done very carefully. Collecting data carelessly can have devastating
results. Surveys mailed to households and then returned may be very biased (they may favora certain group). It is better for the person conducting the survey to select the sample
respondents.
True random sampling is done with replacement. That is, once a member is picked, thatmember goes back into the population and thus may be chosen more than once. However forpractical reasons, in most populations, simple random sampling is done without
replacement. Surveys are typically done without replacement. That is, a member of the
population may be chosen only once. Most samples are taken from large populations and thesample tends to be small in comparison to the population. Since this is the case, sampling
without replacement is approximately the same as sampling with replacement because thechance of picking the same individual more than once with replacement is very low.
Sampling without replacement instead of sampling with replacement becomes amathematical issue only when the population is small. For example, if the population is 25
people, the sample is ten, and you are sampling with replacement for any particularsample, then the chance of picking the first person is ten out of 25, and the chanceof picking a different second person is nine out of 25 (you replace the first person).
If you sample without replacement, then the chance of picking the first person is ten outof 25, and then the chance of picking the second person (who is different) is nine out of 24(you do not replace the first person). Compare the fractions 9/25 and 9/24. To four decimal
places, 9/25 = 0.3600 and 9/24 = 0.3750. So these numbers are not equivalent.
When you analyze data, it is important to be aware of sampling errors and non-sampling
errors. The actual process of sampling causes sampling errors. For example, the sample may
not be large enough. Factors not related to the sampling process cause non-sampling
errors. A defective counting device can cause a non-sampling error.
In reality, a sample will never be exactly representative of the population so there will always
be some sampling error. In general, the larger the sample, the smaller the sampling error.
In Statistics, a sampling bias is created when a sample is collected from a population andsome members of the population are not as likely to be chosen as others (remember, eachmember of the population should have an equally likely chance of being chosen).
When sampling bias happens, there can be incorrect conclusions drawn about the
population that is being studied.

12

Example 1.11
A study is done to determine the average tuition that College of Lake County transfer,career, and adult education students pay per semester. Each student in the following
samples is asked how much tuition they paid for the fall semester. What is the type ofsampling in each case?
a) A sample of 100 College of Lake County students is taken by organizing thestudents’ names by classification (transfer, career or adult education), and then selecting 25
students from each.
b) A random number generator is used to select a student from the alphabetical listing of
all college students in the fall semester. Starting with that student, every 50th student ischosen until 75 students are included in the sample.
c) A completely random method is used to select 75 students. Each undergraduate student
in the fall semester has the same probability of being chosen at any stage of the sampling
process.
d) The transfer, career, and adult education are numbered one, two and threerespectively. A random number generator is used to pick two of those categories. All
students in those two categories are in the sample.
e) An administrative assistant is asked to stand in front of the library one Wednesday andto ask the first 100 students they encounter what they paid for tuition in the fall semester.
Those 100 students are the sample.

Solution 1.11: a stratified; b. systematic; c. simple random; d. cluster; e. convenience
Example 4.10

1.11 You are going to use the random number generator to generate different types ofsamples from the data. This table displays six sets of quiz scores for anelementary Statistics class.
#1
5
10
9
9
7
9
7
8
9
8

#2
7
5
10
10
8
9
7
8
7
8

#3
10
9
8
10
9
9
10
9
8
10

#4
9
8
6
9
5
10
9
10
7
9

#5
8
7
7
8
7
8
8
8
7
8

#6
3
6
9
9
4
7
8
8
8
7

13

1. Create a stratified sample by column. Pick three quiz scores randomly from eachcolumn.
o Number each row one through ten.
o On your calculator, press Math and arrow over to PRB.
o For column 1, select randInt( and enter 1,10. Press ENTER. Record thenumber. Press ENTER 2 more times (even the repeats). Record thesenumbers. Record the three quiz scores in column one that correspond to these threenumbers.
o Repeat for columns two through six.
o These 18 quiz scores are a stratified sample.
2. Create a cluster sample by first randomly picking two of the columns.
Use the column numbers: one through six.
o Press MATH and arrow over to PRB.
o Select randInt and enter 1,6. Press ENTER. Record the number.
o Press ENTER and record that number.
o The two numbers are for two of the columns.
o The quiz scores (20 of them) in these 2 columns are the cluster sample.
3. Create a simple random sample of 15 quiz scores.
o Use the numbering 1 through 60.
o Press MATH. Arrow over to PRB. select randInt and enter 1, 60).
o Press ENTER 15 times and record the numbers.
o Record the quiz scores that correspond to these numbers.
o These 15 quiz scores are the systematic sample.
4. Create a systematic sample of 12 quiz scores.
o Use the numbering one through 60.
o Press MATH. Arrow over to PRB. Press 5: randInt( and enter 1, 60).
o Press ENTER. Record the number and the first quiz score. From that number,count ten quiz scores and record that quiz score. Keep counting ten quiz scores andrecording the quiz score until you have a sample of 12 quiz scores. You may needto wrap around (go back to the beginning).
Example 1.12
Determine type of sampling used (simple random, stratified, systematic, cluster,or convenience).
a. A soccer coach selects six players from a group of boys aged eight to ten, sevenplayers from a group of boys aged 11 to 12, and three from a group of boys aged 13 to 14
to form a recreational soccer team.
b. A pollster interviews all human resource personnel in five different hightech companies.
c. A high school educational researcher interviews 50 high school female teachers and50 high school male teachers.

14

d. A medical researcher interviews every third cancer patient from a list of cancer
patients at a local hospital.
e. A high school counselor uses a computer to generate 50 random numbers and thenpicks students whose names correspond to the numbers.
f. A student interviews classmates in his algebra class to determine how many pairs ofjeans a student owns, on the average.
Solution 1.12
a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f. convenience

1.12 Determine the type of sampling used (simple random, stratified, systematic, cluster,or convenience).
A high school principal polls 50 freshmen, 50 sophomores, 50 juniors, and 50
seniors regarding policy changes for after school activities.
If we were to examine two samples representing the same population, they would not
be exactly the same, even if we used random sampling methods for the samples. Just as
there is variation in data, there is variation in samples. As you become accustomed tosampling, the variability will begin to seem natural.
Example 1.13
Suppose ABC College has 10,000 part-time students (the population). We are interested in
the average amount of money a part-time student spends on books in the fall term. Asking
all 10,000 students is an almost impossible task.
Suppose we take two different samples. First, we use convenience sampling and survey ten
students from a first term organic chemistry class. Many of these students are taking firstterm calculus in addition to the organic chemistry class. The amount of money they spend
on books is as follows:$128; $87; $173; $116; $130; $204; $147; $189; $93; $153
The second sample is taken using a list of senior citizens who take P.E. classes and taking
every fifth senior citizen on the list, for a total of ten senior citizens. They spend:$50; $40; $36; $15; $50; $100; $40; $53; $22; $22
It is unlikely that any student is in both samples.
a. Do you think that either of these samples is representative of (or is characteristic of)the entire 10,000 part-time student population?
b. If these samples are not representative of the entire population, would it be wise to
use the results to describe the entire population?

15

c. Now suppose we take a third sample. We choose ten different part-time students fromthe disciplines of Chemistry, Math, English, Psychology, Sociology, History, Nursing,
Physical Education, Art, and Early Childhood Development. (We assume that these are
the only disciplines in which part-time students at ABC College are enrolled and that anequal number of part-time students are enrolled in each of the disciplines.) Each student
ischosenusing
simplerandom sampling.
Usingacalculator, random
numbers are generated
and a student from a particular discipline is selected if he or she has a corresponding
number. The students spend the following amounts:$180; $50; $150; $85; $260; $75; $180; $200; $200; $150
Is this sample biased? Explain.
Solution 1.13
a. No. The first sample probably consists of science-oriented students. For example, in
addition to the chemistry course, some of them are also Calculus or Biology courses.
Books for these classes tend to be expensive. Most of these students are, more thanlikely, paying more than the average part-time student for their books. The second
sample is a group of senior citizens who are likely taking courses for health andinterest. The amount of money they spend on books is probably much less than the
average part-time student. Both samples are biased. Also, in both cases, not all students
have a chance to be in either sample.
b. No. For these samples, each member of the population did not have an equally likely
chance of being chosen.
c. The sample is unbiased, but a larger sample would be recommended to increase thelikelihood that the sample will be close to representative of the population. However, fora biased sampling technique, even a large sample runs the risk of not being
representative of the population.
Students often ask if it is "good enough" to take a sample, instead of surveying the entire
population. If the survey is done well, the answer is yes.

1.13 A local radio station has a fan base of 20,000 listeners. The station wants to know if its
audience would prefer more music or more talk shows. Since asking all 20,000 listeners would
be impossible, the station uses convenience sampling and surveys the first 200 people theymeet at one of the station’s music concert events. Of those sampled, 24 people said they’dprefer more talk shows, and 176 people said they’d prefer more music. Is this sample
representative of the entire 20,000 listener population?

16

Variation in Data and in Samples
Variation is present in any set of data. For example, 16-ounce cans of beverage may
contain more or less than 16 ounces of liquid. In one study, eight 16 ounce cans were
measured and produced the following amount (in ounces) of beverage:
15.8; 16.1; 15.2; 14.8; 15.8; 15.9; 16.0; 15.5
Measurements of the amount of beverage in a 16-ounce can may vary because
different people make the measurements or because the exact amount, 16 ounces of liquid,
was not put into the cans. Manufacturers regularly run tests to determine if the amount ofbeverage in a 16-ounce can falls within the desired range.
Be aware that as you take data, your data may vary somewhat from the data someone else
is taking for the same purpose. This is completely natural. However, if two or more of you aretaking the same data and get very different results, it is time for you and the others toreevaluate your data selection methods and your accuracy.
It was mentioned previously that two or more samples from the same population, taken
randomly, and having close to the same characteristics of the population will likely be
different from each other. Suppose Doreen and Jung both decide to study the average amountof time students at their college sleep each night. Doreen and Jung each take samples of 500
students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample
will be different from Jung's sample. Even if Doreen and Jung used the same sampling
method, in all likelihood their samples would be different. Neither would be wrong,
however.
Think about what contributes to making Doreen’s and Jung’s samples different.
If Doreen and Jung took larger samples (i.e. the number of data values is increased), theirsample results (the average amount of time a student sleeps) might be closer to the actual
population average. But still, their samples would be, in all likelihood, different from eachother. This variability in samples cannot be stressed enough.
The size of a sample (often called the number of observations) is important. The examples
you have seen in this book so far have been small. Samples of only a few hundredobservations, or even smaller, are sufficient for many purposes. In polling, samples that arefrom 1,200 to 1,500 observations are considered large enough and good enough if the survey
is random and is well done. You will learn why when you study confidence intervals.
Be aware that many large samples are biased. For example, call-in surveys areinvariably biased, because people choose to respond or not.
Critical Evaluation
We need to critically evaluate statistical studies we read about and analyze them before
accepting the results of the studies. Common problems to be aware of include:17

Problems with samples: A sample must be representative of the population. A
sample that is not representative of the population is biased. Biased samples give
results that are inaccurate and not valid.
•

Self-selected samples: Responses only by people who choose to respond, such as callin surveys, are often unreliable.
•

Sample size issues: Samples that are too small may be unreliable. Larger samples
are better, if possible. In some situations, having small samples is unavoidable and canstill be used to draw conclusions. Examples include crash testing cars or medical testing
for rare conditions.
•Undue influence: collecting data or asking questions in a way that influences the
response.
•

Non-response or refusal of subject to participate: The collected responses may nolonger be representative of the population. Often, people with strong positive or
negative opinions may answer surveys, which can affect the results.
•

Causality: A relationship between two variables does not mean that one causes theother to occur. They may be related (correlated) because of their relationship through adifferent variable.
•

Self-funded or self-interest studies: A study performed by a person or organization inorder to support their claim. Is the study impartial? Read the study carefullyto evaluate the work. Do not automatically assume that the study is good, but do not
automatically assume the study is bad either. Evaluate it on its merits and the work
done.
•

Misleading use of data: improperly displayed graphs, incomplete data, or lack ofcontext.
•

Confounding: When the effects of multiple factors on a response cannot be
separated. Confounding makes it difficult or impossible to draw valid conclusions about
the effect of each factor.
•

18

1.3 | Experimental Design and Ethics
Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at
growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol?
Questions like these are answered using randomized experiments. In this module, you willlearn important aspects of experimental design. Proper study design ensures the productionof reliable, accurate data.
The purpose of an experiment is to investigate the relationship between two
variables. When changes in one variable produce changes in another, we call the firstvariable the explanatory variable. The affected variable is called the response variable.
In a randomized experiment, the researcher manipulates values of the explanatory variable
and measures the resulting changes in the response variable. The different values of the
explanatory variable are called treatments. An experimental unit is a single object orindividual to be measured.
You want to investigate the effectiveness of vitamin E in preventing disease. You recruit agroup of subjects and ask them if they regularly take vitamin E. You notice that the subjects
who take vitamin E exhibit better health on average than those who do not. Does this provethat vitamin E is effective in disease prevention? It does not. There are many differences
between the two groups compared in addition to vitamin E consumption. People who takevitamin E regularly often take other steps to improve their health: exercise, diet, othervitamin supplements, choosing not to smoke. Any one of these factors could be influencing
health. As described, this study does not prove that vitamin E is the key to disease
prevention.
Additional variables that can cloud a study are called lurking variables (or confounding
variables). In order to prove that the explanatory variable is causing a change in the response
variable, it is necessary to isolate the explanatory variable. The researcher must design her
experiment in such a way that there is only one difference between groups being compared:
the planned treatments. This is accomplished by the random assignment of experimental
units to treatment groups. When subjects are assigned treatments randomly, all of thepotential lurking variables are spread equally among the groups. At this point the only
difference between groups is the one imposed by the researcher. Different outcomes measured
in the response variable, therefore, must be a direct result of the different treatments. In thisway, an experiment can prove a cause-and-effect connection between the explanatory andresponse variables.
The power of suggestion can have an important influence on the outcome of an experiment.
Studies have shown that the expectation of the study participant can be as important as theactual medication. In one study of performance-enhancing drugs, researchers noted:
Results showed that believing one had taken the substance resulted in [performance] times
almost as fast as those associated with consuming the drug itself. In contrast, taking the drug
without knowledge yielded no significant performance increment.[1]When participation in a study prompts a physical response from a participant, it is difficult
to isolate the effects of the explanatory variable. To counter the power of suggestion,
researchers set aside one treatment group as a control group. This group is given19

a placebo treatment–a treatment that cannot influence the response variable. The control
group helps researchers balance the effects of being in an experiment with the effects of theactive treatments. Of course, if you are participating in a study and you know that you arereceiving a pill which contains no actual medication, then the power of suggestion is no longera factor. Blinding in a randomized experiment preserves the power of suggestion. When aperson involved in a research study is blinded, he does not know who is receiving the active
treatments(s) and who is receiving the placebo treatment. A double-blind experiment is
one in which both the subjects and the researchers involved with the subjects are blinded.
1. McClung, M. Collins, D. “Because I know it will!” Placebo effects of an ergogenic aid on athletic
performance. Journal of Sport & Exercise Psychology. 2007 Jun. 29(3):382-94. Web. April 30, 2013.

Example 1.19
Researchers want to investigate whether taking aspirin regularly reduces the risk of heart
attack. Four hundred men between the ages of 50 and 84 are recruited as participants. The
men are divided randomly into two groups: one group will take aspirin, and the other groupwill take a placebo. Each man takes one pill each day for three years, but he does not knowwhether he is taking aspirin or the placebo. At the end of the study, researchers count the
number of men in each group who have had heart attacks.
Identify the following for this study: population, sample, experimental units,
explanatory variable, response variable, treatments.
Solution 1.19
The population is men aged 50 to 84.
The sample is the 400 men who participated.
The experimental units are the individual men in the study.
The explanatory variable is oral medication.
The treatments are aspirin and a placebo.
The response variable is whether a subject had a heart attack.
Example 1.20
The Smell & Taste Treatment and Research Foundation conducted a study to investigate
whether smell can affect learning. Subjects completed mazes multiple times while wearing
masks. They completed the pencil and paper mazes three times wearing floral-scented masks,
and three times with unscented masks. Participants were assigned at random to wear thefloral mask during the first three trials or during the last three trials. For each trial,
researchers recorded the time it took to complete the maze and the subject’s impression of
the mask’s scent: positive, negative, or neutral.
a. Describe the explanatory and response variables in this study.
b. What are the treatments?
c. Identify any lurking variables that could interfere with this study.
d. Is it possible to use blinding in this study?

20

Solution 1.20
a. The explanatory variable is scent, and the response variable is the time it takes tocomplete the maze.
b. There are two treatments: a floral-scented mask and an unscented mask.
c. All subjects experienced both treatments. The order of treatments was randomly assigned
so there were no differences between the treatment groups. Random assignment eliminates
the problem of lurking variables.
d. Subjects will clearly know whether they can smell flowers or not, so subjects cannot be
blinded in this study. Researchers timing the mazes can be blinded, though. The
researcher who is observing a subject will not know which mask is being worn.

Example 1.21
A researcher wants to study the effects of birth order on personality. Explain why thisstudy could not be conducted as a randomized experiment. What is the main problem in astudy that cannot be designed as a randomized experiment?
Solution 1.21
The explanatory variable is birth order. You cannot randomly assign a person’s birth order.
Random assignment eliminates the impact of lurking variables. When you cannot assign
subjects to treatment groups at random, there will be differences between the groups otherthan the explanatory variable.

1.21 You are concerned about the effects of texting on driving performance. Design a study
to test the response time of drivers while texting and while driving only. How many seconds
does it take for a driver to respond when a leading car hits the brakes?
a. Describe the explanatory and response variables in the study?
b. What are the treatments?
c. What should you consider when selecting participants?
d. Your research partner wants to divide participants randomly into two groups: one todrive without distraction and one to text and drive simultaneously. Is this a good idea?
Why or why not?
e. Identify any lurking variables that could interfere with this study.
f. How can blinding be used in this study?

Ethics
The widespread misuse and misrepresentation of statistical information often gives thefield a bad name. Some say that “numbers don’t lie,” but the people who use numbers to
support their claims often do.

21A recent investigation of famous social psychologist, Diederik Stapel, has led to theretraction of his articles from some of the world’s top journals including Journal of
Experimental Social Psychology, Social Psychology, Basic and Applied Social Psychology,
British Journal of Social Psychology, and the magazine Science. Diederik Stapel is a former
professor at Tilburg University in the Netherlands. Over the past two years, an extensive
investigation involving three universities where Stapel has worked concluded that thepsychologist is guilty of fraud on a colossal scale. Falsified data taints over 55 papers heauthored and 10 Ph.D. dissertations that he supervised.
Stapel did not deny that his deceit was driven by ambition. But it was more complicatedthan that, he told me. He insisted that he loved social psychology but had been frustrated by
the messiness of experimental data, which rarely led to clear conclusions. His lifelong
obsession with elegance and order, he said, led him to concoct sexy results that journals
found attractive. “It was a quest for aesthetics, for beauty—instead of the truth,” he said. He
described his behavior as an addiction that drove him to carry out acts of increasingly
daring fraud, like a junkie seeking a bigger and better high.[2]2. Yudhijit Bhattacharjee, “The Mind of a Con Man,” Magazine, New York Times, April 26, 2013. Available online
at: http://www.nytimes.com/2013/04/28/magazine/diederik-stapels-audacious-academicfraud.html?src=dayp&_r=2& (accessed May 1, 2013).

The committee investigating Stapel concluded that he was guilty of several practices
including:• creating datasets, which largely confirmed the prior expectations,• altering data in existing datasets• changing measuring instruments without reporting the change, and• misrepresenting the number of experimental subjects.
Clearly, it is never acceptable to falsify data the way this researcher did. Sometimes,
however, violations of ethics are not so easy to spot.
Researchers have a responsibility to verify that proper methods are being followed. The
report describing the investigation of Stapel’s fraud states that, “statistical flaws
frequently revealed a lack of familiarity with elementary statistics.”[3] Many of Stapel’s coauthors should have spotted irregularities in his data. Unfortunately, they did not know verymuch about statistical analysis, and they simply trusted that he was collecting and reporting
data properly.
Many types of statistical fraud are difficult to detect. Some researchers simply stop collecting
data once they have just enough to prove what they had hoped to prove. They don’t want totake the chance that a more extensive study would complicate their lives by producing data
contradicting their hypothesis.
Professional organizations, like the American Statistical Association, clearly define
expectations for researchers. There are even laws in the federal code about the use of researchdata.
When a statistical study uses human participants, as in medical studies, both ethics andthe law dictate that researchers should be mindful of the safety of their research subjects.
The U.S. Department of Health and Human Services oversees federal regulations ofresearch studies with the aim of protecting participants. When a university or other22

research institution engages in research, it must ensure the safety of all human subjects.
For this reason, research institutions establish oversight committees known
as Institutional Review Boards (IRB). All planned studies must be approved in advanceby the IRB. Key protections that are mandated by law
include the following:• Risks to participants must be minimized and reasonable with respect to projected
benefits.
• Participants must give informed consent. This means that the risks of participationmust be clearly explained to the subjects of the study. Subjects must consent in writing,and researchers are required to keep documentation of their consent.
• Data collected from individuals must be guarded carefully to protect their privacy.
These ideas may seem fundamental, but they can be very difficult to verify in practice. Is
removing a participant’s name from the data record sufficient to protect privacy? Perhaps the
person’s identity could be discovered from the data that remains. What happens if the study
does not proceed as planned and risks arise that were not anticipated? When is informed
consent really necessary? Suppose your doctor wants a blood sample to check your cholesterollevel. Once the sample has been tested, you expect the lab to dispose of the remaining blood.
At that point the blood becomes biological waste. Does a researcher have the right to take it
for use in a study?
3. “Flawed Science: The Fraudulent Research Practices of Social Psychologist Diederik Stapel,”
Tillburg University,
November28,
2012, http://www.tilburguniversity.edu/upload/064a10cd- bce5-4385b9ff- 05b840caeae6_120695_Rapp_nov_2012_UK_web.pdf (accessed May 1, 2013).

It is important that students of statistics take time to consider the ethical questions thatarise in statistical studies. How prevalent is fraud in statistical studies? You might besurprised—and disappointed. There is a website (www.retractionwatch.com)(http://www.retractionwatch.com) dedicated to cataloging retractions of study articles thathave been proven fraudulent. A quick glance will show that the misuse of statistics is abigger problem than most people realize.
Vigilance against fraud requires knowledge. Learning the basic theory of statistics will
empower you to analyze statistical studies critically.
Example 1.22
Describe the unethical behavior in each example and describe how it could impact thereliability of the resulting data. Explain how the problem should be corrected. A researcher
is collecting data in a community.
a. She selects a block where she is comfortable walking because she knows many of thepeople living on the street.
Example 1.22 continued
b. No one seems to be home at four houses on her route. She does not record theaddresses and does not return at a later time to try to find residents at home.

23

c. She skips four houses on her route because she is running late for an appointment.
When she gets home, she fills in the forms by selecting random answers fromother residents in the neighborhood.
Solution 1.22
a. By selecting a convenient sample, the researcher is intentionally selecting a sample
that could be biased. Claiming that this sample represents the community is
misleading. The researcher needs to select areas in the community at random.
b. Intentionally omitting relevant data will create bias in the sample. Suppose theresearcher is gathering information about jobs and child care. By ignoring people whoare not home, she may be missing data from working families that are relevant to herstudy. She needs to make every effort to interview all members of the target sample.
c. It is never acceptable to fake data. Even though the responses she uses are “real”
responses provided by other participants, the duplication is fraudulent and can create
bias in the data. She needs to work diligently to interview everyone on her route.

1.22 Describe the unethical behavior, if any, in each example and describe how it could
impact the reliability of the resulting data. Explain how the problem should be corrected.
A study is commissioned to determine the favorite brand of fruit juice among teens in
California. The survey is commissioned by the seller of a popular brand of apple
juice. There are only two types of juice included in the study: apple juice and cranberry
juice. Researchers allow participants to see the brand of juice as samples are poured for ataste test. Among the participants, 25% preferred Brand X, 33% preferred Brand Y and
42% had no preference between the two brands. Brand X then references the study in acommercial saying “Most teens like Brand X as much as or more than Brand Y.”

24

KEY TERMS
Average a number that describes the central tendency of the data
Blinding not telling participants which treatment a subject is receiving
Categorical Variable variables that take on values that are names or labels
Cluster Sampling a method for selecting a random sample and dividing the population
into groups (clusters); use simple random sampling to select a set of clusters. Every
individual in the chosen clusters is included in the sample.
Continuous Random Variable a random variable (RV) whose outcomesare measured; e.g. the height of trees in the forest is a continuous RV.
Control Group a group in a randomized experiment that receives an inactive treatment
but is otherwise managed exactly as the other groups
Convenience Sampling a nonrandom method of selecting a sample; this method selects
individuals that are easily accessible and may result in biased data.
Data a set of observations (a set of possible outcomes); most data can be put into twogroups: qualitative (an attribute whose value is indicated by a label) or quantitative (anattribute whose value is indicated by a number). Quantitative data can be separated into
two subgroups: discrete and continuous. Data is discrete if it is the result of counting(such as the number of students of a given ethnic group in a class or the number of books ona shelf). Data is continuous if it is the result of measuring (such as distance traveled orweight of luggage)
Discrete Random Variable a random variable (RV) whose outcomes are counted
Double-blinding the act of blinding both the subjects of an experiment and theresearchers who work with the subjects
Experimental Unit any individual or object to be measured
Explanatory Variable the independent variable in an experiment; the value controlled
by researchers
Informed Consent Any human subject in a research study must be cognizant of any risks
or costs associated with the study. The subject has the right to know the nature of thetreatments included in the study, their potential risks, and their potential benefits. Consent
must be given freely by an informed, fit participant.
Institutional Review Board a committee tasked with oversight of research programs
that involve human subjects
Lurking Variable a variable that has an effect on a study even though it is neither an
explanatory variable nor a response variable25

Non-sampling Error an issue that affects the reliability of sampling data other thannatural variation; it includes a variety of human errors including poor study design,biased sampling methods, inaccurate information provided by study participants, data
entry errors, and poor analysis.
Numerical Variable variables that take on values that are indicated by numbers
Parameter a number that is used to represent a population characteristic
Placebo an inactive treatment that has no real effect on the explanatory variable
Population all individuals, objects, or measurements whose properties are being studied
Proportion the number of successes divided by the total number in the sample
Qualitative Data See Data.
Quantitative Data See Data.
Random Assignment the act of organizing experimental units into treatment groups
using random methods
Random Sampling a method of selecting a sample that gives every member of the
population an equal chance of being selected.
Representative Sample a subset of the population that has the same characteristics as
the population
Response Variable the dependent variable in an experiment; the value that is measured
for change at the end of an experiment
Sample a subset of the population studied
Sampling Bias not all members of the population are equally likely to be selected
Sampling Error the natural variation that results from selecting a sample to represent alarger population; this variation decreases as the sample size increases, so selecting
larger samples reduces sampling error.
Sampling with Replacement Once a member of the population is selected for inclusion ina sample, that member is returned to the population for the selection of the next
individual.
Sampling without Replacement A member of the population may be chosen forinclusion in a sample only once. If chosen, the member is not returned to the population
before the next selection.
Simple Random Sampling a straightforward method for selecting arandom sample; give each member of the population a number. Use a random
number generator to select a set of labels. These randomly selected labels identify themembers of your sample.

26

Statistic a numerical characteristic of the sample; a statistic estimates the corresponding
population parameter.
Stratified Sampling a method for selecting a random sample used to ensure that
subgroups of the population are represented adequately; divide the population into groups(strata). Use simple random sampling to identify a proportionate number of individuals
from each stratum.
Systematic Sampling a method for selecting a random sample; list the members of the
population. Use simple random sampling to select a starting point in the population. Let k= (number of individuals in the population)/(number of individuals needed in the sample).
Choose every kth individual in the list starting with the one that was randomly selected. If
necessary, return to the beginning of the population list to complete your sample.
Treatments different values or components of the explanatory variable applied in
an experiment
Variable a characteristic of interest for each person or object in a population

27

CHAPTER REVIEW
1.1 Basic Definitions
The mathematical theory of statistics is easier to learn when you know the language. Thissection presented important terms that will be used throughout the text.
1.2 Data and Sampling
Data are individual items of information that come from a population or sample. Data may
be classified as qualitative, quantitative continuous, or quantitative discrete.
Because it is not practical to measure the entire population in a study, researchers use
samples to represent the population. A random sample is a representative group from the
population chosen by using a method that gives each individual in the population an equalchance of being included in the sample. Random sampling methods include simple random
sampling, stratified sampling, cluster sampling, and systematic sampling. Conveniencesampling is a nonrandom method of choosing a sample that often produces biased data.
Samples that contain different individuals result in different data. This is true even when
the samples are well-chosen and representative of the population. When
properly selected, larger samples model the population more closely than smaller samples.
There are many different potential problems that can affect the reliability of a sample.
Statistical data needs to be critically analyzed, not simply accepted.
1.2 Experimental Design and Ethics
A poorly designed study will not produce reliable data. There are certain key componentsthat must be included in every experiment. To eliminate lurking variables, subjects must be
assigned randomly to different treatment groups. One of the groups must act as acontrol group, demonstrating what happens when the active treatment is not applied.
Participants in the control group receive a placebo treatment that looks exactly like the active
treatments but cannot influence the response variable. To preserve the integrity of theplacebo, both researchers and subjects may be blinded. When a study is designed properly,
the only difference between treatment groups is the one imposed by the researcher.
Therefore, when groups respond differently to different treatments, the difference must be
due to the influence of the explanatory variable.
“An ethics problem arises when you are considering an action that benefits you orsome because you support, hurts or reduces benefits to others, and violates some rule.”[4]Ethical violations in statistics are not always easy to spot. Professional associations andfederal agencies post guidelines for proper conduct. It is important that you learn basic
statistical procedures so that you can recognize proper data analysis.
4. Andrew Gelman, “Open Data and Open Methods,” Ethics and Statistics,
http://www.stat.columbia.edu/~gelman/research/ published/ChanceEthics1.pdf (accessed
May 1, 2013).

28

EXERCISES FOR CHAPTER 1
Use the following information to answer the next five exercises. Studies are often done bypharmaceutical companies to determine the effectiveness of a treatment program. Supposethat a new AIDS antibody drug is currently under study. It is given to patients once the
AIDS symptoms have revealed themselves. Of interest is the average (mean) length of timein months patients live once they start the treatment. Two researchers each follow a
different set of 40 patients with AIDS from the start of treatment until their deaths. The
following data (in months) are collected.
Researcher A:
3; 4; 11; 15; 16; 17; 22;
14; 24; 25; 15; 26; 27; 33;
13; 21; 22; 10; 12; 8; 40;
31; 34 ; 29; 17; 8; 24; 18;
Researcher B:
3; 14; 11; 5; 16;
14; 14; 26; 25; 21;
23; 21; 21; 16; 12;
33 ; 34; 29; 13; 18;

17;
22;
18;
24;

44; 37; 16;
29; 35; 44;
32; 26; 27;
47; 33; 34.

28; 41; 31; 18;
31; 2; 35; 44;
41; 22; 16; 25;
23; 42; 33; 29

Determine what the key terms refer to in the example for Researcher A.
1. population
2. sample
3. parameter
4. statistic
5. variable
Use the following information to answer the next five exercises: A study was done todetermine the age, number of times per week, and the duration (amount of time) ofresidents using a local park in San Antonio, Texas. The first house in the neighborhood
around the park was selected randomly, and then the resident of every eighth house in theneighborhood around the park was interviewed.
6. The sampling method was:a. simple randomb. systematicc. stratified

d. cluster

7. “Number of times per week” is what type of data?
a. qualitative b. quantitative discretec. quantitative continuous
8. “Duration (amount of time)” is what type of data?
a. qualitativeb. quantitative discretec. quantitative continuous

29

9. The colors of the houses around the park are what kind of data?
a. qualitative b. quantitative discretec. quantitative continuous
10. What is the population for this study?
11. The table below contains the total number of deaths worldwide as a result ofearthquakes from 2000 to 2012.
Year
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
Total

Total Number ofDeaths231
21,357
11,68533,819228,802
88,0036,605
712
88,011
1,790
320,120
21,953768
823,856

Use this data to answer the following questions:a. What is the proportion of deaths between 2007 and 2012?
b. What percent of deaths occurred before 2001?
c. What is the percent of deaths that occurred in 2003 or after 2010?
d. What is the fraction of deaths that happened before 2012?
e. What kind of data is the number of deaths?
f. Earthquakes are quantified according to the Richter scale, which measures theamount of energy they produce (ex. are 2.1, 5.0, 6.7). What type of data is that?
g. What contributed to the large number of deaths in 2010? In 2004? Explain.
For the following four exercises, determine the type of sampling used (simple random,
stratified, systematic, cluster, or convenience).
12. A group of test subjects is divided into twelve groups; then four of the groups are chosen
at random.
13. A market researcher polls every tenth person who walks into a store.

30

14. The first 50 people who walk into a sporting event are polled on their televisionpreferences.
15. A computer generates 100 random numbers, and 100 people whose names correspondwith the numbers on the list are chosen.
Use the following data to answer the next five exercises: A pair of studies was performed tomeasure the effectiveness of a new software program designed to help stroke patients
regain their problem-solving skills. Patients were asked to use the software program twice
a day, once in the morning and once in the evening. The studies observed 200 stroke
patients recovering over a period of several weeks. The first study collected the data
in Table 1.31. The second study collected the data in Table 1.32.
Group
Used Program
Did Not
use Program

Group
Used Program
Did Not
use Program

ShowedImprovement
142
72

No
Improvement
43
110

Deterioration

No
Improvement
74
99

Deterioration

15
18

Table 1.31
ShowedImprovement
105
88

19
12

Table 1.32

16. Given what you know, which study is correct?
17. The first study was performed by the company that designed the software program.
The second study was performed by the American Medical Association. Which study is morereliable?
18. Both groups that performed the study concluded that the software works. Is thisaccurate?
19. The company which makes the software uses the two studies as proof that theirsoftware causes mental improvement in stroke patients. Is this a fair statement?
20. Patients who used the software were also a part of an exercise program whereas
patients who did not use the software were not. Does this change the validity of theconclusions from question #18?
For each of the following eight exercises, identify: a. the population, b. the sample, c. theparameter, d. the statistic, e. the variable, and f. the data. Give examples whereappropriate.

3121. A fitness center is interested in the mean amount of time a client exercises in the center
each week.
22. Ski resorts are interested in the mean age that children take their first ski andsnowboard lessons. They need this information to plan their ski classes optimally.
23. A cardiologist is interested in the mean recovery period of her patients who have hadheart attacks.
24. Insurance companies are interested in the mean health costs each year of their clients,
so that they can determine the costs of health insurance.
25. A politician is interested in the proportion of voters in his district who think he is doinga good job.
26. A marriage counselor is interested in the proportion of clients she counsels who staymarried.
27. Political pollsters may be interested in the proportion of people who will vote fora particular cause.
28. A marketing company is interested in the proportion of people who will buy a particular
product.
Use the following information to answer the next three exercises: A Lake Tahoe Community
College instructor is interested in the mean number of days Lake
Tahoe Community College math students are absent from class during a quarter.
29. What is the population she is interested in?
a. all Lake Tahoe Community College students
b. all Lake Tahoe Community College English students
c. all Lake Tahoe Community College students in her classes
d. all Lake Tahoe Community College math students
30. Let X = number of days a Lake Tahoe Community College math student is absent. In
this case, X is an example of a:a. variable.
b. population.
c. statistic.
d. data.
31. The instructor’s sample produces a mean number of days absent of 3.5 days. This value
is an example of a:a. parameterb. datac. statistic
d. variable
For the following exercises (32 – 40), identify the type of data that would be used to describe
a response (quantitative discrete, quantitative continuous, or qualitative), and give an
example of the data.
32. number of tickets sold to a concert
33. percent of body fat
34. time spent in line to buy groceries

3235. number of students enrolled at Evergreen Valley College
36. most-watched television show
37. brand of toothpaste
38. distance to the closest movie theatre
39. age of executives in Fortune 500 companies
40. number of competing computer spreadsheet software packages
41. Airline companies are interested in the consistency of the number of babies on eachflight, so that they have adequate safety equipment. Suppose an airline conducts a survey.
Over Thanksgiving weekend, it surveys six flights from Boston to Salt Lake City todetermine the number of babies on the flights. It determines the amount of safety
equipment needed by the result of that study.
a. Using complete sentences, list three things wrong with the way the survey was
conducted.
b. Using complete sentences, list three ways that you would improve the survey if itwere to be repeated.
42. Suppose you want to determine the mean number of students per statistics class in yourstate. Describe a possible sampling method in three to five complete sentences. Make
the description detailed.
43. Suppose you want to determine the mean number of cans of soda drunk each month by
students in their twenties at your school. Describe a possible sampling method in three tofive complete sentences. Make the description detailed.
44. List some practical difficulties involved in getting accurate results from atelephone survey.
45. List some practical difficulties involved in getting accurate results from a mailed
survey.
46. With your classmates, brainstorm some ways you could overcome these problems if you
needed to conduct a phone or mail survey.
47. Name the sampling method used in each of the following situations:a. A woman in the airport is handing out questionnaires to travelers asking them toevaluate the airport’s service. She does not ask travelers who are hurrying through the
airport with their hands full of luggage, but instead asks all travelers who are sittingnear gates and not taking naps while they wait.
b. A teacher wants to know if her students are doing homework, so she
randomly selects rows two and five and then calls on all students in row two and all
students in row five to present the solutions to homework problems to the class.

33

c. The marketing manager for an electronics chain store wants information about theages of its customers. Over the next two weeks, at each store location, 100
randomly selected customers are given questionnaires to fill out asking for information
about age, as well as about other variables of interest.
d. The librarian at a public library wants to determine what proportion of thelibrary users are children. The librarian has a tally sheet on which she marks whether
books are checked out by an adult or a child. She records this data for every fourth
patron who checks out books.
e. A political party wants to know the reaction of voters to a debate between thecandidates. The day after the debate, the party’s polling staff calls 1,200 randomly
selected phone numbers. If a registered voter answers the phone or is available to cometo the phone that registered voter is asked whom he or she intends to vote for andwhether the debate changed his or her opinion of the candidates.
48. A “random survey” was conducted of 3,274 people of the “microprocessor generation”(people born since 1971, the year the microprocessor was invented). It was reported that
48% of those individuals surveyed stated that if they had $2,000 to spend, they would use it
for computer equipment. Also, 66% of those surveyed considered themselves
relatively savvy computer users.
a. Do you consider the sample size large enough for a study of this type? Why or why
not?
b. Based on your “gut feeling,” do you believe the percentages accurately reflect the
U.S. population for those individuals born since 1971? If not, do you think thepercentage of the population is actually higher or lower than the sample statistics?
Why?
Additional information: The survey, reported by Intel Corporation, was filled out by
individuals who visited the Los Angeles Convention Center to see the Smithsonian
Institute's road show called “America’s Smithsonian.”c. With this additional information, do you feel that all demographic and ethnic groups
were equally represented at the event? Why or why not?
d. With the additional information, comment on how accurately you think the sample
statistics reflect the population parameters.
49. The Gallup-Healthways Well-Being Index is a survey that follows trends of U.S.
residents on a regular basis. There are six areas of health and wellness covered in thesurvey: Life Evaluation, Emotional Health, Physical Health, Healthy Behavior, Work
Environment, and Basic Access. Some of the questions used to measure the Index are listed
below. Identify the type of data obtained from each question used in this survey
as qualitative, quantitative discrete, or quantitative continuous.
a. Do you have any health problems that prevent you from doing any of the things
people your age can normally do?

34

b. During the past 30 days, for about how many days did poor health keep you fromdoing your usual activities?
c. In the last seven days, on how many days did you exercise for 30 minutes or more?
d. Do you have health insurance coverage?
50. In advance of the 1936 presidential election, a magazine titled Literary Digest released
the results of an opinion poll predicting that the Republican candidate Alf Landon would
win by a large margin. The magazine sent post cards to approximately 10,000,000
prospective voters. These prospective voters were selected from the subscription list of themagazine, from automobile registration lists, from phone lists, and from club membershiplists. Approximately 2,300,000 people returned the postcards.
a. Think about the state of the United States in 1936. Explain why a sample chosenfrom magazine subscription lists, automobile registration lists, phone books, and clubmembership lists was not representative of the population of the United States at thattime.
b. What effect does the low response rate have on the reliability of the sample?
c. Are these problems examples of sampling error or non-sampling error?
d. During the same year, George Gallup conducted his own poll of 30,000 prospective
voters. His researchers used a method they called "quota sampling" to obtain survey
answers from specific subsets of the population. Quota sampling is an example of whichsampling method described in this module?
51. Crime-related and demographic statistics for 47 US states in 1960 were collected fromgovernment agencies, including the FBI's Uniform Crime Report. One analysis of this data
found a strong connection between education and crime indicating that higher levels of
education in a community correspond to higher crime rates. Which of the potential
problems with samples discussed in Section 1.2 could explain this connection?
52. YouPolls is a website that allows anyone to create and respond to polls. One question
posted April 15 asks: “Do you feel happy paying your taxes when members of the Obamaadministration are allowed to ignore their tax liabilities?”[5] As of April 25, 11 people
responded to this question. Each participant answered “NO!” Which of the potential
problems with samples discussed in this module could explain this connection?
53. A scholarly article about response rates begins with the following quote: “Declining
contact and cooperation rates in random digit dial (RDD) national telephone surveys raise
serious concerns about the validity of estimates drawn from such research.”[6] The Pew
Research Center for People and the Press admits: “The percentage of people we interview –out of all we try to interview – has been declining over the past decade or more.” [7]a. What are some possible reasons for the decline in response rate over the past
decade?
b. Explain why researchers are concerned with the impact of the declining response
rate of public opinion polls.

35

REFERENCES
1.1 Basic Definitions
Data and Story
Library, http://lib.stat.cmu.edu/DASL/Stories/CrashTestDummies.html (accessed May 1, 2013).

1.2 Data and Sampling
Gallup-Healthways Well-Being Index. http://www.well-beingindex.com/default.asp (accessed May 1,2013). Gallup-Healthways Well-Being Index. http://www.wellbeingindex.com/methodology.asp (accessed May 1, 2013).
Gallup-Healthways Well-Being Index. http://www.gallup.com/poll/146822/gallup-healthwaysindex-questions.aspx (accessed May 1, 2013).
Data from http://www.bookofodds.com/Relationships-Society/Articles/A0374-How-George-GallupPicked-the-President
Dominic Lusinchi, “’President’ Landon and the 1936 Literary Digest Poll:
Were Automobile and Telephone Owners to Blame?” Social Science History 36, no. 1: 23-54(2012), http://ssh.dukejournals.org/content/36/1/23.abstract (accessed May1, 2013).
“The Literary Digest Poll,” Virtual Laboratories in Probability andStatistics http://www.math.uah.edu/stat/data/ LiteraryDigest.html (accessed May 1, 2013).
“Gallup Presidential Election Trial-Heat Trends, 1936–2008,” Gallup
Politics http://www.gallup.com/poll/110548/gallup- presidential-election-trialheat-trends19362004.aspx#4 (accessed May 1, 2013).
The Data and Story Library, http://lib.stat.cmu.edu/DASL/Datafiles/USCrime.html (accessed May 1,2013).
LBCC Distance Learning (DL) program data in 2010-2011, http://de.lbcc.edu/reports/201011/future/highlights.html#focus (accessed May 1, 2013). Data from San Jose Mercury News

1.3 Experimental Design and Ethics“Vitamin E and Health,” Nutrition Source, Harvard School of Public
Health, http://www.hsph.harvard.edu/nutritionsource/ vitamin-e/ (accessed May 1, 2013).
Stan Reents. “Don’t Underestimate the Power of Suggestion,”
athleteinme.com, http://www.athleteinme.com/ ArticleView.aspx?id=1053 (accessed May 1, 2013).
Ankita Mehta. “Daily Dose of Aspiring Helps Reduce Heart Attacks: Study,” International Business
Times, July 21, 2011. Also available online at http://www.ibtimes.com/daily-dose-aspirin-helpsreduce-heart-attacks-study-300443 (accessed May 1, 2013).
The Data and Story
Library, http://lib.stat.cmu.edu/DASL/Stories/ScentsandLearning.html (accessed May 1, 2013).
M.L. Jacskon et al., “Cognitive Components of Simulated Driving Performance: Sleep Loss effect andPredictors,” Accident Analysis and Prevention Journal, Jan no. 50(2013), http://www.ncbi.nlm.nih.gov/pubmed/22721550 (accessed May 1, 2013).

36

“Earthquake Information by Year,” U.S. Geological
Survey. http://earthquake.usgs.gov/earthquakes/eqarchives/year/ (accessed May 1, 2013).
“Fatality Analysis Report Systems (FARS) Encyclopedia,” National Highway Traffic and Safety
Administration. http://www-fars.nhtsa.dot.gov/Main/index.aspx (accessed May 1, 2013).
Data from www.businessweek.com (accessed May 1, 2013).
Data from www.forbes.com (accessed May 1, 2013).
“America’s Best Small Companies,” http://www.forbes.com/best-small-companies/list/ (accessed May
1, 2013).
U.S. Department of Health and Human Services, Code of Federal Regulations Title 45 Public
Welfare Department of Health and Human Services Part 46 Protection of Human Subjects revised
January 15, 2009. Section 46.111: Criteria for IRB Approval of Research.
“April 2013 Air Travel Consumer Report,” U.S. Department of Transportation, April 11(2013), http://www.dot.gov/ airconsumer/april-2013-air-travel-consumer-report (accessed May 1,2013).
Lori Alden, “Statistics can be Misleading,”
econoclass.com, http://www.econoclass.com/misleadingstats.html (accessed May 1, 2013).
Maria de los A. Medina, “Ethics in Statistics,” Based on “Building an Ethics Module for Business,
Science, and Engineering Students” by Jose A. Cruz-Cruz and WilliamFrey, Connexions, http://cnx.org/content/m15555/latest/ (accessed May 1, 2013).
lastbaldeagle. 2013. On Tax Day, House to Call for Firing Federal Workers Who Owe Back
Taxes. Opinion poll posted online at: http://www.youpolls.com/details.aspx?id=12328 (accessed May
1, 2013).
Scott Keeter et al., “Gauging the Impact of Growing Nonresponse on Estimates from a National RD
D Telephone Survey,” Public OpinionQuarterly70 no. 5 (2006), http://poq.oxfordjournals.org/content/
70/5/759 (accessed May 1, 2013).
Frequently Asked Questions, Pew Research Center for the People & the Press, http://www.peoplepress.org/methodology/frequently-asked-questions/#dont-you-have-trouble-getting-people-to-answeryour-polls (accessed May 1, 2013).

37

CHAPTER 1 SOLUTIONS:
1)AIDS patients; 3) The average length of time (in months) AIDS patients live after
treatment.
5)

X = the length of time (in months) AIDS patients live after treatment;

9)

a;

11)

a. 0.5242

7) b;b. 0.03% c. 6.86% d. 823,088/823,856 e. quantitative discrete

f. quantitative continuous g. In both years, underwater earthquakes produced
massive tsunamis.
13)

systematic; 15) simple random;

17)

There is not enough information given to judge if either one is correct or incorrect.

19)

Yes, because we cannot tell if the improvement was due to the software or the
exercise; the data is confounded, and a reliable conclusion cannot be drawn.

21)

a. all clients for the fitness center b. a smaller, selected group of these clientsc. the population mean number of hours spent each week in the fitness centerd. the sample mean number of hours spent each week in the fitness centere. X = the number of hours spent in the fitness center for a given client in a given
week.
f. values for X, such as 2, 1.7, 3.5, …

23)

a. all patients of the doctor b. a smaller, selected group of these patientsc. the mean recovery period for all patients d. the mean recovery time for the
sample patients e. X = the recovery time of a single patientf. values for X, such as, 9 months, 4.3 months,…

25)

a. all voters in the district b. a smaller, selected group of these votersc. the proportion of all her voters in the district who approve of the politician’s
performance. d. the proportion of the sample who approve of the politician’s job
performance.
e. X = whether or not a voter approves of the politician’s job performance f. yes, no27)

a. all voters in the region (county, state or nation) b. a smaller, selected group ofthese voters c. the proportion of all voters in the region who will vote for the cause.
d. the proportion of the sample who will vote for the cause. e. X = whether or not a
voter will vote for the cause f. yes, no

29) a

31) c

33) quantitative continuous; e.g. 21.2%35) quantitative discrete; e.g. 11,234 students 37) qualitative; e.g. Crest, Colgate

38

39)

quantitative continuous; e.g. 51 yrs, 63.5 yrs

41)

a. The survey was conducted using six similar flights. The survey would not be atrue representation of the entire population of air travelers. Conducting the survey
on a holiday weekend will not produce representative results. b. Conduct the survey
during different times of the year. Conduct the survey using flights to and fromvarious locations. Conduct the survey on different days of the week.

43)

Answers will vary. Sample Answer: You could use a systematic sampling method.
Stop the tenth person as they leave one of the buildings on campus at 9:50 in themorning. Then stop the tenth person as they leave a different building on campus at
1:50 in the afternoon.

45)

Answers will vary. Sample Answer: Many people will not respond to mail surveys. Ifthey do respond to the surveys, you can’t be sure who is responding. In addition,
mailing lists can be incomplete.

47)

a. convenience b. cluster c. stratified d. systematic e. simple random

49)

a. qualitative

51)

Causality: The fact that two variables are related does not guarantee that one
variable is influencing the other. We cannot assume that crime rate impacts
education level or that education level impacts crime rate. Confounding: There aremany factors that define a community other than education level and crime rate.
Communities with high crime rates and high education levels may have otherlurking variables that distinguish them from communities with lower crime rates
and lower education levels. Because we cannot isolate these variables of interest, wecannot draw valid conclusions about the connection between education and crime.
Possible lurking variables include police expenditures, unemployment levels, region,
average age, and size.

53)

a. Possible reasons: increased use of caller id, decreased use of landlines, increased
use of private numbers, voice mail, privacy managers, hectic nature of personalschedules, decreased willingness to be interviewed b. When a large number ofpeople refuse to participate, then the sample may not have the same characteristicsof the population. Perhaps the majority of people willing to participate are doing so
because they feel strongly about the subject of the survey.

b. quantitative discrete c. quantitative discrete d. qualitative

39

This page is purposely left blank.

40

2 | DISPLAYING DATA

Figure 2.1 Photo by Myriam Jessier on Unsplash

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:•••Create and interpret frequency distributions.
Display data graphically and interpret graphs for Qualitative Data (Bar, Pareto, Pie)
Display data graphically and interpret graphs for Quantitative Data(Histogram, Frequency Polygon, and Time Series)

41

Once you have collected data, what will you do with it? Data can be described and presented
in many different formats. For example, suppose you are interested in buying a house in aparticular area. You may have no clue about the house prices, so you might ask your realestate agent to give you a sample data set of prices. Looking at all the prices in the sample
often is overwhelming. A better way might be to look at the median price and the variationof prices. The median and variation are just two ways that you will learn to describe data.
Your agent might also provide you with a graph of the data. In this chapter, you will study
numerical and graphical ways to describe and display your data. This area of statistics iscalled "Descriptive Statistics." You will learn how to calculate, and even more importantly,
how to interpret these measurements and graphs.
A statistical graph is a tool that helps you learn about the shape or distribution of a sample
or a population. A graph can be a more effective way of presenting data than a mass ofnumbers because we can see where data clusters and where there are only a few data values.
Newspapers and the Internet use graphs to show trends and to enable readers to compare
facts and figures quickly. Statisticians often graph data first to get a picture of the data.
Then, more formal tools may be applied.

https://openclipart.org/detail/19980/graphs

Some of the types of graphs that are used to summarize and organize data are the dot plot,
the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken
line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-andleaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs.
Our emphasis will be on histograms and box plots.

422.1 | Frequency Distributions
Once you have a set of data, you will need to organize it so that you can analyze howfrequently each datum occurs in the set. This is done by creating a table of frequencies which
is known as a frequency distribution. There are three frequency distributions to choose
from, depending on if the data is categorical or quantitative. When it is quantitative data,
there are two frequency distributions to choose from. When calculating the frequency, you
may need to round your answers so that they are as precise as possible.

Answers and Rounding Off
A simple way to round off answers is to carry your final answer one more decimal place than
was present in the original data. Round off only the final answer. Do not round off any
intermediate results, if possible. If it becomes necessary to round off intermediate results,
carry them to at least twice as many decimal places as the final answer. For example, theaverage of the three quiz scores four, six, and nine is 6.3, rounded off to the nearest tenth,because the data are whole numbers. Most answers will be rounded off in this manner.

Ungrouped Frequency Distribution
This first type of frequency distribution is for quantitative data sets. It is best when therange of the data is less than 10 units. Range is the difference between the largest data value
and the smallest data value. For example, twenty students were asked how many hours they
worked per day. Their responses, in hours, are as follows:
5; 6; 3; 3; 2; 4; 8; 5; 2; 3; 5; 6; 5; 4; 4; 3; 5; 2; 5; 3.
Since the range is 6, we will keep each data value separate and not group them together. To
create an ungrouped frequency distribution is a simple task. Place the data values fromsmallest to the largest without skipping any values on the first column. Place the frequency,
the count of each data value, in the corresponding row of the second column.
Table 2.1 lists the different data values in ascending order and their frequencies. Notice all
the data values are listed including 7 which is not listed on the original data set.
DATA VALUES FREQUENCY (f)
2
3
3
5
4
3
5
6
6
2
7
0
8
1
Table 2.1 Frequency distribution of Student Work Hours
A frequency is the number of times a value of the data occurs. According to Table 2.1, there
are three students who work two hours, five students who work three hours, and so on. The
sum of the values in the frequency column, 20, represents the total number of students
included in the sample, known as sample size (n).
43A relative frequency is the ratio (fraction or proportion) of the number of times a value ofthe data occurs in the set of all outcomes to the total number of outcomes. To find the relative
frequencies, divide each frequency by the total number of students in the sample, n, in thiscase, 20. Relative frequencies can be written as fractions, decimals, or percent.
The following table is known as a relative frequency distribution because frequencies aretranslated as percent. To find the percent of the first row, take the frequency and divide by
the sample size, n. 3/20 = 15%.
DATA VALUE FREQUENCY (f) RELATIVE FREQUENCY
2
3
15%
3
5
25%
4
3
15%
5
6
30%
6
2
10%
7
0
0%
8
1
5%
Table 2.2 Frequency distribution of Student Work Hours with Relative Frequencies
Cumulative relative frequency is the accumulation of the previous relative frequencies.
To find the cumulative relative frequencies, add all the previous relative frequencies to therelative frequency for the current row, as shown in Table 2.3. The third row undercumulative relative frequency has the number .55 which was found by adding the previous
relative frequencies (.15 + .25 + .15). This represents that 55% of the data has the value of 4
or under.
DATA
VALUE

FREQUENCY RELATIVECUMULATIVE(f)
FREQUENCY RELATIVEFREQUENCY
2
3
15%
.15
3
5
25%
.40
4
3
15%
.55
5
6
30%
.85
6
2
10%
.95
7
0
0%
.95
8
1
5%
1
Table 2.3 Frequency distribution of Student Work Hours with Relative Frequencies

44

Grouped Frequency Distribution
This second type of frequency distribution is also used when there is quantitative data.
However, it is used when the range is large and the data values need to be grouped together.
For example, 28 students were asked how many hours they worked per week. Their
responses, in hours, are as follows:
15; 26; 13; 33; 22; 14; 27; 15; 32; 23; 5; 26; 25; 14; 34; 13; 15; 22; 15; 28; 10; 18; 21; 24; 20; 18;34; 20;Here there are too many different data values to list them separately as in the ungrouped
frequency distribution. Notice the range is 29 (highest – lowest = 34 – 5). Therefore we needto construct a grouped frequency distribution (Table 2.4) to group data values into classes.
A class is an interval where the lowest value of the interval is known as the lower limit andthe highest value of the interval is known as the upper limit.
••••

Guidelines for classes:
There should be between 5 and 20 classes
Classes must be mutually exclusive (no overlap of data values)
Classes must be all inclusive and continuous
Classes must be equal in width

Constructing a Grouped Frequency Distribution:
1. Find Range (highest data value – lowest data value)
2. Determine the number of classes (usually the minimum is 5 classes and a maximum of
20 classes)
3. Find Class Width =NOTE: Round up

4. Choose first lower limit (usually the lowest data value)
5. Create the other lower limits of the classes by adding the class width to the previous
lower limit
6. Create the upper limits by not overlapping the limits

45

Example 2.1
Twenty-eight students were asked how many hours they worked per week. Their responses,
in hours, are as follows: 15; 26; 13; 33; 22; 14; 27; 15; 32; 23; 5; 26; 25; 14; 34; 13; 15; 22; 15;28; 10; 18; 21; 24; 20; 18; 34; 20; construct a grouped frequency distribution using 5 classes.
5; 10; 13; 13; 14; 14; 15; 15; 15; 15; 18; 18; 20; 20; 21; 22; 22; 23; 24; 25; 26; 26; 27; 28; 32; 33;
34;
1.

Range = 34 – 5 = 292.

Use 5 classes (given in Exercise 2.1)

3.

Class Width = 29/5 = 5.8 round up to 6

4.

First lower limit will be 5 which is the minimum data value

5.

The other lower limits will be 11, 17, 23, 29 by adding the class width of 6 to theprevious lower limit

6.

The first upper limit will be 10 since the next class begins at 11. Using class width
again, the other upper limits are 16, 22, 28, 34

Solution 2.1
CLASSES FREQUENCY (f)5 – 102
11 – 168
17 – 227
23 – 287
29 – 34
4
Frequency distribution of Student Work Hours.

2.1 The grade point averages for 40 students are listed below.

Construct a frequency distribution, a relative frequency distribution, and acumulative frequency distribution using eight classes. Include the midpoints of theclasses.

46

Solution ‘Try It’ 2.1:
Range = 4 - .5 = 3.5
Class width = 3.5/8 = .4375 Round up to .5 since the data values are in tenths (one decimalspot) then we round the class width to tenths.
CLASSES FREQUENCY (f).5 - .9
4
1 – 1.42
1.5 – 1.9
7
2 – 2.4
9
2.5 – 2.9
2
3-3.4
10
3.5-3.9
2
4-4.4
4
Table 2.5 Frequency distribution of GPA

Frequency Distribution for Qualitative Data
Qualitative data is non-numeric data. Therefore, when creating a table of data values, place
the category in the first column and the count of each data value, frequency, in the second
column. For example, twenty students are asked their blood type. Their responses are as
follows:
A; B; O; A; AB; O; O; A; O; B; A; A; A; O; O; O; B; O; AB; B
Table 2.5 lists the different data values and their frequencies.
DATA VALUE FREQUENCY (f)
A
6
AB
2
B
4
O
8
Table 2.5 Frequency distribution of Student Blood Types
Frequency distributions (tables) are a good way of organizing and displaying data. Butgraphs can be even more helpful in understanding the data

47

The TI-83/TI-84 Calculator can sort the data values
Press STAT
Choose EDIT

Once you have enter the data into L1,
2nd QUIT
Press STAT

Choose #2 SortA(
Press 2nd L1

To view the
STAT, EDIT

sortedlist,

returnto48

2.2 | Graphs for Qualitative Data
There are no strict rules concerning which graphs to use. However, scale of the bars or slices
should be accurate. For example, if a category is 32% then the slice for the category should
not be smaller than a quarter of the circle or bigger than half the circle. Three graphs thatare used to display qualitative data are pie charts, bar graphs, and Pareto charts.
In a pie chart, categories of data are represented by wedges in a circle and are proportionalin size to the percent of individuals in each category. When creating a pie chart, each slice
should be labeled with the category name and the relative frequency (percent).
In a bar graph, the length of the bar for each category is proportional to the number orpercent of individuals in each category. Bars may be vertical or horizontal. For verticalbars, the categories are on the x-axis and frequency or relative frequency are on the y-axis.
In a Pareto chart consists of bars that are sorted into order by category size (largest to
smallest). Pareto charts are typically used with nominal data. If a Pareto chart were createdfrom ordinal or quantitative data, the values on the x-axis might seem out of order after thebars were rearranged from largest to smallest.

Pie Graphs:
Example 2.2
Below is a table comparing the number of part-time and full-time students at De Anza College
and Foothill College enrolled for the spring 2010 quarter. The tables display counts(frequencies) and percentages or proportions (relative frequencies).

Full-timePart-time
Total

De Anza College
frequency
relative frequency
9,200
40.9%
13,29659.1%
22,496100%

Full-timePart-time
Total

Table 2.6 Fall Term 2007 (Census day)

49

Foothill College
frequency
relative frequency
4,059
28.6%
10,12471.4%
14,183100%

Solution 2.2

Figure 2.1 Pie Graphs

Pie Charts: No Missing Data
The following pie charts have the “Other/Unknown” category included (since thepercentages must add to 100%). The chart in Figure 2.7b is organized by the size of eachwedge, which makes it a more visually informative graph than the unsorted, alphabeticalgraph in Figure 2.7a.

Figure 2.7 a Figure 2.7 b

50

Bar Graphs:
Below is the same data from Example 2.2 displayed as a bar graph.

Figure 2.2 Bar Graph

Look at Figure 2.1 and Figure 2.2 and determine which graph (pie or bar) you thinkdisplays the comparisons better. It is a good idea to look at a variety of graphs to see whichis the most helpful in displaying the data. We might make different choices of what we thinkis the “best” graph depending on the data and the context. Our choice also depends on what
we are using the data for.
Percentages That Add to More (or Less) Than 100%
Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, thepercentages add to more than100% because students can be in more than one category. A bar
graph is appropriate to compare the relative size of the categories. A pie chart cannot be
used. It also could not be used if the percentages added to less than 100%.
Characteristic/Category
Full-Time Students
Students who intend to transfer to a 4-year
educational institution
Students under age 25
Total

Percent
40.9%
48.6%
61%
150.5%

Table 2.7 De Anza College Spring 201051

Figure 2.3 Bar Graph

Omitting Categories/Missing Data
The table displays Ethnicity of Students but is missing the "Other/Unknown” category. This
category contains people who did not feel they fit into any of the ethnicity categories or
declined to respond. Notice that the frequencies do not add up to the total number ofstudents. In this situation, create a bar graph and not a pie chart.
Asian
Black
Filipino
Hispanic
Native American
Pacific Islander
White
TOTAL

Frequency
8,794
1,412
1,298
4,180
146
236
5,978
22,044 out of 24,382

Percent
36.1%
5.8%
5.3%
17.1%
0.6%
1.0%
24.5%
90.4% out of 100%

Table 2.8 Ethnicity of Students at De Anza College Fall Term 2007 (Census Day)

52

Figure 2.4 Bar Graph

The following graph is the same as the previous graph but the “Other/Unknown” percent(9.6%) has been included. The “Other/Unknown” category is large compared to some of theother categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know
when we think about what the data are telling us. This particular bar graph in Figure
2.5 can be difficult to understand visually.

Figure 2.5 Bar Graph with Other/Unknown Category53

Pareto Charts:
As the above data may be more difficult to interpret in a bar graph, a Pareto chart is easierto read and interpret. The graph in Figure 2.6 is a Pareto chart.

Figure 2.6 Pareto Chart

Example 2.3
Twenty-five students are asked their blood type. Their responses are as follows:
A; B; O; A; AB; O; O; A; O; B; A; A; A; O; O; O; B; O; AB; B, O, B, O, A, A
DATA VALUE
A
AB
B
O

FREQUENCY (f)
8
2
5
10

Create a Pareto Chart.
Solution 2.3
Blood type O has the largest frequency therefore it should be placed first on the x-axis andblood type A follows.

542.3 | Graphs for Quantitative Data
Histogram:

For most of the work you do in this book, you will use a histogram to display the data. One
advantage of a histogram is that it can readily display large data sets. A rule of thumb is touse a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) bars. It has both a horizontal axis and a
vertical axis. The horizontal axis is labeled with what the data represents (for instance,
distance from your home to school). Horizontal axis uses the class boundaries. The verticalaxis is labeled either frequency or relative frequency (percent or probability). The graph will
have the same shape with either label. The histogram can give you the shape of the data, thecenter, and the spread of the data.
Below is a histogram of the number of new coronavirus disease (Covid-19) cases reported
daily*† (N = 4,226) – United States, February 12 – March 16, 2020 from the CDC(https://www.cdc.gov/mmwr/volumes/69/wr/mm6912e2.htm)* Includes both COVID-19 cases confirmed by state or local public health laboratories, as well as those testing positive at thestate or local public health laboratories and confirmed at CDC.
† Cases identified before February 28 were aggregated and reported during March 1–3.

Notice that there are no gaps between the bars and that date is in the middle of each bar.
The class boundaries are where the bars meet. The horizontal axis may also be labeled by
the class boundaries.

55

The relative frequency is equal to the frequency for an observed value of the data divided
by the total number of data values in the sample. (Remember, frequency is defined as thenumber of times an answer occurs.)• f = frequency• n = total number of data values (or the sum of the individual frequencies)• RF = relative frequency

RF =For example, if three students in Mr. Ahab's English class of 40 students received from 90%
to 100%, then, f = 3, n = 40, and RF = 3/40 = .075 = 7.5% of the students received 90–100%.
To construct a histogram:
1. Create class boundaries on the grouped frequency distribution. Choose a starting point
for the first interval to be less than the smallest data value. A convenient starting point is alower value carried out to one more decimal place than the value with the most decimal
places. For example, if the value with the most decimal places is 6.1 and this is the smallest
value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more
precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, aconvenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal
places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005= 0.9995). If all the data happen to be integers and the smallest value is two, then aconvenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and otherboundaries are carried to one additional decimal place, no data value will fall on a boundary.
2. Place frequency or relative frequency on the y-axis. Scale is important.
3. Draw bars as high as the frequency for each class interval within each boundary
The next two examples go into detail about how to construct a histogram using continuous
data and how to create a histogram using discrete data.

Example 2.4
The following data are the heights (in inches to the nearest half inch) of 100 malesemiprofessional soccer players. The heights are continuous data, since height is measured.
60; 60.5; 61; 61; 61.5; 63.5; 63.5; 63.5; 64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5;
64.5; 64.5; 64.5; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5;
66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5;
67.5; 67.5; 68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5; 70; 70; 70;
70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71; 72; 72; 72; 72.5; 72.5; 73; 73.5; 74

56

Create a relative frequency histogram using the frequency distribution that follows:
CLASSES FREQUENCY (f)60.0 – 61.9
562.0 – 63.9
364.0 – 65.9
1566.0 – 67.9
40
68.0 – 69.9
1770.0 – 71.9
1272.0 – 73.9
774.0 – 75.9
1
Table 2.8 Frequency distribution of Heights
Solution 2.4
First find the relative frequencies and class boundaries for each class.
BOUNDARIES RELATIVE FREQUENCY
59.95 – 61.95.05
61.95 – 63.95.03
63.95 – 65.95.1565.95 – 67.95.4
67.95 – 69.95.17
69.95– 71.95.12
71.95 – 73.95.07
73.95 – 75.95.01
Second place the boundaries on the x-axis and the relative frequency on the y-axis.

Figure 2.8
57

Example 2.5
The following data are the number of books bought by 50 part-time college students at theCollege of Lake County. The number of books is discrete data, since books are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3
4; 4; 4; 4; 4; 4
5; 5; 5; 5; 5
6; 6;Construct a Histogram.
BOOKS
FREQUENCY (f)
1
11
2
10
3
16
4
6
5
5
6
2
Table 2.9 Ungrouped Frequency distribution of Books
Solution 2.5

BOUNDARIES.5 – 1.5
1.5 – 2.5
2.5 – 3.5
3.5 – 4.5
4.5 – 5.5
5.5 – 6.5

FREQUENCY
11
10
16
6
5
2

58

2.2 The following data are the shoe sizes of 50 male students. The sizes arecontinuous data since shoe size is measured. Construct a histogram and calculate
the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11;
11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 12; 12; 12;
12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14

Using the TI-83, 83+, 84, 84+ Calculator

•
•
•
•

•

Go to Appendix G.
There are calculator instructions for entering data and for creating a customized histogram.
Create the histogram for Example 2.5.
Press Y=. Press CLEAR to delete any equations.
Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR andthen arrow down. If necessary, do the same for L2.
Into L1, enter 1, 2, 3, 4, 5, 6.
Into L2, enter 11, 10, 16, 6, 5, 2.

Press WINDOW. Set Xmin = .5, Xscl = (6.5 – .5)/6, Ymin = –1, Ymax = 20, Yscl =1, Xres = 1.

59

•
•

Press 2nd Y=. Start by pressing 4: Plotsoff ENTER.
Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3rd picture(histogram). Press ENTER.

•

Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2 (2nd 2). Press Graph.
Use TRACE key and the arrow keys to examine the histogram.

Descriptions of Histograms:

There are four descriptions of histograms. These descriptions, describe how the histogram isshaped.
1.

Symmetrical: the largest bar is in the center interval and the bars on each side of the centerdecrease about the same rate.

2. Skewed Left: the largest bar is closer to the right side and the bars start decreasing toward
the left.

60

3. Skewed Right: the largest bar is closer to the left side and the bars start decreasing toward
the right.

4. Uniform: each bar of the histogram is the same size for every interval

NOTE: Histograms will not always be exactly one of the four descriptions; therefore use
adjectives to help describe the histogram. For example, “mostly symmetrical”.

Stem-and-Leaf Graph:

One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory
data analysis. It is a good choice when the data sets are small.
To create the stem-and-leaf plot:
1. Divide each observation of data into a stem and a leaf. The leaf consists of a final
significant digit.
For example:
The number 23 has stem two and leaf three. The number 432 has stem 43 and leaf two.
Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine andleaf three.
2. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the rightof the stems. Then write the leaves in increasing order next to their corresponding stem.

2.6
61

Example 2.6
For Susan Dean's spring pre-calculus class, scores for the first exam were as follows:
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90;
92; 94; 94; 94; 94; 96; 100
Create a Stem and leaf plot.
Solution 2.6:
Stem Leaf
33
4299
5355
61378899
72348
80388890244446
10 0
Table 2.9 Stem-and- Leaf Graph
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31scores or approximately 26% were in the 90s or 100, a fairly high number of A’s.

2.3 For the Park City basketball team, scores for the last 30 games were as follows(smallest to largest):
32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53;
54; 56; 57; 57; 60; 61
Construct a stem plot for the data.
The stem and leaf plot is a quick way to graph data and gives an exact picture of the data.
You want to look for an overall pattern and any outliers. An outlier is an observation ofdata that does not fit the rest of the data. It is sometimes called an extreme value. Whenyou graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due
to mistakes (for example, writing down 50 instead of 500) while others may indicate thatsomething unusual is happening. It takes some background information to explain outliers,
so we will cover them in more detail later.

62

Example 2.7
The data are the distances (in km) from a home to local supermarkets. Create
a stemplot using the data:
1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5;
6.7; 12.3
Do the data seem to have any concentration of values? The leaves are to the right of thedecimal.
Solution 2.7
The value 12.3 may be an outlier. Values appear to concentrate at three and four
kilometers.
Stem Leaf
115
2357
3233584025578
556
657
7
8
9
10
11
12 3
Table 2.10

2.4 The following data show the distances (in miles) from the homes of off-campus
statistics students to the college. Create a stem plot using the data and identify any
outliers:
0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5;
3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0
Example 2.8
A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns.
In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves areto the left and the right of the stems. Table 2.11 and Table 2.12 show the ages of presidents
at their inauguration and at their death. Construct a side- by-side stem-and-leaf plot using
this data.

63

President
Washington
J. Adams
Jefferson
MadisonMonroeJ.Q. Adams
JacksonVan Buren
W. H. Harrison
Tyler
Polk
TaylorFillmore
PierceBuchanan
Lincoln

Age
57
61
57
57
58
57
61
54
68
51
49
64
50
48
65
52

President
A. Johnson
Grant
Hayes
Garfield
ArthurCleveland
B. Harrison
Cleveland
McKinley
T. Roosevelt
Taft
WilsonHarding
Coolidge
HooverF. Roosevelt

Age
56
46
54
49
51
47
55
55
54
42
51
56
55
51
54
51

President
TrumanEisenhower
Kennedy
L. Johnson
Nixon
Ford
CarterReaganG.H.W Bush
Clinton
G.W. Bush
Obama
Trump

Age
60
62
43
55
56
61
52
69
64
47
54
47
70
78

President
Washington
J. Adams
Jefferson
MadisonMonroeJ.Q. Adams
JacksonVan Buren
W. H. Harrison
Tyler
Polk
TaylorFillmore
PierceBuchanan
Lincoln

Age
67
90
83
85
73
80
78
79
68
71
53
65
50
64
77
56

President
A. Johnson
Grant
Hayes
Garfield
ArthurCleveland
B. Harrison
Cleveland
McKinley
T. Roosevelt
Taft
WilsonHarding
Coolidge
HooverF. Roosevelt

Age
66
63
70
49
56
71
67
71
58
60
72
67
57
60
90
63

President
TrumanEisenhower
Kennedy
L. Johnson
Nixon
Ford
CarterReaganG.H.W Bush
Clinton
G.W. Bush
Obama
Trump

Age
88
78
46
64
81
93

Table 2.11 Presidential Ages at Inauguration

Table 2.12 Presidential Ages at Death

Solution 2.8

Biden

Biden

2.3Ages at Inauguration Stem Ages at Death

998777632
877776665555444442111110
954421110
80

64

4
5
6
7
8
9

69
366778003344567778
0011147889
01358
00334

93
94

Line Graphs:
Another type of graph that is useful for specific data values is a line graph. In theparticular line graph shown in Example 2.9 is for an ungrouped frequency distribution,
the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists offrequency points. The frequency points are connected using line segments.
Example 2.9
In a survey, 40 mothers were asked how many times per week a teenager must be reminded
to do his or her chores. The results are shown in Table 2.13 and in Figure 2.10.
# of times
0
1
2
3
4
5
Table 2.13

FREQUENCY (f)
2
5
8
14
7
4

Solution 2.9:.6

Figure 2.10

2.5 In a survey, 40 people were asked how many times per year they had their

car in the shop for repairs. The results are shown in Table 2.14. Construct a linegraph.
# of times
FREQUENCY (f)
0
7
1
10
2
14
3
9

65

Frequency Polygon:
Frequency polygons are analogous to line graphs, and just as line
make continuous data visually easy to interpret, so too do frequency polygons.

graphsTo construct a frequency (relative frequency) polygon:
1. Find the midpoints of each class and place them on the x-axis
2. Place frequency or relative frequency on the y-axis
3. Draw a line graph corresponding to each midpoint and frequency
Example 2.10
Construct a frequency polygon using the distribution of Statistic Final Test Scores
CLASSES FREQUENCY (f)50 - 595
60 - 6910
70 - 7930
80 - 8940
90 - 99
15
Solution 2.10:
First find the midpoints of each class. Midpoints: 54.5, 64.6, 74.5, 84.5, 94.5

The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5.
Since the lowest test score is 54.5, this interval is used only to allow the graph to touch thex-axis. The point labeled 54.5 represents the next interval or the first “real” interval from thetable, and contains five scores. This reasoning is followed for each of the remaining intervals
with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval
contains no data and is only used so that the graph will touch the x-axis. Looking at thegraph, we say that this distribution is skewed because one side of the graph does not mirror
the other side.
66

2.6 Construct a frequency polygon using the ages of President’s inauguration in thefollowing table.

Classes (Ages)
42 - 46
47 - 51
52 - 56
57 - 6162 - 66
67 - 71

Frequency (f)
4
11
14
9
4
2

Frequency polygons are useful for comparing distributions. This is achieved by overlaying
the frequency polygons drawn for different data sets.
Example 2.11
We will construct an overlay frequency polygon comparing the scores from Example 2.8
with the students’ final numeric grade.
CLASSES50 - 59
60 - 69
70 - 79
80 - 89
90 - 99

FREQUENCY (f)
10
10
30
45
5

Solution 2.11:

67

Suppose that we want to study the temperature range of a region for an entire month. Every
day at noon we note the temperature and write this down in a log. A variety of statisticalstudies could be done with this data. We could find the mean or the median temperature forthe month. We could construct a histogram displaying the number of days thosetemperatures reach a certain range of values. However, all of these methods ignore a portionof the data that we have collected.
One feature of the data that we may want to consider is that of time. Since each date is pairedwith the temperature reading for the day, we don‘t have to think of the data as being
random. We can instead use the times given to impose a chronological order on the data. A
graph that recognizes this ordering and displays the changing temperature as the monthprogresses is called a time series graph.

Constructing a Time Series Graph

To construct a time series graph, we must look at both pieces of our paired data set. We startwith a standard Cartesian coordinate system. The horizontal axis is used to plot the date ortime increments, and the vertical axis is used to plot the values of the variable that we aremeasuring. By doing this, we make each point on the graph correspond to a date and ameasured quantity. The points on the graph are typically connected by straight lines in theorder in which they occur.
Example 2.12
The following data shows the Annual Consumer Price Index, each month, for ten years.
Construct a time series graph for the Annual Consumer Price Index data only.
Year
2003 2004 2005 2006 2007
2008
2009
2010
2011
2012
Annual 184 188.9 195.3 201.6 207.342 215.303 214.537 218.056 224.939 229.594
Solution 2.12:

68

2.7 The following table is a portion of a data set from www.worldbank.org. Use the table toconstruct a time series graph for CO2 emissions for the United States.
CO2 Emissions
Ukraine United
Kingdom2003 352,259 540,640
2004 343,121 540,409
2005 339,029 541,990
2006 327,797 542,045
2007 328,357 528,631
2008 323,657 522,247
2009 272,176 474,579

United States
5,681,664
5,790,761
5,826,394
5,737,615
5,828,627
5,656,839
5,299,563

Uses of a Time Series Graph

Time series graphs are important tools in various applications of statistics. When recording
values of the same variable over an extended period of time, sometimes it is difficult to discern
any trend or pattern. However, once the same data points are displayed graphically, somefeatures jump out. Time series graphs make trends easy to spot.

69

KEY TERMSBar Graphs a representation of categorical data where horizontal axis contains thequalitative categories, and the vertical axis represents the frequency of each category. The
height of the bars represents the amount of data in the category.
Frequency Polygon looks like a line graph but uses intervals to display ranges of large
amounts of data.
Frequency Table a data representation in which grouped data is displayed along with thecorresponding frequencies.
Frequency the number of times a value of the data occurs.
Histogram a graphical representation in of the distribution of data in a data set; thehorizontal axis represents the data, and the vertical axis represents the frequencies orrelative frequencies. The graph consists of contiguous rectangles.
Interval also called a class interval; an interval represents a range of data and is used
when displaying large data sets.
Paired Data Set two data sets that have a one-to-one relationship so that:•both data sets are the same size, and•each data point in one data set is matched with exactly one point from the other set.
Pie Graphs a circular graph that shows how large each category is in relation to thewhole. Created from the frequency distribution and using relative frequencies, the central
angles are calculated by multiplying the relative frequency by 360 degrees.
Qualitative Data categorical data consisting of labels or descriptions.
Quantitative Continuous Data measurable data that is numerical and can take on any
value in a given range of numbers.
Quantitative Discrete Data measurable data that is numerical that can only take onparticular values and cannot take on the values in between.
Relative Frequency the ratio of the number of times a value of the data occurs in the setof all outcomes to the number of all outcomes.
Time Series the same variable is measured over a period and plotted on the x-y coordinate
plane where time is on the x-axis and the variable in the y-axis.

70

EXERCISES for CHAPTER 2
For exercises #1 – 4, use the following frequency distribution
Number of times in store
1
2
3
4
5

Frequency
4
10
16
6
4

1. Find the boundaries for the second class.
2. Find the relative frequency for the third class.
3. What is the sample size?
4. What is the cumulative frequency for the fourth class?
5. The following data represent the number of potholes on 35 randomly selected1-mile
stretches of highway around a particular city. Construct an ungrouped frequency
distribution of the data.
143147513612
111716271644
11535232413
For exercises #6 – 11, use the following frequency distribution
Temperature (°F)
32 - 36
37 - 41
42 - 46
47 - 51
52 - 56
57 – 6162 - 66

Frequency
1
3
5
11
7
7

1

6. Find the class width.
7. Find the class midpoints.
8. Find the class boundaries.
9. Find the relative frequencies.
10. Find the cumulative frequencies.

7111. Find the sample size.
For exercises #12 – 17, use the following frequency distribution
Time (minutes)
Frequency
70 - 7814
79 - 8715
88 - 9616
97 - 10518
106 - 11422
115 - 1237
124 - 1323
12. Find the class width.
13. Find the class midpoints.
14. Find the class boundaries.
15. Find the relative frequencies.
16. Find the cumulative frequencies.
17. Find the sample size.
18. Use the given minimum and maximum data entries, and the number of classes, to findthe class width, the lower class limits, and the upper class limits.
a. minimum = 12, maximum = 68, 6 classesb. minimum = 28, maximum = 75, 6 classesc. minimum = 5, maximum = 92, 8 classes
d. minimum = 30, maximum = 83, 8 classes
19. The following data represents the number of miles a full tank of gas allows for a sample
of vehicles inspected. Create a grouped frequency distribution with 7 classes.
260 271 236 244 279 296 284 299 288
288 247 256 338 360 341 333 261 266
287 296 313 311 307 307 299 303 277
283 304 305 288 290 288 289 297 299
332 330 309 328 307 328 285 291 295
298 306 315 310 318 318 320 333 321
323 324 327
20. The data represent the time, in minutes, spent reading a political blog in a day.
Construct a frequency distribution using 5 classes. In the table, include the midpoints,
relative frequencies, and cumulative frequencies. Which class has the greatest frequency
and which has the least frequency?
34 18 10 44 41 47 0 33 49 45
16 10 27 24 44 46 21 25 45 2372

21. Construct a frequency distribution for the given data set using 6 classes. In the table,
include the midpoints, relative frequencies, and cumulative frequencies. Which class has
the greatest frequency and which has the least frequency?
Amount (in dollars) spent on books for a semester
508 488 430 49 278 542 451 366 451 531 196 115
118 362 434 399 148 30 287 236 185 210 178 105
533 236 477 329
22. The students in Ms. Ramirez’s math class have birthdays in each of the four
seasons. The table below shows the four seasons, the number of students who havebirthdays in each season, and the percentage (%) of students in each group. Construct a bar
graph showing the number of students.
Seasons Number ofStudents
Spring8
Summer9
Autumn11
Winter6

Percentage
24%
26%
32%
18%

23. Using the data from Mrs. Ramirez’s math class supplied in Exercise 22, construct a pie
graph showing the percentages.
24. David County has six high schools. Each school sent students to participate in a countywide science competition. The table below shows the percentage breakdown of competitorsfrom each school, and the percentage of the entire student population of the county that
goes to each school. Construct a bar graph and pie graph that shows the
population percentage of competitors from each school.
High
Science Competition
Overall Student Population
SchoolPercentage
Percentage
Alabaster
28.9%
8.6%
Concordia
7.6%
23.2%
Genoa
12.1%
15.0%
Mocksville
18.5%
14.3%
Tynneson
24.2%
10.1%
West End
8.7%
28.8%
25. Use the data from the David County science competition supplied in 24. Construct a bar
graph that shows population percentage of students at each school.

73

For exercises #26 - 29, create a stem/leaf plot for the given data.
26. The miles per gallon rating for 30 cars are shown below (lowest to highest).
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31,
32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
27. The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39,
40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
28. The following data are the prices of different laptops at an electronics store. Round eachvalue to the nearest multiple of ten.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350,
350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
29. The following data are the daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71,
72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95
30. In a survey, 40 people were asked how many times they visited a store before making amajor purchase. Construct a frequency polygon. The results are shown in the table:
Number of times in
store
1
2
3
4
5

Frequency
4
10
16
6
4

31. In a survey, several people were asked how many years it has been since theypurchased a mattress. Construct a histogram. The results are shown below.
Years Since Last Purchase
0
1
2
3
4
5

Frequency
2
8
13
22
16
9

74

32. Several children were asked how many TV shows they watch each day. Construct a
relative frequency polygon. The results of the survey are shown in the table below:
Number of TV Shows
0
1
2
3
4

Frequency
12
18
36
7
2

33. Sixty-five randomly selected car salespersons were asked the number of cars theygenerally sell in one week. Fourteen people answered that they generally sell three cars;
nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars;
eleven generally sell seven cars. Complete the table.
Data Values (# cars)

Frequency

Relative Frequency

Cumulative
Relative Frequency

34. Refer to the table for problem 33. Construct a histogram for the given data.
35. Construct a frequency polygon and histogram for each of the following frequency tables:a.
b.
Pulse Rates for Women Frequency
Actual Speed in a Frequency
60 - 691230 mph Zone
70 - 7914
42 – 4525
80 – 8911
46 – 49
14
90 – 991
50 – 537
100 – 1091
54 – 573
110 – 1190
58 – 611
120 – 1291c.
Tar (in mg) of Nonfiltered Cigarettes
10 – 13
14 – 17
18 – 21
22 – 25
26 – 29

Frequency
1
0
15
7
2

75

36. Construct a relative frequency polygon and relative frequency histogram from thefrequency distribution for the 50 highest ranked countries for depth of hunger.
Depth of Hunger
230 - 259
260 - 289
290 - 319
320 - 349
350 - 379
380 - 409
410 - 439

Frequency
21
13
6
7
1
1
1

37. Use the two frequency tables to compare the life expectancy of men and women from 20
randomly selected countries. Include an overlaid frequency polygon and discuss the shapesof the distributions, the center, the spread, and any outliers. What can we conclude about
the life expectancy of women compared to men?
Life Expectancy at Birth – Women
49 – 55
56 – 62
63 – 69
70 – 76
77 – 83
84 - 90
Life Expectancy at Birth – Men
49 – 55
56 – 62
63 – 69
70 – 76
77 – 83
84 - 90

Frequency
3
3
1
3
8
2
Frequency
3
3
1
1
7
5

38. Construct a times series graph for (a) the number of male births, (b) the number offemale births, and (c) the total number of births.
Sex/Yr 1855
Female 49,545
Male
47,804
Total 93,349

1856
1857
1858
1859
1860
1861
49,582 50,257 50,234 51,915 51,220 52,403
52,239 53,158 53,694 54,628 54,409 54,606
101,821 103,415 104,018 106,543 105,629 107,009

39. The following data sets list full time police per 100,000 citizens along with homicides
per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1972.
Year
1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972
Police260.4 269.8 272 273 272.5 261.3 268.9 296 319.9 341.4 356.6 376.7
Homicides 8.6
8.9
8.5
8.9 13.1 14.6 21.4 28
31.5 37.4 46.3 47.2

a. Construct a double time series graph using a common x-axis for both sets of data.
b. Which variable increased the fastest? Explain.
c. Did Detroit’s increase in police officers have an impact on the murder rate? Explain.

76

REFERENCES

2.2 Graphs for Qualitative Data

Burbary, Ken. Facebook Demographics Revisited – 2001 Statistics, 2011. Available online
at http://www.kenburbary.com/
2011/03/facebook-demographics-revisited-2011-statistics-2/ (accessed August 21, 2013).
“9th Annual AP Report to the Nation.” CollegeBoard, 2013. Available online
at http://apreport.collegeboard.org/goals-and- findings/promoting-equity (accessed September 13,2013).
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control andPrevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13,2013).

2.3 Graphs for Quantitative Data

Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis andits Application’, Marcel Dekker“Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party,
term of office, and more.” Scholastic, 2013. Available online
at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013).
“Presidents.” Fact Monster. Pearson Education, 2007. Available online
at http://www.factmonster.com/ipka/A0194030.html(accessed April 3, 2013).
“Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available
online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013).
“Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics.
Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013).
“CO2 emissions (kt).” The World Bank,
2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed
April 3, 2013).
“Births Time Series Data.” General Register Office For Scotland,
2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/timeseries.html (accessed April 3, 2013).
“Demographics: Children under the age of 5 years
underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013).
Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach.
CRC Press: 1980.
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control andPrevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13,2013).

77

CHAPTER 2 SOLUTIONS:
1) 1.5 – 2.55) frequency distribution# of potholes
1
2
3
4
5
6
7

3)
n = Σf = 40
7) midpoints: 34, 39, 44, 49, 54, 59, 64

f
12
4
5
5
3
3
3

9) r.f: 2.9%, 8.6%, 14.3%, 31.4%, 20%, 20%, 2.9%11) n = Σf = 35
13) midpoints: 74, 83, 92, 101, 110,
119, 12815) r.f: .147, .158, .168, .189, .232, .074, .032
17) n = Σf = 95
19) grouped frequency distribution with class width = 18
Classes
236 – 253
254 – 271
272 – 289
290 – 307
308 – 325
326 – 343
344 – 361

Frequency
3
5
11
17
11
9
1

21) grouped frequency distribution with class width = 86
ClassesFrequency

Midpoint

Rel. Freq.

30 – 115
116 – 201
202 – 287
288 – 373
374 – 459
460 – 545

4
5
5
3
5
6

72.5
158.5
244.5
330.5
416.5
502.5

14.3%
17.9%
17.9%
10.7%
17.9%
21.4%

78

Cum.
Freq.
4
9
14
17
22
28

23)

25)

27)

29)

Stem
2
3
4
5

Leaf
_
57
34445778999
015679003344

Stem Leaf_
1124677789
0001124445556678899

6
7
8
9

31)

5

33)

Cars

79

3
4
5
6
7

Freq.
14
19
12
9
11

Rel Freq
21.5%
29.2%
18.5%
13.8%
16.9%

Cum. Freq
14
33
45
54
65

35)
a. Frequency Polygon

Histogram

b.

c.

80

37)
Both skewed left. Similar life expectancy
pattern for both men and women39) Police variable increased faster. The number of police did not impact the number ofhomicides. Homicides continued to increase.

81

This page is purposely left blank.

82

3 | NUMERICAL DESCRIPTORS

Figure 4.1 Photo by Dayne Topkin on Unsplash

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:•••

Recognize, describe, and calculate the measures of central tendencies: mean,
median, and mode.
Recognize, describe, and calculate the measures of variation: range, variance, andstandard deviation.
Recognize, describe, and calculate the measures of position: quartiles, percentiles,
outliers, Chebyshev’s Theorem, and Empirical Rule83

3.1 | Measures of Central Tendency
The "center" of a data set is also a way of describing location. The two most widely usedmeasures of the "center" of a data set are the mean (average) and the median. To calculate
the mean weight of 50 people, add the 50 weights together and divide by 50. To find the
median weight of the 50 people, order the data and find the number that splits the data
into two equal parts. The median is generally a better measure of the center when there are
extreme values or outliers because it is not affected by the precise numerical values of theoutliers. The mean is the most common measure of the center.
NOTE
The words “mean” and “average” are often used interchangeably. The substitution of one
word for the other is common practice. The technical term is “arithmetic mean” and “average”
is technically a center location. However, in practice among non-statisticians, “average" iscommonly accepted for “arithmetic mean.”The symbol used to represent the sample mean is an x with a bar over it (pronounced “xbar”): x .
The Greek letter m (pronounced "mew") represents the population mean. One of therequirements for the sample mean to be a good estimate of the population mean is forthe sample used to be truly random.
When the values in a data set are not unique, the mean can be calculated by multiplying
each distinct value by its frequency and then dividing the sum by the total number of data
values. To see that both ways of calculating the mean are the same, consider the sample:
1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4x=

1+1+1+ 2 + 2 + 3 + 4 + 4 + 4 + 4 + 4
= 2.7
11

=
x

3(1) + 2(2) + 1(3) + 5(4)
= 2.7
11

n +1
.
2
The letter n is the total number of data values in the sample. If n is an odd number, themedian is the middle value of the ordered data (ordered smallest to largest). If n is an evennumber, the median is equal to the two middle values added together and divided by two
after the data has been ordered. For example, if the total number of data values is 97, then

You can quickly find the location of the median by using the expression84n + 1 97 + 1== 49, so the median is the 49th value in the ordered data. If the total number of2
2
n + 1 100 + 1
data values is 100, then=
= 50.5, so the median occurs midway between the 50th22and 51st values.

In general, the values of the mean and median are not the same. The upper-case letter M isoften used to represent the median. The next example illustrates the location of the medianand the value of the median.

Example 3.1
AIDS data indicating the number of months a patient with AIDS lives after taking a new
antibody drug are as follows (smallest to largest):
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24;
24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
Calculate the mean and the median.
Solution 3.1
The calculation for the mean is:
3 + 4 + (8)(2) + 10 + 11 + 12 + 13 + 14 + (15)(2) + (16)(2) + ... + 35 + 37 + 40 + (44)(2) + 47
x=
= 23.640
To find the median, M, first use the formula for the location. The location isn + 1 40 + 1== 20.5.
2
2
Starting at the smallest value, the median is located between the 20th and 21st values (thetwo 24s); so
24 + 24
M== 24.
2
Using the TI-83, 83+, 84, 84+ Calculator__________________________________________________________________________
To find the mean and the median:
Go to STAT >> EDIT. Clear list L1 by pressing 4: ClrList, and then pressing 2nd 1 for list L1.
Press ENTER. Enter data into the list editor. Go to STAT >> EDIT and enter the data values
into list L1.

85

Press STAT and arrow to CALC. Select 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.
At the top of the screen you will see x = 23.6. Use the arrow keys to scroll down to thesecond output screen, and you will see the median: Med = 24.

3.1 The following data show the number of months patients typically wait on a transplantlist before getting surgery. The data are ordered from smallest to largest. Calculate themean and median.
3; 4; 5; 7; 7; 7; 7; 8; 8; 9; 9; 10; 10; 10; 10; 10; 11;
12; 12; 13; 14; 14; 15; 15; 17; 17; 18; 19; 19; 19; 21 ; 21; 22; 22
23; 24; 24; 24; 24

Example 3.2
Suppose that in a small town of 50 people, one person earns $5,000,000 per year and theother 49 each earn $30,000. Which is the better measure of the "center": the mean or the
median?
Solution 3.2
5,000,000 + 49(30,000)= 129,400 , whereas the median is M = 30,000.
50
The median is a better measure of the "center" than the mean. The mean is distorted
because of one very large value, 5,000,000. Such a value is called an “outlier”, meaning that
it is an extreme value. The 30,000 gives us a better sense of the middle of the data.

The mean is x =3.2 In a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth$280,000, and all the others are worth $315,000. Which is the better measure of the“center”: the mean or the median?
Another measure of the center is the mode. The mode is the most frequent value. Therecan be more than one mode in a data set as long as those values have the same frequency,
and that frequency is the highest. A data set with two modes is called bimodal. There also
can be no mode. No mode exists when each data value has the same frequency.

86

Example 3.3
Statistics exam scores for 20 students are as follows:50; 53; 59; 59; 63; 63; 72; 72; 72; 72; 72; 76; 78; 81; 83; 84; 84; 84; 90; 93
Find the mode.
Solution 3.3
The most frequent score is 72, which occurs five times. Mode = 72.

3.3 The number of books checked out from the library from 25 students are as follows:0; 0; 0; 1; 2; 3; 3; 4; 4; 5; 5; 7; 7; 7; 7; 8; 8; 8; 9; 10; 10; 11; 11; 12; 12
Find the mode.

Example 3.4
When is the mode the best measure of the "center"?
Solution 3.4
Consider a weight loss program that advertises a mean weight loss of six pounds the firstweek of the program. The mode might indicate that most people lose two pounds the first
week, making the program less appealing.

NOTE
The mode can be calculated for qualitative data as well as for quantitative data. For
example, if the data set is: {red, red, red, green, green, yellow, purple, black, and blue}, thenthe mode is red.
Statistical software will easily calculate the mean, the median, and the mode. Some graphingcalculators can also make these calculations. In the real world, people make thesecalculations using software.

87

3.4 Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores
680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The
mode is $25,000 and occurs 150 times out of 301. The median is $50,000 and the mean is$47,500. What would be the best measure of the “center” for each of these situations?

The Law of Large Numbers and the Mean
The Law of Large Numbers says that if you take samples of larger and larger size from any
population, then the mean of the sample is highly likely to get closer and closer to m.
Similarly, if we take larger and larger samples, the sample proportion will approach the
population proportion p. These facts are discussed in more detail later in the text. Recall thata statistic is a number calculated from a sample, whereas a parameter is a corresponding
number calculated from a population. Examples of statistics include the mean, median, modeand sample proportion.
The Law of Large Numbers tells that x is a sample statistic that estimates the parameter m(the population mean).
Similarly, the sample proportion p̂ is a sample statistic that estimates the population
proportion, p.

Calculating the Mean from Frequency Tables
Suppose that thirty randomly selected students were asked the number of movies they
watched the previous week. The results are shown in the frequency table below.
# of movies Frequency
0
5
1
15
2
6
3
3
4
1
Table 3.1
Suppose that we wanted to know the mean number of movies watched for the sample. Of
course, this will be the sum of the data, divided by 30 (the size of the sample):sample mean =data sum.
number of data values88

To do this, we add up the data values, multiplied by their respective frequencies and divideby 30:
x=0(5) + 1(15) + 2(6) + 3(3) + 4(1)≈ 1.3330

Note that this could also be written as:x=0(5) + 1(15) + 2(6) + 3(3) + 4(1) 0 ⋅ 5 1 ⋅15 2 ⋅ 6 3 ⋅ 3 4 ⋅1
=
+
+
+
+
30
30
30
30 30 30115
6
3
1=0 ⋅ + 1 ⋅ + 2 ⋅ + 3 ⋅ + 4 ⋅ ≈ 1.33
30
30
30
30
30

In other words, we can also calculate the mean by multiplying the data values by theirrespective relative frequencies and adding the resulting values. And this can be easily done
using
the
TI-84
familyofcalculators:
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the mean of a data set from a frequency table_

Go to STAT >> EDIT and clear lists L1 and L2.
Enter the data values into L1 and enter the frequencies into L2. Then go to STAT >>CALC. Select 1-VarStats, type L1, L2 and hit ENTER. The mean will appear at the top of
the screen as x .

When only grouped data is available, we do not know the individual data values (we only
know intervals and interval frequencies); therefore, we cannot compute an exact mean forthe data set. However, we can still estimate the sample mean by from the frequency table.
To calculate the mean from a grouped frequency table we can apply the same method as
above; but since we do not know the individual data values, we instead use the midpoint ofeach interval. The midpoint is the average of the lower boundary and upper boundary:
midpoint =upper boundary + lower boundary.
2

89

Example 3.5
A frequency table displaying Professor Blount’s last statistic test is shown below:
Grade Interval Number of Students
50-56.5
1
56.5-62.5
0
62.5-68.5
4
68.5-74.5
4
74.5-80.5
2
80.5-86.5
3
86.5-92.5
4
92.5-98.5
1
Find the best estimate of the class mean.
Solution 3.5
First, find the midpoints for all intervals:
Grade Interval Midpoint Number of Students
50.5-56.5
53.5
1
56.5-62.5
59.5
0
62.5-68.5
65.5
4
68.5-74.5
71.5
4
74.5-80.5
77.5
2
80.5-86.5
83.5
3
86.5-92.5
89.5
4
92.5-98.5
95.5
1
•
•
•
•

Go to STAT >> EDIT, clear lists L1, L2. Enter the midpoints into L1 and thefrequencies into L2.
Go to STAT >> CALC, select 1-VarStats and type L1, L2 (don’t forget the comma for
the TI-83!).
Hit ENTER, and the mean will appear at the top of the output screen: x = 76.86.
For TI-84 (plus) type L1 into FREQ: and L2 into FREQLIST:, then CALCULATE

90

3.5 Maris conducted a study on the effect that playing video games has on memory recall.
As part of her study, she compiled the following data:
Hours Spent on Video Games0 – 3.5
3.5 – 7.5
7.5 – 11.5
11.5 – 15.5
15.5 – 19.5

Number of teenagers
3
7
12
7
9

What is the best estimate for the mean number of hours spent playing video games?

Skewness and the Mean, Median, and Mode
Consider the following data set: 4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10This data set can be represented by following histogram. Each interval has width one, andeach value is in the middle of an interval.

The histogram displays a symmetrical distribution of data. A distribution is symmetrical
if a vertical line can be drawn at some point in the histogram such that the shape to the left
and the right of the vertical line are mirror images of each other. The mean, the median,and the mode are each seven for these data. In a perfectly symmetrical distribution,
the mean and the median are the same. This example has one mode (unimodal), andthe mode is the same as the mean and median. In a symmetrical distribution that has twomodes (bimodal), the two modes would be different from the mean and median.

91

On the other hand, the histogram for the data: 4; 5; 6; 6; 6; 7; 7; 7; 7; 8 is not
symmetrical. The right-hand side seems "chopped off" compared to the left side. A
distribution of this type is called skewed to the left because it is pulled out to the left.

The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is lessthan the median, and they are both less than the mode. The mean and the medianboth reflect the skewing, but the mean reflects it more so.
Finally, consider the histogram for the data: 6; 7; 7; 7; 7; 8; 8; 8; 9; 10.
This graph also is not symmetrical. It is skewed to the right.

The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, themean is the largest, while the mode is the smallest. Again, the mean reflects theskewing the most.

92

To summarize:•••

If the distribution of data is skewed to the left, the mean is less than the median.
If the data is symmetrical, then the median and mean are equal.
If the data is skewed to the right, then the mean is greater than the median.

Skewness and symmetry become important when we discuss probability distributions inlater chapters.

3.6 Discuss the mean, median, and mode for the following two graphs (dot plot andhistogram).
The GPA’s of all students in a business college. Each dot represents up to 4 observations.

933.2 | Measures of Variation
An important characteristic of any set of data is the variation in the data. In some data sets,
the data values are concentrated closely near the mean; in other data sets, the data values
are more widely spread out from the mean. In this section we will discuss three measures ofvariation: the range, the standard deviation, and the variance.
Standard deviation.
The most used measure of variation, or spread, is the standard deviation. The standard
deviation is a non-negative number that measures how far data values are from their mean.
Its importance will become much clearer when we begin studying methods of Inferential
Statistics. For now, we point out two key features of this important statistic.
The standard deviation provides a measure of the overall variation in a data set.
The standard deviation is small when the data are all concentrated close to the mean,
exhibiting little variation or spread. The standard deviation is larger when the data values
are more spread out from the mean, exhibiting more variation.
Suppose that we are studying the amount of time customers wait in line at the checkout at
supermarket A and supermarket B. the average wait time at both supermarkets is fiveminutes. At supermarket A, the standard deviation for the wait time is two minutes; at
supermarket B the standard deviation for the wait time is four minutes. Because
supermarket B has a higher standard deviation, we know that there is more variation in thewait times at supermarket B. Overall, wait times at supermarket B are more spread out fromthe average; wait times at supermarket A are more concentrated near the average, and hence
more predictable.
The standard deviation can be used to determine whether a data value is close toor far from the mean. Suppose that Rosa and Binh both shop at supermarket A. Rosa waits
at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A,
the mean waiting time is five minutes and the standard deviation is two minutes. The
standard deviation can be used to determine whether a data value is close to or far from the
mean.
Rosa’s wait time of seven minutes is two minutes longer than the average of five minutes;
that is, her wait time is one standard deviation above the average of five minutes.
Binh’s wait time is four minutes less than the average of five minute; that is, his wait timeof one minute is two standard deviations below the average of five minutes. A data
value that is two standard deviations from the average is just on the borderline for whatmany statisticians would consider to be “unusual”, or “far” from the average. Considering
data to be far from the mean if it is more than two standard deviations away is more of an
approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the94data affects how much of the data is further away than two standard deviations. (As we will
see in the next section)
As with measures of center, we can have a standard deviation measured from sample data (astatistic) or from population data (a parameter). The lower-case letter s represents a sample
standard deviation and the Greek letter σ (sigma, lower case) represents a population
standard deviation. This is like our notation protocol for the mean, where we used the symbol
x for the sample mean and the Greek letter m for the population mean.
Calculating the Standard Deviation
If x is a data value, then the difference between x and the mean is called its deviation. In adata set, there are as many deviations as there are values in the data set. The deviations areused to calculate the standard deviation; in fact, we intuitively think of the standard
deviation as a sort of “average” of the deviations. If the data belong to a population, in
symbols a deviation is x – μ. For sample data, a deviation will be x – x .
The formula for calculating the standard deviation will also be slightly different for sample
data than for population data. If the sample is reasonably large and representative of the
population, then s should be a good estimate of σ.
Suppose that we wanted to measure the totality of all variation in a data set. Then we mightstart by adding all of the deviations x – x . However, in any data set, there are values bothabove and below the mean x . So, some deviations would be positive, and some will be
negative, and there will be considerable cancellation. In fact, the sum of all the deviations
would always be zero. To prevent deviations from cancelling each other out we look at thesquares of the deviations: terms of the form
(x – x )2.
The variance is the average of the squares of the deviations. For reasons that will
become clear later, we compute these averages slightly differently for population data than
we do for sample data.
The population variance is: σ 2 = ∑(x − µ)2,
N

where N = the number of data values in the population.
The sample variance is: s 2 = ∑(x − x)2,
n −1

where n = the number of data values in the sample.

95

The symbol σ2 represents the population variance - so the population standard deviation
σ is the square root of the population variance. Similarly, the symbol s2 represents the
sample variance, so the sample standard deviation s is the square root of the sample
variance. Taking the square root of the variance can be thought of as reversing the effect ofsquaring the deviations. Thus, we can think of the standard deviation as a sort of “average”
of the deviations. Note also that the standard deviation will always be measured in the sameunits as the data values.
Formulas for the Standard Deviation:σ=∑(x − µ)2Ns=

Population Standard Deviation∑(x − x)2
n −1

Sample Standard Deviation

NOTE
In practice, we will always use a calculator or computer software to calculate a standard
deviation. If you are using a TI-83, 83+, 84+ calculator (or Excel) you need to select theappropriate standard deviation σx or Sx from the summary statistics.

We will concentrate on using and interpreting the information that the standard deviation
gives us. However, it is instructive to do one step-by-step example to help you understandhow the standard deviation measures variation from the mean. The calculator instructions
appear at the end of this example.

Example 3.6
In a fifth-grade class, teacher was interested in the average age and the sample standard
deviation of the ages of her students. The following data are the ages for a sample of n = 20
fifth grade students. The ages are rounded to the nearest half year:
9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5;
11.5;
The average age is 10.53 years, rounded to two places. Find the sample standard deviation,rounded to two decimal places.

96Solution 3.6
The variance may be calculated by using a table. Then the standard deviation is obtainedby taking the square root of the variance. We will explain the parts of the table after
calculating s.
Data
x
9

Frequency
f
1

9.5

2

10

4

10.5

4

11

6

11.5

3

Deviation(x – x bar)
9 – 10.525 = 1.525
9.5 – 10.525 = 1.025
10 – 10.525 = 0.525
10.5 – 10.525 = 0.025
11 – 10.525 =
0.475
11.5 – 10.525 =0.975

Deviation2(x – x bar)2(-1.525)2 =2.325625(-1.025)2 =1.050625(-0.525)2 =0.275625(-0.025)2 =0.000625(0.475)2 =0.225625(0.975)2 =0.950625

Freq * Deviation2f * (x – x bar)2
1 * 2.325625 =2.325625
2* 1.050625 =2.101250
4 * 0.275625 =
1.1025
4 * 0.000625 =0.0025
6 * 0.225625 =
1.35375
3 * 0.950625 =
2.851875
Total = 9.7375

The sample variance, s2, is equal to the sum of the last column (9.7375) divided by the total
number of data values minus one, which is 19.
s2 =9.7375= 0.5125
19

The sample standard deviation s is equal to the square root of the sample variance:s=

0.5125 = 0.715891

Rounded to two decimal places, s = 0.72.
Again, we almost never use the formulas or the procedure above to calculate a standard
deviation. Instead, we will use either software or a calculator to calculate these.

97Using the TI-83, 83+, 84, 84+ Calculator
To find the standard deviation:_

Go to STAT >> EDIT. Clear list L1, and enter the data into the list.
Go to STAT >> CALC and select 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.

In the output screen you will see both Sx and σx:
If the data is from a sample, use Sx as it represents the sample standard deviation, s.
If the data is from a population use σx represents the population standard deviation, σ.

3.7 On a baseball team, the ages of each of the players are as follows: 21; 21; 22; 23; 24;
24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40
Use a calculator or computer to find the mean and standard deviation. Then find the valuethat is two standard deviations above the mean.
Standard deviation of Grouped Frequency Tables
Recall that for grouped data we do not know individual data values, so we cannot describe
the typical value of the data with precision. Just as we could not find the exact mean froma frequency table, neither can we find the exact standard deviation. However, by again
replacing the class intervals by their midpoints, we can estimate the standard deviation
using the same procedure as we used for estimating the mean.

98

Example 3.7
Find the standard deviation for the data in the following table:
Class Frequency
0-2
1
3-5
6
6-8
10
9-11
7
12-14
0
15-17
2
Solution 3.7••••

Go to STAT >> EDIT, and clear lists L1 and L2.
Enter the midpoints of the classes into L1; enter the frequencies into L2:
Next, go to STAT >> CALC and select 1-VarStats:
Finally, type L1, L2 and hit Enter to get:

Here we see both a population standard deviation, σx, and the sample standard deviation Sx
displayed.
The following rules provide a little more insight into what the standard deviation tells us
about the distribution of the data.
Chebyshev's Rule: For any data set, no matter what the distribution of the data is:•••

At least 75% of the data is within 2 standard deviations of the mean.
At least 89% of the data is within 3 standard deviations of the mean.
At least 95% of the data is within 4.5 standard deviations of the mean.

99

Chebyshev’s Theorem
Empirical Rule: For data having a distribution that is bell-shaped and symmetric:• Approximately 68% of the data is within one standard deviation of the mean.
• Approximately 95% of the data is within two standard deviations of the mean.
• Approximately 99.7% of the data is within three standard deviations of the mean.
• It is important to note that this rule applies only when the shape of the distribution of
the data is bell-shaped and symmetric. We will learn more about this when studying"Normal" distribution in later chapters.

Empirical Rule100

Finally, we have seen how the mean and standard deviation can be affected by extreme
high or extreme low values. An outlier is a data value that is far removed from the rest of
the data. Outliers need to be examined carefully - sometimes they are the result of
measurement error and should be removed from the data and not used in the analysis atall. Other times, they hold valuable information about the population under study andshould be included in the data. Now that we have introduced the standard deviation, we
can present a commonly used criterion for classifying a data value as an outlier:
A data value that is more than three standard deviations from the mean is called
an outlier.

Example 3.8
Using Chebyshev’s theorem and Empirical find the percentage of values that lie within 3
standard deviations from the mean; given the heights of 20-year-old males have asymmetrical distribution with a mean of 70.2 inches and a standard deviation of 1.5 inches.
Solution 3.8:
We are given m = 70.2 and s = 1.5. Three standard deviations is represented by 3×s = 3(1.5)= 4.5. Subtract and add 4.5 from the mean, 70.2.
70.2 – 4.5 = 65.7
70.2 + 4.5 = 74.7
Chebyshev’s theorem states that 89% of 20-year-old male heights lie within 65.7 inches and74.7 inches. While Empirical Rule states that 99.7% of 20-year-old male heights lie within
65.7 inches and 74.7 inches. Chebyshev’s theorem is more conservative since it is used forany type of distribution. Empirical Rule is specifically for symmetrical distributions.

101

3.3 | Measures of Relative Position
Measures of relative position are used to describe data values relative to other data values.
Here we will discuss commonly used measures of relative position – in particular, quartiles,
percentiles, and z-scores.

Quartiles
Quartiles are special cases of percentiles. The first quartile, Q1, is the same as the 25th
percentile, and the third quartile, Q3, is the same as the 75th percentile. The median M isboth the second quartile and the 50th percentile. Thus 25% of the data values in any data
set are less than Q1. Similarly, 75% of the data values in the data set are less Q3, and so
25% of data values are above Q3. Finally, 50% of all data values are between Q1 and Q3.
The quartiles separate the data into quarters. To find the quartiles, first find the median orsecond quartile. The first quartile, Q1, is the middle value of the lower half of the data, andthe third quartile, Q3, is the middle value, or median, of the upper half of the data. To get theidea, consider the following data set:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5•••

There are 14 data values, so the median is the average of the 7th and 8th values in thelist. Thus, we have M = (7.2 + 6.8)/2 = 7.
The first quartile will be the median of the lower half of the data: 1, 1, 2, 2, 4, 6, 6.8.
The middle
value of these seven data values is Q1 = 2.
The third quartile will be the median of the upper half of the data: 7.2, 8, 8.3, 9, 10,
10, 11.5.
The middle value of these seven data values is Q3 = 9.

The interquartile range is the difference between the third quartile (Q3) and the firstquartile (Q1):
IQR = Q3 – Q1
The IQR provides a numerical measure of the overall amount of variation in a data set; it isoften used in conjunction with the median. When the median is used to describe the center,the IQR is often used to describe the spread.
The IQR can also be used to determine potential outliers. Recall that an outlier is a data
value that is significantly separated from the other data. Using the IQR, we have thefollowing criteria for classifying a data value as an outlier:
A data value x is considered an outlier if it is either x < Q1 – 1.5(IQR) or x > Q3 +1.5(IQR).

102Example 3.9
The following data represent selling prices for homes in a certain city. Calculate the IQR
and determine if any prices are potential outliers. Prices are in dollars.
389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000;
387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000
Solution 3.9 Order the data from smallest to largest:
114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800;
529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000
There are 13 data values, so the median is the 7th value:

M = 488,800

Q1 will be the median of the lower six values: Q1 = (230,500 + 387,000)/2 = 308,750.
Q3 will be the median of the upper six values: Q3 = (639,000 + 659,000)/2 = 649,000.
Thus, the inter-quartile range is IQR = 649,000 – 308,750 = 340,250=> Q1 – (1.5)(IQR) = 308,750 – (1.5)(340,250) = –201,625=> Q3 + (1.5)(IQR) = 649,000 + (1.5)(340,250) = 1,159,375
No house price is less than –201,625. However, 5,500,000 is more than 1,159,375.
Therefore, the data value 5,500,000 is a potential outlier.

3.8 Two classes take a test. The test Scores for Class A are:69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94
The test Scores for Class B are:90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100
Find the interquartile range for the following two data sets and compare them. Which class
had more variable test scores?

103

Five Number Summary and Boxplots
The quartiles are part of the five-number summary - these five numbers consist of:
The minimum value
The first quartile Q1
The median M
The third quartile Q3
The maximum value
For example, for the real estate data in Example 3.9, the five-number summary would be:
Min = 114,950
Q1 = 308,750
M = 488,800
Q3 = 649,000
Max = 5,500,000
For a small example like this, the quartiles can be easily calculated by hand; but for a large
data set this can be tedious and time-consuming. But most statistical software packages andgraphing calculators will calculate the five-number summary for us. When entering the data
into the calculator or software packages, the data does not have to be in order.
Using the TI-83, 83+, 84, 84+ Calculator_

To find the Five Number Summary:•••

Go to STAT >> EDIT. Clear list L1 and enter the data into the list.
Go to STAT >> CALC and select 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.
Use the down arrow key to scroll down; the 5-number summary appears in the 2nd
output screen.

A box plot is a graphical representation of the five-number summary. These graphs (alsocalled box and whisker plots or box-whisker plots) allow us to see where provide animage of the concentration of the data and show how far the extreme values are from mostof the data. So, the box plot gives a good, quick picture of the data.
A box plot is constructed from the five values: the minimum value, the first quartile, themedian, the third quartile, and the maximum value. To construct a box plot, use ahorizontal number line and mark the five numbers along the axis. We make small verticalmarks above the minimum and maximum values, and larger vertical marks above thequartiles Q1, M, and Q3. Finally, we “box off” the marks above the quartiles.

104

So, the first quartile marks one end of the box and the third quartile marks the other end of
the box. And the middle 50 percent of the data fall inside the box. The "whiskers"
extend from the ends of the box to the smallest and largest data values.
As an example, consider again the dataset:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5.
We calculated the quartiles already, and know that the five number summary is:
Min = 1; Q1 = 2 ; M = 7 ; Q3 = 9; Max = 11.5
Using this information, the boxplot is:

NOTES
1. It is important to start a box plot with a scaled number line. Otherwise, the box plot
may distort the spread of the data.
2. You may encounter box-and-whisker plots that have dots marking outlier values. In
those cases, the whiskers are not extending to the minimum and maximum values.

Example 3.10
The following data are the heights (in inches) of 40 students in a statistics class:59; 60; 61; 62; 62; 63; 63; 64; 64; 64;
65; 65; 65; 65; 65; 65; 65; 65; 65; 66;
66; 67; 67; 68; 68; 69; 70; 70; 70; 70;
70; 71; 71; 72; 72; 73; 74; 74; 75; 77
a)
b)
c)

Find the five-number summary for the data and construct a boxplot.
Find the IQRGiven an interval that has the middle 50% of the data.

105

Solution 3.10:a.) To find the five-number summary, we use the TI-84 calculator.
Go to STAT >> EDIT, and enter the data into L1. Then go to STAT >> CALC and select
1-VarStats. Press 2nd and 1 to select L1, and hit ENTER. Scroll down to the second output
screentoget:
Min = 59,

Q1 = 64.5,

Q2: = M = 66,

Q3 = 70,

Max = 77

Using these values, we get the following graph:b) The interquartile range is IQR = Q3 – Q1 = 70 – 64.5 = 5.5 inches.
c) About 50% of the data will be between Q1 and Q3. That is, about 50% of the students inthe class had heights that were between 64.5 inches and 70 inches.
Finally, note that the calculator can also be used to construct the box plot.
Using the TI-83, 83+, 84, 84+ Calculator
To construct a boxplot:
•
•
•
•
•

_

Go to the STAT PLOT menu by pressing 2nd and then the Y = button.
Turn on Plot 1 (Plot 2, Plot 3 should be turned off).
Use the down arrow key to move to Type, and select the boxplot option (middle ofsecond row).
For Xlist, type 2nd and then 1 to select L1.
Press Zoom and then press 9 to select the ZoomStat graphing option; this willautomatically configure the window so that all of the data are included.

Finally, if we press TRACE, and use the arrow keys to move from left to right, we can see
the quartiles.

1063.9 The following data are the number of pages in 40 books on a shelf. Construct a box plotusing a graphing calculator and state the interquartile range.
136; 140; 178; 190; 205; 215; 217; 218; 232; 234;
240; 255; 270; 275; 290; 301; 303; 315; 317; 318;
326; 333; 343; 349; 360; 369; 377; 388; 391; 392;
398; 400; 402; 405; 408; 422; 429; 450; 475; 512
It is possible that in a data set, two or more numbers in the five-number summary are
equal. For instance, you might have a data set in which the median and the third quartile
are the same. In this case, the diagram would not have a dotted line inside the box
displaying the median. The right side of the box would display both the third quartile andthe median.
As an example, suppose we have data set in which the smallest value and the first quartile
were both 1, the median and the third quartile were both 5, and the maximum value is 7;
then the box plot would look like:In this case, at least 25% of the values are equal to 1. Twenty-five percent of the values are
between 1 and 5, inclusive. At least 25% of the values are equal to 5, and the top 25% of thevalues fall between 5 and 7, inclusive.
Example 3.11
Given the following test scores solve the following questions:
Test scores for a college statistics class held during the day are:99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90

Test scores for a college statistics class held during the evening are:98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5

107

a. Find the five-number summary for the day class.
b. Find the five-number summary for the night class.
c. Create a box plot for each set of data, using the same number line for both box plots.
d. Which box plot has the widest spread for the middle 50% of the data?
What does this mean for that set of data in comparison to the other set of data?
Solution 3.11
a. Min = 32, Q1 = 56, M = 74.5, Q3 = 82.5, Max = 99
b. Min = 25.5, Q1 = 78, M = 81, Q3 = 89, Max = 98c. The upper graph is for the
day class.
The lower graph is for thenight class.

d. The first data set has the wider spread for the middle 50% of the data; so the IQR for
the first data set is greater than the IQR for the second set. This means that there is morevariability in the middle 50% of the first data set.
Interpreting Percentiles, Quartiles, and Median
A percentile indicates the relative standing of a data value when data are sorted into
numerical order from smallest to largest. For example, 15% of data values are less than orequal to the 15th percentile.
• Low percentiles always correspond to lower data values.
• High percentiles always correspond to higher data values.
A percentile may or may not correspond to a value judgment about whether it is "good" or"bad". The interpretation will depend on the context of the situation to which the dataapplies. In some situations, a low percentile would be considered "good;" in other contexts ahigh percentile might be considered "good". In many situations, there is no value judgmentthat applies. Understanding how to interpret percentiles properly is important not only
when describing data, but also when calculating probabilities in later chapters of this text.

108

GUIDELINES
When interpreting a percentile in the context of the given data, the sentence should containthe following information:••••

Information about the context of the situation being considered
the data value (value of the variable) that represents the percentile
the percent of individuals or items with data values below the percentile
the percent of individuals or items with data values above the percentile.

Example 3.12
On a timed math test, the first quartile for time it took to finish the exam was 35 minutes.
Interpret the first quartile in the context of this situation.
Solution 3.12:• Twenty-five percent of students finished the exam in 35 minutes or less.
• Seventy-five percent of students finished the exam in 35 minutes or more.
• A low percentile could be considered good, as finishing more quickly on a timed exam isdesirable. (If you take too long, you might not be able to finish.)3.10 For the 100-meter dash, the third quartile for times for finishing the race was 11.5
seconds. Interpret the third quartile in the context of the situation.

Example 3.13
On a 20 question math test, the 70th percentile for number of correct answers was 16.
Interpret the 70th percentile in the context of this situation.
Solution to 3.13:• Seventy percent of students answered 16 or fewer questions correctly.
• Thirty percent of students answered 16 or more questions correctly.
• A higher percentile could be considered good, as answering more questions correctly isdesirable.

1093.11 On a 60 point written assignment, the 80th percentile for the number of points earned
was 49. Interpret the 80th percentile in the context of this situation.

3.12 During a college basketball season, the 40th percentile for points scored per player in agame is eight. Interpret the 40th percentile in the context of this situation.

Percentiles
We say that a data value x is at the P-th percentile if P% of all data values are less than x.
For example, if a student taking a standardized test scores at the 85th percentile, this meansthat 85% of all students taking the test scored lower than the student, and so 15% of students
taking the test scored better.
Percentiles divide ordered data into hundredths and are useful for comparing values. For
example, universities and colleges use percentiles extensively. One instance in which colleges
and universities use percentiles is when SAT results are used to determine a minimum
testing score that will be used as a criterion for admission. For example, suppose Duke
accepts SAT scores at or above the 75th percentile. That translates into a score of at least
1220.
Percentiles are most often used with very large populations. Therefore, if you were to saythat 90% of the test scores are less (and not the same or less) than your score, it would be
acceptable because removing one particular data value is not significant. We will learn more
about how to calculate percentiles for large data sets in a later chapter. But sometimes wealso need to calculate a percentile for small data sets. The best way to do this is using acumulative relative frequency table.

110

Example 3.14
Fifty statistics students were asked how much sleep they get per school night (rounded to
the nearest hour). The results are as follows:
Hours of Sleep
4
5
6
7
8
9
10

Frequency
2
5
7
12
14
7
3

Relative Frequency
0.04
0.10
0.14
0.24
0.28
0.14
0.06

Cumulative Relative
Frequency
0.04
0.14
0.28
0.52
0.80
0.94
1.00

a. Find the 28th percentile.
b. Find the median.
c. Find the third quartile.
Solution 3.14:
a. Find the 28th percentile. Notice the 0.28 in the "cumulative relative frequency" column.
Twenty-eight percent of the 50 is a total of 14 data values. There are 14 values less thanthe 28th percentile. They include the two 4’s, the five 5’s, and the seven 6’s. The 28th
percentile is between the last six and the first seven. The 28th percentile is 6.5.
b. Find the median. Look again at the "cumulative relative frequency" column and find
0.52. The median is the 50th percentile. Since 50% of 50 is 25, there are 25 data values
less than the median. They include the two 4s, the five 5’s, the seven 6’s, and eleven of
the 7’s. The median is between the 25th and 26th data values, both of which are 7. The
median is M = 7.
c. Find the third quartile. The third quartile is the same as the 75th percentile. You can"eyeball" this answer. If you look at the "cumulative relative frequency" column, you find0.52 and 0.80. When you have all the 4’s, 5’s, 6’s and 7’s, you have 52% of the data. When
you include all the 8’s, you will have 80% of the data. The 75th percentile must be 8.
Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to
38. The third quartile, Q3, is the 38th value, which is again an 8.

111

A Formula for Finding the kth Percentile
In addition to using a frequency distribution, there are several formulas for calculating thekth percentile. Here is one of them.
Suppose that we want the kth percentile. It may or may not be part of the data. Order thedata, and let i be the index of the kth percentile; i.e. this is the position that the percentile
appears in the list of data. Let n be the total number of data values. Then we calculateki=(n + 1)
100
If i is a positive integer, then the kth percentile is the data value in the ith position in theordered set of data. If i is not a positive integer, then round i up and round i down to thenearest integers. Average the two data values in these two positions in the ordered data set.
This is easier to understand in an example.

Example 3.15
Listed below are the ages for 29 Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47;
52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a. Find the 70th percentile.
b. Find the 83rd percentile.
Solution 3.15
k70
(n + 1) =(29 + 1) = 21 .
100
100
This is a whole number, so we want the data value in the 21st position in the ordered data.
The 70th percentile is 64.

a. k = 70, i = the index, n = 29. Calculate i =k83(n + 1) =(29 + 1) = 24.9 .
100
100This is not a whole number, so we round down to get 24, and round up to get 25.
The age in the 24th position is 71, and age in the 25th position is 72. Average these two
numbers to obtain the 83rd percentile, 71.5.

b. k = 83, i = the index, n = 29. Calculate i =112

3.13 Listed below are the ages for 29 Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47;
52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Calculate the 20th percentile and the 55th percentile.
A Formula for Finding the Percentile of a Value in a Data Set
Again, order the data from smallest to largest.
Let x = the number of data values counting from the bottom of the data list up to but notincludingthe data value for which you want to find the percentile.
Let y = the number of data values equal to the data value for which you want to find thepercentile.
Let n = the total number of data values.
Calculatex + 0.5 y⋅ 100 and round to the nearest integer.
n

Example 3.16
Listed below are the ages for 29 Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47;
52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a. Find percentile for 58
b. Find percentile for 25
Solution 3.16
a. Counting from the bottom of the list, there are 18 data values less than 58; and there is
one data value equal to 58. So x = 18 and y = 1.
x + 0.5 y
18 + 0.5(1)=>⋅ 100 =
⋅ 100 = 63.8
n29
Thus an age of 58 would be at the 64th percentile.
b. Counting from the bottom of the list, there are 3 data values less than 25; and there is
one data value equal to 25. So x = 3 and y = 1.
x + 0.5 y
3 + 0.5(1)=>⋅ 100 =
⋅ 100 = 12.07
n
29
Thus an age of 25 would be at the 12th percentile.

113

z-Scores

The standard deviation is useful when comparing data values that come from different data
sets. If the data sets have different means and standard deviations, then comparing thedata values directly can be misleading. But we can tell which data value is farther removed
from the center by finding how many standard deviations away from its mean the value is.
And we have a statistic called a "z-score" for this purpose. Let x be a data value, µ be themean, and σ the standard deviation. Then the z-score corresponding to x is:
z=

x−µ

σ

Solving for x in this equation, we get x = μ + zσ. So, this really does measure how many
standard deviations x is from the mean. Moreover, if z > 0, a positive value, then x is
greater than the mean; and if z < 0, a negative value, then x is less than the mean. When z= 0, the standard deviation is zero and the value is at the mean.

Example 3.17
Two students, John and Ali, from different high schools, wanted to find out who had thehighest GPA when compared to his school.
Student

John
Ali

GPA

School Mean GPA

2.85
77

3.0
80

School Standard Deviation

0.7
10Which student had the highest GPA when compared to his school?
Solution to 3.17:
For each student, we use the z-score to determine how many standard deviations his GPA
is away from the average for his school:
For John, z =For Ali, z =

x−µ

=

σ

x−µ

σ

=

2.85 − 3.0= −0.21 .
0.7

77 − 80= −0.3 .
10John’s GPA is 0.21 standard deviations below his school’s mean, while Ali’s GPA is 0.3
standard deviations below his school’s mean. So, John’s z-score is higher than Ali’s. For
GPA, higher values are better; so, John has the better GPA relative to his school.

114

3.14 Two swimmers, Angie and Beth, from different teams, wanted to find out who had thefastest time for the 50 meter freestyle when compared to her team. The table below showstheir times.
Swimmer Time in Seconds Team Mean Time Team Standard Deviation
Angie
26.2
27.2
0.8
Beth
27.3
30.1
1.4
Which swimmer had the fastest time when compared to her team?

Z- scores will be revisited in Chapter 7: The Normal Distribution. The normal distribution
is an important continuous probability distribution. Z-scores will be on the x-axis and will
also be used in hypothesis testing.
The figure below shows how z-scores relate to position to the mean and to percentile. Recall
that the z-score at the mean is zero. The figure shows the normal distribution for the SAT.
The z-score for Student A was 1, meaning Student A was 1 standard deviation above themean. Thus, Student A performed in the 84.13 percentile on the SAT.

Figure 4.2 (credit: BaumannAshley, Wikimedia Commons)

115

Example 3.18
Three statistics students, Mia, Fiona and Raina, are in different classes and all scored an
82 on their last statistics test. In Mia’s class the mean was 68 with a standard deviation of
6. In Fiona’s class the mean was 73 with a standard deviation of 5. In Raina’s class themean was 80 with a standard deviation of 4. Who performed the best within their own
class?
Solution 3.18
Even though they have the same score we do not know how they compare to theirclassmates so begin by calculating the z-scores for each student.
For Mia: z =x−µ

For Fiona: z =For Raina: z =

σ

=

x−µ

σ
x−µ

σ

82−68
= 2.33
6

=
=

82−73
= 1.8
5

82−80
= 0.5
4

Comparing the graphs of the z-scores, it is easier to see that Mia scored the best within herclass. The z-score for her score is at the highest percentile with the most area shaded to theleft.
Mia: z-score 2.33

Fiona: z-score 1.8

116

Raina: z-score 0.5

KEY TERMS
Box plot a graph that gives a quick picture of the middle 50% of the data
First Quartile the value that is the median of the of the lower half of the ordered data set
Interquartile Range or IQR, is the range of the middle 50 percent of the data values; theIQR is found by subtracting the first quartile from the third quartile.
Mean a number that measures the central tendency of the data; a common name for mean
is 'average.'
The term 'mean' is a shortened form of ‘arithmetic mean.' By definition, the mean for asample is
x = sample mean =sum of data=
number of data values∑xnWhen the data is from a population, we denote the mean by the Greek letter μ.
Median the number that separates ordered data into halves; half the values are the same
number or smaller than the median and half the values are the same number or larger thanthe median. The median may or may not be part of the data.
Midpoint the number that is halfway between the endpoints of an interval
Mode the value that appears most frequently in a set of data
Outlier an observation that falls far outside the rest of the data.
Percentile a number that divides ordered data into hundredths; percentiles may or may not
be part of the data. The median of the data is the second quartile and the 50th percentile.
The first and third quartiles are the 25th and the 75th percentiles, respectively.
Quartiles the numbers that separate the data into quarters; quartiles may or may not be
part of the data. The second quartile is the median of the data.
Standard Deviation: A numerical measure of dispersion in a data set. Specifically, this is
the square root of the variance and measures how far data values are from their mean;
notation: s for sample standard deviation and σ for population standard deviation.
Variance: The mean of the squared deviations from the mean, or the square of the standard
deviation; for a set of data, a deviation can be represented as x – x where x is a data value
and x is the sample mean. The sample variance is equal to the sum of the squares of thedeviations divided by the difference of the sample size and one.

117

FORMULA REVIEW
Mean

mean =

sum of data=
number of data values∑xnFormulas for the Standard Deviation:σ=∑(x − µ)2Ns=

Population Standard Deviation∑(x − x)2
n −1

Sample Standard Deviation

A Formula for Finding the kth Percentile
Suppose that we want the kth percentile. It may or may not be part of the data.
Order the data, and let i be the index of the kth percentile; i.e. this is the position that thepercentile appears in the list of data. Let n be the total number of data values.
kThen we calculate i =(n + 1)
100
If i is a positive integer, then the kth percentile is the data value in the ith position in theordered set of data.
If i is not a positive integer, then round i up and round i down to the nearest integers.
Average the two data values in these two positions in the ordered data set. This is easier tounderstand in an example.
A Formula for Finding the Percentile of a Value in a Data Set
Again, order the data from smallest to largest.
Let x = the number of data values counting from the bottom of the data list up to but notincluding the data value for which you want to find the percentile.
Let y = the number of data values equal to the data value for which you want to find thepercentile.
Let n = the total number of data values.

118

Exercises for Chapter 3
For problems #1 - 4, find the mean, median, standard deviation and five number summaries
for the given data set.
1. The following data shows the mileage (in mpg) for a random sample of 15 new

cars.

30.1
28.2
31.5

32.8
31.5
32.9

29.2
32.8
31.4

32.6
32.8
30.6

29.0
30.2
31.0

2. The following data shows the ages (in years) of a random sample of 22 Boeing 747
airplanes.
17
15

5
5

5
8

12
17

8
5

5
4

8
2

16
22

14
17

12
20

22
23

3. The following data shows the speeds (in mph) for a random sample of 20 cars driving on
I-294 at 1 pm on a single day.
68
75

65
67

50
72

79
58

77
70

60
62

55
67

61
72

78
70

75
74

4. The following data shows the number of cardiograms done each day for a random sample
of 20 days
at an outpatient testing center.
25
36

31
32

20
33

32
32

13
44

14
32

43
52

2
44

57
51

23
45

5. Find the mean and standard deviation for the following frequency tables:
a.

b.
Grade
Frequency
49.5 – 59.5
2
59.5 – 69.5
3
69.5 – 79.5
8
79.5 – 89.5
12
89.5 – 99.5
5

Daily Low Temperature Frequency
49.5 - 59.5
53
59.5 - 69.5
52
69.5 - 79.5
15
79.5 - 89.5
1
89.5 - 99.5
0

119

c.
Points per Game Frequency
49.5 - 59.5
14
59.5 - 69.5
32
69.5 - 79.5
15
79.5 - 89.5
23
89.5 - 99.5
2
6. Sixty-five randomly selected car salespersons were asked the number of cars theygenerally sell in one week. Fourteen people answered that they generally sell three cars;
nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars;
eleven generally sell seven cars. Calculate the following:a. sample mean = ______b) median =_____c. mode = _____

7. When the data are skewed left, what is the typical relationship between the mean andmedian?
8. When the data are symmetrical, what is the typical relationship between the mean andmedian?
9. For each of the following histograms,a) Describe the shape of the distributionb) Describe the relationship between the mean and median.

(i)

(ii)

10. Which is the greatest, the mean, the mode, or the median of the data set?
11; 11; 12; 12; 12; 12; 13; 15; 17; 22; 22; 22120

11. Which is the least, the mean, the mode, and the median of the data set?
56; 56; 56; 58; 59; 60; 62; 64; 64; 65; 67
12. Of the three measures, which tends to reflect skewing the most, the mean, the mode, or
the median? Why?
13. In a perfectly symmetrical distribution, when would the mode be different from themean and median?
14. The following data show the distances in miles between 20 retail stores and a large
distribution center:29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150a) Use a calculator to find the standard deviation and round to the nearest tenth.
b) Find the value that is one standard deviation below the mean.
For the next four exercises, calculate the mean, median and standard deviation:
15. The miles per gallon rating for 30 cars are shown:
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31,
32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
16. The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39,
40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
17. The following data are the prices of different laptops at an electronics store.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350,
350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
18. The following data are the daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71,
72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95

121

19. Fredo and Karl, two baseball players on different teams, wanted to find out who had
the higher batting average when compared to his team.
Player Batting Average
Fredo
0.158
Karl
0.177

Team Batting Average Team Standard Deviation
0.166
0.012
0.189
0.015

Which baseball player had the higher batting average when compared to his team?
20. The ages for 29 Academy Award winning best actors are shown below, in order fromsmallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47;
52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a. Find the 40th percentile.
c. Find the percentile for age 37.

b. Find the 78th percentile.
d. Find the percentile for age 72.

21. Jesse was ranked 37th in his graduating class of 180 students. At what percentile isJesse’s ranking?
22. On an exam, would it be more desirable to earn a grade with a high or low percentile?
Explain.
23. Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32
minutes is the 85th percentile of wait times. Is that good or bad? Write a sentenceinterpreting the 85th percentile in the context of this situation.
24. In a survey collecting data about the salaries earned by recent college graduates, Li
found that her salary was in the 78th percentile. Should Li be pleased or upset by thisresult? Explain.
25. In a study collecting data about the repair costs of damage to automobiles in a certain
type of crash tests, a certain model of car had $1,700 in damage and was in the 90th
percentile. Should the manufacturer and the consumer be pleased or upset by this result?
Explain and write a sentence that interprets the 90th percentile in the context of thisproblem.
26. Suppose that you are buying a house. You and your realtor have determined that themost expensive house you can afford is the 34th percentile. The 34th percentile of housing
prices is $240,000 in the town you want to move to. In this town, can you afford 34% of thehouses or 66% of the houses?

12227. The following data show the lengths of boats moored in a marina. Use this data toconstruct a boxplot.
16; 17; 19; 20; 20; 21; 23; 24; 25; 25; 25; 26; 26; 27;
27; 27; 28; 29; 30; 32; 33; 33; 34; 35; 37; 39; 40
28. Find the five number summary and construct a boxplot for the given data set.
Amount (in dollars) spent on books for a semester
508

488

430

49

278

542

451

366

451

531

196

115

118

362

434

399

148

30

287

236

185

210

178

105

533

236

477

329

Use the following information to answer the next two exercises: Suppose one hundred elevenpeople who shopped in a special t-shirt store were asked the number of t-shirts they own
costing more than $19 each. The results are summarized in the histogram:

29. The percentage of people who own at most three t-shirts costing more than $19 each isapproximately:a. 21%

b. 59%

c. 41%

d. Cannot be determined

30. If the data were collected by asking the first 111 people who entered the store, then thetype of sampling is:a. clusterb. simple random

c. stratified

123

d. convenience31. Given the following box plot:a. Which quarter has the smallest spread of data? What is that spread?
b. Which quarter has the largest spread of data? What is that spread?
c. Find the interquartile range (IQR).
d. Are there more data in the interval [5, 10] or in the interval [10, 13]? Explain
32. The following box plot shows ages of the U.S. population for 1990, the latest available
year.

a. Are there fewer or more children (age 17 & under) than senior citizens (age 65 & over)?
Explain.
b. Given that about 12.6% are age 65 and over, approximately what percentage of the
population are working age adults (above age17 to age 65)?
33. In a survey of 20-year-olds in China, Germany, and the United States, people wereasked the number of foreign countries they had visited in their lifetime. The following box
plots display the results.

a. In complete sentences, describe what the shape of each box plot implies about the
distribution of the data collected.
b. Of the U.S. and Germany, which country has the greater percentage of 20-year olds thathave visited more than eight foreign countries?
c. Compare the three box plots. What do they imply about the foreign travel of 20-year-old
residents of the three countries when compared to each other?

124

34. A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of
new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In the survey,
people were asked the age they were when they purchased their car. The following box plots
display the results.

a. In complete sentences, describe what the shape of each box plot implies about the
distribution of the data collected for that car series.
b. Which group is most likely to have an outlier? Explain how you determined that.
c. Compare the three box plots. What do they imply about the age of purchasing a BMW
from the series when compared to each other?
d. For the BMW 5 series, which quarter has the smallest spread of data? What is the
spread?
e. For the BMW 5 series, which quarter has the largest spread of data? What is the spread?
f. Estimate the interquartile range (IQR) for the BMW 5 series.
g. For the BMW 5 series, are there more data in the interval 31 to 38 or in the interval 45 to
55? How do you know this?
h. For the BMW 5 series, which interval has the fewest data in it? How do you know this?
i. 31 to 35 ii. 38 to 41 iii. 41 to 64
35. Given the following box plot:

What does it mean to have the first and second quartiles so close together, while the second
to third quartiles are far apart?
125

36. The most obese countries in the world have obesity rates that range from 11.5% to83.4%. This data is summarized in the following table:
Percent of Population Obese Number of Countries
11.45–20.45
29
20.45–29.45
13
29.45–38.45
438.45–47.45
047.45–56.45
2
56.45–65.45
165.45–74.45
074.45–83.45
1
a. What is the best estimate of the average obesity percentage for these countries?
b. The United States has an average obesity rate of 33.9%. Is this rate above average or
below?
c. What is the standard deviation for the listed obesity rates?
d. How does the United States compare to other countries? Would it be considered“unusual”?
37. The following table gives the percent of children under five considered to beunderweight.
Percent of Underweight Children Number of Countries
16–21.45
23
21.45–26.9
4
26.9–32.35
932.35–37.8
7
37.8–43.25
6
43.25–48.7
1
a. What is the best estimate for the mean percentage of underweight children?
b. What is the standard deviation?
c. Which interval(s) could be considered unusual? Explain.
38. A music school has budgeted to purchase three musical instruments. They plan topurchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The
mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a
guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with astandard deviation of $100. Which cost is the lowest, when compared to other instrumentsof the same type? Which cost is the highest when compared to other instruments of thesame type? Justify your answer.

12639. Three students were applying to the same graduate school. They came from schools
with different grading systems. Which student had the best GPA when compared to otherstudents at his school? Explain how you determined your answer.
Student GPA School Average GPA
Thuy
2.7
3.2
Vichet87
75
Kamala8.6
8School Std. Deviation
0.8
20
0.4

40. An elementary school class ran one mile with a mean of 11 minutes and a standard
deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A
junior high school class ran one mile with a mean of nine minutes and a standard deviationof two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school classran one mile with a mean of seven minutes and a standard deviation of four minutes.
Nedda, a student in the class, ran one mile in eight minutes.
a. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster thanhe?
b. Who is the fastest runner with respect to his or her class? Explain.
41. The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age
was 33.1 years.
a. What does it mean for the median age to rise?
b. Give two reasons why the median age could rise.
c. For the median age to rise, is the actual number of children less in 1991 than it was in
1980? Why or why not?
42. We are interested in the number of years students in a particular elementary statistics
class have lived in California. The information in the following table is from the entire
section.
Number of years Frequency Number of years Frequency
7
1
22
1
14
3
23
1
15
1
26
1
18
1
40
2
19
4
42
2
20
3
Total = 20
a. What is the IQR?
b. What is the mode?

127

43. Javier and Ercilia are supervisors at a shopping mall. Each was given the task ofestimating the mean distance that shoppers live from the mall. They each randomly
surveyed 100 shoppers. The samples yielded the following information:
Sample mean x
Sample SD, s

Javier
6.0 miles
4.0 miles

Ercilia
6.0 miles
7.0 milesa. How can you determine which survey was more accurate?
b. Explain what the difference in the results of the surveys implies about the data.
c. If the two histograms depict the distribution of values for each supervisor, which one
depicts Ercilia's sample? How do you know?

d. If the two box plots depict the distribution of values for each supervisor, which one
depicts Ercilia’s sample? How do you know?

44. Santa Clara County, CA, has approximately 27,873 Japanese Americans. Their ages
are as follows:
Age Group
0 - 1718 - 24
25 - 34
35 - 4445 - 54
55 - 64
65+

Percent of Community
18.9
8.0
22.8
15.0
13.1
11.9
10.3

128

a. Construct a histogram of the Japanese American community in Santa Clara County, CA.
The bars will not be the same width for this example. Why not? What impact does thishave on the reliability of the graph?
b. What percentage of the community is under age 35?
c. Which box plot most resembles the information above?

45. Forty randomly selected students were asked the number of pairs of sneakers theyowned. Let X = the number of pairs of sneakers owned. The results are as follows:
X = # pairs of sneakers
1
2
3
4
5
6
7

Frequency
2
5
8
12
12
0
1

Relative Frequency

a. Find the sample mean xb. Find the sample standard deviation, s.
c. Construct a histogram of the data.
d. Complete the third column of the chart.
e. Find the first quartile.
f. Find the median.
g. Find the third quartile.
h. Construct a boxplot for the data.
i. What percent of the students owned at least five pairs?
j. Find the 40th percentile.
k. Find the 90th percentile.

12946. Refer to the graphs below determine which of the following are true and which are false.
Explain your solution to each part in complete sentences.

a.
b.
c.
d.

The medians for all three graphs are the same.
We cannot determine if any of the means for the three graphs is different.
The standard deviation for graph b is larger than the standard deviation for graph a.
We cannot determine if any of the third quartiles for the three graphs is different.

47. Following are the published weights (in pounds) of all of the team members of the San
Francisco 49ers from a given year.
177;
184;
245;
285;
265;

205;
174;
259;
286;
290;

210;
185;
278;
200;
276;

210;
242;
270;
215;
228;

232;
188;
280;
185;
265

205;
212;
295;
230;

185;
215;
275;
250;

185;
247;
285;
241;

178;
241;
290;
190;

210;
223;
272;
260;

206;
220;
273;
250;

212;
260;
280;
302;

a. Find the five-number summary for the data.
b. Construct a box plot of the data.
c. The middle 50% of the weights are from ___ to ____.
d. If our population were all professional football players, would the above data be asample of weights or the population of weights? Why?
e. If our population included every team member who ever played for the San Francisco
49ers, would the above data be a sample of weights or the population of weights? Why?
f. Assume the population was this particular year’s San Francisco 49ers. Find:i. the population mean, μ.
ii. the population standard deviation, σ.
iii. the weight that is two standard deviations below the mean.
iv. When quarterback Steve Young played football, he weighed 205 pounds.
How many standard deviations above or below the mean was he?

48. In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced.
Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days.
Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eightdays. One lasted nine days. Let X = the length (in days) of an engineering conference.
130

a. Organize the data in a relative frequency table.
b. Find the median, the first quartile, and the third quartile.
c. Find the 65th percentile.
d. The middle 50% of the conferences last from days to
days.
e. Construct a box plot of the data.
f. Find the 10th percentile.
g. Calculate the sample mean of days of engineering conferences.
h. Calculate the sample standard deviation of days of engineering conferences.
i. Find the mode.
j. If you were planning an engineering conference, which would you choose as the length ofthe conference: mean, median, or mode? Explain why you made that choice.
k. Give two reasons why you think that three to five days seem to be popular lengths ofengineering conferences.
49. A survey of enrollment at 35 community colleges across the United States yielded thefollowing data:
6414; 1550; 2109; 9350; 21,828; 4300; 5944;
2044; 5481; 5200; 5853; 2750; 10,012; 6357;
7681; 3200; 17,500; 9200; 7380; 18,314; 6557;
7493; 2771; 2861; 1263; 7285; 28,165; 5080;

5722; 2825;
27,000; 9414;
13,713; 17,768;
11,622a. Make a frequency table for the data using five intervals of equal width.
b. Construct a histogram of the data.
c. If you were to build a new community college, which piece of information would be morevaluable: the mode or the mean?
d. Calculate the sample mean.
e. Calculate the sample standard deviation.
f. A school with an enrollment of 8000 would be how many standard deviations away fromthe mean?
50. The average height of US men in 2010 is 69.3 inches according to Medical News Today.
Let’s say the standard deviation is 2.8 inches.
a. Using Chebyshev’s theorem, at least 75% of heights will be between what two heights?
b. Using Empirical rule, 68% of heights will be between what two heights?
c. If a randomly chosen male has a height of 78 inches, what is his z-score?
d. Using Empirical rule, what percentage of heights are above 74.9 inches?
51. The average ACT score in 2010 was 21 with a standard deviation of 5.2 according toDigest of Education Statistics.
a. Using Chebyshev’s theorem, at least 89% of scores will be between what two scores?
b. Using Empirical rule, 95% of scores will be between what two scores?
131

c. If a randomly chosen student has a score of 18, what is this student’s z-score?
d. Using Empirical rule, what percentage of scores are below 26.2?
52. Let’s say that the mean return of mutual funds in the last quarter is 2.8% with astandard deviation of 4.5%. Assume the returns are symmetrically distributed. Find thereturn value(s) that would separate thea. top 2.5%b. middle 99.7%
c. bottom 84%
53. Let’s say that the mean return of mutual funds in the last quarter is 2.8% with astandard deviation of 4.5%. Assume the returns are symmetrically distributed.
a. Find the percent of returns that are above 7.3%.
b. Find the percent of returns below 11.8%.
54. Let’s say that the mean return of mutual funds in the last quarter is 5.8% with astandard deviation of 2.1%. Assume the returns are skewed right.
a. Find the percent of returns that are between 1.6% and 10%.
b. At least 89% of returns are between what two values?

132

REFERENCES
3.1 Measures of the Center of the Data
Data from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013).
“Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at
http://www.indexmundi.com/g/
r.aspx?t=50&v=2228&l=en (accessed April 3, 2013).

3.2 Measures of Variation
Data from Microsoft Bookshelf.
King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College.
Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013).
Villines, Z. “What is the average height for men?” Medical News Today. Available online athttps://www.medicalnewstoday.com/articles/318155.php (accessed May 20, 2018).
ACT Scores data from https://nces.ed.gov/programs/digest/d10/tables/dt10_155.asp

3.3 Measures of Relative Position
Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.”
USA Today, 2012. Available online at http://usatoday30.usatoday.com/news/nation/story/2012-0517/minority-birthscensus/55029100/1 (accessed April 3, 2013).
Data from the United States Department of Commerce: United States Census Bureau. Available
online at http://www.census.gov/ (accessed April 3, 2013).
“1990 Census.” United States Department of Commerce: United States Census Bureau. Available
online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013).
Data from San Jose Mercury News.
Data from Time Magazine; survey by Yankelovich Partners, Inc.
Data from West Magazine.
Introductory Statistics, Sawyer Academy, 2012 (https://saylordotorg.github.io/text_introductorystatitstics/s06-05-the-empirical-rule-and-chebysh.html) accessed on 7/15/2021.

133

CHAPTER 3 SOLUTIONS:
1) 𝑥𝑥̅ = 31.1, med = 31.4, s = 1.54, min = 28.2, Q1 = 30.1, Q2 = 31.4, Q3 = 32.8, max = 32.9
3) x = 67.75, med = 69, s = 8.04, min = 50, Q1 = 61.5, Q2 = 69, Q3 = 74.5, max = 79
5) a. x bar = 79.5, s = 11.1

b. x bar = 61.5, s = 7.1 c. x bar = 70.7, s = 11.2

7) The mean is less than the median9) a. symmetricb. skewed to the right11) mode13) If the distribution was bimodal, the mode would be different from the mean andmedian.
15) x bar = 30.7, med = 31.5, s = 7.5
17) x bar = 371.3, med = 350, s = 109.9
19) Fredo, Fredo’s z score = -67 (less negative), Karl’s z score = -.8
21) 94th percentile23) Being at the 85th percentile of wait times is bad as 85 percent of the people have been
waiting less time than Mina. Only 15 percent will be waiting for longer.
25) $1,700 is in the 90% percentile of repair costs so it is comparatively high. The
manufacturer and the consumer would not be pleased. Only 10% of cars in the crash test
had higher repair costs.
27) min = 16, Q1 = 23, med = 27, Q3 = 33, max = 4029) c31) a. Q4 has a spread of 1 b. Q2 has a spread of 8c. 12 –2 = 10 d. There is moredata in [10,13] as it contains half of the data, both Q3 and Q4. [5,10] is within Q2 of thedata.
33) a. From the survey of 20-year-olds from China, all the surveyed people have been
between 0 and 5 countries. The 20-year-olds from Germany data is symmetric where the134middle 50% of the data is between 4 and 8 countries. The 20-year-olds from the US areright skewed with the middle 50% of the data between 0 – 5 countries.
b. Germany, the 4th quartile is 8 countries and above.
c. The survey implies that 20-year-olds from Germany have visited more countries
compared to the US and China. It also implies that 20-year-olds from China have visited
fewer countries comparatively.
35) Since the first and second quartile, so close together a quarter of the data falls withinthat interval. Since the second and third are far apart, the data is more spread out in thatquartile.
37) a. estimated mean = 26.573 b. s = 8.540
c. The last interval could be consideredunusual because it is greater than 2 standard deviations above the mean (26.573 + 2[8.540]= 43.653)
39) ZThuy = -.625, ZVichet = 0.60, ZKamala = 1.5, Kamala has the highest GPA compared to otherstudents in her school.
41) a. When the median age increases the middle age of the US is older. More of the
population is older.
b. Two possible reasons are that the birth rate is not increasing as fast as the death rate orthat people are living longer.
c. Not necessarily, it does not have to be less children in 1991. It just has to 50% of thepeople older than 33.1 and 50% below. It could be that people were living longer.
43) a. Since there are only two samples taken with the same n value, it would be importantto have information about sampling and take more samples. Taking more samples would
help determine which sample is more representative of the population.
b. The mean is the same, but Ercilia’s data implies customers live farther from the mall.
c. (b) since the data has a greater range, and Ercilia has the greater standard deviation.
d. (a) since the data has a greater range, and Ercilia has the greater standard deviation.
45) a. x bar = 3.8

b. s = 1.3

c.

135

d. 2/40 (0.05 or 5%), 5/40 (0.125 or 12.5%), 8/40 (0.2 or 20%), 12/40 (0.3 or 30%), 12/40 (0.3 or30%), 0/40 (0 or 0%), 1/40 (0.025 or 2.5%)
e. Q1 = 3

f. med = 4

g. Q3 = 5h.
I. 13/40 = 0.325 x 100 = 32.5% j. 40th percentile = 4

k. 90th percentile = 547) a. min = 174, Q1 = 205.5, med = 241, Q3 = 272.5, max = 302b.
c. 205.5 to 272.5
d. sample of weights; it’s just one teame. sample of weights; there are only 53 data points given, and there have been more than 53
San Francisco 49ers players since the team began
f. (no d) i. 236.3 ii. 37.5 iii. 161.3 iv. Zsy = -.81, 0.81 standard deviations below the mean.
49) a.

b.

Interval
1 - 6000
6001 - 12000
12001 - 18000
18001 - 24000
24001 - 30000

Frequency
16
12
3
2
2

c. the moded. mean = 8631.6
below the meane. 6944.1

51) a. between 5.4 and 36.6

b. between 10.6 and 31.5

53) a. 16%

b. 97.5%

136

f. Z = -0.091, 0.091 standard deviations
c. Z = -0.58

d. 84%

4 | PROBABILITY TOPICS

Figure 4.1 Photo by Erik Mclean on Unsplash

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:
•
•
•
•
•
•
•

Understand and use the terminology of probability.
Determine whether two events are mutually exclusive.
Determine whether two events are independent.
Calculate probabilities using the Addition Rules and Multiplication Rules.
Construct and interpret Contingency Tables.
Construct and interpret Venn Diagrams.
Construct and interpret Tree Diagrams.

Chapter Objectives137

It is often necessary to "guess" about the outcome of an event to make a decision. Politicians
study polls to guess their likelihood of winning an election. Teachers choose a particular
course of study based on what they think students can comprehend. Doctors choose thetreatments needed for various diseases based on their assessment of likely results. You may
have visited a casino where people play games chosen because of the belief that the likelihoodof winning is good. You may have chosen your course of study based on the probableavailability of jobs.
You have highly likely used probability before, and most people have an intuitive sense of
the concept. Probability deals with the chance of an event occurring. Whenever you weighthe odds of whether to do your homework or to study for an exam, you are using probability.
In this chapter, you will learn how to solve probability problems using a systematic approach.

4.1 | Terminology
Probability is a measure that is associated with how certain we are of outcomes of a particularexperiment or activity. An experiment is a planned operation carried out under controlled
conditions. If the result is not predetermined, then the experiment is said to be a chance
experiment. Flipping a fair coin twice is an example of an experiment.
A result of an experiment is called an outcome. The sample space of an experiment is theset of all possible outcomes. We will discuss three ways to represent a sample space are: list
the possible outcomes, create a tree diagram, or using a Venn diagram. The uppercase letter
S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} whereH = heads and T = tails are the outcomes. Sample spaces can often be written in multiple
ways. For example, if we toss a coin three times in succession, we could describe the sample
space in terms of the number of heads as S = {0, 1, 2, 3} or we can list the possible sequencesof heads and tails as S ={HHH, THH, HTH, HHT, HTT, THT, TTH, TTT}. We will see in a
moment that the second way is more useful, since it has equally likely outcomes. That is,each of the eight outcomes is equally likely to occur. It should be intuitively clear that thisis not true of the first set; i.e. there is only one way for all three tosses to land as heads, butthere is more than one way to get 1 head in three tosses.
An event is any collection of outcomes. Upper case letters, like A and B, represent events.
For example, if the experiment is to flip one fair coin, event A might be getting at most one
head. The probability of an event A is written as P(A).
There are two basic ways to calculate probabilities; Theoretical Probability and Empirical
Probability.
Theoretical Probability: Suppose that all outcomes in the sample space are equally likely.
To calculate the probability of an event A, count the number of outcomes for event A anddivide by the total number of outcomes in the sample space. That is,138

P( A) =number of outcomes in A.
number of outcomes in S

For example, suppose we toss a coin three times in succession. As we saw above, there is a
total of 8 outcomes in the sample space, and three of these outcomes have exactly twoheads. Thus, P(two heads) = 3/8.
Empirical Probability: The probability of any event is the long-term relative
frequency of that event. That is, if we were to do an exceptionally large number of trials(repetitions of our experiment) then the probability of A is:P( A) =

number of times A occurs
total number of trials

For example, suppose that an insurance company wants to find the probability that a
suburban male driver, whose age is between 20 and 25 years, will have an accident in the
coming year. Then they would use data from thousands of such drivers and calculate theproportion of those drivers that had an accident in the past year.
The Law of Large Numbers states that as the number of repetitions of an experiment isincreased, the relative frequency obtained in the experiment will become closer and closer
to the theoretical probability. For example, suppose you roll one fair six-sided die, with thenumbers 1, 2, 3, 4, 5, 6 on its faces. Let E be the event that we roll a number that is at least
five. There are two outcomes (5, 6) in E, so P(E) = 1/3. If you were to roll the die only a few
times, you would not be surprised if your observed results did not match the probability.
However, if we were to roll the die an exceptionally large number of times, the ratio oftimes E occurred would get closer and closer to 1/3.
While a prisoner of war during World War II, the English mathematician John Kerrich(1903–1985) performed a number of probability experiments to demonstrate the Law ofLarge Numbers. The most famous of these consisted of tossing a coin 10,000 times andrecording whether it landed heads or tails. He obtained 5067 heads, so the empiricalprobability of heads was 5067/10,000 or 50.67%.
In many real-world situations, the outcomes are not equally likely. A coin or die may be
unfair, or biased. Two math professors in Europe had their Statistic students test theBelgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of thetime and a tail was obtained 44% of the time. The data seem to show that the coin is not afair coin; more repetitions would be helpful to draw a more accurate conclusion about suchbias. Some dice are also biased. Look at the dice in a game you have at home; the spots oneach face are usually small holes carved out and then painted to make the spots visible.
Your dice may or may not be biased; it is possible that the outcomes may be affected by theslight weight differences due to the different numbers of holes in the faces. Gamblingcasinos make a lot of money depending on outcomes from rolling dice, so casino dice are139made differently to eliminate bias. Casino dice have flat faces; the holes are filled withpaint having the same density as the material that the dice are made out of so that eachface is equally likely to occur. Later we will learn techniques to use to work withprobabilities for events that are not equally likely.
Using either definition of probability, it is easy to obtain the following three generalproperties:
Properties of Probability
1. For any event A, 0 < P(A) < 1. That is, the probability of any event is always anumber between 0 and 1.
2. The probability of any event A is equal to the sum of the probabilities of theindividual outcomes in A.
3. The sum of the probabilities of all outcomes in S must equal 1. This is true whether
S consists of equally likely outcomes.
Note that we can write probabilities as a fraction, a decimal or a percent. Moreover, if P(A)= 0 this means the event A can never happen, whereas P(A) = 1 means the event A must
always occur (A is a certainty).
Often, we deal with events that are described in terms of other events; in particular, events
involving the connectives “and”, “or” and “not”. These are sometimes called compound
events.
An outcome is in the event A or B if the outcome is in A or is in B or is in both A and B.
For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. Then the event A or B = {1, 2, 3, 4,
5, 6, 7, 8}.
Notice that 4 and 5 are NOT listed twice.
The event A and B is the set of outcomes that are common to both A and B. That is, anoutcome is in A and B if it is in both A and B at the same time. For example, let A and B be
{1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A and B = {4, 5}.
The complement of event A consists of all outcomes that are not in A. The complement of
A is denoted as A′ (read "A prime"). Notice that P(A) + P(A′) = 1, since every outcome iseither in A or is in its complement. This observation gives us a valuable rule forcalculating probabilities:
Complement Rule
The probability that A does not occur is P(A′) = 1 – P(A).

140The conditional probability of A given B is written P(A | B). P(A | B) is the probability
that event A will occur given that the event B has already occurred. The additional
information that event B occurs changes the sample space, since now we are interested onlyin those outcomes that are in B. And since A must also occur, we see that

P( A | B) =number of outcomes in (A and B). If we divide the numerator and denominator
number of outcomes in B

by the number of outcomes in S, we get the equivalent formula:
Conditional Probability

The probability that event A occurs, given that event B also occurs is:P( A | B ) =P( A and B).
P( B )For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}.
Let A = {2, 3}, and B = {2, 4, 6}. To calculate P(A | B), first count the number of outcomes
common to the two events; the outcome 2 is in both. Using the first formula, we haveP( A | B)

number of outcomes in (A and B) 1=.
number of outcomes in B
3

We get the same result by using the second formula: P( A | B) =1
P( A and B)
1 6 1= 6= ⋅ = .
3
6 3 3
P( B)
6

Understanding Terminology and Symbols
When working with probability, it is important to read each problem carefully to understandwhat the events are. Understanding the wording is an important first step in solving
probability problems. Reread the problem several times if necessary. Clearly identify theevent of interest. Determine whether there is a condition stated in the wording that would
indicate that the probability is conditional; carefully identify the condition, if any.
Example 4.1
The sample space S is the set of the first 20 whole numbers.
Let event A = the even numbers and event B = numbers greater than 13. Fill in the blanks:a. A =____b. P(A) =c. A and B =,B=, P(B) =,A or B =141

d. P(A and B) =e. A′ =,P(A or B) =, P(A′ ) =f. P(A) + P(A′) =g. P(A | B) =, P(B | A) =. Are the probabilities equal?

Solution 4.1
a. A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}, B = {14, 15, 16, 17, 18, 19, 20}7
10 1
b. P( A) == ; P( B) =20
20 2
c. A and B = {14, 16, 18, 20}; A or B = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19, 20}
413; P( A or B) =d. P( A and B) =
20
20
10 1
e. A′ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}; P( A′) ==20 2
10 10 20
f. P( A) + P( A′) =+== 1.
20 20 20

number of outcomes in (A and B) 4=;
number of outcomes in B
7number of outcomes in (A and B) 4=P( B | A) =.
number of outcomes in A
10g.

P( A | B)

The probabilities are not equal.

Example 4.2
Suppose that we roll two fair dice, one red and one white. Then there are 36 possible
outcomes in the sample space, as shown in the table below:

142Suppose we write the sample space in terms of the number of dots facing up on the twodice. That is, write S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Use the table to answer each of thefollowing:a. Are the outcomes in S equally likely.
b. Find P(4).
c. Find P(6).
d. Let E be the event “number of dots is even”. Find P(E).
e. Let G be the event “number of dots is greater than 9”. Find P(G).
f. Suppose we rolled the dice 7200 times. How many times would we expect to roll a 7?
Solution 4.2a) The events are not equally likely – for example the probability of rolling “snake eyes” isP(2) = 1/36 whereas the probability of rolling a “3” is P(3) = 2/36.
b) There are 3 outcomes in which the dots add to 4, so P(4) = 3/36.
c) There are 5 outcomes in which the dots add to 6, so P(6) = 5/36.
d) The event E = {2, 4, 6, 8, 10, 12}. So we add the probabilities of the outcomes:
1
3
5
5
3
1 18
+
+
+
+
+
=
P(E) = P(2) + P(4) + P(6) + P(8) + P(10) + P(12) =36 36 36 36 36 36 36e) The event G = {10, 11, 12}. So we add the probabilities of the outcomes:
3
2
1
6
+
+
=
P(G) = P(10) + P(11) + P(12) =36 36 36 36

6 1
= .
36 6
By the Law of Large Numbers, we expect the proportion of rolls that equal 7 to be about
7200
1/6. So we would expect the number of 7’s to be about= 1200.
6f) The theoretical probability of rolling a 7 is P(7) =Example 4.3
The table below describes the distribution of a random sample S of 100 individuals,organized by preference for Star Wars or Star Trek and whether they are right- or lefthanded.
Star Wars
Star Trek

Right-handed
43
44

Left-handed
94

Let’s denote the events as W = the subject prefers Star Wars, T = the subject prefers Star
Trek, R = the subject is right-handed, and L = the subject is left-handed. Compute thefollowing probabilities:
a. P(W)b. P(T)c. P(R)d. P(L)143

e. P(W and R)

f. P(W and L)

g. P(T or W )h. P(T or R)

i. P(T or L)

j. P(W' )k. P(R |W)l. P(L | T )Solution 4.3
a. P(W) = 0.52
b. P(T) = 0.48
c. P(R) = 0.87
d. P(L) = 0.13
e. P(W and R) = 0.43
f. P(T and L) = 0.04
g. P(W or T) = 1
h. P(W or R) =43 + 9 + 44
= 0.96100

i. P(T or L) =44 + 4 + 9= 0.57100

j. P(W') = P(F) = 0.48
k. P(R |W) =l. P(L |T ) =number of outcomes in 𝑅𝑅 and 𝑊𝑊
43= = 0.8269
number of outcomes in 𝑊𝑊
52
number of outcomes in 𝐿𝐿 and T
4= = 0.0833
number of outcomes in 𝑇𝑇
48

144

4.2 | Independent and Mutually Exclusive Events
When applying the rules of probability, we will sometimes need to consider the relationshipbetween two events; in particular, we will need to consider whether two events are
independent or whether they are mutually exclusive. Although students often confuse thetwo, they do not mean the same thing.

Independent Events
Two events A and B are independent if the knowledge that one occurred does not affect theprobability that the other occurs. For example, the outcomes of two roles of a fair die are
independent events. The outcome of the first roll does not change the probability for theoutcome of the second roll. To show two events are independent, you must show only one of
the above conditions. If two events are not independent, then we say that they are
dependent. If it is not known whether A and B are independent or dependent, you should
assume they are dependent until you can show otherwise.
To show that two events are independent, we show that any of following statements is true:• P(A | B) = P(A)• P(B | A) = P(B)
• P(A and B) = P(A)P(B)The first two are just restatements of the definition. The third statement follows from firsttwo statements and the formula for conditional probability. If we multiply both sides of theequationP( A | B ) =by P(B), we get P( A and B) = P( B)P( A | B) .

P( A and B)
P( B )

If the events are independent, we substitute P(A)in place of P(A | B) to get P(A and B) = P(A)P(B).
There are some situations where we will have information about the probabilities and/or
conditional probabilities, and we will use the rules above to determine whether the events
are independent. There are other situations where the events are known to be independent,
and then we will use this fact to help compute the probability of P(A and B).

145

Recall that sampling may be done with replacement or without replacement.
•

Sampling with replacement: If each member of a population is replaced after it ispicked, then that member has the possibility of being chosen more than once. Whensampling is done with replacement, then the sample space will not change from one
selection to the other; meaning the result of the first pick will not change theprobabilities for the second pick. Thus, events will be independent from one selection
to the next.

•

Sampling without replacement: When sampling is done without replacement,each member of a population may be chosen only once. In this case, the probabilities
for the second pick are affected by the result of the first pick. The events are dependent
or not independent.

A standard deck of cards is often used in probability problems. To help you visualize thesample space of a standard deck of cards, all the cards are shown below.
•
•
•
•
•
•
•
•
2♥
2♦
2♣
2♠

There are 52 cards in a standard deck of cards.
There are 4 suits: diamonds, hearts, clubs, and spades.
Two of the suits are black, the clubs and spades.
Two of the suits are red, the diamonds and hearts.
There are 13 cards in each suit.
Cards in each suit are 2 to Ace.
There are three face cards in each suit: jack, queen, and king. These along with the
10 and the Aces are called high cards.
The Ace may be used as either a high or low card depending on the situation.
3♥
3♦
3♣
3♠

4♥
4♦
4♣
4♠

5♥
5♦
5♣
5♠

6♥
6♦
6♣
6♠

7♥
7♦
7♣
7♠

8♥
8♦
8♣
8♠

9♥
9♦
9♣
9♠

10♥
10♦
10♣
10♠

J♥
J♦
J♣
J♠

Q♥
Q♦
Q♣
Q♠

K♥
K♦
K♣
K♠

A♥
A♦
A♣
A♠

Example 4.4
Suppose we have a standard, well-shuffled deck of 52 cards. Suppose that we select two
cards. The first card is a Queen of Diamonds.
a. Find the probability of selecting a queen on the second draw if the first card is replaced
before the second is drawn.
b. Find the probability of selecting a queen on the second draw if the first card is not
replaced before the second is drawn.

146

Solution 4.4:
a. If the first card is replaced before the second is drawn, then the second draw is
independent of the first. So, the probability of selecting a queen on the second draw is4.
52

b. If the first card is not replaced before the second is drawn, then the second draw isdependent on the first draw. There are now 51 cards left, of which 3 are queens. So, the3probability of selecting a queen on the second draw is.
51

4.4 Suppose we have a standard, well-shuffled deck of 52 cards. It consists of four suits:clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of Ace, 2, 3,
4, 5, 6, 7, 8, 9, 10, J, Q and K. Suppose that we select two cards. The first card is a Queen of
Diamonds.
a. Find the probability of selecting a diamond on the second draw if the first card is
replaced before the second is drawn.
b. Find the probability of selecting a diamond on the second draw if the first card is not
replaced before the second is drawn.
In Chapter 1 we saw that if we are selecting a sample from an exceptionally large population,
the probability of getting the same individual twice will be very small. As a result, there isvirtually no difference between sampling with replacement and treating without
replacement. Thus, when sampling from an exceptionally large population, we can think of
the individual selections as being independent of one another. This is useful for the next
example:Example 4.5
In a large metropolitan area, it is known that 45% of all voters are registered as Democrats.
Suppose we select two voters at random.
a. What is the probability that both are registered as Democrats?
b. What is the probability that neither are registered as Democrats?

147

Solution 4.5
Let event D1 = Democrat is selected on the first choice, and D2 = Democrat is selected on thesecond. The key observation here is that the draws are independent, so we can use the ruleP(D1 and D2) = P(D1)P(D2).
a. The probability that both are registered as Democrats is:P(D1 and D2) = P(D1) P(D2) = 0.45 x 0.45 = 0.2025.
b. If 45% are registered as Democrats, then the remaining 55% are not registered as
Democrats.
P(D1ʹ and D2ʹ) = P(D1ʹ) P(D2ʹ) = 0.55 x 0.55 = 0.3025.

4.5 Draw two cards from a standard 52-card deck with replacement. Find the probability ofgetting at least one black card.

Example 4.6
Let G = event that a student is taking a math class. Let H = event that a student is taking
a science class. Then, G and H = the event that a student is taking both a math class and ascience class. Suppose that P(G) = 0.6, P(H) = 0.5, and P(G and H) = 0.3. Based on thisinformation, are G and H independent?
Solution 4.6
To show that G and H are independent, we must show ONE of the following:• P(G | H) = P(G)• P(H | G) = P(H)
• P(G and H) = P(G)P(H)The option we choose depends on the information given in the problem. We could choose
any of the methods here because we have the necessary information.
For example, we can show that P(G | H) = P(G):P(G | H ) =P(G and H ) 0.3== 0.6 = P(G )P( H )
0.5

148

Or, we can show P(G and H) = P(G)P(H):
P(G)P(H) = (0.6)(0.5) = 0.3 = P(G and H)
Since G and H are independent, knowing that a person is taking a science class does notaffect the probability that he or she is taking a math class.

4.6 In a bag, there are six red marbles and four green marbles. The red marbles aremarked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with thenumbers 1, 2, 3, 4. One marble is selected. Let R = event that a red marble is drawn,
let G = event that a green marble is drawn, and let O = event that an odd-numberedmarble is selected.
a) Calculate P(G and O).

b) Are P and O independent? Explain

Example 4.7
In a particular college class, 60% of the students are taking math courses. Fifty percent of
all students in the class have graphing calculators. Forty-five percent of the students aretaking math course and have graphing calculators. Of the students taking math courses,
75% have graphing calculators. One student is picked randomly. Let M be the event thata student is taking a math course, and let C be the event that a student has a graphingcalculator. Are the events M and C independent?
Solution 4.7
There are three conditions we can check to determine independence:P(M) = P(M | C) OR P(C) = P(C | M) OR P(C and M) = P(M)P(C)The one we use will depend on the information given in the problem. So, we first write
down the probabilities that are given in the problem:
P(M) = 0.60; P(C) = 0.50;

P(M AND C) = 0.45;

P(C | M) = 0.75.

Based on this information we could use either the second condition or the third.
(We do not know P(M | C) yet, so we cannot use the first condition.)

149

Using the second rule, we check whether P(C |M) equals P(C):We are given that P(C |M) = 0.75, whereas P(C) = 0.50. Since these are not equal, theevents are not independent. This shows that a student who taking a math course is more
likely to have a graphing calculator than a student who is not in a math course.
Using the third rule, we check whether P(M and C) = P(M)P(C):We are given that P(M and C) = 0.45, whereas P(M)P(C) = (0.60)(0.50) = 0.30. Since these
are not equal, the events M and C are not independent.

Mutually Exclusive Events
We say that A and B are mutually exclusive events if they cannot occur at the same time.
This means that A and B do not share any outcomes and so it follows that P(A and B) = 0.
For example, suppose that S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8},
and C = {7, 9}. Then A and B = {4, 5}; so the events A, B are not mutually exclusive. On theother hand, the events B, C are mutually exclusive, since they do not have any outcomes incommon.
If it is not known whether A and B are mutually exclusive, assume they are not until you canshow otherwise. The following examples illustrate these definitions and terms.
Example 4.8
Suppose that we toss two fair coins. The sample space is {HH, HT, TH, TT} where T = tails
and H = heads. The possible outcomes are HH, HT, TH, and TT. Note that the outcomes
HT and TH are different. The outcome HT means that the first coin showed heads and thesecond coin showed tails. The outcome TH means that the first coin showed tails and thesecond coin showed heads.
a. Let A = the event of getting at most one tail. Find the event space.
b. Let B = the event of getting all tails. Find the event space and thecomplement of B.
c. Find the probability of A and the probability of B.
d. Let C = the event of getting all heads. Are B and C mutually exclusive?
e. Let D = the event of getting more than one tail. Find the event space andthe probability of D.
f. Let E = the event of getting a head on the first toss. Find the event space
and the probability of E.
g. Let F = the event of getting at least one (one or two) tail in two flips. Find
the event space and the probability of F.

150

Solution 4.8
a. Let A = the event of getting at most one tail. (At most one tail means zero or one
tail.) Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT
and TH each show one tail.
b. Let B = the event of getting all tails. B can be written as {TT}. Note that Events A
and B are mutually exclusive. In fact, B is the complement of A, so P(A) + P(B) =P(A) + P(A′) = 1.
c. P(A) = 3/4, and P(B) = 1/4.
d. Let C = the event of getting all heads; then B and C are mutually exclusive. Clearly
B, C have no outcomes in common because it is not possible to have all tails and allheads at the same time.
e. Let D = event of getting more than one tail. D = {TT}, so P(D) = 1/4.
f. Let E = event of getting a head on the first toss. This means that we can get either ahead or tail on the second toss, so E = {HT, HH}. Thus, P(E) = 2/4.
g. Find the probability of getting at least one (one or two) tail in two flips. Let F =event of getting at least one tail in two flips. Then F = {HT, TH, TT} and so P(F) =3/4.
Example 4.9
Suppose that we toss two fair coins. Find the probabilities of the events.
a. Let F = the event of getting at most one tail (zero or one tail).
b. Let G = the event of getting two faces that are the same.
c. Let H = the event of getting a head on the first flip followed by a head or tail on thesecond flip.
d. Are F and G mutually exclusive?
e. Let J = the event of getting all tails. Are J and H mutually exclusive?
Solution 4.9:
Refer to the sample space in Example 4.8.
a. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = 3/4.
b. Two faces are the same if HH or TT show up. P(G) = 2/4 = 1/2.
c. A head on the first flip followed by a head or tail on the second flip occurs when HH
or HT show up. So P(H) = 2/4 = 1/2.
d. F and G share the outcome HH so F and G are not mutually exclusive.
e. Getting all tails occurs when tails shows up on both coins (TT). H’s outcomes are HH
and HT. J and H have no outcomes in common, J and H are mutually exclusive.

151

4.8 A box has two balls, one white and one red. We select one ball, put it back in the box,
and then select a second ball. Find the probability of the following events:
a. F = the event of getting the white ball twice.
b. G = the event of getting two balls of different colors.
c. H = the event of getting white on the first pick.
d. Are F and G mutually exclusive?
e. Are G and H mutually exclusive?

Example 4.10
Let event C = taking an English class. Let event D = taking a speech class. Suppose P(C) =0.75, P(D) = 0.3, P(C | D) = 0.75 and P(C and D) = 0.225. Use this information to answer
the following:a. Are C and D independent?
b. Are C and D mutually exclusive?
c. What is P(D | C)?
Solution 4.10
a. Yes, because P(C | D) = P(C).
b. No, because P(C and D) ≠ 0.
c.

P( D | C ) =P(C and D) 0.225== 0.75
P(C )
0.3

152

4.3 | Two Basic Rules of Probability
Recall that in classical probability, the probability of an event is a fraction where thenumerator is the number of outcomes in the event and the denominator is the number ofoutcomes in the sample space. When calculating probability, there are two rules to consider
when determining if two events are independent or dependent and if they are mutually
exclusive or not.

Addition Rule
If Events A and B are defined on the same sample space, then the probability of event A orevent B is written as
If A and B are two events defined on the same sample space, thenP(A or B) = P(A) + P(B) - P(A and B).
If A and B are mutually exclusive which means that event A and event B can’t happen at thesame time; that is, P(A and B) = 0. Therefore the addition rule becomes:
If A and B are mutually exclusive, then:P(A or B) = P(A) + P(B)The Multiplication Rule
Recall that the probability of A given B equals the probability of A and B divided by theP( A and B)
probability of B. That is, P( A | B) =. Multiplying both sides by P(B), we have:P( B )
If A and B are two events defined on the same sample space, then:P(A and B) = P(B)∙P(A|B)

If A and B are independent, then P(A|B) = P(A). So, for independent events, themultiplication rule simplifies to:
If A and B independent, then:P(A and B) = P(A)∙P(B)153

Example 4.11
Klaus is trying to choose where to go on vacation. Klaus can only afford one vacation.
His two choices are: A = New Zealand and B = Alaska. The probability that he chooses A isP(A) = 0.6 and the probability that he chooses B is P(B) = 0.35. P(A and B) = 0, because
Klaus can only afford to take one vacation. What is the probability that Klaus chooses
either New Zealand or Alaska?
Solution 4.11
The probability that he chooses either New Zealand or Alaska is:P(A or B) = P(A) + P(B) = 0.6 + .35 – 0 = .95
Note that the probability that he does not choose to go anywhere on vacation must be 0.05.
Example 4.12
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going toattempt two goals in a row in the next game. Let A = the event Carlos is successful on his
first attempt, so P(A) = 0.65. Let B = the event Carlos is successful on his second attempt.
P(B) = 0.65. Carlos tends to shoot in streaks; the probability that he makes the second goal
given that he made the first goal is 0.90.
a.
b.
c.
d.

What is the probability that he makes both goals?
What is the probability that Carlos makes either the first goal or the second goal?
Are A and B independent?
Are A and B mutually exclusive?

Solution 4.12
a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90:
P(B AND A) = P(B|A)P(A)= (0.90)(0.65)= 0.585
Carlos makes the first and second goals with probability 0.585.
b. The problem is asking you to find P(A OR B).
P(A OR B)= P(A) + P(B) - P(A AND B)= 0.65 + 0.65 - 0.585= 0.715
Carlos makes either the first goal or the second goal with probability 0.715.

154

c. No, they are not independent events, because P(B AND A) = 0.585.
P(B)∙P(A) = (0.65)(0.65) = 0.423
0.423 ≠ 0.585 = P(B AND A) So, P(B AND A) is not equal to P(B)∙P(A).
d. No, they are not mutually exclusive because P(A and B) = 0.585.
To be mutually exclusive P(A and B) = 0.

4.11 Helen plays basketball. For free throws, she makes the shot 75% of the time. Helenmust now attempt two free throws. C = the event that Helen makes the first shot. P(C) =0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helenmakes the second free throw given that she made the first is 0.85. What is the probability
that Helen makes both free throws?

Example 4.13
A community swim team has 150 members. Seventy-five of the members are advanced
swimmers. Forty- seven of the members are intermediate swimmers. The rest are novice
swimmers. Forty of the advanced swimmers’ practice four times a week. Thirty of theintermediate swimmers’ practice four times a week. Ten of the novice swimmers’ practicefour times a week. Suppose one member of the swim team is chosen randomly.
a. What is the probability that the member is a novice swimmer?
b. What is the probability that the member practices four times a week?
c. What is the probability that the member is an advanced swimmer and practices fourtimes a week?
d. What is the probability that a member is an advanced swimmer and an intermediateswimmer? Are being an advanced swimmer and an intermediate swimmer mutually
exclusive? Why or why not?
e. Are being a novice swimmer and practicing four times a week independent events?
Why or why not?

155Solution 4.13
a.
28/150b.
80/150c.
40/150d.
P(advanced and intermediate) = 0, so these are mutually exclusive events. A
swimmer cannot be an advanced swimmer and an intermediate swimmer at thesame time.
e.
No, these are not independent events.
P(novice and practices four times per week) = 0.0667
P(novice)P(practices four times per week) = 0.0996
Since 0.0667 ≠ 0.0996, the events are not independent.

4.12 A school has 200 seniors of whom 140 will be going to college next year. Forty will begoing directly to work. The remainder will be taking a gap year. Fifty of the seniors going
to college play sports. Thirty of the seniors going directly to work play sports. Five of theseniors taking a gap year play sports. What is the probability that a senior is taking a gap
year?
Example 4.14
Felicity attends the College of Lake County in Grayslake, IL. The probability that Felicity
enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65.
The probability that she enrolls in a math class given that she enrolls in speech class is0.25. Let: M = math class, S = speech class.
a. What is the probability that Felicity enrolls in math and speech? i.e. Find P(M andS) = P(M|S)∙P(S).
b. What is the probability that Felicity enrolls in math or speech classes? i.e. Find P(Mor S) = P(M) + P(S) - P(M and S).
c. Are M and S independent? Is P(M | S) = P(M)?
d. Are M and S mutually exclusive? Is P(M and S) = 0?
Solution 4.14
a. 0.1625, b. 0.6875,

c. No,

d. No

156

4.13 A student goes to the library. Let events B = the student checks out a book and D =the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
a. Find P(B and D).
b. Find P(B or D)

Example 4.15
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 willdevelop breast cancer. Suppose that of those women who develop breast cancer, a test isnegative 2% of the time. Also suppose that in the general population of women, the test forbreast cancer is negative about 85% of the time. Let B = woman develops breast cancer andlet N = tests negative. Suppose one woman is selected at random.
a. What is the probability that the woman develops breast cancer? What is the probability
that woman tests negative?
b. Given that the woman has breast cancer, what is the probability that she tests
negative?
c. What is the probability that the woman has breast cancer AND tests negative?
d. What is the probability that the woman has breast cancer or tests negative?
e. Are having breast cancer and testing negative independent events?
f.

Are having breast cancer and testing negative mutually exclusive?

Solution to 4.15
a. P(B) = 0.143; P(N) = 0.85
b. P(N|B) = 0.02
c. P(B and N) = P(B)∙P(N|B) = (0.143)(0.02) = 0.0029
d. P(B or N) = P(B) + P(N) - P(B and N) = 0.143 + 0.85 - 0.0029 = 0.9901
e. No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N).
f. No. P(B and N) = 0.0029. For B, N to be mutually exclusive, P(B and N) must equal 0.

157

4.14 A school has 200 seniors of whom 140 will be going to college next year. Forty will
be going directly to work. The remainder is taking a gap year. Fifty of the seniors going
to college play sports. Thirty of the seniors going directly to work play sports. Five of theseniors taking a gap year play sports. What is the probability that a senior is going tocollege and plays sports?

Example 4.16
Refer to the information in Example 4.15 and let P = tests positive.
a. Given that a woman develops breast cancer, what is the probability that she tests
positive? Find P(P|B) = 1 - P(N|B).
b. What is the probability that a woman develops breast cancer and tests positive?
Find P(B and P) = P(P|B)∙P(B).
c. What is the probability that a woman does not develop breast cancer? Find P(B′).
d. What is the probability that a woman tests positive for breast cancer? Find P(P) = 1- P(N).
Solution 4.16
a. 0.98;b. 0.1401;
c. 0.857;
d. 0.15

4.15 A student goes to the library. Let events B = the student checks out a book and D =the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.
a. Find P(B′).
b. Find P(D and B).
c. Find P(B|D).
d. Find P(D AND B′).
e. Find P(D|B′).

158

Tree Diagrams and Venn Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment.
It consists of "branches" that are labeled with either frequencies or probabilities. Tree
diagrams can make some probability problems easier to visualize and solve. The following
example illustrates how to use a tree diagram.
Example 4.17
In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw twoballs, one at a time, with replacement. "With replacement" means that you put the first ballback in the urn before you select the second ball.
The tree diagram using frequencies that show all the possible outcomes follows.

8/11

3/11

8/11

3/11

64/121

24/121

24/121

9/121

The first set of branches represents the first draw. The second set of branches represents
the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1,
R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR
outcomes can be written as: R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3.
The other outcomes are similar.

159

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement.
There is a total of 11 x 11 = 121 outcomes; this is the size of the sample space.
a.
b.
c.
d.
e.
f.

List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
Using the tree diagram, calculate P(RR)
Using the tree diagram, calculate P(RB or BR)..
Using the tree diagram, calculate P(R on 1st draw and B on 2nd draw).
Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw)
Using the tree diagram calculate P(BB)

Solution 4.17
a. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3;
B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3
b. P(RR) = P(1st Draw Red and 2nd Draw Red) = (3/11)∙(3/11) = (9/121)
Note that we get this probability by “multiplying along the branches”
c. P(RB or BR) = (3/11)∙(8/11) + (8/11)∙(3/11) = (48/121)
d. P(R on 1st draw and B on 2nd draw) = (3/11)∙(8/11) = (24/121)
e. P(R on 2nd draw GIVEN B on 1st draw) = P(R|B) = (24/88) = (3/11)
This problem is a conditional one. The sample space has been reduced to thoseoutcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible
outcomes (24 BR and 64 BB). Twenty-four of those outcomes are BR.
f. P(BB) = (64/121)
If you do the same experiment this time without replacement, then the tree diagram is160

There are a total of 11 balls in the urn. Draw two balls, one at a time, without replacement.
There are 11 × 10 = 110 outcomes; again this is the size of the sample space.
a.
b.
c.
d.
e.

Using the tree diagram, calculate P(RR)
Using the tree diagram, calculate P(RB OR BR)..
Using the tree diagram, calculate P(R on 1st draw and B on 2nd draw).
Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw)
Using the tree diagram calculate P(BB)

Solution for without replacement
a. P(RR) = P(1st Draw Red and 2nd Draw Red) = (3/11)∙(2/10) = (6/110) = 3/55
b. P(RB OR BR) = (3/11)∙(8/10) + (8/11)∙(3/10) = (48/110) = 24/55
c. P(R on 1st draw and B on 2nd draw) = (3/11)∙(8/10) = (24/110) = 12/55
d. P(R on 2nd draw GIVEN B on 1st draw) = P(R|B) = (3/10)
This is a conditional probability. Since a Blue was draw first that leaves only 10balls left. From the 10 balls left, all 3 red are still in the urn.
e. P(BB) = (8/11)∙(7/10) = (56/110) = 28/55

4.16 In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are
not face cards (N). Draw two cards, one at a time, without replacement. Draw a tree diagramis labeled with all possible probabilities.
A Venn Diagram is another visual aid that can be used to display the relationships
between different events for an experiment, and their probabilities. It generally consists ofa box that represents the sample space S together with circles or ovals representing events.
Example 4.18
Forty percent of the students at a local college belong to a club and 50% work part time.
Five percent of the students work part time and belong to a club. Draw a Venn diagram
displaying this information, letting C = “student belongs to a club” and PT = “student
works part time”.
If a student is selected at random, find:a. The probability that the student belongs to a club given that the student works parttime.
b. The probability that the student belongs to a club or works part time.
c. The probability that a student works part-time but does not belong to a club.
d. The probability that a student neither belongs to a club nor works part time.

161

Solution 4.18:
We draw a rectangle representing the sample space S, and then draw two circles, one for C
and the other representing PT. The overlap between the two circles represents theoutcomes that the two events have in common. That is, this region represents the event C
and PT.
Next, we fill in the probabilities, starting with the event C and PT, whose probability is .05.
We are given that P(C) = .40; so the portion of C that does not overlap PT must haveprobability .35. Similarly, the part of PT that does not overlap C will have probability .45.
This gives us the diagram:

S

C
0.35

PT
.05

0.45

C and PT
a. The probability that the student belongs to a club given that the student works parttime is:P(C and PT ) .05
P(C | PT ) === 0.10
P( PT )
.50
b. The probability that the student belongs to a club or works part time can be found using
the addition rule:P(C or PT ) = P(C ) + P( PT ) − P(C and PT ) = .40 + .50 – 0.05 = 0.85.

Or, we can just add the three probabilities that are inside the two circles:
P (C or PT ) = .35 + .05 + .45 = 0.85.

c. The outcomes where a student works part time, but does not belong to a club isrepresented by the portion of the PT circle that does not overlap with C. This probability is
0.45.
d. The event that a student neither belongs to a club nor works part time is represented by
the part of the box that is outside both circles. That is, this event is the complement of theevent C or PT. By the complement rule,P(neither in club nor working part time) = 1 – P(C or PT) = 1 – .85 = 0.15.

162

4.17 Fifty percent of the workers at a factory work a second job, 25% have a spouse who
also works, and 5% work a second job and have a spouse who also works. Draw a Venn
diagram showing the relationships, letting W = works a second job and S = spouse also
works. Find the probability that a randomly selected worker would neither work a second
job nor have a spouse that works.
Example 4.19
A person with type O blood and a negative Rh factor (Rh-) can donate blood to any personwith any blood type. It is known that 4% of African Americans have both type O blood and anegative Rh factor, 8% of African Americans have the Rh- factor, and 51% type O blood.
Make a Venn diagram for this situation, using O for the set of individuals with type O
blood, and Rh- for the individuals with the negative Rh factor.
a. Find the probability that a randomly selected African American has type O blood ornegative Rh factor.
b. Find the probability that a randomly selected African American has negative Rh
factor but does not have type O blood.
c. Find the probability that a randomly selected African American has neither type O
blood nor a negative Rh factor.
Solution 4.19:

a. 55% b. 4% c. 45%

O
.47

Rh.04

.04

O and Rh-

4.18 In a bookstore, the probability that the customer buys a novel is 0.6, and theprobability that the customer buys a non-fiction book is 0.4. Suppose that the probability
that the customer buys both is 0.2.
Draw a Venn diagram representing the situation.
b. Find the probability that the customer buys either a novel or a non-fiction book.
c. Find the probability that a customer buys a novel but does not buy a non-fiction book.

163

4.4 | Contingency Tables
A contingency table is a commonly used way of displaying data that can facilitate
calculating probabilities. The table is particularly helpful for calculating conditional
probabilities. The table displays sample values in relation to two different variables that may
be dependent or contingent on one another. Contingency tables will be used again later inthe course.
Example 4.20
Suppose a study of speeding violations and drivers who use cell phones produced the data:

Cell phone user
Not a cell phone user
Total

Speeding
violationin last year
25
45
70

No speeding
violationin last year
280
405
685

Total
305
450
755

As we see in the bottom right corner, the total number of people in the sample is 755. The
row totals are 305 and 450. The column totals are 70 and 685. Notice that we can get the
total sample size by adding either the row totals or column totals:
305 + 450 = 755 and 70 + 685 = 755.
Let C = event that person is a cell phone user; so Cʹ = event that person is not a cell phoneuser.
Let V = event that person had a speeding violation in the last year, so Vʹ = event that
person had no speeding violation in the last year. Use the table to calculate the following
probabilities:
a. Find P(C).
b. Find P(Vʹ).
c. Find P(Vʹ and C).
d. Find P(Vʹ or C).
e. Find P(C given V)
f. Find P(Vʹ given Cʹ)Solution 4.20
a. P (=
C)

number of outcomes in C 305=≈ 0.404
number of outcomes in S 755

164

′)
b. P (V=number of outcomes in V ′ 685=≈ 0.907
number of outcomes in S 755

number of outcomes in (V ′ and C ) 280=≈ 0.371
number of outcomes in S
755

c. P (V ′ and=C)

d. For part d, we can use the addition rule:number of outcomes in (V ′ or C )number of outcomes in S
685 305 280= P (V ′) + P (C ) − P (V ′ and C ) =+−≈ 0.940755 755 755

P (V ′ or C ) =Or, we can find the number of outcomes and simply add the cells that are in either Vʹ
or C, being careful not to count any outcome twice:=
P (V ′ or C )number of outcomes in (V ′ or C ) 280 + 405 + 25=
≈ 0.940
number of outcomes in S
755

e. To find P(C given V), we use the conditional probability formula:P(C | V=)

number of outcomes in (C and V ) 25=
≈ .357
number of outcomes in V
70

Note that the sample space is limited to the number of individuals who had aviolation.
f. To find P(Vʹ given Cʹ), we again use the conditional probability formula:

P (V ′ | C ′) =number of outcomes in (C ′ and V ′) 405== .90
number of outcomes in C ′450This time the sample space is limited to the number of individuals who were not cell
phone users.

165

4.19 The following table shows the number of athletes who stretch before exercising andhow many injuries within the past year had.
Injury in last year No injury in last year Total
Stretches
55
295
350
Does not stretch
231
219
450
Total
286
514
800
a. Find the probability that an athlete stretches before exercising.
b. Find the probability that an athlete stretches before exercising given that there was
no injury in the last year.

Example 4.21
The table below shows the location preferences for a random sample of 100 boaters.
Boaters
CanoerKayakers
Total

The
Coastline
18

Freshwater
Lakes
16
41

Rivers andStreams
14

Total
45
55

a. Complete the tableb. Let A = event “canoers” and C = event “Prefers the coastline”.
Are A and C independent?
c. Find the probability that a kayaker given that the person prefers freshwater lakes.
d. Find the probability that a canoer or prefers river and streams.
Solution 4.21.
a. We start by filling in the grand total; this is the total sample size, 100. This is also thesum of the two row totals: 45 + 55 = 100. Next, we can fill in the entry in the row for“kayaker” and the column “Freshwater Lakes”. The total for this column is 41, and so
the missing value must be 41 – 16 = 25. Next, we fill in the entry in the row for kayaker
and the column “The Coastline”. The row total is 55, and so this last entry must be 55 –25 – 14 = 16:
Boaters The Coastline Freshwater Lakes River and Streams Total
Canoers
18
16
45
Kayakers
16
25
14
55
Total
41
100

166

Continuing in this way, we get the completed table:
Boater The Coastline Freshwater Lakes River and Streams Total
Canoer18
16
11
45
Kayaker
16
25
14
55
Total
34
41
25
100
When finished, make sure to check that the row totals and the column totals add to 100.
b. To show that A and C are independent, we must show any one of the following:P(C | A) = P(C) or

P(A | C) = P(A) or P(C and A) = P(C)P(A)We will use the last one. From the table we quickly calculate:
34

45

18

𝑃𝑃(𝐶𝐶) = 100 = .34, 𝑃𝑃(𝐴𝐴) = 100 = .45 and 𝑃𝑃(𝐶𝐶 and 𝐴𝐴) = 100 = .18

Since P(C and A) = .18 ≠ (.34)(.45) = P(C)P(A), the events are not independent.
c. Let L be the event “prefers lakes and streams” and let K be “kayakers”. Then we wantto calculate P(K | L):
𝑃𝑃(𝐾𝐾|𝐿𝐿) =number of outcomes in (𝐾𝐾 and 𝐿𝐿)25= ≈ .610
number of outcomes in 𝐿𝐿
41

d. If we let A = “canoer” and P = “prefers rivers and streams”, then we want to calculateP(A or P) = P(A) + P(P) – P(A and P) =

45
25
11
59
+
−
=
= .59.
100
100
100
100

4.20 The following table shows a random sample of 200 cyclists and the routes they prefer.
Bike Type Lake Path Hilly Path Wooded Path Total
Upright
45
38
27
110
Recumbent
26
52
12
90
Total
71
90
39
200
a. Out of the upright cyclists, what is the probability that the cyclist prefers a hilly path?
b. Let R = recumbent and H = hilly path. Are the events R and H independent?

167

Example 4.22
The table below contains the number of crimes per 100,000 inhabitants from 2008 to 2011
in the U.S.
Year
Robbery Burglary Rape Vehicle Total
2008
145.7
732.1
29.7
314.7
2009
133.1
717.7
29.1
259.2
2010
119.3
701
27.7
239.1
2011
113.7
702.2
26.8
229.6
Total:
United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

a. Find P(2009 and Robbery).
b. Find P(2010 and Burglary).
c. Find P(2010 or Burglary).
d. Find P(2011 | Rape).
e. Find P(Vehicle | 2008).
Solution 4.22:
a. 0.0294; b. 0.1551;

c. 0.7165;

d. 0.2365;

e. 0.2575

4.21 The following table relates the vaccines and age ranges of a group of individuals
participating in an observational study.
Vaccines/Ages
21 – 40 Years Old 41 – 60 Years Old 61+ Years Old
Total
Johnson & Johnson 18
28
14
Pfizer20
51
28
Moderna
12
25
9
Total
a. Find the total for each row and column.
b. Find the probability that a randomly chosen individual from this group is 21 – 40years old.
c. Find the probability that a randomly chosen individual from this group had Johnson& Johnson and is 21 – 40 years old.
d. Find the probability that a randomly chosen individual from this group is 21 – 40years old given that the individual has Johnson & Johnson.
e. Find the probability that a randomly chosen individual from this group had Johnson& Johnson given that the individual is 21 – 40 years old.
f. Find the probability a randomly chosen individual from this group is 21 – 40 years
old and had Moderna.
g. Are the events of having Johnson & Johnson and 21 – 40 years old independent?

168

4.5 | Counting and Probability
We now present some tools for counting to better utilize the basic definition of theoreticalprobability. The underlying theme is to represent outcomes in a sample space as “lists”
formed from some collection of symbols. The fundamental questions we will need to answer
are:
1. Are the lists ordered or unordered?
2. Are the lists drawn with or without replacement?
In most cases, if we can explicitly describe the sample space, and can answer these basic
questions, we will be able to count outcomes in our event and in the sample space.
The first counting principle is sometimes called the “Fundamental Principle of Counting”—
as its name suggests, it forms the foundation of all our subsequent counting methods.
Fundamental Principle of Counting
Suppose we have a task that can be broken down into a sequence of k steps. Further
suppose that there are

N 1 ways to do step 1, N 2 ways to do step 2, . . . , and N k ways to do step k. Then there are
N 1 N 2  N k ways to complete the entire task.

Notice that the Basic Counting Principle can be used only when counting an orderedarrangement—the statement explicitly states that the task is broken down into a sequenceof steps.
Example 4.23
A license plate consists of three letters (from the English alphabet) followed by three digits.
How many license plates are possible?
Solution 4.23:
To form a valid license, we must fill in six slots: ___ ___ ___ ___ ___ ___For each of the first three, there are 26 choices. For the last three, there are 10 choices
each. Thus, there are a total of 26 x 26 x 26 x 10 x 10 x 10 = 26 310 3 total possibilities.

169

Example 4.24
a. How many social security numbers are possible?
b. What is the probability that a social security number has no repeated digits?
c. What is the probability that a social security number has at least one repeated digit?
Solution 4.24:
a. To form a valid social security number, we need to fill in nine slots:___ ___ ___ - ___ ___ - ___ ___ ___ ___There are 10 choices for each slot, hence there are 10 9 possible social security numbers.
b. Note that the calculation in part a counted the number of outcomes in the sample space
for this problem. Again, we draw nine slots:___ ___ ___ - ___ ___ - ___ ___ ___ ___There are 10 choices for the first slot. Since we cannot have repeated digits, there are only
9 choices for the second, only 8 choices for the third, etc. Thus, there is a total of
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3,628,800
Possible social security numbers without repeated digits. Thus, the probability that a social
security number has no repeated digits is:# (no repeats)
3,628,800
P(no repeats) =≈ 0.0036 .
=total numbef of SSN' s
10 9c. This problem is easy once the previous problem is completed. Noting that “at least one
repeat” is the complement of “no repeats”, we use the complement rule to get:Pr ( at least one repeat) = 1 – Pr ( no repeats)= 1 – 0.0036 = 0.9964.

4.23 A company uses five-digit identification numbers for their employees. What is theprobability that a randomly selected employee has no repeated digits in his/her ID
number?

170

Factorial Notation
The product of the first n positive integers is often abbreviated by n!; this is read as “nfactorial”. That is,n! = n (n − 1)(n − 2)  3 ⋅ 2 ⋅ 1
Note that n ! = n ⋅ (n − 1)! .
Also, it is convenient to define 0! = 1! = 1.

To understand why n ! = n ⋅ (n − 1)! , we consider a simple example and look at 7!.
By definition, 7! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 . But we can also put parentheses around the last six
factors to get 7! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 7 ⋅ (6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) = 7 ⋅ 6!
As our examples above suggest, factorial notation is very convenient for counting problems
using the basic counting principle; that is, counting ordered arrangements.
Example 4.25:
How many ways can eight people line up for a photograph?
Solution 4.25:
We have eight slots to fill in: ___ ___ ___ ___ ___ ___ ___ ___ .
There are 8 choices for the first, 7 choices for the second, 6 choices for the third, etc. Thus,
there is a total of 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 8! ways to line them up.
The basic counting rules for ordered arrangements can be summarized as follows:
Counting Ordered Arrangements
1. There are k! ways to order k distinct objects.
2. The number of ordered arrangements of k objects, chosen without
replacement from a collection of n objects, isn (n − 1)(n − 2)  (n − k + 1) =n!
.
(n − k )!

3. The number of ordered arrangements of k objects, chosen with replacement.
from a collection of n objects, is n k .

171Justification:

In each of the three cases, we have k slots to fill:___ ___ ___ . . . ___ ___ ___In case (1), there are k choices for the first, k –1 choices for the second..., and only one
choice for the last slot. Hence there are k (k − 1)  2 ⋅ 1 = k! possible orderings.
In case (2), there are n choices for the first, n –1 choices for the second, . . ., andn − (k − 1) = n − k + 1 choices for the last slot. Hence there is a total ofn(n − 1)(n − 2)  (n − k + 1)
possible arrangements. Multiplying and dividing by (n − k )! yields the alternative way ofwriting the expression.
In case (3), we are choosing with replacement, so there are n choices for each slot, hence a
total of n ⋅ n ⋅  n = n k possible arrangements.
NOTES These formulas are recorded mostly for future reference. When counting orderedarrangements, it is best to simply use the basic counting principle directly—write out thecorrect number of slots, fill in the number of choices for each slot, and multiply.
Example 4.26
A club at a local community college has 14 freshman and 18 sophomores. Suppose that theclub will elect a president, vice president and treasurer.
a. What is the probability that the president will be sophomores, and the other two
officers are freshman?
b. What is the probability that all three officers are sophomores?
Solution 4.26:
First we note that there are a total of 32 members in the club; and no person can fill two
different posts, so there will be no repetitions. By the Fundamental Counting Principle,
there is a total of
32 x 31 x 30 = 29,760 ways to choose the officers.
a. There are 18 sophomores to choose from to serve as president. Then there are 14
freshmen, so there are 14 choices for vice president and 13 choices for treasurer. Thus,
there is a total of 18 x 14 x 13 = 3276 ways to get a sophomore president, and freshman vicepresident and treasurer. So, the probability is 3276/29,760 = 0.110.
b. There are 18 sophomores to choose from. So, there are 18 choices for president, 17 choices
for vice president, and 16 choices for treasurer. Thus, there is a total of 18 x 17 x 16 = 4896
ways to get all sophomore officers. So, the probability is 4896/29,760 = 0.165.

172

4.25 Suppose that a club has 20 freshmen and 22 sophomores. If they are to elect apresident, vice president and secretary, what is the probability that the officers will all be
freshmen? What is the probability that at least one sophomore is elected as an officer?

Permutations and Combinations:
An ordered arrangement of k objects, chosen without replacement from a collection of nobjects, is called a permutation. The total number of such arrangements is denoted by nPk,.
And as we saw above, this number can be calculated using the following formula:n Pk =n!
(n − k )!

This formula is not always necessary in practice, because any time we want to count orderedarrangements, we can just use the fundamental counting principle. However, the formula
is necessary to develop a formula for counting unordered arrangements.
An unordered arrangement of k objects, chosen without replacement from a collection of nobjects, is called a combination. The total number of such arrangements is denoted by nCk,,
and is given by the formulan Ck =

n!
(read as “n choose k”)
k!(n − k )!

To understand where this formula comes from, we think about the number of orderedarrangements of k objects chosen without replacement from a set of n objects. We can suchan arrangement in two steps: First, we select a subset of k objects from the set; then we putthem in order. By the Fundamental Principle of Counting, we have:Total # of ordered arrangements = Total # of subsets of size k × Number of ways to order kobjects

We know that the number on the left is n Pk =n!
. And we know that there are k! ways(n − k )!

to order k distinct objects. Finally, the number of subsets of size k is nCk,, which is what wen!
= n C k ⋅ k ! Dividing both sides by
are trying to find. So the equation above becomes:(n − k )!
k!, we get n C k =n!
.
k!(n − k )!

173

NOTE
The numbers nCk are often called binomial coefficients, because they appear in the Binomial
Theorem, which you may remember from your high school algebra class. In the next
chapter they will also be used for an important probability distribution called the Binomial
distribution.
Permutations and Combinations
Permutations:
Order matters so look for key words like position titles, “in order” or “in a line”
n!
Formulan Pk =(n − k )!
Combinations:
Order does not matter so look for key words like “committee” or “group”n!
Formulan Ck =
k!(n − k )!

Note also that the formulas for nPk, and nCk, are built into the TI-83 and TI-84 calculators:

Using the TI-83, 83+, 84, 84+ Calculator
To calculate nPk and nCk we go press the MATH button, and scroll to the PRB menu.
To find nPk,, enter n, then press MATH >> PRB; press 2 (nPr) and then enter k and press
the ENTER key.
To find nCk,, enter n, then press MATH >> PRB; press 3 (nPr) and then enter k and press
the ENTER key.

4.26 Use the calculator to find each of the following:
10P3, 10C3 , 8P4 ,and 8C4.

174

Example 4.27
A shipment of 40 DVD players contains eight defective units. Suppose that a sample of four
players is selected, and that the entire shipment will be rejected if at least one DVD player
in the sample is defective.
a. What is the probability that the sample contains no defective players?
b. What is the probability that the shipment will be rejected?
Solution 4.27:
There are 40C4 = 91,390 ways to choose a sample of four DVD players from the shipment.
a. There are 36 non-defective players, so there are 36C4 = 58,905 ways to choose four nondefective players. Thus, the probability is P(no defective players) = 58,905/91,390 = 0.6445,
or about 64.5%.
b. The probability calculated in part a is the probability that the shipment is not rejected.
So we can use the complement rule to get:P(rejected) = 1 – P(no defective players) = 1 – 0.6445 = 0.3555.

Example 4.28
A school committee of four is to be selected from a class of 7 teachers and 9 administrators.
a.
b.
c.
d.
e.

In how many ways can the committee be selected?
How many ways can the committee be chosen to have 3 administrators?
What is the probability that the committee has 3 administrators?
What is the probability that the committee has at least three administrators?
What is the probability that the committee has at least one teacher?

Solution 4.28:
a. A committee is an unordered arrangement without replacement, so the formula above
applies: We are choosing 4 objects from a set of 16 objects, so the total number ofcommittees is 16C4.Using the calculator, we press 16, then go to MATH >> PRB and selectitem 3 (nCr); then press 4 and ENTER to get 16C4 = 1820.
b. We will choose the committee in two stages: First select three administrators, thenselect one teacher to complete the committee. There are 9C3 = 84 ways to choose threeadministrators from among the nine. There are 7C1 = 7ways to choose the teachers andcomplete the committee. Thus there are (9C3)( 7C1) = 84 x 7 = 588 committees with threeadministrators.

175

c. The probability that a committee has three administrators is
P (three administrators) =#(committees with three administrators)
588== 0.323.
total # committees
1820

d. The probability that a committee has at least three administrators there are two
possibilities: either there are exactly three administrators, or there are exactly four
administrators. So, the probability we want is P (three administrators and one teacher) + P(four administrators).
We have already calculated the probability of getting three administrators on thecommittee, so we must now calculate the probability of getting four administrators on thecommittee. There are 9C4 = 126 ways to choose the committee consisting of alladministrators (verify this), and from part (a), there are1820 committees altogether.
Thus, the probability that the committee consists entirely of administrators is 126/1820 =.06923. Combining this with the result from part c we have:P (three administrators) + P (four administrators) = 0.323 + 0.06923 = 0.39223.
e. We will do this one using the complement rule. If there is not one or more teacher on thecommittee, then the committee must consist entirely of administrators. I.e. thecomplement of the event “at least one teacher” is the event “all administrators”. Thus theprobability that the committee has at least one teacher isP(at least one teacher) = 1 – P(all administrator).
Using the result from part (d), the probability that the committee has at least one teacher is
P (at least one teacher) = 1 – P (all administrator) = 1 – 0.06923 = 0.93077.
Some important things to notice about this example:••

Often, we are using more than one counting principle at a time.
The overall strategy for computing probabilities is more important than theformulas.

176

KEY TERMS
Chance Experiment The result of the experiment is not predetermined
Complement The complement of event A (denoted as Aʹ ) the set of outcomes that are notin A.
Conditional Probability the likelihood that an event will occur given that another event
has already occurred.
Contingency Table A method of displaying a frequency distribution as a table with rows
and columns to show how two variables may be dependent (contingent) upon each other;
the table provides an easy way to calculate conditional probabilities.
Dependent Events If two events are NOT independent, then we say that they are
dependent.
Empirical Probability The long-term relative frequency of an event which approaches
theoretical probability for that event.
Equally Likely Each outcome of an experiment has the same probability.
Event A subset of the set of all outcomes of an experiment; that is, an event is a subset ofthe sample space S. Standard notations for events are capital letters such as A, B, C, and soon.
Experiment a planned activity carried out under controlled conditions.
Independent Events The occurrence of one event has no effect on the probability of theoccurrence of another event. Events A and B are independent if one of the following is true:
1. P(A|B) = P(A)
2. P(B|A) = P(B)3. P(A AND B) = P(A)P(B)Law of Large Numbers As the number of repetitions of an experiment is increased therelative frequency observed in the experiment will become closer and closer to thetheoretical probability.
Mutually Exclusive Two events are mutually exclusive if they have no outcomes incommon. Thus, if events A and B are mutually exclusive, then P(A and B) = 0.
Outcome a particular result of an experiment
Probability a number between zero and one, inclusive, that gives the likelihood that a
specific event will occur.
Sample Space the set of all possible outcomes of an experiment177

Sampling with Replacement means that each member of a population is replaced after it
is picked, so that there is a possibility that the same item is chosen more than once.
Sampling without Replacement means that items are not put back after being chosen,
so each member of a population may be chosen only once.
Theoretical Probability The probability of an event calculated by the number ofoutcomes in the event divided by the total number of outcomes in the sample space.
The AND Event The event A and B consists of all outcomes that are common to both
events. So, an outcome is in the event A and B if the outcome is in both A and in event B.
Tree Diagram A visual representation of a sample space and events in the form of a “tree”
with branches marked by possible outcomes together with associated probabilities(frequencies, relative frequencies)
Venn Diagram A visual representation of a sample space and events in the form of circles
or ovals showing their intersections.

178

FORMULA REVIEW
Theoretical Probability: For a sample space with equally likely outcomes,
number of outcomes in A
P( A) =
number of outcomes in S
Empirical Probability:

P( A) =

number of times A occurs
total number of trials

Conditional Probability: P( A | B) =Complement Rule:P( A and B)
P( B )P ( A′) = 1 − P ( A)

Addition Rules:•

For any events A, B:•

If A, B are mutually exclusive, then P( A or B) = P( A) + P( B)

P( A or B) = P( A) + P( B) − P( A and B)

Multiplication Rules:•

For any events A, B:•

If A, B are independent, then P ( A and B ) = P ( A) P ( B)

P( A and B) = P( A) P( B | A)

Fundamental Principle of Counting.
Suppose we have a task that can be broken down into a sequence of k steps. Further
suppose that there are N 1 ways to do step 1, N 2 ways to do step 2, . . . , N k ways to do
step k.
Then there are N 1 N 2  N k ways to do the entire task.
A permutation is an ordered arrangement of k objects, chosen without replacement from an!
collection of n objects. The total number of such arrangements is: n Pk =.
(n − k )!
A combination is an unordered arrangement of k objects, chosen without replacement fromn!
a collection of n objects. The total number of such arrangements is: n C k =k! (n − k )!

179

Exercises for Chapter 4
1. If the experiment is drawing two colored marbles from a box of 8 blue marbles, 10 green
marbles, and 5 purple marbles, then state the sample space.
2. If the experiment is testing 3 items from a shipment box that contain defective (D) andnon-defective (N) items. State the sample space.
3. A bag with 7 letters (A, B, C, D, E, F, G) is shaken and two letters are drawn without
replacement. State the sample space.
4. In a particular college class, there are sophomore and freshman students. Some students
have a math class, and some students have an English class. Write the symbols for the
probabilities of the events for parts a through j. (Note that you cannot find numerical answers
here. You were not given enough information to find any probability values yet; concentrateon understanding the symbols.)• Let F be the event that a student is freshman.
• Let S be the event that a student is sophomore.
• Let M be the event that a student has a math class.
• Let E be the event that a student has an English class.
a. The probability that a student does not have a math class.
b. The probability that a student is sophomore or has an English class.
c. The probability that a student is a freshman and has a math class.
d.

The probability that a student is sophomore, given that the student has a mathclass.

e. The probability that a student has a math class, given that the student issophomore.
f.

Of all the freshman students, the probability that a student has an English class.

g.

Of all students with a math class, the probability that a student is freshman.

h. The probability that a student is freshman or has a math class.
i.

The probability that a randomly selected student is a sophomore student taking an
English class.

j.

The probability that a student is freshman.

180

5. A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger
traps, and five bags of confetti.
Let H = the event of getting a hat.
Let N = the event of getting a noisemaker. Let F = the event of getting a finger trap.
Let C = the event of getting a bag of confetti.
a.
b.
c.
d.

Find P(H).
Find P(N).
Find P(F).
Find P(C).

6. A jar of 150 jellybeans contains 22 red jellybeans, 38 yellow, 20 green, 28 purple, 26
blue, and the rest are orange.
a. Let B = the event of getting a blue jellybean. Find P(B)b. Let G = the event of getting a green jellybean. Find P(G)c. Let O = the event of getting an orange jellybean. Find P(O)d. Let P = the event of getting a purple jellybean. Find P(P)e. Let R = the event of getting a red jellybean. Find P(R)f. Let Y = the event of getting a yellow jellybean. Find P(Y)7. There are 23 countries in North America, 12 countries in South America, 47 countries in
Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean
region).
a.
b.
c.
d.
e.
f.

Let A = the event that a country is in Asia. Find P(A).
Let E = the event that a country is in Europe. Find P(E).
Let F = the event that a country is in Africa. Find P(F).
Let N = the event that a country is in North America. Find P(N).
Let O = the event that a country is in Oceania. Find P(O).
Let S = the event that a country is in South America. Find P(S).

8. A card is drawn from a standard deck. Find the following probabilities.
a. What is the probability of a red card is drawn?
b. What is the probability of a club is drawn?
c. What is the probability of a 9 or diamond is drawn?
d. What is the probability of a 4 or king is drawn?
e. What is the probability of a 6 and heart is drawn?
9. The experiment is rolling a fair, six-sided die numbered one through six.
a. What is the probability of rolling an even number of dots?
b. What is the probability of rolling a prime number of dots?

18110. You see a game at a local fair. You must throw a dart at a color wheel. Each section on
the color wheel is equal in area.
a. Let B = the event of landing on blue. Find P(B)b. Let R = the event of landing on red. Find P(R′)
c. Let G = the event of landing on green. Find P(G′)
d. Let Y = the event of landing on yellow. Find P(Y).
11. On a baseball team, there are infielders and outfielders. Some players are great hitters,
and some players are not great hitters. Let I = the event that a player in an infielder.
Let O = the event that a player is an outfielder. Let H = the event that a player is agreat hitter. Let N = the event that a player is not a great hitter.
a. Write the symbols for the probability that a player is not an outfielder.
b. Write the symbols for the probability that a player is an outfielder or is a great
hitter.
c. Write the symbols for the probability that a player is an infielder and is not a
great hitter.
d. Write the symbols for the probability that a player is a great hitter, given that theplayer is an infielder.
e. Write the symbols for the probability that a player is an infielder, given that theplayer is a great hitter.
f. Write the symbols for the probability that of all the outfielders, a player is not a
great hitter.
g. Write the symbols for the probability that of all the great hitters, a player is an
outfielder.
h. Write the symbols for the probability that a player is an infielder or is not a great
hitter.
i. Write the symbols for the probability that a player is an outfielder and is a great
hitter.
j. Write the symbols for the probability that a player is an infielder.
12. What is the word for the set of all possible outcomes?
13. What is conditional probability?
14. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different
book with a unique title. The fiction books are numbered one to eight. The nonfiction
books are numbered one to four. Randomly select one book. Let F = event that book is
fiction. Let N = event that book is nonfiction. What is the sample space?
15. What is the sum of the probabilities of an event and its complement?

182

16. Use the following information to answer the next two exercises. You are rolling a fair,six-sided number cube. Let E = the event that it lands on an even number. Let M = theevent that it lands on a multiple of three.
a. What does P(E|M) mean in words?
b. What does P(E OR M) mean in words?
17. Let E and F be mutually exclusive events. P(E) = 0.4; P(F) = 0.5. Find P(E∣F).
18. Let J and K be independent events. P(J|K) = 0.3. Find P(J).
19. Let U and V be mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find:a. P(U AND V) =b. P(U|V) =c. P(U OR V) =20. Let Q and R be independent events. P(Q) = 0.4 and P(Q AND R) = 0.1. Find P(R).
21. Forty-eight percent of all Californians registered voters prefer life in prison without
parole over the death penalty for a person convicted of first-degree murder. Among Latino
California registered voters, 55% prefer life in prison without parole over the death
penalty for a person convicted of first-degree murder. 37.6% of all Californians are Latino.
Let C = Californians (registered voters) preferring life in prison without parole over thedeath penalty for a person convicted of first-degree murder. Let L = Latino Californians.
Suppose that one Californian is randomly selected.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.

Find P(C).
Find P(L).
Find P(C|L).
In words, what is C|L?
Find P(L AND C).
In words, what is L AND C?
Are L and C independent events? Show why or why not.
Find P(L OR C).
In words, what is L OR C?
Are L and C mutually exclusive events? Show why or why not.

22. The following table shows a random sample of musicians and when they learned to playtheir instruments.

Elementary School
Middle School
Total

Self-taught
12
19
31

Studied in School
38
24
62

183

Private Instruction
22
15
37

Total
72
58
130

a.
b.
c.
d.

Find P(musician learned in elementary school).
Find P(musician learned in middle school AND had private instruction).
Find P(musician learned in elementary school OR is self-taught).
Are the events “learned in elementary school” and “learning music in school”
mutually exclusive events?

23. The probability that a man develops some form of cancer in his lifetime is 0.4567. The
probability that a man has at least one false positive test result (meaning the test comesback for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in
his lifetime; P = man has at least one false positive. Construct a tree diagram of thesituation.
24. An article in the New England Journal of Medicine, reported about a study of smokers in
California and Hawaii. In one part of the report, the self-reported ethnicity and smoking
levels per day were given. Of the people smoking at most ten cigarettes per day, there
were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 JapaneseAmericans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there
were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 JapaneseAmericans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there
were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 JapaneseAmericans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there
were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 JapaneseAmericans, and 3,970 Whites.
SmokingLevel
1 – 10
11 – 20
21- 3131+
TOTALS

AfricanAmericanNativeHawaiian

LatinoJapanese
Americans

White

Totalsa. Complete the table using the data provided. Suppose that one person from the study
is randomly selected. Find the probability that person smoked 11 to 20 cigarettes perday.
b. Suppose that one person from the study is randomly selected. Find the probability
that person smoked 31+ cigarettes per day.
c. Find the probability that the person was Latino.
d. In words, explain what it means to pick one person from the study who is “JapaneseAmerican AND smokes 21 to 30 cigarettes per day.” Also, find the probability.
e. In words, explain what it means to pick one person from the study who is “JapaneseAmerican OR smokes 21 to 30 cigarettes per day.” Also, find the probability.

184

f.

In words, explain what it means to pick one person from the study that is “JapaneseAmerican GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find theprobability.
g. Prove that smoking level/day and ethnicity are dependent events.
25. The graph in Figure 4.11 displays the sample sizes and percentages of people in different
age and gender groups who were polled concerning their approval of Mayor Ford’s actions
in office. The total number in the sample of all the age groups is 1,045.

a.
b.
c.
d.
e.
f.
g.
h.
i.
j.

Define three events in the graph.
Describe in words what the entry 40 means.
Describe in words the complement of the entry in question 2.
Describe in words what the entry 30 means.
Out of the males and females, what percent are males?
Out of the females, what percent disapprove of Mayor Ford?
Out of all the age groups, what percent approve of Mayor Ford?
Find P(Approve|Male).
Out of the age groups, what percent are more than 44 years old?
Find P(Approve|Age < 35).

26. Explain what is wrong with the following statements. Use complete sentences.
a. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday,
then there is a 130% chance of rain over the weekend.
b. The probability that a baseball player hits a home run is greater than theprobability that he gets a successful hit.

185

27. The graph shown is based on more than 170,000 interviews done by Gallup that took
place from January through December 2012. The sample consists of employed Americans
18 years of age or older. The Emotional Health Index Scores are the sample space. We
randomly sample one Emotional Health Index Score.

79.6
80
80.1
80.2
80.3
80.7
80.9
81.5
81.7
81.8
82.3
82.7
82.7
83.7

a.
b.
c.
d.
e.

Find the probability that an Emotional Health Index Score is 82.7.
Find the probability that an Emotional Health Index Score is 81.0.
Find the probability that an Emotional Health Index Score is more than 81?
Find the probability that an Emotional Health Index Score is between 80.5 and 82?
If we know an Emotional Health Index Score is 81.5 or more, what is the probability
that it is 82.7?
f. What is the probability that an Emotional Health Index Score is 80.7 or 82.7?
g. What is the probability that an Emotional Health Index Score is less than 80.2 giventhat it is already less than 81?
h. What occupation has the highest emotional index score?
i. What occupation has the lowest emotional index score?
j. What is the range of the data?
k. Compute the average EHIS.
l. If all occupations are equally likely for a certain individual, what is the probability
that he or she will have an occupation with lower-than-average EHIS?
28. The data from Gallup’s annual Values and Beliefs pill conducted on May 3 – 18, 2021,
70% support same-sex marriage. Republicans show majority support in 2021 first time at
55%. Among older adults (55+) surveyed 6 out of 10 say they favor same-sex marriage.
Assuming the surveyed Republicans who support same-sex marriage, 42% of them areolder adults.

186

a.
b.
c.
d.
e.
f.
g.
h.
i.
j.

• C = Surveyed adults who support same-sex marriage.
• B = Older adults surveyed who favor same-sex marriage.
• A = Surveyed Republicans who support same-sex marriage.
Find P(C).
Find P(B).
Find P(C|A).
Find P(B|C).
In words, what is C|A?
In words, what is B|C?
Find P(C AND B).
In words, what is C AND B?
Find P(C OR B)
Are C and B mutually exclusive events? Show why or why not.

29. After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late2021, the Forum Research polled 1,046 people to measure the mayor’s popularity.
Everyone polled expressed either approval or disapproval. These are the results theirpoll produced:• In early 2021, 60 % of the population approved of Mayor Ford’s actions in office.
• In mid-2021, 57 % of the population approved of his actions.
• In late 2021, the percentage of popular approval was measured at 42 percent.
a. What is the sample size for this study?
b. What proportion in the poll disapproved of Mayor Ford, according to the results fromlate 2021?
c. How many people polled responded that they approved of Mayor Ford in late 2021?
d. What is the probability that a person supported Mayor Ford, based on the datacollected in mid-2021?
e. What is the probability that a person supported Mayor Ford, based on the datacollected in early 2021?
30. Two spinners are spun, and the sum recorded. One spinner is numbered 1 – 6 where eachnumber is equally possible, and the even numbers are gray. The other spinner isnumbered 1- 5 where each number is equally possible, and the numbers 1 – 3 are gray.
a. Draw out the sample space shading the values where both spinners are gray.
Spinners
1
2
3
4
5
6
1
2
3
4
5

187

b.
c.
d.
e.
f.
g.

Find the probability of getting a 5, P(5).
Find the probability of getting a shaded box, P(shaded).
Find the probability of getting an even number, P(even number).
Is getting an even number the complement of getting an odd number? Why?
Find two mutually exclusive events.
Are the shaded events and the odd values independent?

31. Using the information from #30, compute the probability of the following events:a. P(an odd number)b. P(sum is 8)
c. P(sum is less than 5)
d. P(an even number and shaded)e. P(multiple of 3)f. P(greater than or equal to 9)g. P(multiple of 4 or shaded multiple of 3)
32. Using the information from #30, compute the probability of the following events:a. P(multiple of 4)b. P(an odd number and shaded)c. P(sum is 10)
d. P(prime number and shaded)e. P(greater than 7 or even shaded)f. P(odd shaded or less than 6)
33. Suppose that you have eight cards. Five are green and three are yellow. The five green
cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3.
The cards are well shuffled. You randomly draw one card.
a. List the sample space.
b. P(G) =c. P(G|E) =d. P(G AND E) =e. P(G OR E) =f. Are G and E mutually exclusive? Justify your answer numerically.
34. Roll two fair dice. Each die has six faces.
a. List the sample space.
b. Let A be the event that either a three or four is rolled first, followed by an evennumber. Find P(A).
c. Let B be the event that the sum of the two rolls is at most seven. Find P(B).
d. In words, explain what “P(A|B)” represents. Find P(A|B).
e. Are A and B mutually exclusive events? Explain your answer in one to three complete
sentences, including numerical justification.
f. Are A and B independent events? Explain your answer in one to three complete
sentences, including numerical justification.
188

35. A special deck of cards has ten cards. Four are green, three are blue, and three are red.
When a card is picked, its color of it is recorded. An experiment consists of first picking acard and then tossing a coin.
a. List the sample space.
b. Let A be the event that a blue card is picked first, followed by landing a head on thecoin toss. Find P(A).
c. Let B be the event that a red or green is picked, followed by landing a head on the cointoss. Are the events A and B mutually exclusive? Explain your answer in one to threecomplete sentences, including numerical justification.
d. Let C be the event that a red or blue is picked, followed by landing a head on the cointoss. Are the events A and C mutually exclusive? Explain your answer in one to threecomplete sentences, including numerical justification.
36. An experiment consists of first rolling a die and then tossing a coin.
a. List the sample space.
b. Let A be the event that either a three or a four is rolled first, followed by landing ahead on the coin toss. Find P(A).
c. Let B be the event that the first and second tosses land on heads. Are the events A
and B mutually exclusive? Explain your answer in one to three complete sentences,including numerical justification.
37. An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side
the coin lands on.
a. List the sample space.
b. Let A be the event that there are at least two tails. Find P(A).
c. Let B be the event that the first and second tosses land on heads. Are the events A
and B mutually exclusive? Explain your answer in one to three complete sentences,including justification.
38. Consider the following scenario: Let P(C) = 0.4. Let P(D) = 0.5. Let P(C|D) = 0.6.
a. Find P(C AND D).
b. Are C and D mutually exclusive? Why or why not?
c. Are C and D independent events? Why or why not?
d. Find P(C OR D).
e. Find P(D|C).
39. Let Y and Z are independent events.
a. Rewrite the basic Addition Rule P(Y OR Z) = P(Y) + P(Z) - P(Y AND Z) using theinformation that Y and Z are independent events.
b. Use the rewritten rule to find P(Z) if P(Y OR Z) = 0.71 and P(Y) = 0.42.
40. Let G and H are mutually exclusive events. P(G) = 0.5 P(H) = 0.3
a. Explain why the following statement MUST be false: P(H|G) = 0.4.
b. Find P(H OR G).
c. Are G and H independent or dependent events? Explain in a complete sentence.
18941. Approximately 281,000,000 people over age five live in the United States. Of these people,
55,000,000 speak a language other than English at home. Of those who speak anotherlanguage at home, 62.3% speak Spanish. Let: E = speaks English at home; E′ = speaks
another language at home; S = speaks Spanish; Finish each probability statement bymatching the correct answer.
a.
b.
c.
d.

Probability Statements
P(E′)
P(E)
P(S and E′)
P(S | E′)

i.
ii.
iii.
iv.

Answers
0.80430.623
0.19570.121942. In 2018, the U.S. government issue 695,525 Family Green Cards (permits for non-citizens
to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately
2.7 million total applications. Let G = awarded green card.
a. What was Renate’s chance of being awarded a Family Green Card? Write your answer
as a probability statement.
b. In the Fall of 2017, Renate received a letter stating she was one of 950,000 finalists
chosen. Once the finalists were chosen, assuming that each finalist had an equalchance to win, what was Renate’s chance of being awarded a Family Green Card?
Write your answer as a conditional probability statement. Let F = was a finalist.
c. Are G and F independent or dependent events? Justify your answer numerically.
d. Are G and F mutually exclusive events? Justify your answer numerically and explain
why.
43. Three professors at George Washington University did an experiment to determine if
economists are more selfish than other people. They dropped 64 stamped, addressedenvelopes with $10 cash in different classrooms on the George Washington campus. 44%
were returned overall. From the economics classes 56% of the envelopes were returned.
From the business, psychology, and history classes 31% were returned. Let: R = money
returned; E = economics classes; O = other classesa. Write a probability statement for the overall percent of money returned.
b. Write a probability statement for the percent of money returned out of the economics
classes.
c. Write a probability statement for the percent of money returned out of the otherclasses.
d. Is money being returned independent of the class? Justify your answer numerically.
e. Based upon this study, do you think that economists are more selfish than otherpeople? Explain why or why not. Include numbers to justify your answer.

190

44. The following table of data obtained from www.baseball-almanac.com shows hit
information for four players. Suppose that one hit from the table is randomly selected.
NameSingleDoubleTripleTotal Hits

136
54

Home
Run
714
137

Babe Ruth
JackieRobinson
Ty Cobb
Hank Aaron
Total

1517
1054

506
273

3603
2294
8471

174
624
1577

295
98
583

114
755
1720

4189
3771
12,351

2873
1518

Are "the hit being made by Hank Aaron" and "the hit being a double” independent events?
45. United Blood Services is a blood bank that serves more than 500 hospitals in 18 states.
According to their website, a person with type O blood and a negative Rh factor (Rh-) can
donate blood to any person with any blood type. Their data show that 43% of people havetype O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor.
a. Find the probability that a person has both type O blood and the Rh- factor.
b. Find the probability that a person does NOT have both type O blood and the Rhfactor.
46. At a college, 72% of courses have final exams and 46% of courses require research papers.
Suppose that 32% of courses have a research paper and a final exam. Let F be the eventthat a course has a final exam. Let R be the event that a course requires a research paper.
a. Find the probability that a course has a final exam or a research project.
b. Find the probability that a course has NEITHER of these two requirements.
47. In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8%
contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.
a. Find the probability that a cookie contains chocolate or nuts (he can't eat it).
b. Find the probability that a cookie does not contain chocolate or nuts (he can eat it).
48. A college finds that 10% of students have taken a distance learning class and that 40% of
students are part time students. Of the part time students, 20% have taken a distancelearning class. Let D = event that a student takes a distance learning class and E = eventthat a student is a part time student.
a. Find P(D AND E).
b. Find P(E|D).
c. Find P(D OR E).
d. Using an appropriate test, show whether D and E are independent.
e. Using an appropriate test, show whether D and E are mutually exclusive.

191

49. The table shows the political party affiliation of each of 67 members of the US Senate inJune 2012, and when they are up for reelection.
Up for
Democratic
Republican
IndependentTotal
reelection:
Party
Party
2016
10
24
0
2018
23
8
2
Total
a. What is the probability that a randomly selected senator has an “Independent”
affiliation?
b. What is the probability that a randomly selected senator is up for reelection in
November 2016?
c. What is the probability that a randomly selected senator is a Democrat and up for
reelection in November 2016?
d. What is the probability that a randomly selected senator is a Republican or is up for
reelection in November 2018?
e. Suppose that a member of the US Senate is randomly selected. Given that the
randomly selected senator is up for reelection in November 2016, what is theprobability that this senator is a Democrat?
f.

Suppose that a member of the US Senate is randomly selected. What is the probability
that the senator is up for reelection in November 2018, knowing that this senator is a
Republican?

g. The events “Republican” and “Up for reelection in 2016” arei. mutually exclusive.
ii. independent.
iii. both mutually exclusive and independent.
iv. neither mutually exclusive nor independent.
h. The events “Independent” and “Up for reelection in November 2016” arei. mutually exclusive.
ii. independent.
iii. both mutually exclusive and independent.
iv. neither mutually exclusive nor independent.

192

50. The table below gives the number of suicides estimated in the U.S. for a recent year byage, race (black or white), and sex. We are interested in possible relationships between
age, race, and sex. We will let suicide victims be our population.
Race and Sex
1 – 1415 – 24
25 – 64
Over 64
Totalswhite, male210
3360
13,61022,050white, female
80
580
3380
4930
black, male
10
460
1060
1670
black, female
0
40
270
330
all others
Totals310
4650
18,78029,760
Do not include "all others" for parts f and g.
a. Fill in the column for the suicides for individuals over age 64.
b. Fill in the row for all other races.
c. Find the probability that a randomly selected individual was a white male.
d. Find the probability that a randomly selected individual was a black female.
e. Find the probability that a randomly selected individual was blackf. Find the probability that a randomly selected individual was male.
g. Out of the individuals over age 64, find the probability that a randomly selectedindividual was a black or white male.
51. The table of data obtained from www.baseball-almanac.com shows hit information for
four well known baseball players. Suppose one hit from the table is randomly selected.
a. Find P(hit was made by Babe Ruth).

1518
2873
2873
ii.
12351
583
iii.
12351
4189
iv.
12351
i.

Name
Babe
Ruth
JackieRobinson
Ty Cobb
Hank
Aaron
Total

Single Double Triple HR
1517 506
136
714

Total
2873

1054

273

54

137

1518

3603
2294

174
624

295
98

114
755

4189
3771

8471

1577

583

1720 12351

b. Find P(hit was made by Ty Cobb|The hit was a Home Run).

4189
12351
114
ii.
1720
1720
iii.
4189
114
iv.
12351
i.

193

52. The following table identifies a group of children by one of four hair colors, and by type of
hair.
Hair Type
Wavy
Straight
Totals

Brown
20
80

Blond

Black
15

15
20

Red
3
12

Totals43
215

a. Complete the table.
b. What is the probability that a randomly selected child will have wavy hair?
c. What is the probability that a randomly selected child will have either brown or blondhair?
d. What is the probability that a randomly selected child will have wavy brown hair?
e. What is the probability that a randomly selected child will have red hair, given thathe or she has straight hair?
f. If B is the event of a child having brown hair, find the probability of the complementof B.
g. In words, what does the complement of B represent?
53. In a previous year, the weights of the members of the San Francisco 49ers and the Dallas
Cowboys were published in the San Jose Mercury News. The factual data were compiled
into the following table. For the following, suppose that you randomly select one playerfrom the 49ers or Cowboys.
Shirt #
1 - 3334 - 66
67 - 99

≤ 210
21
6
6

211 - 250518
12

251 - 290
0
7
22

> 290
0
4
5

a. Find the probability that his shirt number is from 1 to 33.
b. Find the probability that he weighs at most 210 pounds.
c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210
pounds.
d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210
pounds.
e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at
most 210 pounds.

19454. This tree diagram shows the tossing of an unfair coin followed by drawing one bead froma cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, P(H)=1
2and P(T) = where H is heads and T is tails.
33

a. Find P(tossing a Head on the coin AND a Red bead)b. Find P(Blue bead).
55. A box of cookies contains three chocolate and seven butter cookies. Miguel randomly
selects a cookie and eats it. Then he randomly selects another cookie and eats it. (Howmany cookies did he take?)a. Draw the tree that represents the possibilities for the cookie selections. Write theprobabilities along each branch of the tree.
b. Are the probabilities for the flavor of the SECOND cookie that Miguel selects
independent of his first selection? Explain.
c. For each complete path through the tree, write the event it represents and find theprobabilities.
d. Let S be the event that both cookies selected were the same flavor. Find P(S).
e. Let T be the event that the cookies selected were different flavors. Find P(T) by two
different methods: by using the complement rule and by using the branches of thetree. Your answers should be the same with both methods.
f. Let U be the event that the second cookie selected is a butter cookie. Find P(U).
56. A previous year, the weights of the members of the San Francisco 49ers and the Dallas
Cowboys were published in the San Jose Mercury News. The factual data are compiled
into the following table:
Shirt #
≤ 210
211-250
251 – 290290 ≤1 – 3321
50
0
34 – 666
18
7
4
67 – 99
6
12
22
5

195

For the following, suppose that you randomly select one player from the 49ers or Cowboys.
If having a shirt number from one to 33 and weighing at most 210 pounds wereindependent events, then what should be true about P(Shirt# 1–33|≤ 210 pounds)?
57. The probability that a male develops some form of cancer in his lifetime is 0.4567. The
probability that a male has at least one false positive test result (meaning the test comesback for cancer when the man does not have it) is 0.51. Some of the following questions
do not have enough information for you to answer them. Write “not enough information”
for those answers. Let C = a man develops cancer in his lifetime and P = man has at least
one false positive.
a. P(C) =b. P(P|C) =c. P(P|C') =d. If a test comes up positive, based upon numerical values, can you assume that man
has cancer? Justify numerically and explain why or why not.
58. Given events G and H: P(G) = 0.43; P(H) = 0.26; P(H AND G) = 0.14
a. Find P(H OR G).
b. Find the probability of the complement of event (H AND G).
c. Find the probability of the complement of event (H OR G).
59. Given events J and K: P(J) = 0.18; P(K) = 0.37; P(J OR K) = 0.45
a. Find P(J AND K).
b. Find the probability of the complement of event (J AND K).
c. Find the probability of the complement of event (J or K).
60. Suppose that you have eight cards. Five are green and three are yellow. The cards arewell shuffled. Suppose that you randomly draw two cards, one at a time, withreplacement. Let G1 = first card is green. Let G2 = second card is green
a. Draw a tree diagram of the situation.
b. Find P(G1 AND G2).
c. Find P(at least one green).
d. Find P(G2|G1).
e. Are G2 and G1 independent events? Explain why or why not.
61. Suppose that you have eight cards. Five are green and three are yellow. The cards arewell shuffled. Suppose that you randomly draw two cards, one at a time, without
replacement. G1 = first card is green G2 = second card is green
a. Draw a tree diagram of the situation.
b. Find P(G1 AND G2).
c. Find P(at least one green).
d. Find P(G2|G1).
e. Are G2 and G1 independent events? Explain why or why not.

196

Use the following information to answer the next two exercises. The percent of licensed U.S.
drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 andunder; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers,
5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.
62. Complete the following.
a. Construct a table or a tree diagram of the situation.
b. Find P(driver is female).
c. Find P(driver is age 65 or over|driver is female).
d. Find P(driver is age 65 or over AND female).
e. In words, explain the difference between the probabilities in part c and part d.
f. Find P(driver is age 65 or over).
g. Are being age 65 or over and being female mutually exclusive events? Explain.
63. Suppose that 10,000 U.S. licensed drivers are randomly selected.
a. How many would you expect to be male?
b. Using the table or tree diagram, construct a contingency table of gender versus age
group.
c. Using the contingency table, find the probability that out of the age 20–64 group, arandomly selected driver is female.
64. Approximately 86.5% of Americans commute to work by car, truck, or van. Out of thatgroup, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to workand approximately 5.3% take public transportation.
a. Construct a table or a tree diagram of the situation. Include a branch for all othermodes of transportation to work.
b. Assuming that the walkers walk alone, what percent of all commuters travel alone towork?
c. Suppose that 1,000 workers are randomly selected. How many would you expect totravel alone to work?
d. Suppose that 1,000 workers are randomly selected. How many would you expect todrive in a carpool?
65. When the Euro coin was introduced in 2002, two math professors had their statistics
students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather
than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110
showed a tail (event T). On that basis, they claimed that it is not a fair coin.
a. Based on the given data, find P(H) and P(T).
b. Use a tree to find the probabilities of each possible outcome for the experiment of
tossing the coin twice.
c. Use the tree to find the probability of obtaining exactly one head in two tosses of thecoin.
d. Use the tree to find the probability of obtaining at least one head.
197

66. Use the following information to answer the next two exercises. The following are real
data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059
documented cases of AIDS in the county. They were grouped into the following categories:
Homosexual/Bisexual IV Drug
Heterosexual
Other Totals
User*
Contact
Female 0
70
136
49
Male
2146
463
60
135
Totals
a.
b.
c.
d.
e.
f.
g.

Find P(Person is female).
Find P(Person has a risk factor heterosexual contact).
Find P(Person is female OR has a risk factor of IV drug user).
Find P(Person is female AND has a risk factor of homosexual/bisexual).
Find P(Person is male AND has a risk factor of IV drug user).
Find P(Person is female GIVEN person got the disease from heterosexual contact).
Construct a Venn diagram. Make one group females and the other group heterosexual
contact.

67. Answer these questions using probability rules. Do NOT use the contingency table. Three
thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, througha certain date. Those cases will be our population. Of those cases, 6.4% obtained thedisease through heterosexual contact and 7.4% are female. Out of the females with thedisease, 53.3% got the disease from heterosexual contact.
a. Find P(Person is female).
b. Find P(Person obtained the disease through heterosexual contact).
c. Find P(Person is female GIVEN person got the disease from heterosexual contact)
d. Construct a Venn diagram representing this situation. Make one group females andthe other group heterosexual contact. Fill in all values as probabilities.
68. Suppose 22% of the population are 65 or older, 28% of those 65 or older have loans, and
55% of those younger than 65 have loans. Find the probabilities that a person fits into
the following categories.
a. 65 or older and has a loan.
b. Has a loan.
c. Younger than 65 and does not have a loan.
d. Are the events that a person is 65 or older and that the person has a loan
independent.
69. In an election with 4 candidates for one office and 8 candidates for another office, howmany different ballots may be printed?

198

70. How many different 4-letter call letters for a student radio station can be made if
a. The first letter must be a M or a K and no letter may be repeated?
b. Repeats are allowed but the first letter is a M or K?
71. How many different 6 space license plates can be made if
a. The first three positions must be letters and the last three positions digits.
b. The first three positions must be letters that do not repeat, and the last threepositions are digits.
72. A financial advisor gives her client 9 potential investments and asked her to select andrank her top 5. In how many ways can she do this?
73. An internet installer is given 12 customers that need installation to be completed by theend of the day. How many different possible ways can the list of customers be arranged?
74. There are 5 defective parts in a crate of 28 parts.
a. How many samples of 3 parts can be drawn from the crate?
b. How many samples of 3 parts can be drawn in which all three are not defective?
c. How many samples of 3 parts can be drawn in which 2 are not defective and 1
is defective?
75. In a club with 12 varsity and 10 junior varsity members, a 5-member committee will berandomly chosen. Find the probability that the committee contains the following,a. All junior varsity membersb. 3 varsity and 2 junior varsity membersc. At least 4 junior varsity members

199

REFERENCES
4.1 Terminology“Countries List by Continent.” Worldatlas, 2013. Available online at
http://www.worldatlas.com/cntycont.htm(accessed May 2, 2013).

4.2 Independent and Mutually Exclusive Events
Lopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.”
Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggleworkplace.aspx (accessed May 2, 2013).
Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).

4.3 Two Basic Rules of Probability
DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at
http://www.field.com/
fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013).
Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available
online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html(accessed May 2, 2013).
“Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at
http://www.forumresearch.com/
forms/News
Archives/News Releases/74209_TO_Issues__Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).
“Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2,2013).
Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States
Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS12.pdf (accessed May 2, 2013).
Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed
May 2, 2013). Data from U.S. Census Bureau.
Data from the Wall Street Journal.
Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available
online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013).
Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed
May 2, 2013).

200

4.4 Contingency Tables“Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learnabout-blood/blood- types (accessed May 3, 2013).
Data from the National Center for Health Statistics, part of the United States Department of Healthand Human Services. Data from United States Senate. Available online at www.senate.gov (accessed
May 2, 2013).
Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel,
Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related
Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at
http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013).
“Human Blood Types.” Unite Blood Services, 2011. Available online at
http://www.unitedbloodservices.org/learnMore.aspx (accessed May 2, 2013).
Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at
http://www.ehow.com/
facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013).
“United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center.
Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013).

Tree and Venn Diagrams
Data from Clara County Public H.D. Data from the American Cancer Society.
Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/(accessed May 2, 2013). Data from the Federal Highway Administration, part of the United States
Department of Transportation.
Data from the United States Census Bureau, part of the United States Department of Commerce.
Data from USA Today.
“Environment.” The World Bank, 2013. Available online at
http://data.worldbank.org/topic/environment (accessed May 2, 2013).
“Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013.
Available online at http://www.ropercenter.uconn.edu/data_access/data/search_for_datasets.html(accessed May 2, 2013).

201

CHAPTER 4 SOLUTIONS:
1) S = {BB, BG, BP, GG, GB, GP, PP, PB, PG}
3) S = {AB, AC, AD, AE, AF, AG, BA, BC, BD, BE, BF, BG, CA, CB, CD, CE, CF, CG, DA,
DB, DC, DE, DF, DG, EA, EB, EC, ED, EF, EG, FA, FB, FC, FD, FE, FG, GA, GB, GC, GD,
GE, GF}
5) P(H) = 12/42 = 2/7
P(N) = 15/42 = 5/14 P(F) = 10/42 = 5/21 P(C) = 5/42
7) P(A) = 22/97
P(E) = 47/194 P(F) = 27/97
P(N) = 23/194 P(O) = 7/97
P(S) = 6/97
9) P(even) = ½
P(prime) = ½
11) a. P(O’) b. P(O or H) c. P(I and H’) d. P(H | I)
13) Probability of an event knowing that a previous event already occurred15) P(E ) + P(E’ ) = 1
17) P(E | F) = P (E and F)/ P(F) = 0/.5 = 019) a. 0
b. 0 c. 0.63
21) a. P(C) = .48 b.P(L) = .376 c. P(C | L ) = .55 d. Choosing a Californian registeredvoter who prefer life in prison…knowing that they are Latino Californian. e. P(L and C) =.376(.55) = .207
f. a person who is Latino Californian and who is a registered voter that prefers life in prison…
g. L and C are not independents because P(C | L)  P(C)h. P(L or C) = .376 + .48 - .207 = .649
i. a person who is Latino Californian or who isregistered voter that prefers life in prison…
j. L and C are not mutually exclusive
because P( L and C)  0
23)25) a. Answers vary (i.e. individual is male, individual is between 18-34, ..)b. 40% of total approve of Mayor Ford’s action in officec. 60% of total disapprove …
d. 30% of the 18-34 years old approve of Mayor …
e. 45.7%
f. 63%
g. 40%
h. 44%
i. 78.9%
j. 30%
27) a. 1/7 b. 0 c. ½ d. 5/14 e. 2/7 f. 3/14 g. 5/7 h. Physician i. Service j. 4.1 k. 81.3 l.
½

20229) a. 1046 b. 58% c. 439 d. 57% e. 60%31) a. 1/2 b. 4/30 = 2/15 c. 6/30 = 1/5 d. 3/30 = 1/10 e. 10/30 = 1/3 f. 6/30 = 1/5
g. 10/30 = 1/3
33) a. S = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
b. P(G) = 5/8 c. P(G | E) = 2/3
d. P( G and E) = ¼ e. P(G or E) = ¾
f. Not mutually exclusive since P(G and E) ≠ 035) a. S = {GH, GT, BH, BT, RH, RT}
b. 3/10*1/2 = 3/20
since they can’t happen at the same time.

c. Yes, mutually exclusive37) a. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
b. P(A) = ½
c. Not mutually exclusive because can’t happen at the same time.
39) a. P(Y or Z) = P(Y) + P(Z) – P(Y)*P(Z) b. 0.541) a. iii
b. i
c. iv d. ii43) a. P(R) = .44
b. P(R | E) = .56
c. P(R | O) = .31
d. Not independent
because P(R | E) ≠ P(R)e. No because P(R | E) > P(R)45) a. .52 = .43 + .15 – P(O and Rh-); P(O and Rh-) = .06 b. complement of part a: .94
47) a. P(C or N) = .36 + .12 - .08 = .4
b. complement of part a: .649) a. P(independent) = 2/67
b. 34/67
c. 10/67
d. 8/67
e. 10/34
f. 8/32
g. iv h. i51) a. ii
b. ii53) a. 26/106 b. 33/10655) a.

c. 21/106

d. 38/106

e. 21/33

b. Not independent since there is no replacement(eaten)c. P(CC) = 6/90; P(CB) = 21/90; P(BC) = 21/90
P(BB) = 42/90
d. 48/90
e. 42/90
f. 63/90

57) a. .4567 b. 0 base on tree diagram from problem 23
c. 0.51 based on tree
diagramd. No because at least 1 false positive happens over 50% of time.
59) a. 0.1

b. 0.90c. P((J or K) ′) = .55

20361) a.
b. 5/14c. 25/28
d. 4/7e. Not independent because
P(G2 | G1) = P(G)63) a. 5140b.

c. .4857

FemaleMale
Totals< 20
0.02440.02590.050365) a. P(H) = 140/250b.

20 – 640.39540.41860.8140>64
0.0661
0.06950.1356

Totals
0.486
0.514
1

P(T) = 110/250
c. 308/625
d. 504/62567) a. 0.074

b. 0.064

69) 32 ballots
71) a. 17,576,00073) 12!

C522 C575) a. 10

b. 12

c. 0.6163

d.

b. 15,600,000

C3 ⋅ 10 C2C ⋅ CCc. 10 4 12 1 + 10 5
22 C5
22 C5
22 C5

204

5|DISCRETE RANDOM VARIABLES

Figure 5.1 Photo by Wilhelm Gunkel on Unsplash

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:• Recognize and understand discrete probability distribution functions, in general.
• Calculate and interpret expected values.
• Recognize the binomial probability distribution and apply it appropriately.
• Recognize the Poisson probability distribution and apply it appropriately.
• Recognize the geometric probability distribution and apply it appropriately.
• Recognize the hypergeometric probability distribution and apply it appropriately.
• Classify discrete word problems by their distributions.
A student takes a ten-question, true-false quiz. Because the student had such a busy
schedule, he or she could not study and guesses randomly at each answer. What is theprobability of the student passing the test with at least a 70%?

205

Small companies might be interested in the number of long-distance phone calls theiremployees make during the peak time of the day. Suppose the average is 20 calls. What isthe probability that the employees make more than 20 long-distance phone calls during thepeak time?
These two examples illustrate two different types of probability problems involving discreterandom variables. Recall that discrete data are data that you can count. A random variable
is a numerical variable that describes the outcomes of a statistical experiment in words. The
values of a random variable can vary with each repetition of an experiment. Thus, a randomvariable is a variable whose values are determined by chance.

Random Variable Notation
We will use upper case letters such as X or Y to denote a random variable. Lower case letterslike x or y denote the value of a random variable. If X is a random variable, then X is
described in words, and the value x is given as a number.
For example, let X = the number of heads you get when you toss three fair coins. The sample
space for the toss of three fair coins is {TTT; THH; HTH; HHT; HTT; THT; TTH; HHH}. So
the values of X are x = 0, 1, 2, 3. Notice that for this example, the x values are countableoutcomes. Because you can count the possible values that X can take on and the outcomesare random (the x values 0, 1, 2, 3), X is a discrete random variable.

5.1 | Probability Distribution Function (PDF) for a Discrete
Random Variable

If we have a random variable that has only finitely many outcomes, then we can make a tablethat shows the values x in one column and the corresponding probabilities in another column.
Such a table is called a probability distribution function (PDF), and is very similar to a
relatively frequency distribution, like those we saw in Chapters 2 and 3. In particular, a
discrete probability distribution function has two key characteristics:
1. Each probability is between zero and one, inclusive.
2. The sum of the probabilities is one.
Example 5.1
A child psychologist is interested in the number of times a newborn baby's crying wakes its
mother after midnight. For a random sample of 50 mothers, the following information was
obtained.

206

Let X = the number of times per week a newborn baby's crying wakes its mother after
midnight. For this example, x = 0, 1, 2, 3, 4, 5. Let P(x) = probability that X takes on avalue x. Then the following is the probability distribution for X :Solution 5.1

x
0
1
2
3
4
5

P(x)
2/50
11/50
23/50
9/50
4/50
1/50

This is a discrete distribution function because:a. Each P(x) is between zero and one, inclusive.
b. The sum of the probabilities is one, that is,

2 11 23 9
4
1
+
+
+
+
+
=1
50 50 50 50 50 50

5.1 A hospital researcher is interested in the number of times the average post-op patient
will ring the nurse during a 12-hour shift. Let X = the number of times a patient rings thenurse during a 12-hour shift. For this exercise, the possible values are x = 0, 1, 2, 3, 4, 5.
P(x) = the probability that X takes on value x. For a random sample of 50 patients, thefollowing information was obtained:
Why is this a discrete probability distribution?
x
0
1
2
3
4
5

P(x)
4/50
8/50
16/50
14/50
6/50
2/50

These two examples show that we can think of a relative frequency distribution as an
approximation to a probability distribution. But using ideas from Chapter 4, we can calculate
the actual PDF for many probability experiments.

207

Example 5.2
Suppose that we toss a coin 4 times in succession. Then we can write the sample space asfollows:
TTTT, HTTT, THTT, TTHT, TTTH, HHTT, THHT, TTHH,
THTH, HTHT, HTTH, THHH, HTHH, HHTH, HHHT, HHHH.
Let X = the number of heads in four tosses; so the values of X are x = 0, 1, 2, 3, 4.
Find the PDF for the random variable X.
Solution 5.2:

X P(x)
0
1
2
3
4

1
16
4
16
6
16
4
16
1
16

5.2 Suppose that a fair coin is tossed three times in succession. List the outcomes in thesample space as sequences of H’s and T’s. Let X = the number of heads in three tosses.
Construct the PDF for X.

Example 5.3
Suppose that we toss a pair of fair dice. Then we can represent the sample space using thetable:(1, 1)(2, 1)(3, 1)(4, 1)(5, 1)(6, 1)(1, 2)(2, 2)(3, 2)(4, 2)(5, 2)(6, 2)(1, 3)(2, 3)(3, 3)(4, 3)(5, 3)(6, 3)(1, 4)(2, 4)(3, 4)(4, 4)(5, 4)(6, 4)(1, 5)(2, 5)(3, 5)(4, 5)(5, 5)(6, 5)(1,6)(2, 6)(3, 6)(4, 5)(5, 6)(6, 6)Let X = number of dots facing up in a single toss. Find the PDF for the random variable X.
208

Solution 5.3:
The possible values of X are x = 2, 3, …, 12.
Using the sample space above, we see that P ( x = 2) =table:
x
P(x)

2

3

1
36

2
36

4

3
36

5

6

4
36

7

5
36

6
36

8

5
36

1
2
, P ( x = 3) = , etc. and we get the
36
36
9

4
36

10

3
36

11

2
36

12

1
36

Finally, we can represent a PDF graphically using a histogram. For example, thehistogram for the PDF obtained by rolling two fair dice would be:

X = sum of dice
0.1946
0.16680.139
0.11120.08340.05560.0278
0

P(X)

1

2

3

4

5

6

7

8

9

10 11 12

5.2 | Mean, Expected Value and Standard Deviation
The expected value or mean of a random variable can be thought of as the "long-term"
average of the variable. That is, if we performed the probability experiment over and over,
we would expect the average of the numerical values to be this amount.
For example, suppose that we toss a fair coin and record the result. Although the probabilityof getting heads is 1/2, this does not mean that in multiple trials that exactly half the tosses
would land as heads. I.e. if you flip a coin two times, does probability tell you that these flips
will result in one heads and one tail? You might toss a fair coin ten times and record nineheads. As we learned in Chapter 3, probability does not describe the short-term results of anexperiment. Instead, it gives information about what can be expected in the long term. To
demonstrate this principle, the British statistician Karl Pearson once tossed a fair coin 24,000
times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment,
Pearson illustrated the Law of Large Numbers.

209

The Law of Large Numbers states that, as the number of trials in a probability experiment
increases, the difference between the theoretical probability of an event and the relative
frequency approaches zero (the theoretical probability and the relative frequency get closer
and closer together). When evaluating the long-term results of statistical experiments, we
often want to know the “average” outcome. This “long-term average” is known as the mean
or expected value of the experiment and is denoted by the Greek letter μ. In other words,
after conducting many trials of an experiment, you would expect this average value.
As we showed in Chapter 2, we can find the mean of a relative frequency distribution bymultiplying each data value by its relative frequency and adding all of these products up.
Given the close relationship of relative frequency distributions and probability
distributions, it is not surprising that the same basic procedure:
Mean/Expected Value of a Discrete Random Variable
Let X be random variable, and x1, x2, . . . xn, the list of possible outcomes for X.
Then the mean of the distribution and expected value of X are the same quantity, given by:
μ = E(X) = Σ xi P(xi)That is, to find the expected value or long-term average, μ, simply multiply each value ofthe random variable by its probability and add the products.

Example 5.4
A men's soccer team plays soccer zero, one, or two days a week. The probability that they playzero days is 0.2, the probability that they play one day is 0.5, and the probability that theyplay two days is 0.3. Find the long-term average or expected value, μ, of the number of days
per week the men's soccer team plays soccer.
Solution 5.4:
To do the problem, first let the random variable X = the number of days the men's soccer
team plays soccer per week. Then X takes on the values 0, 1, 2. Construct a PDF table,
and add an extra column labeled x*P(x). In this column, you will multiply each x value by
its probability:
x
0
1
2

P(x)
x*P(x)
0.2 0*(0.2) = 0
0.5 1*(0.5) = 0.5
0.3 2*(0.3) = 0.6

210This table is called an expected value table, and helps you calculate the expected value orlong-term average. In this case, we have
E(X) = μ = 0*(0.2) + 1*(0.5) + 2*(0.3) = 0 + 0.5 + 0.6 = 1.1.
This means that the men’s soccer team would, on average, expect to play 1.1 days per week.
We can also calculate the variance and standard deviation of a random variable; again, thiswill be similar to the way we found these statistics for a relative frequency distribution in
Chapter 3. That is, we will take a weighted average of the squared deviations from the mean.
Variance and Standard Deviation of a Discrete Random Variable
Let X be random variable, and x1, x2, . . . xn, the list of possible outcomes for X.
•

The variance of the random variable is:•

The standard deviation is the square root of the variance: σ =𝜎𝜎 2 = ∑(𝑥𝑥𝑖𝑖 − 𝜇𝜇)2 𝑃𝑃(𝑥𝑥𝑖𝑖 )∑ ( x − µ ) P( x )i

2i

Example 5.5
Refer to Example 5.1 Find the expected value of the number of times a newborn baby's
crying wakes its mother after midnight. The expected value is the expected number oftimes per week a newborn baby's crying wakes its mother after midnight. Calculate thestandard deviation of the variable as well.
Solution 5.5:
We start with the PDF from Example 5.1; then add columns column showing xP(x), and(x – μ)2P(x):
x
0
1
2
3
4
5

P(x)
0.04
0.22
0.46
0.18
0.08
0.02

xP(x)
0
0.22
0.92
0.54
0.32
0.10

(x – μ)2P(x)(0 – 2.1)2(0.04) = 0.1764(1 – 2.1)2(0.22) = 0.2662(2 – 2.1)2(0.46) = 0.0046(3 – 2.1)2(0.18) = 0.1458(4 – 2.1)2(0.08) = 0.2888(5 – 2.1)2(0.02) = 0.1682

Summing the entries in the third column, we get μ = E(X) = 2.1. So, on average, we expect
a newborn to wake its mother after midnight 2.1 times per week. We then use this value211to complete the fourth column. Add the values in the fourth column of the table to get thevariance:
σ2 = 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05
The standard deviation of X is the square root of this sum: σ ≈ 1.0247.
While doing an example using these tables is instructive, we can do these calculations quickly
and efficiently using the TI-84. The procedure is exactly as we did for calculating the meanand standard deviation of a frequency distribution in Chapter 3:_
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the mean and standard deviation of a discrete probability distribution.
Go to STAT >> EDIT and clear lists L1 and L2.
Enter the x values into L1 and enter the probabilities P(x) into L2.
Then go to STAT >> CALC. Select 1-VarStats, type L1, L2 and hit ENTER.
The mean will appear at the top of the screen as x .
The standard deviation will appear as σx.

5.5 A hospital researcher is interested in the number of times the average post-op patient
will ring the nurse during a 12-hour shift. For a random sample of 50 patients, thefollowing information was obtained.
x
0
1
2
3
4
5

P(x)
0.08
0.16
0.32
0.28
0.12
0.04

Find the expected value and the standard deviation of this random variable.

212

Example 5.6
Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6,
7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You
pay $2 to play and could profit $100,000 if you match all five numbers in order (you getyour $2 back plus $100,000). Over the long term, what is your expected profit of playing
the game?
Solution 5.6
Let X = the amount of money you profit.
(The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.) Since you are interested in your profit (orloss), the possible values of x are 100,000 dollars and −2 dollars.
To win, you must get all five numbers correct, in order. The probability of choosing one

1
because there are ten numbers. We are sampling without
10
replacement, so the selections are independent of one another; thus the probability ofcorrect number is

1 5
10choosing all five correctly is � � =0.00001. Therefore, the probability of winning is

0.00001 and the probability of losing is 1 − 0.00001 = 0.99999. Thus, we have the PDF:
Win
Loss

x
100,000-2P(x)0.00001
0.99999

From the calculator, we have μ = Expected Value = -0.99998.
Since –0.99998 is about –1, on average you should expect to lose approximately $1 for eachgame you play. Note that we will never actually lose $1; the only potential outcomes are
winning $100,000 or losing $2. But if we play the game many, many times, the average loss
will be about $1 per game.

5.5 You are playing a game of chance in which four cards are drawn from a standard deckof 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in thedeck on each draw. You pay $1 to play. If you guess the right suit every time, you get yourmoney back and $256. What is your expected profit of playing the game over the long term?

213

Example 5.7
Toss a pair of fair six-sided dice. Recall from Example 5.3 that the sample space has 36
outcomes:(1, 1)(2, 1)(3, 1)(4, 1)(5, 1)(6, 1)(1, 2)(2, 2)(3, 2)(4, 2)(5, 2)(6, 2)(1, 3)(2, 3)(3, 3)(4, 3)(5, 3)(6, 3)(1, 4)(2, 4)(3, 4)(4, 4)(5, 4)(6, 4)(1, 5)(2, 5)(3, 5)(4, 5)(5, 5)(6, 5)(1,6)(2, 6)(3, 6)(4, 5)(5, 6)(6, 6)Let X = the number of faces that show an even number.
Calculate the mean μ and standard deviation σ of X.
Solution 5.7:
Using the sample space, we have the following PDF:
X P(x)
0
1
2

9
36
18
36
9
36

From the calculator, we get μ = 1, and σ = 0.7071.

Example 5.8
According to a YouGov survey of 1354 US Adults on Dec 27, 2019, 21% of all respondents
clean their cell phone daily.
While at the mall suppose you make a bet with a friend that if you ask a random person at
the mall if the person clean their cell phone daily. If you win the bet, you win $50. If you
lose the bet, you pay $20. Let X = the amount of profit from a bet.
If you bet many times, will you come out ahead? Explain your answer in a complete
sentence using numbers. What is the standard deviation of X?
Solution 5.8:
We are given that P(win) = P(the person cleans their cell phone daily) = .21

214

Thus, we also have P(loss) = P(the person does not clean their cell phone daily) = 1 – .21 =.79
And if we win, then we win $50, so x = 50. If we lose, we pay $20, so x = -20.
Thus, we have the PDF:
Win
Loss
From the calculator, we have:

x
50
-20

P(x)
0.21
0.79

μ = Expected Value = -5.3, and the standard deviation is σ = 28.512
If you make this bet many times under the same conditions, the long-term outcome will be
an average loss of $5.30 per bet.
The standard deviation is σ ≈ $28.51.

5.8 According to the YouGov survey on Dec 27, 2019, of 1354 US Adults, 8% of thosesurveyed never wash their cell phone. As in Example 5.8, you make a bet with your friendthat a randomly surveyed person at the mall never washes their cell phone. If you win thebet, you win $100. If you lose the bet, you pay $10. Let X = the amount of profit from a bet.
Find the mean and standard deviation of X.

So far, we have presented discrete probability distribution functions in table form. But forsome distributions, we can come up with a formula that describes the probabilities in the
distribution.
In the next few sections, we will investigate some well-known discrete
probability functions: the binomial, geometric, and Poisson distributions.
A probability distribution function is a pattern. Each distribution has its own special
characteristics. Learning the characteristics each of these enables us to distinguish among
the different distributions, and to match a given probability problem into the correct pattern.

215

5.3 | Binomial Distribution
Binomial distribution is a special discrete probability distribution. There are four
conditions that the experiment has to meet to be considered a binomial experiment:
Conditions for a Binomial Experiment
1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The
letter n denotes the number of trials.
2. There are only two possible outcomes, called "success" and "failure," for each trial.
3. The n trials are independent and are repeated using identical conditions. Because then trials are independent, the outcome of one trial does not help in predicting theoutcome of another trial.
4. The letter p denotes the probability of a success on one trial, and q denotes theprobability of a failure on one trial, so p + q = 1. Since the trials are independent, premains the same for each trial.
Any experiment that has characteristics two, three, four, and where n = 1 is called a
Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied themextensively). A binomial experiment takes place when the number of successes is counted in
one or more Bernoulli Trials.

Example 5.9
Randomly guessing at a multiple-choice question with 4 possible answers has only two
outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose
there are 6 multiple choice questions. Joe guesses on each question with no pattern.
Determine if this is a binomial experiment.
Solution 5.9
Condition 1 is met since there are a fixed number of trials: n = 6 questions
Condition 2 is met since there are two outcomes: each question is either correct or incorrect.
Condition 3 is met since each trial is independent: Joe guessing with no pattern on eachquestion does not help in predicting the outcome of another question.
Condition 4 is met since the probability of success (correct) is constant for each trial: p = .25
is the probability of guessing a question correct and remains constant for each question.

216

Example 5.10
Randomly guessing at a multiple-choice question has only two outcomes. If a success isguessing correctly, then a failure is guessing incorrectly. Suppose there are 6 multiple choicequestions. The first 3 questions have 4 choices, and the last 3 questions have 5 choices each.
Cesar guesses on each question with no pattern. Determine if this is a binomial experiment.
Solution 5.10
Condition 1 is met since there are a fixed number of trials: n = 6 questions.
Condition 2 is met since there are two outcomes: each question is either correct or incorrect.
Condition 3 is met since each trial is independent: Cesar guessing with no pattern on eachquestion does not help in predicting the outcome of another question.
Condition 4 is not met since the probability of success (correct) is not constant for each trial:
p = .25 (1 out of the 4 choices is correct) is the probability of guessing a question correct for
the first 3 question but changes to p = .20 (1 out of 5 choices is correct) for the last 3 questions.
The probability of success (correct) does not remain constant for each trial.

5.9 Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50
individuals who have taken the driver’s exam is randomly selected. Give two reasons why
this is a binomial problem.

The outcomes of a binomial experiment fit a binomial probability distribution. The
random variable X = the number of successes obtained in the n independent trials. To findthe probability of a specific value of the random variable, the binomial formula is used.
Notation and Properties of the Binomial Distribution•

X ~ B(n, p) This notation states the random variable X is a binomial distribution
with n trials and the probability of success , p.

•••P(x = k) = nCk∙pk∙qn-k Binomial formula
The mean of a binomial probability distribution, μ = n∙p, andVariance of a binomial probability distribution, σ2 = n∙p∙q.

•

The standard deviation, σ = �𝑛𝑛𝑛𝑛𝑛𝑛

217

Using the TI-83, 83+, 84, 84+ Calculator
To calculate binomial formula.
Go into 2nd DISTR:_

P(x = k) = binompdf(n, p, k)

The above screenshots are for the following situation:
X~B(6, .25) which means that the random variable is Binomial with n = 6, p = .25,
and the formula finds P(x = 1) = 6C1∙.251∙.755 = .356
Using an Excel function, the binomial formula is used for both calculating x values andcumulative x values (see Example 5.13). The Excel function is =BINOM.DIST(number_s,
trials, probability, true/false). Notice that the x value is first, and the number of trials is
second, which is different than the TI 83/84. Select FALSE to calculate just the x value.

Example 5.11
Randomly guessing at a multiple-choice question with 4 possible answers has only two
outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose
there are 6 multiple choice questions. Mayra guesses on each question with no pattern.
Create a binomial probability distribution. Draw a histogram.
Solution 5.11

n = 6, p = .25 which means q = .75
X
0
1
2
3
4
5
6

P(X)
Binompdf(6, .25, 0) = .178
Binompdf(6, .25, 1) = .356
Binompdf(6, .25, 2) = .297
Binompdf(6, .25, 3) = .132
Binompdf(6, .25, 4) = .033
Binompdf(6, .25, 5) = .00439
Binompdf(6, .25, 6) = .00024

218

0.4
0.3
0.2
0.1
0

.5

1.5

2.5

3.5

4.5

5.5

6.5

Example 5.12
Suppose you play a game that you can only either win or lose. The probability that you win
any game is 55%, and the probability that you lose is 45%. Each game you play is
independent.
a. If you play the game 20 times, write the function that describes the probability thatyou win 15 of the 20 times.
b. Find the mean number of wins.
c. Find the standard deviation of wins.
Solution 5.12
a. Let X be the number of wins (0, 1, 2, 3, ..., 20). The probability of a success is p =0.55. The number of trials is n = 20. The probability question can be statedmathematically as P(x = 15) = Binompdf(20, .55, 15) = .0365
b. μ = n∙p = 20∙.55 = 11 wins is the mean number of wins for 20 trials.
c. σ = �𝑛𝑛𝑛𝑛𝑛𝑛 = 2.22

Example 5.13
A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully
performs the trick is 35%, and the probability that the dolphin does not successfully
perform the trick is 65%. Out of 10 attempts, you want to find the probability that thedolphin succeeds at most 5 times. State the probability question mathematically.
Solution 5.13
Here, if you define X as the number of successful performances, then X takes on the values
0, 1, 2, 3, ..., 10. The probability of success is p = .35. The probability question can be statedmathematically as P(x ≤ 5). Here we want to add all the probabilities from 0 to 5 as seen in
the table below:P(x ≤ 5) = .0135 + .0725 + .176 + .252 + .238 + .154 = .906

219

X
0
1
2
3
4
5
6
7
8
9
10

P(X)
Binompdf(10, .35, 0) = .0135
Binompdf(10, .35, 1) = .0725
Binompdf(10, .35, 2) = .176
Binompdf(10, .35, 3) = .252
Binompdf(10, .35, 4) = .238
Binompdf(10, .35, 5) = .154
Binompdf(10, .35, 6) = .0689
Binompdf(10, .35, 7) = .0212
Binompdf(10, .35, 8) = .00428
Binompdf(10, .35, 9) = .00051
Binompdf(10, .35, 10) = .000028

Using the TI-83, 83+, 84, 84+ Calculator_

A simpler way of finding this same probability is to use the cumulative binomial function on
the calculator. The function binomcdf (n, p, x) calculates the cumulative probability. In
this case the calculator automatically sums up the probabilities from x = 0 to x = 5.

Using an Excel function, the binomial formula is used for both calculating x values andcumulative x values. The Excel function is =BINOM.DIST(number_s, trials, probability,
true/false). Notice that the x value is first, and the number of trials is second, which isdifferent than the TI 83/84. Select TRUE to calculate the cumulative x value.

220

TI 83/84 and Excel Function Recap where k is the value of x, n is the number of trials, andp is the probability. Remember to include the equal sign, =, when using the Excel function.

P(x = k)Exact value
Not cumulative
P(x≤ k)Cumulative including kArea to the left
P(x < k)Cumulative NOT
including kArea to the left
P(x > k) = 1 – P(x ≤ k)
Cumulative NOT
including kArea to the rightP(x ≥ k) = 1 – P(x > k)
Cumulative including kArea to the rightCalculator
Function

Excel Function

binompdf(n, p, k)

BINOM.DIST(k, n, p, FALSE)

binomcdf(n, p, k)

BINOM.DIST(k, n, p, TRUE)

binomcdf(n, p, k – 1)

BINOM.DIST(k – 1, n, p, TRUE)

1 – binomcdf(n, p, k)

1 – BINOM.DIST(k, n, p, TRUE)

1 – binomcdf(n, p, k – 1)

1 – BINOM.DIST(k – 1, n, p, TRUE)

Example 5.14
A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting
more than ten heads? Let X = the number of heads in 15 flips of the fair coin. X takes on
the values 0, 1, 2, 3, ..., 15. Since the coin is fair, p = 0.5 and q = 0.5. The number of trials is
n = 15. State the probability question mathematically. Find the probability.
Solution 5.14
P(x > 10) = P(x = 11) + P(x = 12) + P(x = 13) + P(x = 14) + P(x = 15)= 1 – P(complement)= 1 – P(x ≤ 10)= 1 – binomcdf(15, .5, 10)= 1 - .941 = .059

5.10 A fair, six-sided die is rolled ten times. Each roll is independent. You want to find theprobability of rolling a one more than three times. State the probability question
mathematically. Find the probability.
221

Example 5.15
It has been stated that about 41% of adult workers have a high school diploma but do not
pursue any further education. If 20 adult workers are randomly selected, find the probability
that at most 12 of them have a high school diploma but do not pursue any further education.
How many adult workers do you expect to have a high school diploma but do not pursue any
further education? Create a histogram.
Solution 5.15
Let X = the number of workers who have a high school diploma but do not pursue any
further education. X takes on the values 0, 1, 2, ..., 20 where n = 20, p = 0.41, and q = 1 –0.41 = 0.59.
X ~ B(20, 0.41)
P(x ≤ 12) = binomcdf(20, .41, 12) = 0.9738.
To answer the question how many adult workers you expect to have a high school diploma
but do not pursue any further education:
μ = n∙p = 20(.41) = 8.2 adult workers.
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0

1 5 ..... 20
x=01234222

5.11 About 32% of students participate in a community volunteer program outside of school.
If 30 students are selected at random, find the probability that at most 14 of them participate
in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator tofind the answer.

Example 5.16
In the 2020 the Paper Clip business supply catalog had 560 pages. Eight of the pages feature
different types of ink pens. Suppose we randomly sample 100 pages. Let X = the number ofpages that feature different types of ink pens.
a. What values does x take on?
b. What is the probability distribution?
c. Find the following probabilities:
i. the probability that two pages feature different types of ink pens
ii. the probability that at most six pages feature different types of ink pensiii. the probability that more than three pages feature different types of ink pens.
d. Using the formulas, calculate the mean and standard deviation.
Solution 5.16
a. x = 0, 1, 2, 3, 4, 5, 6, 7, 8
b. X ~ B(100, 8/560)
c. probabilities:
i. P(x = 2) = binompdf(100, 8/560, 2) = .2466
ii. P(x ≤ 6) = binomcdf(100, 8/560, 6) = .9994
iii. P(x > 3) = 1 – P(x ≤ 3) = 1 – binomcdf(100, 8/560, 3) = .0557
d. μ = n∙p = 100(8/560) ≈ 1.43 pages
8552
� � � ≈ 1.19560
560

σ = �𝑛𝑛𝑛𝑛𝑛𝑛 = �100 �223

5.12 According to a Gallup poll, 60% of American adults prefer saving money over spending.
Let X = the number of American adults out of a random sample of 50 who prefer saving tospending.
a. What is the probability distribution for X?
b. Use your calculator to find the following probabilities:
i. the probability that 25 adults in the sample prefer saving money over spendingii the probability that at most 20 adults prefer savingiii. the probability that more than 30 adults prefer saving money.
c. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.

Example 5.17
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we
randomly sample 200 people. Let X = the number of people who will develop pancreaticcancer.
a. What is the probability distribution for X?
b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
c. Use your calculator to find the probability that at most eight people develop
pancreatic cancer.
d. Is it more likely that five or six people will develop pancreatic cancer? Justify youranswer numerically.
Solution 5.17
a. X ∼ B(200, 0.0128)
b. Mean and standard deviation
i. μ = np = 200(0.0128) = 2.56
ii. σ = �𝑛𝑛𝑛𝑛𝑛𝑛 = (200)(0.0128)(0.9853) ≈ 1.5897

c. P(x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988
d. P(x = 5) = binompdf(200, 0.0128, 5) = 0.0707
e. P(x = 6) = binompdf(200, 0.0128, 6) = 0.0298 So P(x = 5) > P(x = 6); it is more likely thatfive people will develop cancer than six.

224

5.13 During the 2020 - 2021 regular NBA season, DeAndre Jordan of the Brooklyn Nets hada field goal completion rate of 76.3%. Suppose you choose a random sample of 80 shots madeby DeAndre during the 2020 - 2021 season. Let X = the number of shots that scored points.
a.
b.
c.
d.

What is the probability distribution for X?
Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
Use your calculator to find the probability that DeAndre scored with 60 of these shots.
Find the probability that DeAndre scored with more than 50 of these shots.

Example 5.18
The following example illustrates a problem that is not binomial. It violates the condition of
independence. ABC College has a student advisory committee made up of ten staff members
and six students. The committee wishes to choose a chairperson and a recorder. What is theprobability that the chairperson and recorder are both students?
Solution 5.18
The names of all committee members are put into a box, and two names are drawn without
replacement. The first name drawn determines the chairperson and the second name therecorder. There are two trials. However, the trials are not independent because the outcomeof the first trial affects the outcome of the second trial. The probability of a student on thefirst draw is 6/16. The probability of a student on the second draw is 5/15, when the first draw
selects a student. The probability is 6/15, when the first draw selects a staff member. The
probability of drawing a student's name changes for each of the trials and, therefore, violates
the condition of independence.

5.14 A lacrosse team is selecting a captain. The names of all the seniors are put into a hat,and the first three that are drawn will be the captains. The names are not replaced once they
are drawn (one person cannot be two captains). You want to see if the captains all play thesame position. State whether this is binomial or not and state why.

225

5.4 | The Geometric Distribution
There are four main characteristics of a geometric experiment:
1. There are one or more Bernoulli trials with all failures except the last one, which isa success. In other words, we keep repeating the experiment until the first success.
Then we stop.
2. The trials are identical and independent.
3. The probability of a success, p, is the same for each trial. Thus, the probability of a
failure, q, the same for each trial. Of course, q = 1 − p.
4. The random variable is X = the number of independent trials until the first success.
For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit
the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit
the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success,
STOP.
There must be at least one trial; in theory, the number of trials could go on forever.
As another example, suppose we roll one fair die; then the probability of rolling a three is1  . This is true no matter how many times you roll the die. Suppose you want to know the6
probability of getting the first three on the fifth roll. On rolls one through four, you do not get
5a three. The probability for each of the rolls is q =   . And since the trials are independent,
6
5 5 5 5 1
the probability of getting a three on the fifth roll is       = 0.0804.
 6  6  6  6  6 
Example 5.19
You play a game of chance that you can either win or lose until you lose. Your probability oflosing is p = 0.57. What is the probability that it takes five games until you lose? Let X = thenumber of games you play until you lose (includes the losing game). Then X takes on thevalues 1, 2, 3, ... (could go on indefinitely). The probability question is P(x = 5).
Solution 5.19
The probability of losing is p = 0.57, so the probability of winning is q = 1 – 0.57 = 0.43.
If the first loss occurs on the fifth play, then the first four plays must be wins. Thus,
P(x = 5) = (0.43)(0.43)(0.43)(0.43)(0.57 ) = 0.0195.

226

5.15 You throw darts at a board until you hit the center area. Your probability of hitting
the center area is p = 0.17. Find the probability that it takes eight throws until you hit thecenter.
Example 5.20
A safety engineer feels that 25% of all industrial accidents in her plant are caused by failureof employees to follow instructions. She decides to look at the accident reports (selected
randomly and replaced in the pile after reading) until she finds one that shows an accident
caused by failure of employees to follow instructions. On average, how many reports would
the safety engineer expect to look at until she finds a report showing an accident caused by
employee failure to follow instructions? What is the probability that the safety engineer will
have to examine at least three reports until she finds a report showing an accident caused by
employee failure to follow instructions?
Solution 5.20
Let X = the number of accidents the safety engineer must examine until she finds a reportshowing an accident caused by employee failure to follow instructions. X takes on thevalues 1, 2, 3, ....
The first question asks us to find the expected value or the mean. If the probability ofsuccess is 0.25, then on average we would expect to check about 4 reports before we had asuccess. I.e. the mean is μ = 4 =

1
.
0.25

The second question asks us to find P(x ≥ 3). (At least three means three or more.)
We can calculate this using the complement rule:P(x ≥ 3) = 1 – P(x < 2) = 1 – P(x = 2) – P(x = 1)
= 1 − (0.75)(0.25) − 0.25 = 0.5625.
The calculations for the geometric distribution can be done very easily using the calculator.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate probabilities involving Geometric Distribution.
Go into 2nd DISTR. The syntax is as follows:
To calculate P(X = x):geometpdf(p, x)

To calculate P(X < x):geometcdf(p, x)227_

5.16 An instructor feels that 15% of students get below a C on their final exam. She decides
to look at final exams (selected randomly and replaced in the pile after reading) until she
finds one that shows a grade below a C. We want to know the probability that the
instructor will have to examine at most ten exams until she finds one with a grade below a
C. Use the calculator to find P(x < 10).

Notation and Properties of the Geometric Distribution•

We write X ~ G(p) to indicate that the random variable X has a geometric
distribution.
The parameter is p = probability of success for a single trial.

•

The distribution function is P(X = x) = (1 − p ) x −1 p•

The mean of the distribution is μ =• 1  1

The variance is σ2 =   − 1 p  p•

The standard deviation is σ =•1p 1  1  − 1 p  p

Example 5.20
Assume that the probability of a defective computer component is 0.02. Components are
randomly selected and tested until one component fails.
a. Find the probability that the first defect is caused by the seventh component tested.
b. How many components do you expect to test until one is found to be defective?
Solution 5.20
Let X = the number of computer components tested until the first defect is found. Then X
takes on the values 1, 2, 3, .... And the probability of a defective component is p = 0.02, so X
~ G(0.02).
a. Here we want the mean: μ =1
1== 50 . I.e. we expect that we would have to
p .02

inspect 50 components before we found a defective one.
228

b. Here we want to find P(x = 7). Go to 2nd DISTR, and select the geometpdf :P(x = 7) = geometpdf(.02, 7) = 0.0177.
The graph of X ~ G(0.02) is:

The y-axis represents the probability of x, where X = the number of computercomponents tested.

5.17 The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find
the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84
calculator to find the answer.

229

Example 5.21
The lifetime risk of developing pancreatic cancer is about one in78 (1.28%). Let X = thenumber of people you ask until one says he or she has pancreatic cancer. Then X is adiscrete random variable with a geometric distribution: X ~ G(0.0128).
a. What is the probability of that you ask ten people before one says he or she haspancreatic cancer?
b. What is the probability that you must ask 20 people?
c. Find the mean and standard deviation of X.

Solution 5.21
a. P(x = 10) = geometpdf(0.0128, 10) = 0.0114
b. P(x = 20) = geometpdf(0.0128, 20) = 0.01
c. The mean is μ =

1
1
= 78
=
p .0128

The standard Deviation is σ = 1  1  − 1 = 78(78 − 1) = 77.6234.
 p  p

5.18 The literacy rate for a nation measures the proportion of people age 15 and over who
can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the numberof Afghani women you ask until one says that she is literate.
a.
b.
c.
d.

What is the probability distribution of X?
What is the probability that you ask five women before one says she is literate?
What is the probability that you must ask ten women?
Find the mean and standard deviation of X.

230

5.5 | Poisson Distribution
Both Binomial and Poisson are discrete probability distributions. In Binomial the goal was
to look for the probability of a specific value of success in n trials. Now we want to look forthe specific number of occurrences in a specific amount of time or space.
Conditions of a Poisson Experiment
1. The experiment consists of counting the number of events occurring in a fixed intervalof time or space if these events happen with a known average rate and independentlyof the time since the last event.
2. The probability of the event remains constant for each interval of equal length.
3. The number of occurrences in one fixed interval is independent of the number ofoccurrences in other fixed intervals.
4. The random variable X = the number of occurrences in the interval of interest.
For example, a book editor might be interested in the number of words spelled incorrectly ina particular book. It might be that, on the average, there are five words spelled incorrectly in
100 pages. The interval is the 100 pages.
Notation and Properties for the Poisson Distribution••

X ~ P(μ) where μ is the mean number of occurrences per fixed interval.
μ = np•

σ= µ•The probability of exactly x occurrences in an interval is𝑃𝑃(𝑥𝑥) =𝜇𝜇 𝑥𝑥 𝑒𝑒 −𝜇𝜇𝑥𝑥!

Where e is an irrational number approximately 2.718

Example 5.22
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12.
Of interest is the number of loaves of bread put on the shelf in five minutes. The time intervalof interest is five minutes. What is the average number of loaves of bread put on a shelf in 5
minutes?

231

Solution 5.22
Let X = the number of loaves of bread put on the shelf in five minutes. If the average numberof loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves

 12  ( 5 )= 2 loaves of bread. X ~ P(2) is Poisson 30 

put on the shelf in 5 minutes is µ= λ= distribution notation.

5.19 The average number of fish caught in an hour is eight. Of interest is the number of fishcaught in 15 minutes. The time interval of interest is 15 minutes. What is the average
number of fish caught in 15 minutes?

Using the TI-83, 83+, 84, 84+ Calculator
To calculate poisson distribution.

_

Press 2nd VARS, Arrow down to poissonpdf(. Press ENTER
P(x) = poissonpdf(λ, x); P(X < x) = poissoncdf(λ, x)Example 5.23
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12.
Of interest is the number of loaves of bread put on the shelf in five minutes. The time intervalof interest is five minutes. What is the probability that the number of loaves, selected
randomly, put on the shelf in 5 minutes is three?
Solution 5.23
The probability question asks you to find P(x = 3).
From the previous example, we found μ = 2 for the average number of loaves to put on the
shelf in 5 minutes.
P(x = 3) = poissonpdf(2, 3) = .180

232

Example 5.24
A bank expects to receive six bad checks per day, on average. What is the probability of thebank getting fewer than five bad checks on any given day? Of interest is the number of checks
the bank receives in one day, so the time interval of interest is one day. Let X = the numberof bad checks the bank receives in one day. If the bank expects to receive six bad checks perday, then the average is six checks per day. Write the correct notation for the Poisson
distribution. Write a mathematical statement for the probability question.
Solution 5.24
X ~ P(6). We want to find P(x < 5) = P(x < 4) = poisssoncdf(6, 4) = 0.2851.

5.24 An electronics store expects to have ten returns per day on average. The manager wants
to know the probability of the store getting fewer than eight returns on any given day. Statethe probability question mathematically.

Example 5.25
You notice that a news reporter says "uh," on average, two times per broadcast. What is theprobability that the news reporter says "uh" more than two times per broadcast?
This is a Poisson problem because you are interested in knowing the number of times thenews reporter says "uh" during a broadcast.
a. What is the interval of interest?
b. What is the average number of times the news reporter says "uh" during one
broadcast?
c. What does X represent? Write the correct notation for Poisson distribution.
d. Write a mathematical statement for the probability question.
e. Find the probability that the news reporter says "uh" more than two times per
broadcast.
Solution 5.25
a. One broadcast is the fixed interval.
b. μ = 2
c. Let X = the number of times the news reporter says "uh" during one broadcast.
x = 0, 1, 2, 3, …
X ~ P(2)d. P(x > 2)
e. P(x > 2) = P(x ≥ 3) = 1 – P(Complement) = 1 – P(x ≤ 2)
= 1 – poissoncdf(2, 2) = .677

233

Using the TI-83, 83+, 84, 84+ Calculator_

2nd VARS, scroll up to poissoncdf(, Press enter
NOTE: similar to binomcdf, poissoncdf(λ, x) gives the cumulative probability from 0 to x.

5.25 An emergency room at a particular hospital gets an average of five patients per hour. A
doctor wants to know the probability that the ER gets more than five patients per hour. Givethe reason why this would be a Poisson distribution.

Example 5.26
Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m.
What is the probability that Leah receives more than one call in the next 15 minutes?
Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15
minutes or ¼ hour)
Solution 5.26
x = 0, 1, 2, 3, ...
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15
minute intervals in two hours, then Leah receives1  ( 6 ) =0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem.
8
X ~ P(0.75)
The probability that Leah receives more than one telephone call in the next 15 minutes isabout:

234

Find P(x > 1).
P(x > 1) = 1 – P(complement) = 1 – P(x ≤ 1) = 1 – poissoncdf(.75, 1) = 0.1734
The graph of X ~ P(0.75) is:
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0

x = 0, 1, 2, 3, ...

The y-axis contains the probability of x where X = the number of calls in 15 minutes.

5.26 A customer service center receives about ten emails every half-hour. What is theprobability that the customer service center receives more than four emails in the next six
minutes? Use the TI-83+ or TI-84 calculator to find the answer.

Example 5.27
According to Baydin, an email management company, an email user gets, on average, 147
emails per day. Let X = the number of emails an email user receives per day. The discreterandom variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson
distribution: X ~ P(147). The mean is 147 emails.
a. What is the probability that an email user receives exactly 160 emails per day?
b. What is the probability that an email user receives at most 160 emails per day?
c. What is the standard deviation?
Solution 5.27
a. P(x = 160) = poissonpdf(147, 160) ≈ 0.0180
b. P(x ≤ 160) = poissoncdf(147, 160) ≈ 0.8666
c. Standard Deviation = σ = μ = 147 ≈ 12.1244

235

5.27 According to a recent poll by the Pew Internet Project, girls between the ages of 14 and
17 send an average of 187 text messages each day. Let X = the number of texts that a girlaged 14 to 17 sends per day. The discrete random variable X takes on the values x = 0, 1, 2…. The random variable X has a Poisson distribution: X ~ P(187). The mean is 187 text
messages.
a. What is the probability that a teen girl sends exactly 175 texts per day?
b. What is the probability that a teen girl sends at most 150 texts per day?
c. What is the standard deviation?

Example 5.28
Text message users receive or send an average of 41.5 text messages per day.
a. How many text messages does a text message user receive or send per hour?
b. What is the probability that a text message user receives or sends two messages perhour?
c. What is the probability that a text message user receives or sends more than two
messages per hour?
Solution 5.28:
a. Let X = the number of texts that a user sends or receives in one hour. The average
number of texts received per hour is

41.5≈ 1.7292.
24

b. X ~ P(1.7292), so P(x = 2) = poissonpdf(1.7292, 2) ≈ 0.2653
c. P(x > 2) = 1 – P(x ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505

5.28 Atlanta’s Hartsfield-Jackson International Airport is the busiest airport in the world.
On average there are 2,500 arrivals and departures each day.
a. How many airplanes arrive and depart the airport per hour?
b. What is the probability that there are exactly 100 arrivals and departures in one hour?
c. What is the probability that there are at most 100 arrivals and departures in one hour?

236

Example 5.29
On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 24hours in Alaska was reported as about 1.02%. Use this information for the next 200 days tofind the probability that there will be low seismic activity in ten of the next 200 days. Useboth the binomial and Poisson distributions to calculate the probabilities. Are they close?
Solution 5.29
Let X = the number of days with low seismic activity. Using the binomial distribution:• P(x = 10) = binompdf(200, .0102, 10) ≈ 0.000039
Using the Poisson distribution:• Calculate μ = np = 200(0.0102) ≈ 2.04
• P(x = 10) = poissonpdf(2.04, 10) ≈ 0.000045
We expect the approximation to be good because n is large (greater than 20) and p is small(less than 0.05). The results are close—both probabilities reported are almost 0.

5.29 On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity forthe next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%.
Use this information for the next 100 days to find the probability that there will be low
seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions
to calculate the probabilities. Are they close?

237

KEY TERMS
Bernoulli Trials an experiment with the following characteristics:
1. There are only two possible outcomes called “success” and “failure” for each trial.
2. The probability p of a success is the same for any trial. So, the probability q = 1 − pof a failure is also the same for any trial.
Binomial Experiment a statistical experiment that satisfies the following four conditions:
1. There are a fixed number of identical trials, n.
2. There are only two possible outcomes, called "success" and, "failure," for each trial.
3. The probability p of a success on one trial remains the same for all trials. Similarly,
the probability q of a failure on one trial remains the same for all trials.
4. The n trials are independent of one another.
Binomial Distribution a discrete random variable that arises from Bernoulli trials; there
are a fixed number, n, of independent trials. “Independent” means that the result of any
trial (for example, trial one) does not affect the results of the following trials, and all trials
are conducted under the same conditions. Under these circumstances the binomial randomvariable X is defined as the number of successes in n trials.
•

The notation is: X ~ B(n, p).

•

The mean is μ = np and the standard deviation is σ = npq = np (1 − p ) .

•The probability of exactly x successes in n trials is P ( X = x)= n C k p qx

n− x

Expected Value: The expected arithmetic average when an experiment is repeated manytimes; also called the mean of the distribution.
••

The expected value is denoted by either E(X) or μ.
For a discrete random variable with probability distribution function P(x), we
calculate the expected value by E(X) = μ = ∑ xP(x).

Geometric Experiment a statistical experiment with the following properties:
1. There are one or more Bernoulli trials with all failures except the last one, which isa success.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, p, of a success and the probability, q, of a failure do not change fromtrial to trial.
4. The trials are independent of one another.

238

Geometric Distribution a discrete random variable (RV) that arises from the Bernoulli
trials; the trials are repeated until the first success. The geometric variable X is defined as
the number of trials until the first success.
•Notation: X ~ G( p).

•

The mean is µ =•The probability of exactly x failures before the first success is given by the formula:1
and the standard deviation is σ =
p


1 1
⋅  − 1 .
p p
P( X = x) = p(1 − p) x −1 .
Mean of a Probability Distribution: The long-term average of the random variable overmany trials of a statistical experiment. The mean is also called the expected value.
Poisson Probability Distribution: A discrete random variable that counts the numberof times a certain event will occur in a specific interval. Characteristics of the variable:•••

The probability p that the event occurs in a given interval is the same for all
intervals of equal length.
The events occur with a known mean and independently of the time since the last
event.
The distribution is defined by the mean λ of the event in the interval.
o Notation: X ~ P(μ).
ooThe mean is μ = np and the standard deviation is σ = µ .

The probability of having exactly x successes in a fixed time interval isP( X = x) = e − µ

µx
x!

.

The Poisson distribution is often used to approximate the binomial distribution, when n is“large” and p is “small” (a general rule is that n should be greater than or equal to 20 and p
should be less than or equal to 0.05)
Probability Distribution Function (PDF): A mathematical description of a discreterandom variable, given either in the form of an equation (formula) or in the form of a table
listing all the possible outcomes of an experiment and the probability associated with eachoutcome.
Random Variable. A characteristic of interest in a population being studied; common
notation for variables are upper case Latin letters X, Y, Z,.... Common notation for a
specific value from the domain (set of all possible values of a variable) are lower case Latin
letters x, y, and z. For example, if X is the number of children in a family, then x represents
a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in algebra in thetwo following ways.

239

•

•

The domain of the random variable is not necessarily a numerical set; the domain
may be categorical in nature and expressed in words; for example, if X = hair color
then the domain is {black, blond, gray, green, orange}.
The values of the random variable are determined by chance. So we can tell what
specific value x the random variable X takes only after performing the experiment.

Standard Deviation of a Probability Distribution: A numerical measure of how
dispersed the outcomes of a statistical experiment are from the mean of the distribution.
The Law of Large Numbers: As the number of trials in a probability experiment
increases, the difference between the theoretical probability of an event and the relative
frequency probability approaches zero.

FORMULA REVIEWLet X be a discrete random variable, and x1, x2, . . . xn, list of possible outcomes for X.
•

The mean of the random variable is: μ = Σ xi P(xi)•The variance of the random variable is:•

The standard deviation is the square root of the variance: σ =σ2 = Σ(xi – μ)2 P(xi)∑ ( x − µ ) P( x )i

2i

Binomial Distribution:
Notation:

X ~ B(n, p).

PDF: P ( X = x)= n C k p x q n − x

Mean: is μ = np
Standard deviation: σ = npq = np (1 − p )Geometric Distribution:1pNotation: X ~ G( p).

Mean: µ =PDF: P ( X = x) = p (1 − p ) x −1

Standard deviation: σ =1 1⋅  − 1 .
p pPoisson Distribution:
Notation: X ~ P(μ).
PDF: P ( X = x) = e − µMean: μ = npµxx!

Standard deviation: σ = µ240

Exercises for Chapter 5
Use the following information to answer the next five exercises:
A company wants to evaluate its attrition rate, in other words, how long new hires staywith the company. Over the years, they have established the following probability
distribution.
Let X = the number of years a new hire will stay with the company.
Let P(x) = the probability that a new hire will stay with the company x years.
1. Complete the table using the data provided:
x
0
1
2
3
4
5
6

P(x)
0.12
0.18
0.30
0.15
0.10
0.05

2. P(x = 4) =3. P(x ≥ 5) =4. On average, how long would you expect a new hire to stay with the company?
5. What does the column “P(x)” sum to?
Use the following information to answer the next four exercises:
A baker is deciding how many batches of muffins to make to sell in his bakery. He wants tomake enough to sell every muffin and no fewer. Through observation, the baker has
established a probability distribution.
X
1
2
3
4

P(x)
0.15
0.35
0.40
0.10

6. Define the random variable X.
7. What is the probability the baker will sell more than one batch? P(x > 1) =241

8. What is the probability the baker will sell exactly one batch? P(x = 1) =9. On average, how many batches should the baker make?
Use the following information to answer the next two exercises:
Ellen has music practice three days a week. She practices for all of the three days 85% of
the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One
week is selected at random.
10. Define the random variable X.
11. Construct a probability distribution table for the data.
12. We know that for a probability distribution function to be discrete, it must have two
characteristics. One is that the sum of the probabilities is one. What is the othercharacteristic?
Use the following information to answer the next five exercises:
Javier volunteers in community events each month. He does not do more than five events ina month. He attends exactly five events 35% of the time, four events 25% of the time, three
events 20% of the time, two events 10% of the time, one event 5% of the time, and no events
5% of the time.
13. Define the random variable X.
14. What values does x take on?
15. Construct a PDF table.
16. Find the probability that Javier volunteers for less than three events each month:That is, calculate P(x < 3).
17. Find the probability that Javier volunteers for at least one event each month.
That is, calculate P(x > 1).
18. Complete the expected value table:
x
0
1
2
3

P(x) x* P(x)
0.2
0.2
0.4
0.2

242

19. Suppose that the PDF for the number of years it takes to earn a Bachelor of Science(B.S.) degree is given in
x
3
4
5
6
7

P(x)
0.05
0.40
0.30
0.15
0.10

a. In words, define the random variable X.
b. What does it mean that the values zero, one, and two are not included for x in the
PDF?
c. On average, how many years do you expect it to take for an individual to earn a
B.S.?
20. Find the mean and standard deviation of the probability distributions:
a.

b.
x
2
4
6
8
10

P(x)
0.1
0.3
0.3
0.2
0.1

x
1
2
3
4
5

P(x)
0.15
0.25
0.30
0.20
0.10

Use the following information to answer the next five exercises:
A physics professor wants to know what percent of physics majors will spend the next
several years doing post-graduate research. He has the following probability distribution.
x
1
2
3
4
5
6

P(x)
0.35
0.20
0.15
0.10
0.05

21. Define the random variable X.
22. Define P(x), or the probability of x.

243

23. Find the probability that a physics major will do post-graduate research for four years,
P(x = 4).
24. Find the probability that a physics major will do post-graduate research for at mostthree years. I.e. find P(x ≤ 3).
25. On average, how many years would you expect a physics major to spend doing postgraduate research?
Use the following information to answer the next five exercises:
A ballet instructor is interested in knowing what percent of each year's class will continue
on to the next, so that she can plan what classes to offer. Let X = the number of years a
student will study ballet with the teacher.
Over the years, she has established the following probability distribution:
x
1
2
3
4
5
6
7

P(x)
0.10
0.05
0.10
0.30
0.20
0.10

26. Complete the table using the data provided.
27. Calculate P(x > 5).
28. Calculate P(x = 4) =29. Calculate P(x < 4) =30. On average, how many years would you expect a child to study ballet with this teacher?
31. What does the column “P(x)” sum to and why?
32. You are playing a game by drawing a card from a standard deck and replacing it. If thecard is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards
in a deck of 52 cards. What is the expected value of playing the game?
33. You are playing a game by drawing a card from a standard deck and replacing it. If thecard is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards
in a deck of 52 cards. Should you play the game?

244

Use the following information to answer the next nine exercises:
The Higher Education Research Institute at UCLA collected data from 203,967 incoming
first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. Of
those surveyed, 71.3% of those students replied that, yes, they believe that same-sex
couples should have the right to legal marital status. Suppose that you randomly pick eight
first-time, full-time freshmen from the survey. You are interested in the number that
believes that same sex-couples should have the right to legal marital status.
34. In words, define the random variable X.
35. Identify the distribution of X:

X~

(

,

)

36. What values does the random variable X take on?
37. Construct the probability distribution function (PDF).
38. On average, how many would you expect to answer yes?
39. What is the standard deviation σ?
40. What is the probability that at most five of the freshmen reply “yes”?
41. What is the probability that at least two of the freshmen reply “yes”?
42. What is the probability that exactly three of the freshmen reply “yes”?
43. A game involves selecting a card from a regular 52-card deck and tossing a coin. The
coin is a fair coin and is equally likely to land on heads or tails.
• If the card is a face card, and the coin lands on Heads, you win $6• If the card is a face card, and the coin lands on Tails, you win $2• If the card is not a face card, you lose $2, no matter what the coin shows.
a. Find the expected value for this game (expected net gain or loss).
b. Explain what your calculations indicate about your long-term average profits andlosses on this game.
c. Should you play this game to win money?
44. You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets
available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes,
and four $25 prizes. Find your expected gain or loss.

245

45. A venture capitalist, willing to invest $1,000,000, has three investments to choose from.
The first investment, a software company, has a 10% chance of returning $5,000,000 profit,
a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars.
The second company, a hardware company, has a 20% chance of returning $3,000,000
profit, a 40% chance of returning $1,000,000 profit, and a 40% chance of losing the million
dollars. The third company, a biotech firm, has a 10% chance of returning $6,000,000 profit,
a 70% of no profit or loss, and a 20% chance of losing the million dollars.
a.
b.
c.
d.
e.

Construct a PDF for each investment.
Find the expected value for each investment.
Which is the safest investment? Why do you think so?
Which is the riskiest investment? Why do you think so?
Which investment has the highest expected return, on average?

46. Suppose that 20,000 married adults in the United States were randomly surveyed as tothe number of children they have. Let X = the number of children married people have.
The results of the survey are compiled and are used as theoretical probabilities:
x
0
1
2
3
4
5
6 or more
a.
b.
c.
d.

P(x)
0.10
0.20
0.30
0.10
0.05
0.05

Find the probability that a married adult has three children.
In words, what does the expected value in this example represent?
Find the expected value.
Is it more likely that a married adult will have two to three children or four to six
children? How do you know?

47. A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope froma box containing 100 seemingly identical envelopes. However, each envelope contains a
coupon for a free gift.
••••

Ten of the coupons are for a free gift worth $6.
Eighty of the coupons are for a free gift worth $8.
Six of the coupons are for a free gift worth $12.
Four of the coupons are for a free gift worth $40.

246

Based upon the financial gain or loss over the long run, should you play the game?
a. Yes, I expect to come out ahead in money.
b. No, I expect to come out behind in money.
c. It doesn’t matter. I expect to break even.
48. Florida State University has 14 statistics classes scheduled for its Summer 2013 term.
One class has space available for 30 students, eight classes have space for 60 students, oneclass has space for 70 students, and four classes have space for 100 students.
a. What is the average class size assuming each class is filled to capacity?
b. Space is available for 980 students. Suppose that each class is filled to capacity andselect a statistics student at random. Let the random variable X equal the size of thestudent’s class. Define the PDF for X.
c. Find the mean of X.
d. Find the standard deviation of X.
49. In a lottery, there are 250 prizes of $5, 50 prizes of $25, and ten prizes of $100.
Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to
break even?
50. According to a recent article the average number of babies born with significant hearing
loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The
number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Supposethat 1,000 babies from healthy baby nurseries were randomly surveyed.
Find the probability that exactly two babies were born deaf.
Use the following information to answer the next four exercises.
Recently, a nurse commented that when a patient calls the medical advice line claiming tohave the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only
about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in
how many actually have the flu.
51. Define the random variable and list its possible values.
52. State the distribution of X.
53. Find the probability that at least four of the 25 patients actually have the flu.
54. On average, for every 25 patients calling in, how many do you expect to have the flu?
55. People visiting video rental stores often rent more than one DVD at a time. The
probability distribution for DVD rentals per customer at Video To Go is given in the tablebelow. There is five-video limit per customer at this store, so nobody ever rents more than
five DVDs.

247

x
P(x)
a.
b.
c.
d.

0
0.03

1
0.50

2
0.24

3

4
0.07

5
0.04

Describe the random variable X in words.
Find the probability that a customer rents three DVDs.
Find the probability that a customer rents at least four DVDs.
Find the probability that a customer rents at most two DVDs.

56. A school newspaper reporter decides to randomly survey 12 students to see if they willattend Tet (Vietnamese New Year) festivities this year. Based on past years, she knowsthat 18% of students attend Tet festivities. We are interested in the number of students
who will attend the festivities.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~( , )d. How many of the 12 students do we expect to attend the festivities?
e. Find the probability that at most four students will attend.
f. Find the probability that more than two students will attend.
Use the following information to answer the next two exercises:
The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming
monthly schedule contains 12 games. Let X = the number of games won in that upcoming
month.
57. Find the expected number of wins for that upcoming month.
58. What is the probability that the San Jose Sharks win six games in that upcoming
month?
a. 0.1476

b. 0.2336

c. 0.7664

d. 0.8903

59. What is the probability that the San Jose Sharks win at least five games in theupcoming month?
a. 0.369
b. 0.526c. 0.473
d. 0.230
60. A student takes a ten-question true-false quiz but did not study and randomly guesses
each answer. Find the probability that the student passes the quiz with a grade of at least
70% of the questions correct.

248

61. A student takes a 32-question multiple-choice exam but did not study and randomly
guesses each answer. Each question has three possible choices for the answer. Find theprobability that the student guesses more than 75% of the questions correctly.
62. Six different colored dice are rolled. Of interest is the number of dice that show a one.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ ( , )d. On average, how many dice would you expect to show a one?
e. Find the probability that all six dice show a one.
f. Is it more likely that three or that four dice will show a one? Justify your answer
numerically.
63. More than 96 percent of the very largest colleges and universities (more than 15,000
total enrollments) have some online offerings. Suppose you randomly pick 13 suchinstitutions. We are interested in the number that offer distance learning courses.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ (,)
d. On average, how many schools would you expect to offer such courses?
e. Find the probability that at most ten offer such courses.
f. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify youranswer numerically and answer in a complete sentence.
64. Suppose that about 85% of graduating students attend their graduation. A group of 22
graduating students is randomly chosen.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ ( ,)d. How many are expected to attend their graduation?
e. Find the probability that 17 or 18 attend.
f. Based on numerical values, would you be surprised if all 22 attended graduation?
Justify your answer numerically.
65. At The Fencing Center, 60% of the fencers use the foil as their main weapon. We
randomly survey 25 fencers at The Fencing Center. We are interested in the number offencers who do not use the foil as their main weapon.
a.
b.
c.
d.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ ( , )How many are expected to not to use the foil as their main weapon?
249

e. Find the probability that six do not use the foil as their main weapon.
f. Based on numerical values, would you be surprised if all 25 did not use foil as theirmain weapon? Justify your answer numerically.
66. Approximately 8% of students at a local high school participate in after-school sports all
four years of high school. A group of 60 seniors is randomly chosen. Of interest is thenumber who participated in after-school sports all four years of high school.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~(,)
d. How many seniors are expected to have participated in after-school sports all four yearsof high school?
e. Based on numerical values, would you be surprised if none of the seniors participated inafter-school sports all four years of high school? Justify your answer numerically.
f. Is it more likely that four or that five of the seniors participated in after-school sports all
four years of high school? Justify your answer numerically.
67. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per
year. We are interested in the expected number of audits a person with that income has ina 20-year period. Assume each year is independent.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~(,)
d. How many audits are expected in a 20-year period?
e. Find the probability that a person is not audited at all.
f. Find the probability that a person is audited more than twice.
68. It has been estimated that only about 30% of California residents have adequate
earthquake supplies. Suppose you randomly survey 11 California residents. We areinterested in the number who have adequate earthquake supplies.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~(,)
d. What is the probability that at least eight have adequate earthquake supplies?
e. Is it more likely that none or that all of the residents surveyed will have adequate
earthquake supplies? Why?
f. How many residents do you expect will have adequate earthquake supplies?
69. There are two similar games played for Chinese New Year and Vietnamese New Year.
In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with aboard with those numbers. In the Vietnamese version, fair dice with pictures of a gourd,
fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also.
250

We will play with bets being $1. The player places a bet on a number or object. The “house”
rolls three dice. If none of the dice show the number or object that was bet, the house keeps
the $1 bet. If one of the dice shows the number or object bet (and the other two do not showit), the player gets back his or her $1 bet, plus $1 profit. If two of the dice show the number
or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus
$2 profit. If all three dice show the number or object bet, the player gets back his or her $1
bet, plus $3 profit. Let X = number of matches and Y = profit per game.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~(,)
d. List the values that Y may take on. Then, construct one PDF table that includes both X
and Y and their probabilities.
e. Calculate the average expected matches over the long run of playing this game for theplayer.
f. Calculate the average expected earnings over the long run of playing this game for theplayer.
g. Determine who has the advantage, the player or the house.
70. According to The World Bank, only 9% of the population of Uganda had access toelectricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = thenumber of people who have access to electricity.
a. What is the probability distribution for X?
b. Using the formulas, calculate the mean and standard deviation of X.
c. Use your calculator to find the probability that 15 people in the sample have access toelectricity.
d. Find the probability that at most ten people in the sample have access to electricity.
e. Find the probability that more than 25 people in the sample have access to electricity.
71. The literacy rate for a nation measures the proportion of people aged 15 and over thatcan read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people
in Afghanistan at random. Let X = the number of people who are literate.
a. Sketch a graph of the probability distribution of X.
b. Using the formulas, calculate the mean and standard deviation of X.
c. Find the probability that more than five people in the sample are literate. Is it is more
likely that three people or four people are literate?

251

5.4 Geometric Distribution
Use the following information to answer the next three exercises:
The Higher Education Research Institute at UCLA collected data from 203,967 incoming
first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3%
of those students replied that, yes, they believe that same-sex couples should have the rightto legal marital status. Suppose that you randomly select freshman from the study until you
find one who replies “yes.” You are interested in the number of freshmen you must ask.
72. In words, define the random variable X and identify the distribution of X.
73. On average, how many freshmen would you expect to have to ask until you found one
who replies "yes?
74. What is the probability that you will need to ask fewer than three freshmen?
75. Suppose that the probability that an adult in America will watch the Super Bowl is40%. Each person is considered independent. We are interested in the number of adults in
America we must survey until we find one who will watch the Super Bowl.
a. In words, define the random variable X.
b. Identify the distribution of X. X ~( , )c. How many adults in America do you expect to survey until you find one who will watch
the Super Bowl?
d. Find the probability that you must ask seven people.
e. Find the probability that you must ask three or four people.
76. It has been estimated that only about 30% of California residents have adequate
earthquake supplies. Suppose we are interested in the number of California residents wemust survey until we find a resident who does not have adequate earthquake supplies.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ ( , )d. What is the probability that we must survey just one or two residents until we find a
California resident who does not have adequate earthquake supplies?
e. What is the probability that we must survey at least three California residents until wefind a California resident who does not have adequate earthquake supplies?
f. How many California residents do you expect to need to survey until you find a
California resident who does not have adequate earthquake supplies?
g. How many California residents do you expect to need to survey until you find a
California resident who does have adequate earthquake supplies?

25277. In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog
pages. Suppose we randomly survey 20 pages. We are interested in the number of pages
that advertise footwear. Each page may be picked more than once.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ (,)
d. How many pages do you expect to advertise footwear on them?
e. Is it probable that all twenty will advertise footwear on them? Why or why not?
f. What is the probability that fewer than ten will advertise footwear on them?
g. Reminder: A page may be picked more than once. We are interested in the number ofpages that we must randomly survey until we find one that has footwear advertised on it.
Define the random variable X and give its distribution.
h. What is the probability that you only need to survey at most three pages in order to find
one that advertises footwear on it?
i. How many pages do you expect to need to survey in order to find one that advertises
footwear?
80. The World Bank records the prevalence of HIV in countries around the world. According
to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are
infected with HIV.”[1] In South Africa, the prevalence of HIV is 17.3%. Let X = the numberof people you test until you find a person infected with HIV.
a.
b.
c.
d.

Sketch a graph of the distribution of the discrete random variable X.
What is the probability that you must test 30 people to find one with HIV?
What is the probability that you must ask ten people?
Find the mean and standard deviation of the distribution of X.

79. According to a recent Pew Research poll, 75% of millennials (people born between 1981
and 1995) have a profile on a social networking site. Let X = the number of millennials you
ask until you find a person without a profile on a social networking site.
a. Describe the distribution of X.
b. Find the mean and standard deviation of X.
c. What is the probability that you must ask ten people to find one person without a social
networking site?
d. What is the probability that you must ask 20 people to find one person without a social
networking site?
e. What is the probability that you must ask at most five people?
80. A consumer looking to buy a used red Miata car will call dealerships until she finds a
dealership that carries the car. She estimates the probability that any independent
dealership will have the car will be 28%. We are interested in the number of dealerships
she must call.

253

a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ (,)
d. On average, how many dealerships would we expect her to have to call until she finds
one that has the car?
e. Find the probability that she must call at most four dealerships.
f. Find the probability that she must call three or four dealerships.
5.5 Poisson Distribution
Use the following information to answer the next six exercises:
On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result,states across the country are debating raising the driving age.
81. Assume the event occurs independently in any given day. Define the random variable
X.
82. X ~(,)

83. What values does X take on?
84. For the given values of the random variable X, fill in the corresponding probabilities.
85. Is it likely that there will be no teens killed from motor vehicle injuries on any given
day in the U.S? Justify your answer numerically.
86. Is it likely that there will be more than 20 teens killed from motor vehicle injuries on
any given day in the U.S.? Justify your answer numerically.
Use the following information to answer the next six exercises:
On average, a clothing store gets 120 customers per day.
87. Assume the event occurs independently in any given day. Define the random variable X
and identify the distribution.
88. What is the probability of getting 150 customers in one day?
89. What is the probability of getting 35 customers in the first four hours? Assume the store
is open 12 hours each day.
90. What is the probability that the store will have more than 12 customers in the firsthour?
91. What is the probability that the store will have fewer than 12 customers in the first twohours?

254

92. The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines
is one of the busiest in the world with an average of 60 births per day. Let X = the numberof births in an hour.
a. Find the mean and standard deviation of X.
b. Sketch a graph of the probability distribution of X.
c. What is the probability that the maternity ward will deliver three babies in one hour?
d. What is the probability that the maternity ward will deliver at most three babies in one
hour?
e. What is the probability that the maternity ward will deliver more than five babies in
one hour?
93. The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls
during the noon hour on Mondays. Experience shows that the existing staff can handle up
to six calls in an hour. Let X = the number of calls received at noon.
a. Find the mean and standard deviation of X.
b. What is the probability that the office receives at most six calls at noon on Monday?
c. Find the probability that the law office receives six calls at noon. What does this mean
to the law office staff who get, on average, 5.5 incoming phone calls at noon?
d. What is the probability that the office receives more than eight calls at noon?
94. The average number of children a Japanese woman has in her lifetime is 1.37. Supposethat one Japanese woman is randomly chosen.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ (,)
d. Find the probability that she has no children.
f. Find the probability that she has more children than the Japanese average.
95. The average number of children a Spanish woman has in her lifetime is 1.47. Supposethat one Spanish woman is randomly chosen.
a. In words, define the Random Variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~ (,)
d. Find the probability that she has no children.
e. Find the probability that she has fewer children than the Spanish average.
f. Find the probability that she has more children than the Spanish average.

25596. Fertile, female cats produce an average of three litters per year. Suppose that one
fertile, female cat is randomly chosen. In one year, find the probability she produces:a. In words, define the random variable X.
b. List the values that X may take on.
c. Give the distribution of X. X ~d. Find the probability that she has no litters in one year.
e. Find the probability that she has at least two litters in one year.
f. Find the probability that she has exactly three litters in one year.
97. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per
year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We areinterested in the number of people audited in one year. Use a Poisson distribution toanswer the following questions.
a. In words, define the random variable X.
b. List the values that X may take on.
c. Identify the distribution of X. X ~( , )d. How many are expected to be audited?
e. Find the probability that no one was audited.
f. Find the probability that at least three were audited.
98. An article about directory assistance stated that “internal surveys paid for by directory
assistance providers show that even the most accurate companies give out wrong numbers
15% of the time.” Assume that you are testing such a provider by making 20 directory
assistance requests and also assume that the provider gives the wrong phone number 15%of the time.
a. Give the distribution of X. X ~ (,)
b. Find the probability of getting 3 wrong numbersc. Find the probability of getting at most 7 wrong numbers
d. Find the probability of getting at least 5 wrong numbers.
99. A company prices its natural disaster insurance using the following assumptions. In one
calendar year, there is at most 1 natural disaster. Let’s say the probability of a natural
disaster is .15. Lastly, the number of natural disasters in any calendar year is independentof the number of natural disasters in any other calendar year.
a. Give the distribution of X. X ~

(

,

)

b. Find the probability that there are fewer than 4 natural disasters in a 10-year period.
c. Find the probability that there are at least 6 natural disasters in a 10-year period.

256

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5.2 Mean, Expected Value and Standard Deviation

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5.3 Binomial Distribution“Access to electricity (% of population),” The World Bank, 2013. Available online at
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Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013).
“One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline.
Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planetaverages-60-babies-day-mothers- bed.html (accessed May 15, 2013).
Vanderkam, Laura. “StopChecking Your Email, Now.” CNNMoney, 2013. Available online at
http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15,2013).
“World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012.
http://www.world- earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013).

1. ”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at
http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value
2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May15, 2013).

259

CHAPTER 5 SOLUTIONS:
1)

x
0
1
2
3
4
5
6

P(x)
0.12
0.18
0.30
0.15
0.10
0.10
0.05

3) P(x > 5) = 0.10 + 0.05 = 0.155) 1
7) P(x > 1) = 0.35 + 0.40 + 0.10 = 0.85
9) E(X) = 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45
11)

x
0
1
2
3

P(x)
0.03
0.04
0.08
0.85

13) Let X = the number of events Javier volunteers for each month.
xP(x)0 0.05
1 0.05
2 0.103 0.204 0.25
5 0.3517) P(x > 0) = 1 – P(x = 0) = 0.95.
19) a. X = number of years needed to complete B.S. degree.
b. No students completed B.S. degrees in two years or less.
c. E[x] = 4.85 years21) X = the number of years a physics major will spend doing post-graduate research.
23) 0.15
25) E(X) = 1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 2.6 years
27) P(x > 5) = .30 + .20 + .10 = 0.60.
29) P(x < 4) = 0.10 + 0.05 + 0.10 = 0.25
31) The sum of the probabilities sum to one because it is a probability distribution.
33) E(X) = 30(12/52) – 2(40/52) = $5.38.
35) X ~B(8, .713)
x
0
1
2
3
4
5
6
7
8
37) the table to the rightP(x) .000046 .00091 .008 .0395 .123 .244 .303 .215 .067
39) σ =�8(. 713)(.287) = 1.28

15)

260

41) P(x ≥ 2) = 1- binomcdf(8, .713, 1) = 0.999.
43) E[x] = 6(6/52) + 2(6/52) – 2(40/52) = -$.62 No you should not play this game to win
money.
45) a.
Software company

b. E[x] for software = $200,000

x
5,000,000
1,000,000-1,000,000

E[x] for hardware = $600,000

P(x)
.10
.30
.60

E[x] for Biotech = $400,000
c. 3rd investment because it has the lowest
probability of loss

Hardware company
x
3,000,000
1,000,000-1,000,000

d. 1st investment because it has the highest
probability of loss

P(x)
.20
.40
.40

e. 2nd investment

Biotech company
x
6,000,000
1,000,000-1,000,000

P(x)
.10
.70
.20

47) b
49) Let X = the amount of money to be won on a ticket. The following table shows the PDF
for X.
X
0
5
25
100

P(x)
0.969
0.025
0.005
0.001

E(X) = 0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35. So a fair price for a ticket is$0.35.
Any price over $0.35 will enable the lottery to raise money.
51) X = the number of patients calling in claiming to have the flu, who actually have theflu. Possible values of X are x = 0, 1, 2, ...25
53) 0.0165

261

55) a. X = the number of DVDs a Video to Go customer rentsc. P(x > 4) = 0.07 + 0.04 = 0.11 d. P(x < 2) = 0.77

b. P(x = 3) = 0.12

57) 4.4359) c
61) X = number of questions answered correctly; X ~ B(32, 0.333). We want theprobability that more than 75% of 32 questions correct; so we want P(x > 24). The event"more than 24" is the complement of "less than or equal to 24."So P(x > 24) = 1 – P(x < 24) =1 – binomcdf(32, .333, 24) ≈ 0.00000026.
63) a. X = the number of college and universities that offer online offerings.
b. Possible values are x = 0, 1, 2, …, 13.
c. X ~ B(13, 0.96) d. 12.48 e. 0.013
f. P(x = 12) = 0.3186; P(x = 13) = 0.5882. So it is more likely to get 13.
65) a. X = the number of fencers who do not use the foil as their main weaponb. x = 0, 1, 2, 3,... 25 c. X ~ B(25, 0.40) d. 10 e. 0.0442
f. The probability that all 25 not use the foil is almost zero. Therefore, it would be very
surprising.
67) a. X = the number of audits in a 20-year period b. x = 0, 1, 2, …, 20 c. X ~ B(20,0.02) d. 0.4 e. 0.6676 f. 0.0071
69) a. X = the number of matches b. 0, 1, 2, 3 c. X ~ B(3, 1/6) d. In dollars: x = −1,
1, 2, 3 e. ½ f. Multiply each Y value by the corresponding X probability from the PDF
table to get -0.0787. You lose about eight cents, on average, per game. g. The house has
the advantage.
71) a. X ~ B(15, 0.281)

b. μ = np = 15(0.281) = 4.215, σ =npq = 15(0.281)(719) =1.7409

c. P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246
P(x = 3) = binompdf(15, 0.281, 3) = 0.1927
P(x = 4) = binompdf(15, 0.281, 4) = 0.2259
It is more likely that four people are literate than three people are.
73) 1.4

26275) a. X = the number of adults surveyed until one says he or she will watch the Super
Bowl. b. X ~ G(0.40) c. μ = 2.5
d. 0.0187 e. 0.2304
77) a. X = the number of pages that advertise footwear b. X takes on the values 0, 1, 2,..., 20 c. X ~ B(20, 29/192) d. 3.02 e. No f. 0.9997 g. X = the number of pages wemust survey until we find one that advertises footwear. X ~ G(29/192) h. P(x < 3) =geometcdf(29/192, 3) = 0.3881 i. μ = 192/29 = 6.6207 pages.
79) a. X ~ G(0.25) b. μ = 1/0.25 = 4. c. P(x = 10) = geometpdf(0.25, 10) = 0.0188
d. P(x = 20) = geometpdf(0.25, 20) = 0.0011 e. P(x ≤ 5) = geometcdf(0.25, 5) =0.7627.
81) X = the number of U.S. teens who die from motor vehicle injuries per day.
83) 0, 1, 2, 3, ...
85) No87) X = number of customers per day; X ~ P(120).
89) If the store averages 120 customers per day, then it will average 10 customers perhour, or 40 per each four hours. So P(x = 35) = poissonpdf(40, 35) = 0.0485
91) The store averages 120 customers per day, so it will average 10 customers per hour, or
20 per each two hours. So P(x < 12) = P(x < 11) = poissonpdf(20, 11) = 0.0214.
93) X ~ P(5.5) a. μ = 5.5 and σ = 5.5 b. P(x < 6) = poissoncdf(5.5, 6) ≈ 0.6860 c. P(x= 6) = poissonpdf(5.5, 6). There is a 15.7% probability that the law staff will receive morecalls than they can handle. d. P(x > 8) = 1 – P(x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – 0.8944 =
0.105695) a. X = the number of children for a Spanish woman; b. 0, 1, 2, 3,.. ;.c. X ~
P(1.47) d. P(x = 0) = poissonpdf(1.47, 0) = 0.2299;
e. P(x < 1.47) = poissoncdf(1.47, 1) =0.5679 f. P(x > 1.47) = 1 – P(x < 1.47) = 1 – 0.5679 = 0.4321
97) a. X = the number of people audited in one year
b. 0, 1, 2, ..., 100
c. X ~ P(2)d. 2 e. P(x = 0) = poissonpdf(2, 0) = 0.1353; f. P(x > 3) = 1 – P(x < 2) = 1 – poissoncdf(2,2) = 0.3233.
99) a. X~B(10, .15)

b. .95 c. 0.00138

263

This page is purposely left blank.

264

6|CONTINUOUS RANDOM VARIABLES

Figure 6.1 The heights of these radish plants are continuous random variables. (Credit: Rev Stan)Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:•••

Recognize and understand continuous probability density functions in general.
Recognize the uniform probability distribution and apply it appropriately.
Recognize the exponential probability distribution and apply it appropriately.

Continuous random variables have many applications. Baseball batting averages, IQ scores,
the length of time a long distance telephone call lasts, the amount of money a person carries,
the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability
depends on a variety of continuous random variables.

265

6.1|Continuous Probability Distributions
NOTE
The values of discrete and continuous random variables can be unclear. For example, if X isequal to the number of miles (to the nearest mile) you drive to work, and then X is a discreterandom variable. You count the miles. If X is the distance you drive to work, then you
measure values of X and X is a continuous random variable. For a second example, if X isequal to the number of books in a backpack, then X is a discrete random variable. If X is theweight of a book, then X is a continuous random variable because weights are measured. How
the random variable is defined is very important.
Recall Discrete Probability Distributions have two rules
1. Each probability is between zero and one, inclusive.
2. The sum of the probabilities is one.
Example 6.1
A child psychologist is interested in the number of times a newborn baby's crying wakes its
mother after midnight. For a random sample of 50 mothers, the following information was
obtained.
Let X = the number of times per week a newborn baby's crying wakes its mother after
midnight. For this example, x = 0, 1, 2, 3, 4, 5. Let P(x) = probability that X takes on avalue x. Then the following is the probability distribution for X:

Recall P(x < 2) =

2 11
+
50 50

x
0
1
2
3
4
5

P(x)
2/50
11/50
23/50
9/50
4/50
1/50

For continuous random variables, the values of X can’t be placed in a table. The outcomes aremeasured, not counted.

266

Properties of Continuous Probability Distributions
The graph of a continuous probability distribution is a curve. Probability is represented byarea under the curve.
The curve is called the probability density function (abbreviated as pdf). We use thesymbol f(x) to represent the curve. f(x) is the function that corresponds to the graph; we use
the density function f(x) to draw the graph of the probability distribution.
Area under the curve is given by a different function called the cumulative distributionfunction (abbreviated as cdf). The cumulative distribution function is used to evaluate
probability as area.

Area under the curve
•
•
•

•

The entire area under the curve and above the x-axis is equal to one.
Probability is found for intervals of x values rather than for individual x values.
P(c < x < d) is the probability that the random variable X is in the interval between
the values c and d. P(c < x < d) is the area under the curve, above the x-axis, to theright of c and the left of d. P(c < x < d) is the same as P(c ≤ x ≤ d) because probability
is equal to area.
P(x = c) = 0. The probability that x takes on any single individual value is zero. The
area below the curve, above the x-axis, and between x = c and x = c has no width, andtherefore no area (area = 0). Since the probability is equal to the area, the probability
is also zero.

We will find the area that represents probability by using geometry, formulas, technology, or
probability tables. In general, calculus is needed to find the area under the curve for many
probability density functions. When we use formulas to find the area, the formulas were
found by using the techniques of integral calculus. However, we will not be using calculus inthis textbook.

267

There are many continuous probability distributions. When using a continuous probability
distribution to model probability, the distribution used is selected to model and fit theparticular situation in the best way. In this chapter and the next, we will study thecontinuous uniform distribution, the exponential distribution, and the normal
distribution. The following graphs illustrate these distributions respectively.

Figure 6.2 Graph shows a Uniform Distribution with the shaded area between x = 3 and x =6 to represent the probability that the value of the random variable X is in the interval
between 3 and 6.
Remember that the x-axis has the values of the continuous random variable.

Figure 6.3 Graph above shows an Exponential Distribution with the area between x = 2 andx = 4 shaded to represent the probability that the value of the random variable X is in theinterval between 2 and 4.

Figure 6.4 Graph shows the Standard Normal Distribution with the area between x = 1 andx = 2 shaded to represent the probability that the value of the random variable X is in theinterval between 1 and 2.
268

We begin by defining a continuous probability density function. We use the function
notation f(x). Intermediate algebra may have been your first formal introduction to functions.
In the study of probability, the functions we study are special. We define the function f(x) sothat the area between it and the x-axis is equal to a probability. Since the maximum
probability is one, the maximum area is also one. For continuous probability distributions,
PROBABILITY = AREA.

Example 6.2
Consider the function f(x) =1
for 0 ≤ x ≤ 20. x = a real number.
20

1
is a horizontal line. However, since 0 ≤ x ≤ 20, f(x) is restricted to the20
portion between x = 0 and x = 20, inclusive.
The graph of f(x) =100%

Notice that the distance on the x-axis is 20 units. Also, the height on the y-axis is 1/20.
The area of the created rectangle (A = Base · Height) is 1 which represents 100%.

Example 6.3
Consider the function f(x) = .1e −.1x for x ≥ 0. x = a real number.
The graph of f(x) = .1e −.1x is an exponential function that decreases starting at x = 0.
f(x)Note: the area under
the curve is 100%.
Continuous probability
distributions have asimilar notion that the“sum” of theprobabilities is 1.

0.1

100%
x

0
269

6.2 | Uniform Distribution
The uniform distribution is a continuous probability distribution and is concerned with
events that are equally likely to occur. When working out problems that have a uniform
distribution, be careful to note if the data is inclusive or exclusive.
Notation and Properties of the Uniform Distribution•

X ~ U(a, b) where a = lowest value of x and b = highest value of x.

•

The probability density function is f ( x ) =•The mean of a uniform probability distribution, µ =

•
•
•

1
b−a

a+b
, and
2
(b − a) 2
Variance of a uniform probability distribution, σ 2 =12(b − a)12
Probability = Area = Base ∙ heightThe standard deviation, σ =The probability density function for Uniform distribution is a horizontal line from a starting
x value to an ending x value. The area under the horizontal line creates a rectangle.

Example 6.4
Consider the function f(x) =1
for 0 ≤ x ≤ 20. x = a real number.
20

a. State the probability distribution in correct notation.
b. Find the mean and standard deviation.
c. Find P(x < 2).
Solution 6.4
a. Here we have a uniform distribution: U(0, 20) where a = 0 and b = 20b.=µ

a + b 0 + 20=
= 10 ;
22=σ(b − a) 20 − 0= = 5.77
12
12

270

c. P(x < 2) = area of rectangle = base ∙ height = 2 ⋅Suppose we want to find the area between f(x) =1=.1 =10%20

1
and the x-axis where 4 < x < 15.
20

Area = (Base)(Height)

 1 
 = 0.55
 20 

Area = (15 – 4) 

The area corresponds to the probability P(4 < x < 15) = 0.55.
NOTE: Suppose we want to find P(x = 15). On an x-y graph, x = 15 is a vertical line. A vertical

 1 
= 0
 20 

line has no width (or zero width). Therefore, P(x = 15) = (base)(height) = (0) 271

P(x ≤ k) (can be written as P(x < k) for continuous distributions) is called the cumulative
distribution function or CDF. Notice the "less than or equal to" symbol. We can use the
CDF to calculate P(x > k). The CDF gives "area to the left" and P(x > k) gives "area to theright." We calculate P(x > k) for continuous distributions as follows:
P(x > k) = 1 – P (x < k)

6.1 The data that follow are the number of passengers on 35 different charter fishing boats.
The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform
distribution where all values between and including zero and 14 are equally likely. State thevalues of a and b. Write the distribution in proper notation, and calculate the theoreticalmean and standard deviation.
1
6

12 4
10 4
14 11 7
11 4
13 2
4
6
9
10 5
13 4
10 14 12 11 6
10 11 03
10 0
11 13 2

12

Example 6.5
Let the smiling times of an 8-week old baby, in seconds; follow a uniform distribution between
0 and 23 seconds inclusive.
a. State the probability density function.
b. Write the distribution in notation form X~______c. What is the probability that a randomly chosen eight-week-old baby smiles between
two and 18 seconds?
d. Find the 90th percentile for an 8-week old baby’s smiling time.
e. Find the probability that a random eight-week-old baby smiles more than 12 seconds
knowing that the baby smiles more than eight seconds.

Solution 6.5

1
for 0 ≤ x ≤ 23
23
b. X~U(0, 23)a. f(x) =272

c. Find P(2 < x < 18).

 1   16 
= 
 23   23 

P(2 < x < 18) = (base)(height) = (18 – 2) d. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90

P(x < k) = 0.90
Base ∙ height = 0.90

 1
 =0.90
 23 

(k – 0) 

 1
 =0.90 solve for k by multiplying both sides by 23
 23 

k

k = 20.7 minutes
e. This probability question is a conditional. You are asked to find the probability thatan eight-week-old baby smiles more than 12 seconds when you already know the baby
has smiled for more than eight seconds.
Find P(x > 12|x > 8) There are two ways to do the problem:273

For the first way, use the fact that this is a conditional and changes the sample space. The
graph illustrates the new sample space. You already know the baby smiled more than eight
seconds.
Write a new f(x):f(x) =

1
1
for 8 < x < 23=
23 − 8 15

P(x > 12|x > 8)

= base ∙ height 1  11=
 15  15= (23 – 12) 

For the second way, use the conditional formula from Probability Topics with the original
distribution X ~ U(0, 23):

P( A | B) =P( A and B)
P( B)For this problem, A is (x > 12) and B is (x > 8).
So, P(x > 12|x > 8)=

P ( x > 12 AND x > 8 )P( x > 8)1 
23 − 12 )  (
P ( x > 12 )
 15  11===
P ( x > 8)1
 15 

( 23 − 8) 

274

15

6.2 A distribution is given as X ~ U (0, 20). What is P(2 < x < 18)? Find the 90th percentile.

Example 6.6
The amount of time, in minutes, that a person must wait for a bus is uniformly distributed
between zero and 15 minutes, inclusive.
a. What is the probability that a person waits fewer than 12.5 minutes?
b. On the average, how long must a person wait? Find the mean, μ, and the standard
deviation, σ.
c. Ninety percent of the time, the time a person must wait falls below what value?
Solution 6.6
a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0,11
for 0 ≤ x ≤ 15.
15). Write the probability density function. f(x) ==15 − 0 15
Find P (x < 12.5). Draw a graph.

1
 = 0.8333
 15 

P(x < k) = (base)(height) = (12.5 - 0) 

The probability a person waits less than 12.5 minutes is 0.8333.
b. μ = (a + b)/2 = (15 + 0)/2 = 7.5 minutes. On the average, a person must wait 7.5 minutes.
=σ(b − a ) 15 − 0 = 4.3. The Standard deviation is 4.3 minutes.
=
12
12

275

c. This asks for the 90th percentile. Draw a graph. Let k = the 90th percentile.

P(x < k) = (base)(height)

1 15 

0.9 = (k − 0) 1 15 

0.9 = k 

k = (0.90)(15) = 13.5 minutes
k is sometimes called a critical value.
The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most
13.5 minutes.

6.3 Based on a Covid 19 incubation study, a researcher looked at the incubation period as a
uniform distribution from 2.1 to 11.1 days to determine the quarantine period.
a.
b.
c.
d.
e.

Find a and b and describe what they represent.
Write the distribution as X~.
Find the mean and the standard deviation.
What is the probability that the incubation period for Covid-19 is less than 4 days?
What is the 70th percentile for the incubation period of the Covid-19 virus?

276

Example 6.7
Suppose the amount of gasoline distributed to a company’s fleet of trucks every week isuniformly distributed. The minimum distributed is 2500 gallons. The maximum distributed
is 3200 gallons.
a. What is the probability that a randomly selected week, the distribution is more than3000 gallons?
b. Find the probability that another week’s distribution is more than 3000 gallons, giventhat distribution has already exceeded 2800 gallons.
Solution 6.71
� = 0.286
700

a. P(x > 3000) = Base · Height = 200 �b. The second question has a conditional probability. You are asked to find the probability
that another week’s distribution is more than 3000 gallons, given that distribution has
already exceeded 2800 gallons.
First way: Since you know the distribution has already exceeded 2800 gallons, the starting
point is no longer a = 2500. The new starting point is 2800.
Write a new f(x):f(x) =

1
1
for 2800 ≤ x ≤ 3200.
=
3200 − 2800 400

Find P(x > 3000|x > 2800). Draw a graph.

P(x > 3000|x > 2800) = (base)(new height) = (3200 – 3000)(1/400) = .5
The probability that another week’s distribution is more than 3000 gallons, given that
distribution has already exceeded 2800 gallons is .5, 50%277

Second way: Draw the original graph for X ~ U (2500, 3200).

NOTE: P(x > 3000 and x > 2800) is
the area that they have incommon. The base of the common
area is 200. The height is 1/700.

Use the conditional formula

 1 
200 

P( x > 3000 and x > 2800)
 700  .5
P(x > 3000|x > 2800) == =P( x > 2800)
 1 
400 

 700 Suppose the time it takes a student to finish a quiz is uniformly distributed between six and
15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. X ~ U(6, 15). Find the probability that a randomly selected student needs at least eight minutes tocomplete the quiz. Then find the probability that a different student needs at least eightminutes to finish the quiz given that she has already taken more than seven minutes.

Example 6.8
Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs
to fix a furnace is uniformly distributed between 1.5 and four hours. Let X = the time needed
to fix a furnace. Then X ~ U (1.5, 4).
a. Find the 30th percentile of furnace repair times.
b. The longest 25% of furnace repair times take at least how long? (In other words: findthe minimum time for the longest 25% of repair times.) What percentile does thisrepresent?
c. Find the mean and standard deviation
Solution 6.8
a. Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left,
representing the shortest 30% of repair times.

278

P(x < k) = (base)(height) = (k – 1.5)(0.4)
0.3 = (k – 1.5) (0.4); Solve to find k:
0.75 = k – 1.5, obtained by dividing both sides by 0.4
k = 2.25, obtained by adding 1.5 to both sides
Interpretation:
The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or
less.
b. Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right
representing the longest 25% of repair times.
P(x > k) = (base)(height)
0.25 = (4 – k)(0.4); Solve for k:
0.625 = 4 − k, obtained by dividing both sides by 0.4−3.375 = −k, obtained by subtracting four from both sides: k = 3.375
The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).
Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair
times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.
c. μ = (a + b)/2 = (1.5 + 4)/2 = 2.75 hours=σ(b − a ) 4 − 1.5= = 0.7217
12
12

279

The amount of time a service technician needs to change the oil in a car is uniformly
distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car.
a. Write the random variable X in words.

.

b. Write the distribution.
c. Graph the distribution.
d. Find P (x > 19).
e. Find the 75th percentile.

280

6.3 | The Exponential Distribution
The exponential distribution is often concerned with the amount of time until some specific
event occurs. For example, the amount of time (beginning now) until an earthquake occurs
has an exponential distribution. Other examples include the length, in minutes, of long
distance business telephone calls, and the amount of time, in months, a car battery lasts. It
can be shown, too, that the value of the change that you have in your pocket or purse
approximately follows an exponential distribution.

This Photo by Unknown Author is licensed under CC BY-ND

Values for an exponential random variable occur in the following way. There are fewer large
values and more small values. For example, the amount of money customers spend in one
trip to the supermarket follows an exponential distribution. There are more people who spend
small amounts of money and fewer people who spend large amounts of money.
The exponential distribution is widely used in the field of reliability. Reliability deals withthe amount of time a product lasts.

Notation and Properties of the Exponential Distribution•

X ~ Exp(m) where m = is the rate of decay (m > 0)•

The probability density function is f ( x) = me•

The mean of an exponential probability distribution, µ =•The standard deviation, σ = µ••P(x > k) = e − mx
P(x < k) = 1 - e − mxln(area to right )kth percentile =
−m

•

281

− mx

1
m

Graphing an exponential probability density function, f ( x) = me

− mx

, you will have a y-intercept of (0, m). Since m is positive and there is a negative sign in front of it, as x increases
the function decreases. The number e = 2.71828182846... It is a number that is used often inmathematics. Scientific calculators have the key ex; if you enter one for x, the calculator will
display the value e.

Using the TI-83, 83+, 84, 84+ Calculator

2nd lnExample 6.9
Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time
is known to have an exponential distribution with the average amount of time equal to four
minutes.
X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To
do any calculations, you must know m, the decay parameter.

µ=

1
m



m=

1

µ

Therefore m = .25
The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25).

f ( x) = .25e −.25 x ; where x is at least zero. For example, f(5) = 0.25e−(0.25)(5) = 0.072.

282

The standard deviation, σ, is the same as the mean. μ = σ.

The amount of time spouses shop for anniversary cards can be modeled by an exponential
distribution with the average amount of time equal to eight minutes. Write the distribution,state the probability density function, and graph the distribution.

Example 6.10
a. Using the information in Example 6.9, find the probability that a clerk spends less thanfour minutes with a randomly selected customer.
b. Half of all customers are finished within how long? (Find the 50th percentile).

Solution 6.10
a. Find P(x < 4).
The cumulative distribution function (CDF) gives the area to the left.
.25

x
4

283

P(x < k) = 1 – e–mkP(x < 4) =1 – e(–0.25)(4) = 0.6321
The probability that a postal clerk spends less than four minutes with a randomly selectedcustomer is 63.21%.
NOTE: µ = 4 is the mean and 63.21% is below it. It is unlike uniform distribution where µ
is the median too.
b. Here it is asking us to find the 50th percentile (the median).
.25

Shaded area
represents probabilityP(x < k) = 0.5

xkP(x < k) = 0.50
kth percentile =ln(area to right )where ln is the natural logarithm.
−m50th percentile =ln(.5)−.25

k = 2.8 minutes
Interpretation: Half of all customers are finished within 2.8 minutes.
NOTE: The mean is larger. The mean is 4 minutes and the median is 2.8 minutes.

The number of days ahead travelers purchase their airline tickets can be modeled by an
exponential distribution with the average amount of time equal to 15 days. Find theprobability that a traveler will purchase a ticket fewer than ten days in advance. How manydays do half of all travelers wait?

284

Example 6.11
On the average, a certain computer part lasts ten years. The length of time the computer partlasts is exponentially distributed.
a. What is the probability that a computer part lasts more than 7 years?
b. On the average, how long would five computer parts last if they are used one after another?
c. Eighty percent of computer parts last at most how long?
d. What is the probability that a computer part lasts between nine and 11 years?

Solution 6.11
a. Let X = the amount of time (in years) a computer part lasts.
μ = 10 so m = 0.1
Find P(x > 7). Draw the graph.

P(x > k) = e-mkP(x > 7) = e(–0.1)(7) = 0.4966.

b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they
are used one right after the other would last, on the average, (5)(10) = 50 years.

285

c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.

Solve for k: 80% is below k and 20% is above k.
kth percentile =k=ln(area to right )−mln(.2)= 16.1 years−.1

Eighty percent of the computer parts last at most 16.1 years.
d. Find P(9 < x < 11). Draw the graph.

P(9 < x < 11) = P(x < 11) – P(x < 9)P(x < 11) = 1 – e(–0.1)(11) = 0.6671
P(x < 9) = 1 – e(–0.1)(9) = 0.5934
P(9 < x < 11) =0.6671 – 0.5934 = 0.0737.
The probability that a computer part lasts between nine and 11 years is 7.37%

286

On average, a pair of running shoes can last 18 months if used every day. The length of timerunning shoes last is exponentially distributed. What is the probability that a pair of runningshoes last more than 15 months? On average, how long would six pairs of running shoes last
if they are used one after the other? Eighty percent of running shoes last at most how long if
used every day?

Example 6.12
The time spent waiting between events is often modeled using the exponential distribution.
For example, suppose that an average of 30 customers per hour arrive at a store and the time
between arrivals is exponentially distributed.
a. On average, how many minutes elapse between two successive arrivals.
b. When the store first opens, how long on average does it take for three customers toarrive?
c. After a customer arrives, find the probability that it takes less than one minute for the
next customer to arrive.
d. After a customer arrives, find the probability that it takes more than five minutes forthe next customer to arrive.
e. Seventy percent of the customers arrive within how many minutes of the previous
customer?
f. Is an exponential distribution reasonable for this situation?
Solution 6.12
a. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one
customer to arrive every two minutes on average.
b. Since one customer arrives every two minutes on average, it will take six minutes on
average for three customers to arrive.
c. Let X = the time between arrivals, in minutes. By part a, μ = 2, so m = ½ = .5
X ∼ Exp(0.5). Exponential distribution property states P(x < k) = 1 – e(–0.5k)

Therefore P(x < 1) = 1 – e(–0.5)(1) ≈ 0.3935.

d. Exponential distribution property states and P(x > k) = e(–0.5k)
Therefore P(x > 5) = e(–5)(0.5) ≈ 0.0821.

287

e. We want to solve for the 70th percentileln(area to right )kth percentile =−mk=ln(.3)= 2.41−.5

Thus, seventy percent of customers arrive within 2.41 minutes of the previous
customer.

f.

This model assumes that a single customer arrives at a time, which may not bereasonable since people might shop in groups, leading to several customers arriving at
the same time. It also assumes that the flow of customers does not change throughout
the day, which is not valid if some times of the day are busier than others.

Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per
minute. Assume that the duration of time between successive cars follows the exponential
distribution.
a. On average, how many seconds elapse between two successive cars?
b. After a car passes by, how long on average will it take for another seven cars to pass
by?
c. Find the probability that after a car passes by, the next car will pass within the next
20 seconds.
d. Find the probability that after a car passes by, the next car will not pass for at least
another 15 seconds.

288

Memorylessness of the Exponential Distribution
In Example 6.9, recall that the amount of time between customers is exponentiallydistributed with a mean of two minutes (X ~ Exp (0.5)). Suppose that five minutes haveelapsed since the last customer arrived. Since an unusually long amount of time has now
elapsed, it would seem to be more likely for a customer to arrive within the next minute. With
the exponential distribution, this is not the case–the additional time spent waiting for the
next customer does not depend on how much time has already elapsed since the lastcustomer. This is referred to as the memoryless property.
Memoryless property says thatP (x > r + t | x > r) = P (x > t) for all r ≥ 0 and t ≥ 0

For example, if five minutes has elapsed since the last customer arrived, then the probability
that more than one minute will elapse before the next customer arrives is computed by usingr = 5 and t = 1 in the foregoing equation.
P(x > 5 + 1 | x > 5) = P(x > 1) = e ( – 0.5)(1) ≈ 0.6065.
This is the same probability as that of waiting more than one minute for a customer to arrive
after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical ormechanical device. In Example 6.11, the lifetime of a certain computer part has the
exponential distribution with a mean of ten years (X ~ Exp(0.1)). The memoryless property
says that knowledge of what has occurred in the past has no effect on future probabilities. In
this case it means that an old part is not any more likely to break down at any particulartime than a brand new part. In other words, the part stays as good as new until it suddenly
breaks. For example, if the part has already lasted ten years, then the probability that it lasts
another seven years is P(x > 17|x > 10) = P(x > 7) = 0.4966.

Example 6.13
Refer to Example 6.9 where the time a postal clerk spends with his or her customer has an
exponential distribution with a mean of four minutes. Suppose a customer has spent four
minutes with a postal clerk. What is the probability that he or she will spend at least an
additional three minutes with the postal clerk?
Solution 6.13
The decay parameter of X is m = .25 so X ∼ Exp(0.25).
289

Exponential distribution property states P(x > k) = e–0.25k.
We want to find P(x > 7|x > 4). The memoryless property says that P(x > 7|x > 4) = P (x >3), so we just need to find the probability that a customer spends more than three minuteswith a postal clerk.
This is P(x > 3) = e–0.25⋅3 ≈ 0.4724.

Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years.
If a bulb has already lasted 12 years, find the probability that it will last a total of over 19
years.

Relationship between the Poisson and the Exponential Distribution
There is an interesting relationship between the exponential distribution and the Poisson
distribution. Suppose that the time that elapses between two successive events follows the
exponential distribution with a mean of μ units of time. Also assume that these times are
independent, meaning that the time between events is not affected by the times between
previous events. If these assumptions hold, then the number of events per unit time follows
a Poisson distribution with mean λ =1/μ. Recall from the chapter on Discrete Random
Variables that if X has the Poisson distribution with mean λ, then P(x = k) =λ𝑘𝑘 𝑒𝑒 −λ𝑘𝑘!

Conversely, if the number of events per unit time follows a Poisson distribution, then theamount of time between events follows the exponential distribution.
Example 6.14
At a police station in a large city, calls come in at an average rate of four calls per minute.
Assume that the time that elapses from one call to the next has the exponential distribution.
Take note that we are concerned only with the rate at which calls come in, and we are
ignoring the time spent on the phone. We must also assume that the times spent between
290calls are independent. This means that a particularly long delay between two calls does notmean that there will be a shorter waiting period for the next call. We may then deduce thatthe total number of calls received during a time period has the Poisson distribution.
a. Find the average time between two successive calls.
b. Find the probability that after a call is received, the next call occurs in less thanten seconds.
c. Find the probability that exactly five calls occur within a minute.
d. Find the probability that less than five calls occur within a minute.
e. Find the probability that more than 40 calls occur in an eight-minute period.
Solution 6.14
a. On average there are four calls occur per minute, so 15 seconds, or 15/60 = 0.25
minutes occur between successive calls on average.
b. Let T = time elapsed between calls. From part a, μ = 0.25, so m =T ∼Exp(4).

1= 4. Thus,µThe cumulative distribution function is P(t < k) = 1 – e–4t.

The probability that the next call occurs in less than ten seconds (ten seconds =1

−4⋅ 
 1
1/6 minute) is P  t <  =−1 e  6  ≈ 0.4866
 6

c. Let X = the number of calls per minute. As previously stated, the number of calls
per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, X ∼ Poisson(4), and so
P(x = 5) =45 𝑒𝑒 −4
= poissonpdf(4, 5) ≈ 0.1563
5!

(NOTE: 5! = 5∙4∙3∙2∙1 = 120)d. Keep in mind that X must be a whole number, so P(x < 5) = P(x ≤ 4).To compute
this, we could take P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4).
Using technology, we see that P(x ≤ 4) = poisssoncdf(4, 4) = 0.6288
e. Let Y = the number of calls that occur during an eight minute period.
Since there is an average of four calls per minute, there is an average of (8)(4) =32 calls during each eight minute period. Hence, Y ∼ Poisson(32). Therefore, P(y >40) = 1 – P (y ≤ 40) = 1 – poissoncdf(32, 40). = 0.0707

291

KEY TERMS
Uniform Distribution a continuous random variable (RV) that has equally likely outcomesover the domain, a < x < b; it is often referred as the rectangular distribution because thea+b
andgraph of the pdf has the form of a rectangle. Notation: X ~ U(a,b). The mean is μ =2
the standard deviation is σ =b−a1. The probability density function is f(x) =for a < x <b−a12b or a ≤ x ≤ b. The cumulative distribution is P(x ≤ k) =

k −a
.
b−a

Decay parameter: The decay parameter describes the rate at which probabilities decay tozero for increasing values of x.
It is the value m in the probability density function f(x) = me(-mx) of an exponential random
1variable. It is also equal to m =, where μ is the mean of the random variable.

µExponential Distribution a continuous random variable (RV) that appears when we areinterested in the intervals of time between some random events, for example, the length oftime between emergency arrivals at a hospital; the notation is X ~ Exp(m). The mean is μ =11
and the standard deviation is σ =. The probability density function is f ( x) = me− mx , x ≥ 0mm
and the cumulative distribution function is P(x ≤ k) = 1 − e−mx.
Memoryless property for an exponential random variable X, the memoryless property is
the statement that knowledge of what has occurred in the past has no effect on futureprobabilities. This means that the probability that X exceeds x + k, given that it has exceeded
x, is the same as the probability that X would exceed k if we had no knowledge about it.
In symbols we say that P(x > r + k|x > r) = P(x > k).
Poisson distribution If there is a known average of λ events occurring per unit time, andthese events are independent of each other, then the number of events X occurring in one
unit of time has the Poisson distribution. The probability of k events occurring in one unit
time is equal to P(x = k) =λ𝑘𝑘 𝑒𝑒 −λ
.
𝑘𝑘!

292

Formula Review
6.2 The Uniform Distribution•

X ~ U(a, b) where a = lowest value of x and b = highest value of x.

•

The probability density function is f ( x) =•1
for a ≤ x ≤ bb−aa+bThe mean of a uniform probability distribution, µ =2(b − a)2• Variance of a uniform probability distribution, σ =12(b − a)• The standard deviation, σ =
12
2

1
b−a

•

P(x < k) = ( k − a )•P(c < x < d) = ( d − c )Probability = Area

1b−aArea = Base ∙ Height

6.3 The Exponential Distribution•

X ~ Exp(m) where m = is the rate of decay•

The probability density function is f ( x) = me•

The mean of an exponential probability distribution, µ =•The standard deviation, σ = µ•••P(x > k) = e − mx
P(x < k) = 1 - e − mxln(area to right )kth percentile =−mMemoryless Property: P(x > r + k|x > r) = P(x > k)•Poisson Probability: P(x = k) =•k! = k∙(k-1)∙(k-2)∙…∙3∙2∙1•

λ𝑘𝑘 𝑒𝑒 −λ
.
𝑘𝑘!

293

− mx

where x ≥ 0 and m > 01
mEXERCISES FOR CHAPTER 6
For each probability and percentile problem, draw the picture.
1. Write the notation for the following distribution.

2. Write the notation for the following distribution.
.205

3. What does the shaded area represent? P(<x<)

4. What does the shaded area represent? P(

<x<

)

294

5. For a continuous probability distribution, 0 ≤ x ≤ 15. What is P(x > 15)?
6. What is the total area under f(x) if the function is a continuous probability density
function?
7. For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)?
8. A continuous probability function is restricted to the portion between x = 0 and 7.
What is P(x = 10)?
1
9. f(x) for a continuous probability function is, and the function is restricted to 0 ≤ x
5≤ 5. What is P(x < 0)?
1
10. f(x), a continuous probability function, is equal toand the function is restricted12
to 0 ≤ x ≤ 12. What is P (0 < x < 12)?
11. For the following distribution

a.
b.
c.
d.

State the distribution X~ _____________
State the continuous probability function, f(x) = ______
State the shaded region in probability notation.
Find the probability that x falls in the shaded area.

12. Find the probability that x falls in the shaded area:295

13. Find the probability that x falls in the shaded area:

14. f(x), a continuous probability function, is equal to1
and the function is restricted to33)2
15. The uniform distribution X ~ U(1.5, 4.5) describes the square footage (in 1000 feetsquared)a. What is the height of f(x) for the continuous probability distribution?
b. What are the constraints for the values of x?
c. Graph P(2 < x < 3).
d. What is P(2 < x < 3)?
e. What is P(x < 3.5| x < 4)?
f. What is P(x = 1.5)?
g. What is the 90th percentile of square footage for homes?
h. Find the probability that a randomly selected home has more than 3,000 square
feet given that you already know the house has more than 2,000 square feet.
1 ≤ x ≤ 4. Describe P(x >16. A distribution is given as X ~ U(0, 12)
a. What is a? What does it represent?
b. What is b? What does it represent?
c. What is the probability density function?
d. What is the theoretical mean?
e. What is the theoretical standard deviation?
f. Draw the graph of the distribution for P(x > 9).
g. Find P(x > 9).
h. Find the 40th percentile.
17. The age of cars in the staff parking lot of a suburban college is uniformly distributed
from six months (0.5 years) to 9.5 years.
a. What is being measured here?
b. In words, define the random variable X.
c. Are the data discrete or continuous?
d. The interval of values for x is
e. The distribution for X is296

f. Write the probability density function.
g. Graph the probability distribution.
h. Find the probability that a randomly chosen car in the lot was less than four
years old.
i. Considering only the cars less than 7.5 years old, find the probability that arandomly chosen car in the lot was less than four years old.
j. What changed in the previous two problems that made the solutions different?
k. Find the third quartile of ages of cars in the lot.
18. A customer service representative must spend different amounts of time with eachcustomer to resolve various concerns. The amount of time spent with each customer
can be modeled by the following distribution: X ~ Exp(0.2)
a. What type of distribution is this?
b. Are outcomes equally likely in this distribution? Why or why not?
c. What is m? What does it represent?
d. What is the mean?
e. What is the standard deviation?
f. State the probability density function.
g. Graph the distribution.
h. Find P(2 < x < 10).
i. Find P(x > 6).
j. Find the 70th percentile.
19. A distribution is given as X ~ Exp(0.75).
a. What is m?.
b. What is the probability density function?
c. What is the cumulative distribution function?
d. Draw the distribution.
e. Find P(x < 4).
f. Find the 30th percentile.
g. Find the median.
h. Which is larger, the mean or the median?
20. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is
said to decay exponentially. The decay rate is 0.000121. We start with one gram ofcarbon-14. We are interested in the time (years) it takes to decay carbon-14.
a.
b.
c.
d.
e.

What is being measured here?
Are the data discrete or continuous?
In words, define the random variable X.
What is the decay rate (m)?
The distribution for X is
297

f.

Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years.
This means, find P(x < 5,730).
g. Find the percentage of carbon-14 lasting longer than 10,000 years.
h. Thirty percent (30%) of carbon-14 will decay within how many years?

21. Consider the following experiment. You are one of 100 people enlisted to take part ina study to determine the percent of nurses in America with an R.N. (registerednurse) degree. You ask nurses if they have an R.N. degree. The nurses answer “yes”
or “no.” You then calculate the percentage of nurses with an R.N. degree. You givethat percentage to your supervisor.
a. What part of the experiment will yield discrete data?
b. What part of the experiment will yield continuous data?
22. When age is rounded to the nearest year, do the data stay continuous, or do they
become discrete? Why?
23. Births are approximately uniformly distributed between the 52 weeks of the year.
They can be said to follow a uniform distribution from one to 53 (spread of 52
weeks).
a. X ~
b. Graph the probability distribution.
c. f(x) =d. μ =e. σ =
f. Find the probability that a person is born at the exact moment week 19 starts.
That is, find P(x = 19) =g. P(2 < x < 31) =h. Find the probability that a person is born after week 40.
i. P(12 < x|x < 28) =j. Find the 70th percentile.
k. Find the minimum for the upper quarter.
24. A random number generator picks a number from one to nine in a uniform manner.
a. X ~
b. Graph the probability distribution.
c. f(x) =d. μ =e. σ =
f. P(3.5 < x < 7.25) =g. P(x > 5.67)
h. P(x > 5|x > 3) =i. Find the 90th percentile.
298

25. According to a study by Dr. John McDougall of his live-in weight loss program at St.
Helena Hospital, the people who follow his program lose between six and 15 pounds
a month until they approach trim body weight. Let’s suppose that the weight loss is
uniformly distributed. We are interested in the weight loss of a randomly selectedindividual following the program for one month.
a. Define the random variable. X =b. X ~
c. Graph the probability distribution.
d. f(x) =e. μ =
f. σ =
g. Find the probability that the individual lost more than ten pounds in a month.
h. Suppose it is known that the individual lost more than ten pounds in a month.
Find the probability that he lost less than 12 pounds in the month.
i. P(7 < x < 13|x > 9) =. State this in a probability question, similarly to parts g
and h, draw the picture, and find the probability.

26. A subway train on the Red Line arrives every eight minutes during rush hour. We
are interested in the length of time a commuter must wait for a train to arrive. The
time follows a uniform distribution.
a. Define the random variable. X =b. X ~
c. Graph the probability distribution.
d. f(x) =e. μ =
f. σ =
g. Find the probability that the commuter waits less than one minute.
h. Find the probability that the commuter waits between three and four minutes.
i. Sixty percent of commuters wait more than how long for the train? State this in
a probability question, similarly to parts g and h, draw the picture, and find theprobability.
27. The age of a first grader on September 1 at Garden Elementary School is uniformly
distributed from 5.8 to 6.8 years. We randomly select one first grader from the class.
a. Define the random variable. X =b. X ~
c. Graph the probability distribution.
d. f(x) =e. μ =
f. σ =
g. Find the probability that she is over 6.5 years old.
299

h. Find the probability that she is between four and six years old.
i. Find the 70th percentile for the age of first graders on September 1 at Garden
Elementary School.
28. The Sky Train from the terminal to the rental–car and long–term parking center is
supposed to arrive every eight minutes. The waiting times for the train are known tofollow a uniform distribution.
a. What is the average waiting time (in minutes)?
b. Find the 30th percentile for the waiting times (in minutes).
c. The probability of waiting more than seven minutes given a person has waited
more than four minutes is?
29. The time (in minutes) until the next bus departs a major bus depot follows a
1
distribution with f(x) =where x goes from 25 to 45 minutes.
20
a. Define the random variable. X =b. X ~
c. Graph the probability distribution.
d. The distribution is _______ (name of distribution). Determine if it is discrete orcontinuous.
e. μ =
f. σ =
g. Find the probability that the time is at most 30 minutes. Sketch and label a graphof the distribution. Shade the area of interest. Write the answer in a probability
statement.
h. Find the probability that the time is between 30 and 40 minutes. Sketch and labela graph of the distribution. Shade the area of interest. Write the answer in aprobability statement.
i. P(25 < x < 55) = . State this in a probability statement, similarly to parts g andh, draw the picture, and find the probability.
j. Find the 90th percentile. This means that 90% of the time, the time is less thanminutes.
k. Find the 75th percentile. In a complete sentence, state what this means. (See partj.)l. Find the probability that the time is more than 40 minutes given (or knowing that)
it is at least 30 minutes.
30. Suppose that the value of a stock varies each day from $16 to $25 with a uniform
distribution.
a. Find the probability that the value of the stock is more than $19.
b. Find the probability that the value of the stock is between $19 and $22.

300

c. Find the upper quartile, which means 25% of all days the stock is above whatvalue? Draw the graph.
d. Given that the stock is greater than $18, find the probability that the stock is morethan $21.
31. A fireworks show is designed so that the time between fireworks is between one andfive seconds, and follows a uniform distribution.
a. Find the average time between fireworks.
b. Find probability that the time between fireworks is greater than four seconds.
32. The number of miles driven by a truck driver falls between 300 and 700 miles, andfollows a uniform distribution.
a. Find the probability that the truck driver goes more than 650 miles in a day.
b. Find the probability that the truck drivers goes between 400 and 650 miles in a
day.
c. At least how many miles does the truck driver travel on the furthest 10% of days?
33. Suppose that the length of long distance phone calls, measured in minutes, is known
to have an exponential distribution with the average length of a call equal to eightminutes.
a. Define the random variable. X =b. X ~c. Determine if the variable is discrete or continuous.
d. μ =e. σ =
f. Draw a graph of the probability distribution. Label the axes.
g. Find the probability that a phone call lasts less than nine minutes.
h. Find the probability that a phone call lasts more than nine minutes.
i. Find the probability that a phone call lasts between seven and nine minutes.
j. If 25 phone calls are made one after another, on average, what would you expect
the total to be? Why?
34. Suppose that the useful life of a particular car battery, measured in months, decays
with parameter 0.025. We are interested in the life of the battery.
a. Define the random variable. X =.
b. Is X continuous or discrete?
c. X ~
d. On average, how long would you expect one car battery to last?
e. On average, how long would you expect nine car batteries to last, if they are used
one after another?
f. Find the probability that a car battery lasts more than 36 months.
g. Seventy percent of the batteries last at least how long?

301

35. The percent of persons (ages five and older) in each state who speak a language athome other than English is approximately exponentially distributed with a mean of
9.848. Suppose we randomly pick a state.
a. Define the random variable. X =.
b. Is X continuous or discrete?
c. X ~
d. μ =e. σ =
f. Draw a graph of the probability distribution. Label the axes.
g. Find the probability that the percent is less than 12.
h. Find the probability that the percent is between eight and 14.
i. The percent of all individuals living in the United States who speak a language athome other than English is 13.8.
i. Why is this number different from 9.848%?
ii. What would make this number higher than 9.848%?
36. The time (in years) after reaching age 60 that it takes an individual to retire isapproximately exponentially distributed with a mean of about five years. Suppose we
randomly pick one retired individual. We are interested in the time after age 60 toretirement.
a. Define the random variable. X =.
b. Is X continuous or discrete?
c. X ~
d. μ =e. σ =
f. Draw a graph of the probability distribution. Label the axes.
g. Find the probability that the person retired after age 70.
h. Do more people retire before age 65 or after age 65?
i. In a room of 1,000 people over age 80, how many do you expect will NOT haveretired yet?
37. The cost of all maintenance for a car during its first year is approximately
exponentially distributed with a mean of $150.
a. Define the random variable. X =.
b. X ~
c. μ =
d. σ =
e. Draw a graph of the probability distribution. Label the axes.
f. Find the probability that a car required over $300 for maintenance during its first
year.

302

38. The average lifetime of a certain new cell phone is three years. The manufacturer will
replace any cell phone failing within two years of the date of purchase. The lifetime ofthese cell phones is known to follow an exponential distribution.
a. The decay rate is:b. What is the probability that a phone will fail within two years of the date ofpurchase?
c. What is the median lifetime of these phones (in years)?
39. Let X ~ Exp(0.1).
a. decay rate =b. μ =
c. Graph the probability distribution function.
d. On the graph, shade the area corresponding to P(x < 6) and find the probability.
e. Sketch a new graph, shade the area corresponding to P(3 < x < 6) and find theprobability.
f. Sketch a new graph, shade the area corresponding to P(x < 7) and find theprobability.
g. Sketch a new graph, shade the area corresponding to the 40th percentile and findthe value.
h. Find the average value of x.
40. Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight
years.
a. Find the probability that a light bulb lasts less than one year.
b. Find the probability that a light bulb lasts between six and ten years.
c. Seventy percent of all light bulbs last at least how long?
d. A company decides to offer a warranty to give refunds to light bulbs whose lifetimeis among the lowest two percent of all bulbs. To the nearest month, what shouldbe the cutoff lifetime for the warranty to take place?
e. If a light bulb has lasted seven years, what is the probability that it fails within
the 8th year.
41. At a 911 call center, calls come in at an average rate of one call every two minutes.
Assume that the time that elapses from one call to the next has the exponential
distribution.
a. On average, how much time occurs between five consecutive calls?
b. Find the probability that after a call is received, it takes more than three minutes
for the next call to occur.
c. Ninety-percent of all calls occur within how many minutes of the previous call?
d. Suppose that two minutes have elapsed since the last call. Find the probability
that the next call will occur within the next minute.
e. Find the probability that less than 20 calls occur within an hour.
303

42. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't
give up any hits throughout the game. No-hitters occur at a rate of about three perseason. Assume that the duration of time between no-hitters is exponential.
a. What is the probability that an entire season elapses with a single no-hitter?
b. If an entire season elapses without any no-hitters, what is the probability that
there are no no-hitters in the following season?
c. What is the probability that there are more than 3 no-hitters in a single season?
d. What is the median?
e. What is the variance?
43. Assume that the life span of Android batteries is exponentially distributed. The
average time the batteries last is 4.2 years.
a. What is the probability that an Android battery lasts less than 5 years?
b. What is the probability that an Android battery lasts less than 6 years given thatit lasted already 5 years?
c. What is variance?
d. What is the median?
e. The middle 50% of batteries last between which 2 years?
44. According to the American Red Cross, about one out of nine people in the U.S. haveType B blood. Suppose the blood types of people arriving at a blood drive are
independent. In this case, the number of Type B blood types that arrive roughly
follows the Poisson distribution.
a. If 100 people arrive, how many on average would be expected to have Type B blood?
b. What is the probability that over 10 people out of these 100 have type B blood?
c. What is the probability that more than 20 people arrive before a person with type
B blood is found?
45. A web site experiences traffic during normal working hours at a rate of 12 visits perhour. Assume that the duration between visits has the exponential distribution.
a. Find the probability that the duration between two successive visits to the website is more than ten minutes.
b. The top 25% of durations between visits are at least how long?
c. Suppose that 20 minutes have passed since the last visit to the web site. What isthe probability that the next visit will occur within the next 5 minutes?
d. Find the probability that less than 7 visits occur within a one-hour period.

304

46. At an urgent care facility, patients arrive at an average rate of one patient every sevenminutes. Assume that the duration between arrivals is exponentially distributed.
a. Find the probability that the time between two successive visits to the urgent carefacility is less than 2 minutes.
b. Find the probability that the time between two successive visits to the urgent carefacility is more than 15 minutes.
c. If 10 minutes have passed since the last arrival, what is the probability that thenext person will arrive within the next five minutes?
d. Find the probability that more than eight patients arrive during a half-hour
period.
e. Find the median time between two successive visits.
f. Find the 75th percentile time between two successive visits.
g. Find the variance.

305

REFERENCES
6.2 The Uniform Distribution
McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.
Backer, Jantien A, et al. “Incubation Period of 2019 Novel Coronavirus (2019-NCoV)
Infections among Travellers from Wuhan, China, 20-28 January 2020.” Euro Surveillance :Bulletin Europeen Sur Les Maladies Transmissibles = European Communicable Disease
Bulletin, European Centre for Disease Prevention and Control (ECDC), Feb. 2020,www.ncbi.nlm.nih.gov/pmc/articles/PMC7014672/.

6.3 The Exponential Distribution
Data from the United States Census Bureau.
Data from World Earthquakes, 2013. Available online at http://www.worldearthquakes.com/ (accessed June 11, 2013).
“No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseballreference.com/bullpen/No-hitter(accessed June 11, 2013).
Zhou, Rick. “Exponential Distribution lecture slides.” Available online atwww.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf (accessed June 11, 2013).

306

CHAPTER 6 SOLUTIONS:
1) X ~ U(3, 8)3) P(2 < x < 5)
5) 0
7) 0
11) a. X~U(0, 9) b. f(x) = 1/9 for 0 ≤ x ≤ 9c. P( 6 < x < 8)
2(1/9) = 2/99) 0
d. P( 6 < x < 8) =13) P(2.5 < x < 5.5) = 3(1/10) = 0.3
15) a. 1/3
b. a = 1.5, b = 4.5
c. Graph:

1
4

.5

1

1.5 2

2.5 3

3.5 4

4.5

d. P(2 < x < 3) = base*height = 1(1/3) = 1/3
e. P(x < 3.5 | x < 4) =𝑃𝑃(𝑥𝑥<3.5 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥<4)
𝑃𝑃(𝑥𝑥<3.5)
2∗1/3
=
=
= 0.8
𝑃𝑃(𝑥𝑥<4)
𝑃𝑃(𝑥𝑥<4)2.5∗1/3

f. P(x = 1.5) = 0
g. P90 = k , where 0.90 = (k – 1.5)(1/3) => k = 4.2
h. P(x > 3 | x > 2) = 0.6

17) a. Age of carsc. continuous
g. Graph:b. X = The age (in years) of cars in the staff parking lot
d. 0.5 to 9.5
e. X ~ U(.5, 9.5) f. f( x) = 1/9

19h. P(x < 4) = .389
i. P(x < 4 | x < 7.5) = 0.5
j. The second problem had a given situation
k. P75 = k, where .75 = (k - .5)(1/9) => k = 7.25.

30719) a. m = .75 b. f(x) = -.75e-.75x
d. Graph:e. P(x < 4) = 1 – e-.75(4) = .95
g. median = ln(.50)/-.75 = .92

c. P(x < k) = 1 – e-.75k

f. k = ln(.70)/-.75 = .48
h. µ = 1.33 which is larger than the median.

21) a. The number of nurses that said yes.

b. The percentage of nurses that said yes.

23) a. X ~ U(1, 53) b. Graph is a uniform graph spanning from 1 to 52 with a height of
1/52.
c. f(x) = 1/9
d. µ = 54/2 = 27
e. σ = 52/sqrt(12) = 15.01
f. P(x = 19) = 0
g. P(2 < x < 31) = base*height = 29(1/52) = .558
h. P(x > 40) = 13 (1/52) = .25
i. P(12 < x | x < 28) = P(12 < x < 28)/ P(x < 28) = .308/.519 =.593
j. P70 = k, where 0.70 = (k – 1)(1/52) => k = 37.4
k. Q3 = k, where 0.75 = (k – 1)(1/52) => k = 40.
25) a. X = weight loss in a monthc. Graph:

1
9b. X ~U(6, 15)

2 4 6 8 10 12 14 16
d. f(x) = 1/9
e. µ = 21/2 = 10.5
f. σ = 9/sqrt(12) = 2.6
g. P(x > 10) = base*height = 5(1/9) = .556
h. P(x < 12 | x > 10) = P(10 < x < 12) / P(x > 10) = .222/.556 = .399
i. P(7< x < 13 | x > 9) = P(9 < x < 13) / P(x > 9) = .444/.667= .666

30827) a. X = age of first grade on Sept. 1st
c. Graph:b. X ∼U(5.8, 6.8)

d. f(x) = 1
e. µ = 12.6/2 = 6.3
f. σ = 1/sqrt(12) = .289
g. P(x > 6.5) = base*height = .3(1) = 0.3 h. P(4 < x < 6) =P(x < 6) =.2(1) = 0.2
i. P70 = k, where .70 = (k – 5.8)(1) => k = 6.5
29) a. X = minutes until the next bus departs
b. X ∼U(25, 45)
c. Graph: a rectangle that spans from 25 to 45 on x –axis with a height of 1/20
d. Uniform. It is continuous. e. µ = 70/2 = 35 f. σ = 20/sqrt(12) = 5.77
g. P(x ≤ 30) = base* height = 5(1/20) = .25

120

5 10 15 20 25 30 35 40 45

h. P(30 < x < 40) = base*height = 10(1/20) = .50
i. P(25 < x < 55) = P(x > 25) = 1
j. P90 = k, where .90 = (k – 25)(1/20) => k = 43.
k. P75 = k, where .75 = (k – 25)(1/20) => k = 40
l. P( x > 40 | x ≥ 30) = P(x > 40) / P(x ≥ 30) = .25/.75 = 1/3
31)a. µ = 3

b. P(x > 4) = .25

33)

a. X = minutes on a long distance phone callb. X ∼ exp(.125)
c. Continuousd. µ = 8 (given in the problem)e. σ = 8
f. Graph:

.125

309

g. P(x < 9) = 1 – e(-.125*9) = .675
h. P(x > 9) = e(-.125*9) = .325
i. P(7 < x < 9) = e(-.125*7) - e(-.125*9) = .0922 j. 25*8 = 200 minutes since average is 8
minutes.
35) a. X = percent of persons in each state who speak a language at home other thanEnglishb. Continuousc. X ∼ exp(.101) d. µ = 9.848 (given in the problem) e. σ =9.848
g. Graph:.101

h. P(x < 12) = 1 - e(-.101*12) = .702 i. P(8 < x < 14) = e(-.101*8) - e(-.101*14) = .203
j. The number is different because the population is different.
It makes it higher because of the younger age children are included.
37) a. X = Cost of car during first yearc. µ = 150 (given in the problem)e. Graph:b. X ∼ exp(1/150)
d. σ = 150

1
150

f. P(x > 300) = e(-2) = .135
39) a. Decay rate = 0.1
c. Graph:b. µ = 10.1

310

d. P(x < 6) = 1 - e(-.10*6) = .451
f. P(x < 7) = 1 - e(-.10*7) = .503

g. P40 = ln(.60)/-.1 = 5.10841) a. µ = 2

e. P(3 < x < 6) = e(-.10*3) - e(-.10*6) = .192

h. Average = µ = 10.

b. P(x > 3) = e(-.5*3) = .223

c. P90 = ln (.10)/-.5) = 4.605

d. P(x < 3 | x > 2) = P(2 < x < 3)/P(x >2) =𝑒𝑒 −.5∗2 −𝑒𝑒 −.5∗3=.393
𝑒𝑒 −.5∗2e. Poisson distribution; poissoncdf(30, 19) = .0219
43) a. P(x < 5) = 1 – e-.283(5) = .696
b. P(x < 6 | x > 5) = P(5 < x < 6)/ P(x >5 )=c. 4.22 = 17.64

𝑒𝑒 −.238∗5 −𝑒𝑒 −.238∗6=.212
𝑒𝑒 −.238∗5

d. P50 = ln(.50)/-.238 = 2.91 years

e. P25 = ln(.75)/-.238 = 1.21 years and P75 = ln(.25)/-.238 = 5.82 years
b. P75 = ln(.25)/-.2 = 6.93 minutes45) a. P(x > 10) = e-.2(10) =.135. m = 12/60 = .2
c. P(x < 25 | x > 20) = P(20 < x < 25)/P(x > 20) = .632
d. P(x < 7 visits) = poissoncdf(12, 6) = .0458

311

This page is purposely left blank.

312

7 | THE NORMAL DISTRIBUTION

Figure 7.1 If you ask enough people about their shoe size, you will find that your graphed data isshaped like a bell curve and can be described as normally distributed. (credit: Ömer Ünlϋ)

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:••••

Recognize the normal probability distribution and apply it appropriately.
Recognize the standard normal probability distribution and apply it appropriately.
Compare normal probabilities by converting to the standard normal distribution.
Find value of the random variable given area under the curve.

The normal distribution is a continuous distribution, and is the single most important of all
the distributions discussed in this text. It is widely used and even more widely abused. Itsgraph is bell-shaped, and we see the bell curve in many disciplines. Some of these include
psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of313

your instructors may use the normal distribution to help determine your grade. Most IQ
scores are normally distributed, as are many standardized test scores. In this chapter, we
will study the normal distribution, the standard normal distribution, and applications
associated with them.
The normal distribution has two parameters (two numerical descriptive measures of the
population), the mean, μ, and the standard deviation, σ. If X is a quantity to be measured
that has a normal distribution with mean μ and standard deviation σ, we designate this by
writing X ~ N(μ, σ). The probability density function is a rather complicated function. Do
not memorize it. It is not necessary for calculations and is included only for completeness:
𝑓𝑓(𝑥𝑥) =𝑒𝑒

𝑥𝑥−𝜇𝜇 2−.5��𝜎𝜎𝜎𝜎 √2𝜋𝜋The graph of this function exhibits the bell-shape of the normal distribution:

The cumulative distribution function is P(X < x). It can be calculated by a calculator,
statistical software, or by using a normal distribution table. Modern technology has made thetables virtually obsolete, and we will focus mainly on the use of the TI-84 family of calculatorsfor calculations. But there is still some insight to be gained from working with the tables so
we will discuss these briefly.
The curve is perfectly symmetrical about a vertical line drawn through the mean, μ. Thus
the mean is the same as the median. As the notation indicates, the normal distribution
depends only on the mean and the standard deviation. Since the area under the curve must
equal one, a change in the standard deviation, σ, causes a change in the shape of the curve;
the curve becomes fatter or skinnier depending on σ. The smaller the standard deviation is,
the narrower the normal curve appears. A change in μ causes the graph to shift to the left or
right. This means there are an infinite number of normal probability distributions.
Drawing the Normal Distribution Curve can be easily done. Firstly, we know that it is bell
shaped. Secondly, we know that it is symmetric about the mean, μ. Therefore, draw the x314

axis and place the mean at the center. Let the scale be the standard deviation, σ. The first
turning points of the bell shape occur one standard deviation from the mean. The second
turning points of the bell occur two standard deviations from the mean.

Example 7.1
Suppose X ~ N(30, 4). Draw the normal distribution.

Solution 7.1
The mean of the normal distribution is 30, μ = 30. The standard deviation of the
distribution is 4, σ = 4.
One standard deviation below the mean is 30 – 4 = 26. One standard deviation above themean is 30 + 4 = 34. The first turning points of the bell occur at 26 and 34.
Two standard deviations below the mean is 30 – 2 · 4 = 22. Two standard deviations above
the mean is 30 + 2 · 4 = 38. The second turning points of the bell occur at 22 and 38.
Note: Top of the bell occur atthe mean, µ. The 2 turning
points are symmetrically
about the mean

7.1 | The Standard Normal Distribution
The standard normal distribution is a normal distribution of standardized values called
z-scores. Recall from Chapter 3 that z-score is measured in units of the standard deviation.
For example, if the mean of a normal distribution is five and the standard deviation is two,the value 11 is three standard deviations above (or to the right of) the mean, sincex = μ + z· σ = 5 + 3· 2 = 11
So the z-score corresponding to x = 11 is z = 3. If the value x comes from a normal
x−µdistribution with mean μ and standard deviation, σ, then the transformation z =σproduces the distribution Z ~ N(0, 1).
315

That is, the mean for the standard normal distribution is zero, and the standard deviationis one. Therefore, zero is in the center.

-2

-1

0

1

2

z-Scoresx−µ.
σ
The z-score tells you how many standard deviations the value x is above (to the right of) or
below (to the left of) the mean μ. Values of x that are larger than the mean have positive zscores, and values of x that are smaller than the mean have negative z-scores. If x equals themean, then x has a z-score of zero.
If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is z =Example 7.2
Suppose X ~ N(5, 6). This says that x is a normally distributed random variable with mean
μ = 5 and standard deviation σ = 6. Suppose x = 17; then the corresponding z-score is:

z=

x−µ

σ

=

17 − 5=2
6This means that x = 17 is two standard deviations above the mean of μ = 5.
Now suppose x = 1; then the corresponding z-score is:

z=

x−µ

σ

=

1− 5
= −0.67 .
6This means that x = 1 is 0.67 standard deviations below the mean μ = 5.
Summarizing,- When z is positive, x is above, or to the right of μ
- When z is negative, x is to the left of, or below μ.

316Graphical interpretation of Example 7.2. The first graph displays x = 17 on X ~ N(5, 6). The
second graph displays z = 2 on X ~ N(0, 1).

-7

-1

5

11

17

Normal Distribution

-2

-1

0

1

2

Standard Normal Distribution

7.1 What is the z-score of x, when x = 7 and X ~ N(12,3)? Display the x value on a normal
distribution graph.

Example 7.3
Some doctors believe that a person can lose five pounds, on average, in a month by reducing
his or her fat intake and by exercising consistently. Suppose weight loss has a normal
distribution. Let X = the amount of weight lost (in pounds) by a person in a month. Use astandard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.
a. Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z =2.5 (verify).
This z-score tells you that x = 10 isor left) of the mean of.

standard deviations to theb. Suppose a person gained three pounds (a negative weight loss).
Then z =.
This z-score tells you that x = –3 isstandard deviations to theor left) of μ, the mean.

(right(rightSolution 7.3
a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean of_5_, μ = 5.
b. 𝑧𝑧 =

𝑥𝑥−𝜇𝜇
−3−5
=
= −4.
𝜎𝜎
2

This z-score tells you that x = –3 is four standard deviations to theleft of the mean.

317

One of the most important aspects of z-scores is that they allow us to compare data that are
scaled differently. To understand the concept, suppose X ~ N(5,6) represents weight gains
for one group of people who are trying to gain weight in a six week period and Y ~ N(3, 1)
measures the same weight gain for a second group of people. A negative weight gain would
be a weight loss. Since x = -1 and y = 2 are each one standard deviations to the left of theirmeans, they represent the same, standardized weight gain relative to their means.

y

x
-7

-1

5

11

1

17

X~N(5, 6)

2

3

4

5

Y~N(3, 1)

7.2 Jerome averages 16 points a game with a standard deviation of four points. Suppose
the number of points in a game is normally distributed: X ~ N(16,4). Jerome scores ten
points in a game. The z–score when x = 10 is –1.5. This score tells you that x = 10 is ___standard deviations to the (right or left) of the mean (What is the mean?).

Example 7.4
The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm witha standard deviation of 6.28 cm. Male heights are known to follow a normal distribution.
Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010, so X ~ N(170,
6.28).
a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010.
Calculate and interpret the z-score for this individual.
b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010
has a z-score of z = 1.27. What was the male’s height?
Solution 7.4
a. z = -0.32. The individual was 0.32 standard deviations to the left of the mean µ = 170.
b. x = µ + z· σ = 170 + 1.27(6.28) = 177.98 cm.

318

7.3 Use the information in Example 7.3 to answer the following questions.
a. Suppose a 15 to 18-year-old male from Chile in this time period was 176 cm tall.
What does that tell us about the individual’s height relative to the population?
b.

Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010
has a z-score of z = -2. What was the individual’s height?

Standard Normal distribution is also known as the Z-distribution. The total area under the
curve is 100%. As in the previous chapter, we can find the probability using area under the
curve. We will be using technology to find the probability of standard normal distribution.
The notation P(z < k) represents the probability of a z-score less than a particular k value
in the standard normal distribution. Below is a graphical representation of P(z < k):To find the probability (area), we can use the Ti-83 or Ti-84 (plus) or Excel.
Using the TI-83, 83+, 84, 84+ Calculator and Excel
To calculate standard normal probabilities_

Go to 2nd DISTR, and select item 2: normalcdf.
The syntax is: normalcdf (lower bound, upper bound, 0, 1).
NotationP(z < k)P(z > k)P(k1 < z < k2)

Calculator Function
Normalcdf (-10^99, k, 0, 1)
Normalcdf(k, 10^99, 0, 1)
Normalcdf(k1, k2, 0, 1)

Excel Function=Norm.s.dist(k, true)=1 – Norm.s.dist(k, true)=Norm.s.dist(k2, true) – Norm.s.dist(k1, true)319

Example 7.5
Use the Calculator or Excel to find the following probabilities
a. P( z < 1.4)
b. P( z > 0.74)
c. P(1.21 < z < 2.43)
Solution 7.5
a. Using the calculator, P(z < 1.4) = Normalcdf(-10^99, 1.4, 0, 1) = 0.9192

Using an Excel function, you must type an equal sign in an empty cell first.
P(z < 1.4) = norm.s.dist(1.4, true)

Go to an empty cell and type equal sign,then start typing Norm.s.dist

After you press Enter, the answer
will appear.

b. Using the calculator, P(z > 0.74) = Normalcdf(.74, 10^99, 0, 1) = 0.2296

320

Using an Excel function, you must type an equal sign in an empty cell first. Since
norm.s.dist(z, true) function gives the area to the left. Therefore, to find the area tothe right, we would use the complement: 1 – norm.s.dist(z, true)

Go to an empty cell and type equal sign,then start typing 1 - norm.s.dist(

After you press Enter, the answer willappear.

c. Using the calculator, P(1.2 < z < 2.4) = normalcdf (1.2, 2.4, 0, 1) = 0.1069

Using an Excel function of norm.s.dist(z, true) gives the area to the left. We want
the area in between. To find the area in between, we take the difference of the areafor z = 2.4 and the area for z = 1.2.
Type in an empty cell: =norm.s.dist(2.4, true) – norm.s.dist(1.2, true)3217.2 | Using Normal Distribution
Recall that the probability that the random variable X falls into any interval will be the area
under the curve corresponding to that interval. For example, scores for the Stanford-Binet
IQ test are normally distributed with a mean of µ = 100 and standard deviation of σ = 15.
Then the probability that a randomly selected individual has an IQ between 95 and 120 would
be about 54%, as shown in the graph:

Distribution Plot

Normal, Mean=100, StDev=15
0.030
0.025

0.5393

Density

0.020
0.015
0.010
0.005
0.000

95 100X

120

To calculate probabilities like the problem above, we need to work with the cumulative
distribution function, P(X < x). The shaded area in the graph below shows this probability;
it is the area under the normal curve that is to the left of x. To calculate this area manuallywould require Calculus; but there is a table of values that shows these areas for the standard
normal distribution, and many calculators and statistical software packages also willcalculate cumulative normal probabilities.

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As we found the area under the curve for the Z-distribution, we will again use Normalcdf tofind the area under the curve for the normal distribution that is not standard normal.

Using the TI-83, 83+, 84, 84+ Calculator and Excel
To calculate standard normal probabilities_

Go to 2nd DISTR, and select item 2: normalcdf.
The syntax is: normalcdf (lower bound, upper bound, µ, σ)
NotationP(x < k)P(x > k)P(k1 < x < k2)

Calculator Function
Normalcdf (-10^99, k, µ, σ)
Normalcdf(k, 10^99, µ, σ)
Normalcdf(k1, k2, µ, σ)

Excel Function=Norm.dist(k, µ, σ, true)=1 – Norm.dist(k, µ, σ, true)=Norm.dist(k2, µ, σ, true) – Norm.dist(k1,µ, σ, true)
Note: For Excel function here we are using Norm.dist instead of Norm.s.dist

Example 7.6
Let’s say in the year 2020, a finance researcher took the daily price of XYZ stock and foundthe daily return for it. The daily return (day to day return) for XYZ stock is normally
distributed with a mean of 0.12% return. This means on average from one day to the next
the stock went up .012%. The standard deviation of the course of the year, day to day, was
1.81%. Let x = 1.93%a.) What does x = 1.93% represent in this situation?
b.) What is the z-score for x = 1.93%?
c.) Find the P(z < 1) using normalcdf on calculator.
d.) Find the P(x < 1.93%) using normalcdf on calculator.
e.) Compare the answers in parts c and d.

Solution 7.6
a.) x = 1.93% represents a particular daily return of XYZ stock had a return of 1.93%b.) 𝑧𝑧 =𝑥𝑥−𝜇𝜇1.93−0.12
=
= 1.
𝜎𝜎
1.81

This means that the x-value is one standard deviation above the mean. Below is thegraphical representation using the normal distribution curve and converting it to
standard normal distribution by finding the z-score of the x value.
323Normal Distribution
μ = 0.11%; σ = 1.81%

Standard Normal Distribution
μ=0σ=1c.) P(z < 1) = normalcdf(-10^99, 1, 0, 1) = 0.8413d.) P(x < 1.93%) = normalcdf(-10^99, 1.93, 0.12, 1.81) = 0.8413e.) Probability for a continuous probability distribution is the area under the respective
curve. The area of the normal distribution curve that is represented by P(x < 1.93%)
with X~N(0.12%, 1.81%) is the same area of the standard normal distribution curvethat is represented by P(z < 1) with X~N(0, 1).

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Example 7.7
The final exam scores in a large statistics class were normally distributed with a mean of
63 points and a standard deviation of 5 points.
a. Find the probability that a randomly selected student scored more than 65 points on theexam without finding the z-score.
b. Find the probability that a randomly selected student scored less than 80 points on theexam without find the z-score.
Solution 7.7
a. Let X = score on the final exam. X ~ N(63, 5) is the notation used to state that the
distribution is normally distributed with μ = 63 and σ = 5.
We start by drawing a graph to represent P(x > 65):

Then, find P(x > 65) by using the calculator, normalcdf(lower, upper, µ, σ). The lower boundof the shaded area is at x = 65, the upper bound of the shaded area is positive infinity which
we represent using 10^99 on the calculator.
The solution: P(x > 65) = normalcdf(65, 10^99, 63, 5) = 0.3446.
b. Here we want to find P(x < 85). The shaded area is on the left side of 85. The lower
bound of the shaded area is negative infinity which we represent using -10^99 on thecalculator. The upper bound of the shaded area is 85.
The solution: P(x < 85) = normalcdf(-10^99, 85, 63, 5) = 0.999995. This tells us that
virtually all of the students in the class scored below 85.

7.4 In 2019, over 2.2 million students took the SAT exam. The distribution of total scores of
the SAT had a mean µ = 1059 and a standard deviation σ = 210. Let X = a SAT exam scorein 2019. Let X ~ N(1059, 210). Find the z-scores for x1 = 933 and x2 = 1220. State theinterpretation. Find the P(x < 933) and P(x > 1220).

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Now, what if we are given the area under the curve and want to find the value on the xaxis. This process is considered doing the inverse operation of what we were doing in thelast example. We can find the percentiles in the Z-distribution, X~N(0,1), or normal
distribution, X~N(µ, σ) using the calculator.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate percentiles in a normal distribution_

Go to 2nd DISTR, and select item 3: invNorm.
The syntax is: invNorm(probability to the left, mean, standard deviation).
i.e., k = invNorm(p, µ, σ) value k so that P(x < k) = p.
NOTE: Newer versions of the calculator allow you to specify which area is given; if you are
given the left area, the center area, or the right area. The above instructions are for older
calculators where the area that needs to be entered is the left area as seen in the next
example.
NotationP(x < k) = p
find kP(z < k) = p
find kCalculator Function
k = invnorm(p, µ, σ)

Excel Function
k = norm.inv(p, µ, σ)

k = invnorm(p, 0, 1)

k = norm.s.inv(p)Example 7.8
The final exam scores in a large statistics class were normally distributed with a mean of
63 points and a standard deviation of 5 points. Find the 90th percentile.

Solution 7.8
Let the random variable X represent scores on the final exam. X ~ N(63, 5) is the notation
used to state that the distribution is normally distributed with μ = 63 and σ = 5. The
variable k on the x-axis is a specific score that has 90% of final exam scores below it and
10% of scores above it.

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To find k using the calculator: invnorm(area to the left, µ, σ)
k = invnorm(.90, 63, 5) = 69.4

So 90% of the test scores were at or below 69.4 points.
Remember that Excel is a different company than the TI calculators. Excel function has thesame input into its function but the name of the function is different.
Using Excel: k = norm.inv(.90,63, 5) = 69.4

There is a third way of finding this percentile. It is using the z-score formula. Recall z-score
gives us the number of standard deviations a value is from the mean.
𝑧𝑧 =𝑥𝑥 − 𝜇𝜇𝜎𝜎

We can isolate the x value: x = µ + z· σ. We know the mean and standard deviation. We canfind z using the calculator. z = invnorm(.9, 0, 1) = 1.28. Remember that the area under the
standard normal curve and the area under the normal curve are the same.

k is the x-value that we are trying to find: k = µ + z· σ = 63 + 1.28(5) = 69.4

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Example 7.9
A personal computer is used for office work at home, research, communication, personal
finances, education, entertainment, social networking, and a myriad of other things. Supposethat the average number of hours a household personal computer is used for entertainmentis two hours per day. Assume the times spent on entertainment are normally distributed andthe standard deviation for the times is half an hour.
a. Find the probability that a household personal computer is used for entertainment
between 1.8 and 2.75 hours per day.
b. Find the maximum number of hours per day that the bottom quartile of households
uses a personal computer for entertainment.
Solution 7.9
Let X = the amount of time (in hours) a household personal computer is used forentertainment. We are told that µ = 2 and σ = 1/2, so X ~ N(2, 0.5)
a. We wish to find P(1.8 < x < 2.75). This probability is the area between x = 1.8 andx = 2.75:

Using the calculator, P(1.8 < x < 2.75) = normalcdf(1.8, 2.75, 2, 0.5) = 0.5886.
b. To find the maximum number of hours per day that the bottom quartile of households
uses a personal computer for entertainment, find the 25th percentile, k. This is thevalue so that P(x < k) = 0.25:

k =?

2

Again using the calculator, k = invNorm(0.25, 2, 0.5) = 1.66.

328

Thus, the maximum number of hours per day that the bottom quartile of households uses a
personal computer for entertainment is 1.66 hours.

7.5 The golf scores for a school team were normally distributed with a mean of 68 and astandard deviation of three.
Find the probability that a randomly selected golfer scored less than 65.
Find the 80th percentile

Example 7.10
There are approximately one billion smartphone users in the world today. In the UnitedStates the ages of smartphone users are approximately normally distributed withapproximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
a. Determine the probability that a randomly selected smartphone user is between 23 and64.7 years old.
b. Determine the probability that a randomly selected smartphone is at most 50.8 years old.
c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.
Solution 7.10
Let X = the age of a randomly selected smartphone user. We are told that µ = 36.9 and σ =13.9, so X ~ N(36.9, 13.9).
a. P(23 < x < 64.7) = normalcdf(23, 64.7, 36.9, 13.9) = 0.8186.
b. P(x < 50.8) = normalcdf( -10^99, 50.8, 36.9, 13.9) = 0.8413.
c. The 80th percentile is given by: invNorm(0.80, 36.9, 13.9) = 48.6.
Thus, 80% of the smartphone users in the U.S. are 48.6 years old or less.

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Example 7.11
There are approximately one billion smartphone users in the world today. In the UnitedStates the ages of smartphone users approximately follow a normal distribution with meanand standard deviation of 36.9 years and 13.9 years respectively.
Use this to answer the following questions, rounding answers to one decimal place:a. Calculate the interquartile range (IQR) of the distribution.
b. What is the cutoff for the top 40% of ages for smartphone users?
Solution 7.11
a. Recall that IQR = Q3 – Q1

Since Q3 is the 75th percentile, we have Q3 = invNorm(0.75, 36.9, 13.9) = 46.2754.
Similarly, Q1 is the 25th percentile, so Q1 = invNorm(0.25,36.9,13.9) = 27.5246. Thus,
IQR = Q3 – Q1 = 46.2754 – 27.5246 = 18.7508, which we round to 18.8.
b. We want to find k so that P(x > k) = 0.40.
40% is the area to the right of k, and so the area to the left of k will be 1 – 0.40 = 0.60.
In other words, we are looking for the 60th percentile. This value is given by k =invNorm(0.60, 36.9, 13.9) = 40.4215.
Rounding, we get k = 40.4. So, forty percent of the ages are at least 40.4 years.

7.6 The golf scores for a school team are normally distributed with a mean of 68 and astandard deviation of 3.
a. Find the probability that a randomly selected golfer scores between 66 and 70.
b. Find the score for a golfer that is at the third quartile.

330

Calculating Probabilities using a Table (optional)
Probabilities are calculated using technology. There are instructions given as necessary for
the TI-83+ and TI-84 calculators. To calculate normal probabilities without technology we
use the normal tables provided in the Appendix. The tables include instructions for how to
use them; but we will review the steps briefly.
The normal distribution table in the Appendix shows the area under the standardized
normal curve to the left of x. I.e. it shows the probability P(z < x). We reproduce a sectionof the table here:

7.7 Use the tables to find P( z < 0.85),

P( z > 1.36) and P(0.45 < z < 1.45)

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To use the table for non-standard normally distributed variables, we first rewrite the desired
probability in terms of z-scores, and then use the table as demonstrated above. For example,
if we have a normally variable X ~ N(50, 4) and wanted to find P(46 < x < 55), we would findthe z-scores corresponding to X = 46 and 55, which are z = (46 – 50)/4 = -1 and z = (55 – 50)/4= 1.25, respectively. From there we could rewrite P(46 < x < 55) = P(-1 < z < 1.25) and use
the table.
Example 7.12
A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges
harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and astandard deviation of 0.24 cm.
a. Find the probability that a randomly selected mandarin orange from this farm has adiameter larger than 6.0 cm by finding z-score and using table.
b. Find the 90th percentile for the diameters of mandarin oranges, and interpret.
Solution 7.12
a. P(x > 6) = normalcdf(6, 10^99, 5.85, 0.24) = 0.2660 but if using the z-chart we would
need to find z first. z = (6 – 5.85)/0.24 = .625 which is halfway between 0.62 and 0.63
When you look up 0.62 and 0.63 on the table, these are the areas to the left.

We want the areas to the right, which can be found by taking the complement: 1 0.7324 = 0.2676 and 1 – 0.7357 = 0.2643. If you average these two areas we get thesame answer as the calculator.

332

b. The 90th percentile is invNorm(0.90, 5.85, 0.24) = 6.16 cm. If wanted to use the z
table. We would need to find the z-score associated with 90% to the left.

The z associated to 90% is z = 1.28. We would plug that into z = µ + zσ.
z = 5.85 + 1.28(0.24) = 6.16 cm.
Thus, ninety percent of mandarin oranges have a diameter that is at most 6.16 cm.

7.8 Using the information from Example 7.12, answer the following:a. Find P(x < 5.5)b. Find the 56th percentile and interpret it in a complete sentence.

333

The Empirical Rule Review

Let X be a random variable that has a normal distribution with mean µ and standard
deviation σ. Then the Empirical Rule says the following:• About 68% of the x values lie within one standard deviation of the mean. That is, about
68% of all data values lie between µ – σ and µ + σ.
• About 95% of the x values lie within two standard deviations of the mean. That is, about
95% of all data values lie between µ – 2σ and µ + 2σ.
• About 99.7% of the x values lie within three standard deviations of the mean. That is,about 99.7% of all data values lie between µ – 3σ and µ + 3σ.
Notice that almost all the x values lie within three standard deviations of the mean. The
empirical rule is also known as the 68-95-99.7 rule.

Since the Empirical rule is used for symmetric distributions and normal distributions are
symmetric, we can use normalcdf to calculate the Empirical rule values.
Example 7.13
Suppose x has a normal distribution with mean 50 and standard deviation 6.
• Then µ – σ = 44 and µ + σ = 56.
So about 68% of the x values lie between 44 and 56.

Check with the calculator. P(44 < x < 56) = normalcdf(44, 56, 50, 6) = .6827 = 68.27%
334

• Similarly, µ – 2σ = 38 and µ + 2σ = 62.
So about 95% of the x values lie between 38 and 62.
• Finally, µ – 3σ = 32 and µ + 3σ = 68.
So about 99.7% of the x values lie between 32 and 68.

7.9 Suppose X has a normal distribution with mean 25 and standard deviation five.
Between what values of x do 68% of the values lie?

Example 7.14
From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm,
and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in1984 to 1985, so Y ~ N(172.36, 6.34).
a. About 68% of the y-values fall between what two values?
b. About 95% of the y values lie between what two values?
c. About 99.7% of the y values lie between what two values?
Solution 7.14
a. Using the Empirical rule, 68% lies within 1 standard deviation below and above themean. 172.36 – 1(6.34) = 166.02 and 172.36 + 1(6.34) = 178.7. Therefore, about 68% of thevalues lie between 166.02 and 178.7 cm.
b. Using the Empirical rule, 95% lies within 2 standard deviation below and above themean. 172.36 – 2(6.34) = 159.68 and 172.36 + 2(6.34) = 185.04. Therefore, about 95% of thevalues lie between 159.68 and 185.04 cm.
c. Using the Empirical rule, 99.7% lies within 3 standard deviation below and above themean. 172.36 – 3(6.34) = 153.34 and 172.36 + 3(6.34) = 191.38. Therefore, about 99.7% of
the values lie between 153.34 and 191.38 cm.

335

Normal Approximation to the Binomial
Historically, being able to compute binomial probabilities was an important application of
the normal distribution. Binomial probabilities with a small value for n (say, 20) weredisplayed in a table in a book. To calculate the probabilities with large values of n, you had
to use the binomial formula, which could be very complicated and time consuming. Using
the normal approximation to the binomial distribution simplified the process. The
graphs below are for four different binomial distributions with varying sample sizes:
Distribution PlotDistribution Plot

Binomial, n=10, p=0.2

Binomial, n=20, p=0.2
0.25

0.30

0.20

Probability

Probability

0.25
0.20
0.15

0.15

0.10

0.10
0.05

0.05
0.00

0

1

2

3
X

4

5

6

0.00

7

0

2

Distribution Plot

4

X

6

8

10

12

30

35

Distribution Plot

Binomial, n=50, p=0.2

Binomial, n=100, p=0.2

0.14

0.10

0.12

0.08
Probability

Probability

0.10
0.08
0.06

0.06
0.04

0.04
0.02

0.02
0.00

0

5

10
X

15

0.00

20

10

15

20

X

25

We see that as the sample size increases, the histogram for the distribution takes on a
distinctive bell shape, which means that a normal distribution will provide a good
approximation.
Recall the necessary conditions for a binomial distribution:• There is a fixed number n of identical, independent trials• The only outcomes for a given trial are success or failure• Each trial has the same probability of a success p
Recall that if X is the binomial random variable, then we write X ~ B(n, p). The shape of thebinomial distribution needs to be similar to the shape of the normal distribution. To ensure
this, the parameters n and p should satisfy the inequalities np > 10 and n(1 – p) > 10. Thenthe binomial can be approximated by the normal distribution with mean μ = np and standard
deviation σ =np (1 − p ) .

336

In order to get the best approximation, we add 0.5 to x or subtract 0.5 from x (use x + 0.5 or
x – 0.5). The number 0.5 is called the continuity correction factor and is used in thefollowing example.
Example 7.15
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the
population favor a charter school for grades K through 5. A simple random sample of 300 issurveyed.
a. Find the probability that at least 150 favor a charter school.
b. Find the probability that at most 160 favor a charter school.
c. Find the probability that more than 155 favor a charter school.
d. Find the probability that fewer than 147 favor a charter school.
e. Find the probability that exactly 175 favor a charter school.
Solution 7.15
Let X = the number that favor a charter school for grades K through 5. Then X follows abinomial distribution with n = 300 and p = 0.53; i.e. X ~ B(300, 0.53). Since np > 10 andnq > 10, we can use the normal approximation to the binomial. The mean and standard
deviation of the distribution are μ = np = 159 and σ =300(.53)(.47) = 8.6447.

Thus, the random variable for the corresponding normal distribution is Y ~ N(159, 8.6447).
For part a, we want P(X ≥ 150), which means that x = 150 is included; this means in thehistogram, we want the rectangle that is centered at x = 150; and the left endpoint of thisrectangle is 149.5.
So we must find the area to the right of x = 149.5, and so we write:P(X ≥ 150) ≈ P(Y ≥ 149.5) = normalcdf(149.5, 10^99, 159, 8.6447) = 0.8641.
Similarly, for part b, we P(X ≤ 160), which means we want to include X = 160. In thehistogram, the rectangle for this value has a right endpoint of 160.6, and soP(X ≤ 160) ≈ P(Y ≤ 160.5) = normalcdf(0, 160.5, 159, 8.6447) = 0.5689.
For part c, we want to exclude 155, so
P(X > 155) ≈ P(y > 155.5) = normalcdf(155.5, 10^99, 159, 8.6447) = 0.6572.

337

For part d, we want to exclude 147 soP(X < 147) ≈ P(Y < 146.5) = normalcdf(0, 146.5, 159, 8.6447) = 0.0741.
For part e, P(X = 175) has normal approximation
P(174.5 < Y < 175.5) = normalcdf(174.5, 175.5, 159, 8.6447) = 0.0083.
Modern calculators and computer software allow us to easily binomial probabilities for any
values of n and p, making the normal approximation to the binomial distribution obsolete forcalculation purposes. However, in Chapters 7, 8 and 9, we will see that this approximation
is important for theoretical reasons.
For the previous example, the probabilities can of course be calculated using the binomcdf
and binompdf functions in the TI-84 calculator, with n = 300 and p = 0.53. Compare thebinomial and normal distribution answers:
a. P(X ≥ 150) = 1 - binomcdf(300, 0.53, 149) = 0.8641
b. P(X ≤ 160) = binomcdf(300, 0.53, 160) = 0.5684
c. P(X > 155) = 1 - binomcdf(300, 0.53, 155) = 0.6576
d. P(X < 147) = binomcdf(300, 0.53, 146) = 0.0742
e. P(X = 175) = binompdf(300, 0.53, 175) = 0.0083

7.10 In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A
simple random sample of 500 is taken. Using the continuity correction factor, find theprobability that at least 250 favor Dawn Morgan for mayor.

338

KEY TERMS
Normal Distribution. A random variable with pdf 𝑓𝑓(𝑥𝑥) =

𝑒𝑒

−.5�

𝑥𝑥−𝜇𝜇 2
�𝜎𝜎𝜎𝜎 √2𝜋𝜋, where µ is the mean ofthe distribution and σ is the standard deviation. The graph of this pdf is a bell-shaped curvethat is perfectly symmetric about the line x = µ. Notation: X ~ N(μ, σ).

µStandard Normal Distribution: A normal distribution with μ = 0 and σ = 1. I.e., this is acontinuous random variable X ~ N(0, 1); since this is the distribution obtained by replacing
data values by their respective z-scores, it is often noted as Z ~ N(0, 1).

0x−µ. The z-score allows us to compareσdata that are normally distributed but scaled differently. If this transformation is applied to
any normal distribution X ~ N(μ, σ) the result is the standard normal distribution Z ~ N(0,1).
z-score: If x is any data value, then its z-score is z =339

FORMULA REVIEW
z-score: z =x−µσ

For Z ~ N(0, 1):
NotationP(z < k)P(z > k)P(k1 < z < k2)

Calculator Function
Normalcdf (-10^99, k, 0, 1)
Normalcdf(k, 10^99, 0, 1)
Normalcdf(k1, k2, 0, 1)

Excel Function=Norm.s.dist(k, true)=1 – Norm.s.dist(k, true)=Norm.s.dist(k2, true) – Norm.s.dist(k1, true)k = invnorm(probability to left, 0, 1)

For X ~ N(µ, σ):NotationP(x < k)P(x > k)P(k1 < x < k2)

Calculator Function
Normalcdf (-10^99, k, µ, σ)
Normalcdf(k, 10^99, µ, σ)
Normalcdf(k1, k2, µ, σ)

Excel Function=Norm.dist(k, µ, σ, true)=1 – Norm.dist(k, µ, σ, true)=Norm.dist(k2, µ, σ, true) – Norm.dist(k1,µ, σ, true)
Note: For Excel function here we are using Norm.dist instead of Norm.s.distk = invnorm(probability to left, µ, σ)

340

EXERCISES FOR CHAPTER 7
1. A bottling plant produces bottles of water whose volumes are normally distributed with amean of 12.05 fluid ounces and a standard deviation of 0.01 ounces. Describe the randomvariable X symbolically: X ~ ____.
2. A normal distribution has a mean of 61 and a standard deviation of 15. What is themedian?
3. Given that X ~ N(1, 2), σ = ____.
4. A company manufactures rubber balls. The diameters of the balls are approximately
normally distributed with mean 12 cm and a standard deviation of 0.2 cm. Describe therandom variable X symbolically: X = ____.
5. Given that X ~ N(–4, 1), what is the median?
6. Given that X ~ N(3, 5), σ = ____.
7. Given that X ~ N(–2, 1), μ = ____.
8. What is the z-score of x = 12, if it is two standard deviations to the right of the mean?
9. What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean?
10. Find the z-score of x = –2 if X ~ N(0, 1.5)
11. Find the z-score of x = 7, if X ~ N(5, 1.2)
12. Suppose X ~ N(–1, 2). What is the z-score of x = 2?
13. Suppose X ~ N(12, 6). What is the z-score of x = 2?
14. Suppose X ~ N(9, 3). What is the z-score of x = 9?
15. Suppose a normal distribution has a mean of 6 and a standard deviation of 1.5.
What is the z-score of x = 5.5?
16. Suppose X ~ N(2, 6). What value of x has a z-score of z = 3?
17. Suppose X ~ N(9, 5). What value of x has a z-score of -0.5?
18. Suppose X ~ N(2, 3). What value of x has a z-score of -0.67?
19. Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?
20. Suppose X ~ N(4, 2). What value of x is 2.8 standard deviations to the right of the mean?

341

21. Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?
22. Suppose X ~ N(15, 2.5). Graph the normal distribution.
23. Suppose X ~ N(-2, 1). Graph the normal distribution
24. Suppose X ~ N(60, 5.9). Graph the normal distribution
25. Suppose X ~ N(520, 35). Graph the normal distribution
26. In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 isdeviations to the(right or left) of the mean.

standard

27. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 isdeviations to the(right or left) of the mean.

standard

28. In a normal distribution, x = –2 and z = 6. This tells you that x = –2 isdeviations to the(right or left) of the mean.

standard

29. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 isstandard deviations to the (right or left) of the mean.
30. In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 isdeviations to the(right or left) of the mean.

standard

31. About what percent of x values from a normal distribution lie within one standard
deviation (left and right) of the mean of that distribution?
32. About what percent of the x values from a normal distribution lie within two standard
deviations (left and right) of the mean of that distribution?
33. About what percent of x values lie between the second and third standard deviations(both sides)?
34. Suppose X ~ N(15, 3). Between what x values does the middle 68.26% of the data lie?
35. Suppose X ~ N(–3, 1). Between what x values does the middle 95.44% of the data lie?
36. Suppose X ~ N(–3, 1). Between what x values does 34.14% of the data lie?
37. About what percent of x values lie between the mean and three standard deviations?
38. About what percent of x values lie between the mean and one standard deviation?
39. About what percent of x values lie between the first and second standard deviations
from the mean (both sides)?
40. About what percent of x values lie between the first and third standard deviations (bothsides)?
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41. How would you represent the area to the left of one as a probability statement?

42. Represent the area to the right of x = 1 as a probability statement.

43. In a normal distribution, is P(x < 1) equal to P(x ≤ 1)? Why or why not?
44. How would you represent the area to the left of x = 3 as a probability statement?

45. Represent the area to the right of x = 3 as a probability statement.

46. Draw and find P(z < 2.16).
47. Draw and find P(z < -2.78).

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48. If the area to the left of x in a normal distribution is 0.123, what is the area to the rightof x?
49. If the area to the right of x in a normal distribution is 0.543, what is the area to the leftof x?
50. X ~ N(54, 8). Find the probability that x > 56.
51. X ~ N(54, 8). Find the probability that x < 30.
52. X ~ N(54, 8). Find the 80th percentile.
53. X ~ N(54, 8). Find the 60th percentile.
54. Given that X ~ N(6, 2), find the probability that x is between 3 and 9.
55. Given that X ~ N(–3, 4), find the probability that x is between 1 and 4.
56. Given that X ~ N(4, 5), find the maximum value of x for the bottom quartile.

General Word Problems
57. The life span of gaming consoles is normally distributed with mean of 7.5 years and astandard deviation of 1.3 years. We are interested in the length of time a console lasts. The
guarantee period is 5 years.
a. Draw the shape of the distribution, label and scale the axes
b. Describe the random variable symbolically: X ~

(

,

)

c. What is the likelihood that a gaming console will break down during theguarantee period?
d. Find the probability that a console last more than 10 years. Is it unusual?
58. The life span of gaming consoles is normally distributed with mean of 7.5 years and astandard deviation of 1.3 years. We are interested in the length of time a console lasts.
Find the 70th percentile of the distribution for the time a gaming console last.
59. The life span of car tires is normally distributed with a mean of 5 years and a standard
deviation of 0.85 years.
a. Draw the shape of the distribution, label and scale the axes
b. Describe the random variable symbolically: X ~

(

,

)

c. What is the likelihood that a car tire will last longer than 7 years?
d. Find the probability that a tire will last between 4 to 5 years.
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Use the following information to answer the next two exercises: The patient recovery timefrom a particular surgical procedure is normally distributed with a mean of 5.3 days and astandard deviation of 2.1 days.
60. What is the median recovery time?
61. What is the z-score for a patient who takes ten days to recover?
62. Clotting time of blood is normally distributed. The mean clotting time of blood is 7.45
seconds, with a standard deviation of .3 seconds. What is the probability that the bloodclotting time will be less than 7 seconds?

63. The average size of the fish in a lake is 12.1 inches, with a standard deviation of 3.8
inches. The size of fish in a lake are normally distributed. Find the probability of catchinga fish that isa. longer than 17 inches
b. shorter than 8 inches
c. between 14 and 17 inches
64. The systolic blood pressure (given in millimeters) of males has an approximately normal
distribution with mean µ =125 and standard deviation σ = 14.
a. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
b. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard
deviations below the mean, but that he believed his blood pressure was between 100 and
150 millimeters, what would you say to him?

65. The monthly income of trainees at a local technology center is normally distributed. The
average income is $1200 a month. The standard deviation is $150.
a. Find the probability that the trainee earns less than $900 a month?
b. Find the 90th percentile of monthly income of trainees.
66. According to Eye safe Nielsen study in 2020, the average time spent on App/Web on asmartphone is 4.5 hours for adults in US. Let’s assume the time spent on App/Web on asmartphone is normally distributed with standard deviation of 1.2 hours.
a. Find the probability that an adult spends less than 1 hour on App/Web on asmartphone.

345

b. Find the probability that an adult spends between 4 to 6 hours on App/Web on asmartphone.
c. Find the middle 50% of hours used by adults.
67. In 2019, 2,220,087 students heading to college took the SAT. The distribution of scores in
the math section of the SAT follows a normal distribution with mean µ = 528 and standard
deviation σ = 117.
a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence.
b. What math SAT score is 1.5 standard deviations above the mean? What can you say
about this SAT score?
c. What is the probability that a randomly selected score is greater than 750? Is it
unusual?
Use the following information to answer the next two exercises:
The patient recovery time from a particular surgical procedure is normally distributed witha mean of 5.3 days and a standard deviation of 2.1 days.
68. What is the probability of spending more than two days in recovery?
69. What is the 90th percentile for recovery times?
Use the following information to answer the next three exercises:
The length of time it takes to find a parking space at 9 am follows a normal distribution
with a mean of five minutes and a standard deviation of two minutes.
70. Based upon the given information, would you be surprised if it took less than one
minute to find a parking space?
a. Yesb. No

c. Unable to determine

71. Find the probability that it takes at least eight minutes to find a parking space.
72. Seventy percent of the time, what is the most number of minutes needed to find aparking space?
73. According to a study done by De Anza students, the height for Asian adult males isnormally distributed with an average of 66 inches and a standard deviation of 2.5 inches.
Suppose one Asian adult male is randomly chosen. Let X = height of the individual.
a. X ~

(

,

)

346

b. Find the probability that the person is between 65 and 69 inches. Include a sketch of
the graph, and write the probability in terms of the random variable x.
c. Would you expect to meet many Asian adult males over 72 inches? Explain why or
why not, and justify your answer numerically.
d. The middle 40% of heights fall between what two values? Sketch the graph, andwrite the probability statement.
74. IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose
one individual is randomly chosen. Let X = IQ of an individual.
a. X ~(,)
b. Find the probability that the person has an IQ greater than 120. Include a sketch of
the graph, and write a probability statement.
c. MENSA is an organization whose members have the top 2% of all IQs. Find theminimum IQ needed to qualify for the MENSA organization. Sketch the graph, andwrite the probability statement.
d. The middle 50% of IQs fall between what two values? Sketch the graph and write
the probability statement.
75. The percent of fat calories that a person in America consumes each day is normally
distributed with a mean of about 36 and a standard deviation of 10. Suppose that one
individual is randomly chosen. Let X = percent of fat calories.
a. X ~(,)
b. Find the probability that the percent of fat calories a person consumes is more than
40. Graph the situation. Shade in the area to be determined.
c. Find the maximum number for the lower quarter of percent of fat calories. Sketch
the graph and write the probability statement.
76. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally
distributed with a mean of 250 feet and a standard deviation of 50 feet.
a. If X = distance in feet for a fly ball, then X ~(,).
b. If one fly ball is randomly chosen from this distribution, what is the probability that
this ball traveled fewer than 220 feet? Sketch the graph, label the horizontal axis X andshade the region corresponding to the probability. Find the probability.
c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write
the probability statement.

347

77. The corn yield in a given area is normally distributed. The average corn yield in a given
area is 210 bushels per acre with a standard deviation of 47 bushels per acre.
a. In words, define the random variable X.
b. X ~(,)
c. Find the probability that a given area produces more than 275 bushels per acre.
d. What percent of given areas produced less than 200 bushels per acre?
e. Seventy percent of given areas produce less than how many bushels per acre of corn?
78. The average weekly income of call center analyst is $600 with a standard deviation of$20. The weekly income of call center analyst is normally distributed. Let X = weekly incomeof call center analyst.
a. Find the probability that a randomly selected analyst has a weekly income less than
$530. Sketch the graph and write the probability statement.
b. Find the probability that a randomly selected analyst has a weekly income between$610 and $630.
c. Find the third quartile for weekly income.
79. Suppose that the duration of a particular type of criminal trial is known to be normally
distributed with a mean of 21 days and a standard deviation of seven days.
a. In words, define the random variable X.
b. X ~(,)
c. If one of the trials is randomly chosen, find the probability that it lasted at least 24
days. Sketch the graph and write the probability statement.
d. Sixty percent of all trials of this type are completed within how many days?
80. Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in aseven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times
is normally distributed. We are interested in one of her randomly selected laps.
a. In words, define the random variable X.
b. X ~(,)
c. Find the percent of her laps that are completed in less than 130 seconds.
d. Find the percent of her laps that are completed in more than 140 seconds.
d. The fastest 3% of her laps are under.
e. The middle 80% of her laps are fromseconds toseconds.
thf. Find the 60 percentile.

348

81. Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as
the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school
average. In complete sentences, explain why each of the following statements may be false.
a. Ricardo’s actual GPA is lower than Anita’s actual GPA.
b. Ricardo is not passing because his z-score is zero.
c. Anita is in the 70th percentile of students at her college.
82. A NUMMI assembly line, which has been operating since 1984, has built an average of
6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off theassembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the numberof defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empiricalrule (one standard deviation, two standard deviations and three standard deviations fromthe mean are being referred to)? Assume a normal distribution for the defective cars in thesample.
83. We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of
the time. The mean and standard deviation for the number of times the coin lands on heads
is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following:a. There is about a 68% chance that the number of heads will be somewhere between
and .
b. There is about achance that the number of heads will be somewhere between 12
and 28.
c. There is about achance that the number of heads will be somewhere between eightand 32.
Business Applications
84. The life of a coffee dispenser machine that uses regular-sized k-cups is normally
distributed with a mean of 4.88 years with a standard deviation of .62 years.
a. What is the probability that the coffee dispenser lasts less than 3 years?
b. If the company wants less than 2% of the coffee dispensers to fail while under warranty,
what should be the guarantee period?
85. A machine that fills quart bottles with fruit juice is normally distributed with a mean of
31.6 oz per bottle, with a standard deviation of 1.2 oz.
a. What is the probability that the amount of juice in a bottle is less than 1 quart?
b. What is the probability that the amount of juice in a bottle is at least 2 ounces
more than a quart?

349

c. What are the largest and smallest amounts dispensed by the middle 50%?
86. Customers at a certain cosmetic store spend an average of $45.80, with a standard
deviation of $1275. Assume the amount spent is normally distributed.
a. What is the probability that a customer spends more than $60?
b. What is the probability that a customer spends between $30 and $40?
c. What are the largest and smallest amounts spend by the middle 60%?
d. Find the 70th percentile.
87. Suppose in a quarter, a group of 25 mutual funds return of 2.8% with a standard
deviation of 4.8%. Assume the returns are normally distributed.
a. What percent of funds would you expect to have returns of 7% or more?
b. What percent of funds would you expect to have returns of 2% or less?
c. What percent of funds would you expect to have returns between 5% and 10%?
d. Find the 10th percentile.
88. Human resource departments look at job satisfaction scores. Let’s say job satisfaction
scores are normally distributed with a mean of 90 and a standard deviation of 12.
a. What percent of scores are less than 70?
b. What percent of scores are between 100 and 120?
c. Human resource departments are concerned with job satisfaction scores thatdrops below a specific score. What score is considered unusual?

350

REFERENCES
7.1 The Standard Normal Distribution“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at
http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).
“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational
resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical
Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14,2013).
“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard 2012. Available online at
http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14,2013).
“Digest of Education Statistics: ACT score average and standard deviations by sex andrace/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned
fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics.
Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14,2013).
Data from the San Jose Mercury News.
Data from The World Almanac and Book of Facts.
“List of stadiums by capacity.” Wikipedia. Available online athttps://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).
Data from the National Basketball Association. Available online at www.nba.com (accessed May 14,2013).
7.2 Using the Normal Distribution“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele's_rule (accessed
May 14, 2013). “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at
http://www.thisamericanlife.org/radio- archives/episode/403/nummi (accessed May 14, 2013).
“Scratch-Off
Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at
http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).
“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smartphone-users-numbers (accessed May 14, 2013).
Eyesafe. “COVID-19: Screen Time Spikes to over 13 Hours per Day According to Eyesafe Nielsen
Estimates.” Eyesafe®, 30 Mar. 2020, eyesafe.com/covid-19-screen-time-spike-to-over-13-hoursper-day/.

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CHAPTER 7 SOLUTIONS:
1) X ~N(12.05, 0.01) 3) σ = 2 5) Median = µ = -4 7) µ = -2
9) Z = -1.5 11) z = 1.67 13) z = -1.67 15) z ≈ –0.33
17) x = 6.5
19) x = 1 21) x = 1.97
23)

25)

27) 0.67, right 29) 3.14, left31) about 68% 33) about 4% 35) between –5 and –1 37) about 50% 39) about 27%41) P(x < 1)43) Yes, because they are the same in a continuous distribution: P(x = 1) = 0.
45) P(x > 3) or 1 – P(x < 3)47) P(z < 2.16) = .00272

49) 1 – 0.543 = 0.457 51) 0.0013 53) 56.03 55) 0.1186
57) a. Check student’s solution. b. X~N(7.5, 1.3)
c. 0.0272
d. 0.0272, unusual59) a. Check student’s solution. b. X~N(5, .85) c. 0.0093
d. 0.38061) z = 2.24
63) a. P(x > 17) = normalcdf(17, 10^99, 12.1, 3.8)b. P(x < 8) = normalcdf(-10^99, 8, 12.1, 3.8)c. P(14 < x < 17) = normalcdf(14, 17, 12.1, 3.8)65) P(x < 900) = normalcdf(-10^99, 900, 1200, $150)
67) a. z = (720 – 528)/117 = 1.64. This student scored 1.64 standard deviations above themean.
b. x = µ + 1.5σ = 528 + 1.5(117) = 703.5. This student scored 1.5 standard deviations
above the mean.
c. P(x > 750) = normalcdf(750, 10^99, 528, 117) = .0289. Yes, unusual.
69)

7.99 days71) 0.0668

73) a. X ~ N(66, 2.5) b. 0.5404

c. No, the probability that an Asian male is over 72 inches tall is 0.0082

352

75) a. X ~ N(36, 10) b. P(x > 40)= 0.3446.
c. Approximately 25% of people consume less than 29.26% of their calories as fat.
77) a. X = number of bushels per acre of corn in a given areab. X ~ N(210) , 47 c. P(x > 275) = 0.0833. d. P(x < 200) = 0.4158.
bushelse. 234.679) a. X = the number of days a particular type of criminal trial will take b. X ~ N(21, 7)
c. The probability that a randomly selected trial will last more than 24 days is 0.3336.
d. 22.77
81) a. mean = 5.51, s = 2.15 e. X ~ N(5.51, 2.15) f. 0.6029
g. The cumulative frequency for less than 6.1 minutes is 0.64.
h. The answers to part f and part g are not exactly the same, because the normal
distribution is only an approximation to the actual data.
i. The answers to part f and part g are close, because a normal distribution is an
excellent approximation when the sample size is greater than 30.
j. The approximation would have been less accurate, because the smaller sample sizemeans that the data does not fit normal curve as well.
83) µ = n*p = 100*.2 = 20; σ = sqrt(npq) = sqrt(100*.2*.8) = 4
a. 16 and 24
b. 95%
c. 99.7%85) a. 0.631 b. 0.0228

c. 30.8 oz and 32.4

87) a. 0.191

c. 0.257

b. 0.434

d. -3.4%

353

This page is purposely left blank.

354

8 | CENTRAL LIMIT THEOREM

This Photo by Unknown Author is licensed under CC BY-SA

Figure 8.1 If you want to figure out the distribution of fish size in fish farm pools to maximize
profit, using the central limit theorem and assuming your sample is large enough, you will find thatthe distribution is normal and bell-shaped.

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:••••

Recognize problems involving the Central Limit Theorem.
Classify continuous word problems by their distributions.
Apply and interpret the Central Limit Theorem for means and proportions.
Apply and interpret the Central Limit Theorem for sums.

355

The Central Limit Theorem (CLT for short) is one of the most powerful and useful ideas inall of Statistics. There are two alternative forms of the theorem, and both describe the center,spread and shape of a certain sampling distribution. In general, the sampling distributionof a statistic (such as a sample mean or sample proportion) is the distribution of values ofthat statistic when all possible samples of the same size are taken from the same population.
Sampling distributions form the foundation for almost all methods in inferential statistics,
and the Central Limit Theorem allows us to explicitly describe the sampling distribution fora sample mean x and the sampling distribution for a sample proportion p.
If we select multiple random samples of size n from a population and calculate the mean foreach sample, then the sample mean x will vary from sample to sample.

Population
Distribution:

Random sample of size n:µ = mean of population

x1 = mean of sample

σ = standard deviation

That is, x will be a random variable, and so it has a probability distribution. If we were able
to collect all possible random samples of size n from the population and calculate x for eachsample, then the resulting distribution is called the sampling distribution for the mean.
What would this distribution look like? Well, if n is reasonably large, we would expect mostof the sample means x to be pretty close to the true population mean, µ. Thus, we would
expect the mean of all sample means to be equal to µ. There will of course be some variation
among the sample means; however, we would expect most of the differences between a sample
mean and population mean to be fairly small, with large deviations quite rare. Thus, wewould expect the pdf of the distribution to approach zero as we move away from the center.
Finally, since a sample mean x is as likely to underestimate as it is to underestimate the true
population mean, we would expect the positive and negative deviations from the mean tooccur with about the same proportion. Thus, we would expect the distribution to besymmetric.
Sampling distribution
Collection of all possible random samples of size n356

When we put these three observations together, would expect the sampling distribution forx to be a bell-shaped, symmetric distribution – so it is not unreasonable to think that the
sampling distribution is a normal distribution. Moreover, the mean of this sampling
distribution is the mean, µ of the population from which we are sampling. The Central Limit
Theorem validates our intuition; specifically, the CLT states that if we collect samples of“sufficiently large” size n then the sampling distribution will be approximately normal.
Similarly, the sampling distribution for 𝑝𝑝̂ will be formed and the CLT can be used for thissampling distribution too. Another version of the theorem says that if we again collectsamples of size n that are "large enough," calculate the sum of each sample, then the sampling
distribution for the sum will again be approximately normal.
In any case, it does not matter what the distribution of the original population is,or whether we even know it. The important fact is that the distribution of sample
means and the sums tend to follow the normal distribution.
The size of the sample, n, that is required in order to be "large enough" depends on theoriginal population from which the samples are drawn. If we are sampling from a normal
distribution, the sampling distribution will exhibit a bell-shape even for small n. But if weare sampling from a skewed distribution, we will need the sample size to be at least 30.

Original Distribution
Uniform
Exponential
SkewedNormalSampling Distribution
Needs the sample size to be at least
30. This distribution is approximately
normal.
Sample size can be less than 30. This
distribution is normal.





357

8.1 | The Central Limit Theorem for Sample Means and Sample
Proportions
Suppose X is a random variable with mean µ and standard deviation σ. Suppose that weselect random samples of size n and denote the corresponding sample means as X . Then wedenote the mean and standard deviation of the sampling distribution for X as µ X and σ Xrespectively.

Central Limit Theorem for Sample Means
Suppose X is a random variable with mean µ and standard deviation σ. Suppose that weselect random samples of size n. Then the following are true:i) The mean of the sampling distribution for X is µ X = µ .
ii) The standard deviation of the sampling distribution for X is σ X =σn.

iii) For n > 30, the sampling distribution for the sampling distribution for X isapproximately normal.

The first two parts of the theorem tell us that the sampling distribution has the same mean
as the original distribution and a variance that equals the original variance divided by thesquare root of the sample size. To state the third part more precisely, if we draw random
samples of size n, the distribution of the random variable X approaches a normal distribution
as the sample size n increases. We can summarize all three parts of the theorem succinctly
by saying:
X ~ N(µ,σn)

Note that the standard deviation for the sampling distribution, σ X =σn, is often called thestandard error of the mean. Note also that as n grows larger, the standard error getssmaller. That is as the sample size gets larger there is less variation among the sample
means – this reinforces our intuition that larger samples give more reliable results when weuse a sample mean x to estimate µ.
Also recall the Law of Large numbers, which says that if you take samples of larger andlarger size from any population, then the sample mean x tends to get closer and closer to the
population mean μ. We now see that this is a direct result of the Central Limit Theorem: we358

know that as n gets larger and larger, the sample means follow a normal distribution withmean μ. Moreover, as the sample size increases, the standard deviation for the sampling
distribution decreases. So n becomes large, the sample means x really will get closer andcloser to the population mean μ.
Example 8.1
An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of sizen = 36 are drawn randomly from the population.
a.) Find the mean of the sampling distributionb.) Find the standard error of meanc.) State if the sampling distribution is approximately normal or normal.
Solution 8.1
a. This is asking for 𝜇𝜇𝑋𝑋̄ . According to the Central Limit Theorem, µ X = µ . Therefore,𝜇𝜇𝑋𝑋̄ = 90.

b. The formula for standard error of mean is σ X =σn, 𝜎𝜎𝑋𝑋̄ =

15
= 2.5
√36

c. Since the original distribution is unknown, the sampling distribution isapproximately normal.
The fact that the sampling distribution of x is approximately normal or normal means that
we can use the techniques learned in Chapter 7 for the calculations. However, we must be
careful to use the standard deviation for the sampling distribution. For example, if
we want to calculate the z-score for a particular sample mean x , then we would use theformula:z=x − µX(σ X )

=

x−µ

σ/ n

And when we calculate probabilities, we will make sure that we will always write theprobability statement in terms of the correct random variable.
Using the TI-83, 83+, 84, 84+ Calculator_
To calculate the probabilities involving the sampling distribution for x , we go to 2nd
DISTR and use the normalcdf function:σ)P(a < x < b) = normalcdf(a, b, 𝜇𝜇𝑋𝑋̄ , 𝜎𝜎𝑋𝑋̄ ) = normalcdf(a, b, µ,nHere, µ = the mean of the original distribution we are sampling, σ = the standard
deviation of the same distribution, and n is the sample size.
359

Example 8.2
Continued from Example 8.1. Recall, an unknown distribution has a mean of 90 and astandard deviation of 15. Samples of size n = 36 are drawn randomly from the population.
a. Find the probability that a randomly selected sample mean is between 85 and 92.
b. Find the value that is two standard deviations above the expected value of the sample
mean.
Solution 8.2
a. This question asks you to find a probability involving the sample mean: P(85 <x < 92).
We first draw a graph:

We from the CLT that the distribution of x is approximately normal, with meanσ
15and standard deviation µ X = µ = 90 and σ X === 2.5 . Since the36ndistribution is normal, we can use the normalcdf function in the calculator to findthe probability:
P(85 < x < 92) = normalcdf(85, 92, 90, 2.5) = 0.7654.
b. To find the value of x that is two standard deviations above the expected value 90,
use the formula:x = µ X + 2 σ X = 90 + 2(2.5) = 95.
The value of x that is two standard deviations above the expected value is 95.

8.1 An unknown distribution has a mean of 45 and a standard deviation of eight. Samples ofsize n = 30 are drawn randomly from the population. Find the probability that the sample
mean is between 42 and 50.

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Example 8.3
The length of time, in hours, it takes an "over 40" group of people to play one soccer match isnormally distributed with a mean of two hours and a standard deviation of 0.5 hours. A
sample of size n = 50 is drawn randomly from the population. Find the probability that the
sample mean is between 1.8 hours and 2.3 hours.
Solution 8.3
Let X = the time, in hours, it takes to play one soccer match.
We are asked to find a probability involving the sample mean time (in hours) needed toplay one soccer match.
We are told that the original distribution has µ = 2, σ = .50, n = 50, and X ~ N(2, 0.5).

From the Central Limit Theorem, the sampling distribution of x is normal, with meanµ X = µ = 2, σ X = σ = .5 ≈ 0.0707 and X ~ N(2, 0.0707)
50
nThus, P(1.8 < x < 2.3) = normalcdf(1.8, 2.3, 2, 0.0707) = 0.9977. Notice the x-axis for both
distributions. Between 1.8 and 2.3 covers the majority of the 2nd graph compared to the firstgraph.

361

8.2 The length of time taken on the SAT for a group of students is normally distributed witha mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn
randomly from the population. Find the probability that the sample mean is between twohours and three hours.
Recall that percentiles separate values We can also calculate percentiles in the sampling
distribution using the techniques from Chapter 7:

Using the TI-83, 83+, 84, 84+ Calculator_nd
To calculate percentiles in the sampling distribution for x , go to 2 DISTR and selectthe invNorm function:σ)
The p-th percentile is k = invNorm(p, µ,nHere, µ = the mean of the distribution we are sampling, σ = the standard deviation, and n is
the sample size.

Example 8.4
According to NOAA (National Oceanic and Atmospheric Administration) Fisheries website,
the average size of the Atlantic Salmon (farmed) is about 29 inches. Let’s assume thestandard deviation is 3 inches. If we take sample sizes of 100 salmons,a. What are the mean and standard deviation for the sampling distribution of average size
salmons?
b. What does the sampling distribution look like?
c. Find the probability that a sample mean length is more than 30 inches?
d. Find the 95th percentile for the sample mean length (to one decimal place).
Solution 8.4
a. The mean of the sampling distribution is µ X = µ = 29 inches and the standard deviationof the sampling distribution is 𝜎𝜎𝑋𝑋̄ =

𝜎𝜎

√𝑛𝑛

=

3

√100

= 0.3 inches

b. From the Central Limit Theorem, we would expect that the sampling distribution will beapproximately normal since the original distribution is unknown.

362

c. The probability that a sample mean length is more than 30 inches is given by P(𝑋𝑋� > 30)= normalcdf(30, 10^99, 29, 0.3) = 0.000429. It is very unlikely that an average of 100
salmons will be more than 30 inches.
d. Let k = the 95th percentile; then k = invNorm(0.95, 29, 0.3) = 29.493, or about 29.5
inches. 95% of sample means are less than 29.5 inches.

8.3 In the same NOAA site, another farmed fish that is monitored is the sablefish. The
average length is 36 inches. Assume the standard deviation is 3.2 inches. If samples of size
49 are taken, find the middle 80% of sample means. Interpret the results in terms of farmed
sablefish.

Example 8.5
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose
the standard deviation is one minute. We select a random sample of 60 tablet users andmeasure the mean time spent on each app.
a. What are the mean and standard deviation for the sample mean number of minutes forapp engagement by a tablet user? Draw the sampling distribution.
b. What is the standard error of the mean?
c. Find the 90th percentile for the sample mean time for app engagement for a tablet user.
Interpret value in a complete sentence.
d. Find the probability that the sample mean is between eight minutes and 8.5 minutes.
Solution 8.5
a. The mean of the sampling distribution is µ X = µ = 8.2 minutes and the standard
deviation of the sampling distribution is σ X =

σ

363

n

=

1
60

= 0.1291 .

b. Recall that the standard error of the mean is just σ X , the standard deviation of thesampling distribution for x . If we select many samples of size 60, this statistic describes
the spread of the sample means about the population mean.
c. The 90th percentile is given by invNorm(.90, 82, .1291) = 8.37 minutes.
This indicates that 90% of average app engagement times are less than 8.37 minutes.
d. P(8 < x < 8.5) = normalcdf(8, 8.5, 8.2, 0.1291) = 0.9293.

8.4 Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample aremeasured and the statistics are n = 34, x = 16.01 ounces. If the cans are filled so that μ =16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 canswill have an average amount greater than 16.01 ounces. Do the results suggest that cans arefilled with an amount greater than 16 ounces?

364

Suppose X is a random variable with population proportion p. Suppose that we selectrandom samples of size n and denote the corresponding sample proportion as 𝑝𝑝̂ . Then wedenote the mean and standard deviation of the sampling distribution for 𝑝𝑝̂ as 𝜇𝜇𝑝𝑝� and 𝜎𝜎𝑝𝑝�
respectively.

Central Limit Theorem for Sample Proportions
Suppose X is a random variable with population proportion p. Suppose that we selectrandom samples of size n. Then the following are true:i) The mean of the sampling distribution for 𝑝𝑝̂ is 𝜇𝜇𝑝𝑝� = 𝑝𝑝.

𝑝𝑝𝑝𝑝ii) The standard deviation of the sampling distribution for 𝑝𝑝̂ is 𝜎𝜎𝑝𝑝� = � .
𝑛𝑛iii) The sampling distribution for the sampling distribution for 𝑝𝑝̂ isapproximately normal when n is large and p is not too near 0 or 1. X ~ N(𝜇𝜇𝑝𝑝� , 𝜎𝜎𝑝𝑝� )Let’s look at how CLT works for a sampling distribution for sample proportion. We open up
a regular bag of colored candy and notice that there are 23 red candies out of 100 pieces.
Therefore 𝑝𝑝 =23
= 0.23.
100

Now we will take samples of size 9, which the notation is n = 9.

Note the properties are approximate when the population is finite and no more than 10% of
the population is included in the sample. In each sample we find the proportion of red
candies:𝑥𝑥
9𝑝𝑝̂ = .

x is the count of red candies in that sample of 9 candies.

Population
Distribution:

Random sample of size n:

𝑝𝑝̂1 =

p = proportion of
population𝑥𝑥

𝑛𝑛

x is the # of red in sample 1

365

The list of 50 sample proportions of size 9:
0
2�
9
3�
9
2�
9
5�
9

1�
9
3�
9
2�
9
1�
9
3�
9

3�
9
2�
9
1�
9
3�
9
2�
9

3�
9
0
5�
9
4�
9
1�
9

2�
9
1�
9
2�
9
1�
9
0

2�
9
3�
9
3�
9
4�
9
3�
9

1�
9
1�
9
0
3�
9
2�
9

4�
9
3�
9
4�
9
2�
9
3�
9

2�
9
4�
9
2�
9
1�
9
3�
9

3�
9
2�
9
3�
9
1�
9
2�
9

If we look at the frequency distribution of this situation, we can see that the histogram issymmetric around p = .23. We take the mean of all the 𝑝𝑝̂ drawn which we denote 𝜇𝜇𝑝𝑝� . CLT
𝑝𝑝𝑝𝑝

states that 𝜇𝜇𝑝𝑝� = p and 𝜎𝜎𝑝𝑝� = � .
𝑛𝑛

𝑝𝑝̂

0
1
9
2
9
3
9
4
9
5
9
6
9
7
9
8
9
1

f

4
10
14
15
5
2
0
0
0
0

𝜇𝜇𝑝𝑝� = .23

Example 8.6
Suppose that 68% of students like donuts. Take samples of size 40. The sampling
distributions of proportions are formed.
a. State the mean and standard deviation of sampling distribution of 𝑝𝑝̂ .
b. If you randomly survey 40 students, what is the probability your survey says at least
80% like donuts?

366

Solution 8.6
a. 𝜇𝜇𝑝𝑝� = 𝑝𝑝 = .68
𝑝𝑝𝑝𝑝

𝜎𝜎𝑝𝑝� = � 𝑛𝑛 =.68 ⋅ .32
= 0.0738
40

b. P(𝑝𝑝̂ ≥ .80) = ? Since we are asking the probability of a sample proportion, we areusing central limit theorem for proportions. P(𝑝𝑝̂ ≥ .80) = normalcdf(.80, 1E99, .68,
.0738) = 0.052

Example 8.7
Based on past experience, a bank claims that 7% of the people who receive student loans
will not make the payments on time. The bank recently approved 300 student loans.
a. What is the mean of the proportion of students in this group who may not maketimely payments?
b. What is the standard deviation of the proportion of students in this group who may
not make timely payments?
c. What’s the probability that over 10% of these students will not make the payments
on time?
Solution 8.7
P(𝑝𝑝̂ > .10) = ? Since we are asking the probability of a sample proportion, we are using
central limit theorem for proportions.

a. 𝜇𝜇𝑝𝑝� = 𝑝𝑝 = .07
𝑝𝑝𝑝𝑝

b. 𝜎𝜎𝑝𝑝� = � 𝑛𝑛 =c.

.07 ⋅ .93
= 0.015300

P(𝑝𝑝̂ ≥ .10) = normalcdf(.10, 1E99, .07, .015) = 0.0228

367

8.2 | Using the Central Limit Theorem
It is important to understand when to use the central limit theorem. If you are being asked
to find the probability of the mean, use the CLT for the mean. If you are being asked to findthe probability of a sum or total, use the CLT for sums. This also applies to percentiles formeans and sums.
NOTE
If you are being asked to find the probability of an individual value, do not use the CLT.
Use the distribution of its random variable.

Example 8.8
On the average, a certain computer part lasts 8 years. The length of time the computer partlasts is exponentially distributed. Let X = the time an INDIVIDUAL computer part lasts,
and find the following probabilities:
a. Find the probability that the mean time the certain part lasts for 40 is shorter than 4
years.
b. Suppose that one computer part is randomly selected. Find the probability that thisindividual computer part lasts shorter than 4 years. This is asking us to find P(x < 4).
c. Explain why the probabilities in parts a and b are different.
d. Find the 95th percentile for the mean time for samples of 40 computer parts lasts.
Solution 8.8
a. This is asking us to find P( x < 4). Since the original distribution is exponential, themean of the exponential distribution is µ = 8, the standard deviation is also σ = 8. The
sampling distribution of sample means is approximately normal with deviation µ X = µ and𝜎𝜎𝑋𝑋̄ =

𝜎𝜎.
√𝑛𝑛

By the CLT, X ~ N(8,

8
).
√40

Using the calculator, P( x < 4) = normalcdf(-10^99, 4, 8,368

8) = 0.000783.

√40

b. Find P(x < 4). Since we are looking at an individual computer part, we use the original
distribution which is the exponential distribution. µ = 8 and to find the rate of decay, m, we
1

1
8

take the reciprocal of mu. X ~ Exp� �. Thus P(x < 4) = 𝑒𝑒 −8(4) = 0.6065

c. The probabilities are not equal because they involve different random variables, andhence use different distributions. Part a uses the sampling distribution of sample means.
The CLT states that this distribution is approximately normal. Part b uses exponential
distribution.
d. To find the 95th percentile for the mean time for samples of 40 computer parts lasts, weagain use the CLT, and the fact that X ~ N(8,This value is given by k = invNorm(.95, 8,

8

8
).
√40

Let k be the 95th percentile.

). = 10.08. Thus, about ninety five percent of√40such samples would have a mean under 10.08 years; only five percent of such samples
would have a mean above 10.08 years.
Example 8.9
A study was done on the market value of homes in a certain city. The market value of homes
is uniformly distributed between $175 thousand to $250 thousand.
a. In a sample of 30 homes, what is the probability of the mean value being greater than$230 thousand.
b. What is the probability that an individual home has a value greater than $230
thousand?
c. Find the 95th percentile for the mean value of 30 homes.
369

Solution 8.9
The original distribution is uniform. In Uniform distribution, µ = (a + b)/2 = ($175 +$250)/2 = $212.5 thousand and the standard deviation σ =𝑏𝑏−𝑎𝑎
250−175=√12√12= $21.65 thousand.

The sampling distribution of sample means is approximately normal with 𝜇𝜇𝑥𝑥̅ = 𝜇𝜇 = $212.5

thousand and 𝜎𝜎𝑥𝑥̅ =

𝜎𝜎

√𝑛𝑛

=

$21.65= $3.95 thousand. We use the CLT to get√30

X ~ N($212.5

thousand,$3.95 thousand) .

a. P( x > $230) = normalcdf(230, 10^99, 212.5, 3.95) = 0.00000471
1
� = 0.267$75

b. P(x > $230) = area of rectangle = base(height) = $20�c. The 95th percentile = invNorm(0.95, 212.5, 3.95) = $219 thousand. This indicates that in
95% all samples of size 30, the average market value of homes in the sample was less $219
thousand.

8.5 Let X~U(10, 40) and Let Y~exp(0.2)
a.
b.
c.
d.

Find P(22 < x < 25).
Find P(2 < 𝑦𝑦� < 4).
Find 25th percentile for X distribution and for 𝑋𝑋� distribution
Find 75th percentile for Y distribution and for 𝑌𝑌� distribution370

8.3 | The Central Limit Theorem for Sums
Suppose X is a random variable with a distribution that may be known or unknown (it canbe any distribution) and suppose that we select samples of size n. Recall that ΣX represents

the sum of all data values from the sample.

Again, ΣX is a random variable, as this sum

will vary from sample to sample; thus, we can discuss the sampling distribution for ΣX as
well. We denote the standard deviation of this sampling distribution as μΣX and σΣΧrespectively. Then we have the following variant of the CLT:

Central Limit Theorem for Sample Sums
Suppose X is a random variable with mean µ and standard deviation σ. Suppose that weselect random samples of size n. Then the following are true:i.

The mean of the sampling distribution for ΣX is µ ∑ X = nµ .

ii.

The standard deviation of the sampling distribution for ΣX is σ ∑ X = σ n .

iii.

For n > 30, the sampling distribution for the sampling distribution for ΣX isapproximately normal.

In other words, if we draw random samples of size n, the random variable ΣX consisting ofsums tends to be normally distributed and ΣX ~ N(nµ, n σ). Again, this means that we

can apply the techniques from Chapter 7 for calculating z-scores and probabilities involving
sample sums.
Example 8.10
An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size
80 is drawn randomly from the population.
a. Find the probability that the sum of the 80 values is more than 7,500.
b. Find the sum that is 1.5 standard deviations above the mean of the sums.
Solution 8.10
Let X = one value from the original unknown population. The probability question asks you
to find a probability for the sum (or total of) 80 values.

ΣX = the sum of 80 values. Since μ = 90, σ = 15, and n = 80, Σ X ~ N((80)(90), 80 (15)).
That is, Σ X ~ N(7200, 134.16).
371

a. Find P(Σx > 7,500).

It is helpful to draw a graph:

To calculate the probability we use the normalcdf function:
P(Σx > 7,500) = normalcdf(7500, 10^99, 7200, 134.16) = 0.0127.
b. Find Σx where z = 1.5. This sum is 1.5 standard deviations above the mean, so

Σx = 7200 + 1.5(134.16) = 7401.2.

8.6 An unknown distribution has a mean of 45 and a standard deviation of eight. A sample
size of 50 is drawn randomly from the population. Find the probability that the sum of the
50 values is more than 2,400.

Example 8.11
According to NOAA (National Oceanic and Atmospheric Administration) Fisheries website,
the average size of the Atlantic Salmon (farmed) is about 29 inches. Let’s assume thestandard deviation is 3 inches. If we take sample sizes of 100 salmon,
a. What are the mean and standard deviation for the sum of the lengths of salmon? Whatis the distribution?
b. Find the probability that the sum of the lengths is between 2850 and 2950 inches.
c. Find the 80th percentile for the sum of the 100 lengths.
Solution 8.11
a. μΣx = nμ = 100(29) = 2,900 and σΣx =n σ = √100 (3) = 30

The distribution is approximately normal for sums by the central limit theorem.
b. P(2850 < Σx < 2950) = normalcdf (2850, 2950, 2900, 30) = 0.9044
c. Let k = the 80th percentile. Then k = invNorm(0.80, 2900, 30) = 2925.2 inches372

8.7 According to NOAA (National Oceanic and Atmospheric Administration) Fisheries
website, the average size of the Atlantic Salmon (farmed) is about 29 inches. Let’s assume
the standard deviation is 4 inches. If we take sample sizes of 50 salmon,
a. What are the mean and standard deviation for the sum of the lengths of the salmon?
b. Find the probability that the sum of the lengths is between 1,400 and 1,500 inches.
c. Find the 90th percentile for the sum of the 50 salmon.

Example 8.12
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose
the standard deviation is one minute. Suppose that we select a sample of size 70.
a. What are the mean and standard deviation for the sums?
b. Find the 95th percentile for the sum of the sample. Interpret this value in a complete
sentence.
c. Find the probability that the total time for the sample is at least ten hours.
Solution 8.12
a. μΣx = nμ = 70(8.2) =574 and σΣx =nσ=70 (1) = 8.37 minutes

b. Let k = the 95th percentile. Then k = invNorm (0.95, 574, 8.37) = 587.76 minutes
Ninety five percent of the app engagement times are at most 587.76 minutes.
c. Note that 10 hours = 600 minutes. P(Σx ≥ 600) = normalcdf(600, 10^99, 574, 8.37) =0.0009.

8.8 The mean number of minutes for app engagement by a table use is 8.2 minutes.
Suppose the standard deviation is one minute. We select a random sample size of 70.
a. What is the probability that the total time for the sample is between seven hours and ten
hours? What does this mean in context of the problem?
b. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values incontext.

373

Example 8.13
A study involving stress is conducted among the students on a college campus. The stressscores follow a uniform distribution with the lowest stress score equal to one and the highest
equal to five. Using a sample of 50 students, find:a. The probability that the mean stress score for the 50 students is less than 2.5.
b. The 90th percentile for the mean stress score for the 50 students.
c. The probability that the total of the 50 stress scores is less than 125.
d. The 90th percentile for the total stress score for the 50 students. Let X = one stressscore.
Solution 8.13
Problems a and b ask you to find a probability or a percentile for a mean.
Problems c and d ask you to find a probability or a percentile for a total or sum.
The sample size is n = 50.
--------------------------------------------------------------------------------------------------------------------Since the individual stress scores follow a uniform distribution, X ~ U(1, 5); thus the meanand standard deviation for uniform distribution are µ = (1 + 4)/2 = 3 and σ = (5 – 1)/ 12 ≈1.155.
Since µ = 3 and σ = 12 ≈ 1.155 are for the original distribution, the sampling distribution
for sample means is approximately normal. Using the central limit theorem for sample
1.155
σmeans, µ X = µ = 3 and σ X === 0.163. Therefore, X ~ N( 3, 0.163).
√50
na. We want to find P( x < 2.5). It is useful to draw a graph.

Since X ~ N( 3, 0.163), P( x < 2.5) = normalcdf(-10^99, 2.5, 3, 0.163) = 0.001. The
probability that the mean stress score is less than 2.5 is virtually zero.
b. Let k = the 90th percentile; that is, P( x < k) = 0.90. Again, a graph is helpful:374

Using the calculator, and using the parameters from part a, we see that the 90th percentile
for the mean of 50 scores is k = invNorm(.90, 3, 0.163), which is about 3.21. This tells us
that, among all samples of size 50, 90% of the mean stress scores are at most 3.21, and that
10% are at least 3.21.
----------------------------------------------------------------------------------------------------------------------------The sampling distribution for sample sums, µ ∑ X = nµ and σ ∑ X = σ n . Therefore, ΣX ~N[(50)(3), √50 (1.155)]; that is, ΣX ~ N(150, 8.167).

c. Since ΣX ~ N(150, 8.167), P(Σx < 125) = normalcdf(-10^99, 125, 150, 8.167) = 0.0011.

d. Let k = the 90th percentile; that is, P( x < k) = 0.90.

Using the calculator, and using the parameters from part c, we see that the 90th
percentile for the mean of 50 scores is k = invNorm(.90, 150, 8.167), which is about
160.5. This tells us that, among all samples of size 50, 90% of the totals of the stressscores are at most 160.5, and 10% are at 160.5 or more.

8.9 According to NOAA (National Oceanic and Atmospheric Administration) Fisheries
website, the average size of the Atlantic Salmon (farmed) is about 29 inches. Let’s assume
the standard deviation is 3.5 inches. If we take sample sizes of 49 salmon,a. Find P( x < 28).

b. Find P(Σx > 1500).
c. Find the 80th percentile for the mean of 49 salmon.
d. Find the 85th percentile for the sum of 49 salmon.

375

KEY TERMS
Central Limit Theorem for Sample Means: Given a random variable X with known
mean µ and standard deviation σ, we select random samples of size n and examine thesampling distribution of the sample mean x . This sampling distribution has the following
characteristics:i) The mean of the sampling distribution for X is µ X = µ .
ii) The standard deviation of the sampling distribution for X is σ X =σ.
niii) For n > 30, the sampling distribution for the sampling distribution for X isapproximately normal. That is, X ~N(µ,σ).
nCentral Limit Theorem for Sample Proportions: Suppose X is a random variable withpopulation proportion p. Suppose that we select random samples of size n. The sampling
distribution has the following characteristics:i) The mean of the sampling distribution for 𝑝𝑝̂ is 𝜇𝜇𝑝𝑝� = 𝑝𝑝.

𝑝𝑝𝑝𝑝ii) The standard deviation of the sampling distribution for 𝑝𝑝̂ is 𝜎𝜎𝑝𝑝� = � .

iii) The sampling distribution for the sampling distribution for 𝑝𝑝̂ isapproximately normal. X ~ N(𝜇𝜇𝑝𝑝� , 𝜎𝜎𝑝𝑝� )𝑛𝑛

Central Limit Theorem for Sample Sums: Given a random variable X with known
mean µ and standard deviation σ, we select random samples of size n and examine thesampling distribution of the sample sum, ΣX. This sampling distribution has the following
characteristics:i.

The mean of the sampling distribution for ΣX is µ ∑ X = nµii.

The standard deviation of the sampling distribution for ΣX is σ ∑ X = σ n

iii.

For n > 30, the sampling distribution for the sampling distribution for ΣX isapproximately normal. That is, ΣX ~ N(nµ, n σ).

Sampling Distribution: The sampling distribution of a statistic (such as a sample meanor sample proportion) is the distribution of values of that statistic when all possible samplesof the same size are taken from the same population.

376

Standard Error of the Mean: The standard deviation of the sampling distribution for x :σ.
n

FORMULA REVIEW
z-score: z =

x−µ
σ
n

For 𝑋𝑋� ~ N(µ,

σ
n

z=

p̂ − ppqn)

σP(𝑋𝑋� < k) = normalcdf(-10^99, k, µ,)nσ)P(𝑋𝑋� > k) = normalcdf(k, 10^99, µ,n
σP(k1 < 𝑋𝑋� < k2) = normalcdf(k, k2, µ,)nσk = invnorm(probability to left, µ,
)
n

𝑝𝑝𝑝𝑝

For 𝑝𝑝̂ ~ N(p, � )
𝑛𝑛

𝑝𝑝𝑝𝑝

P(𝑝𝑝̂ < k) = normalcdf(-10^99, k, p, � )
𝑛𝑛

𝑝𝑝𝑝𝑝

P(𝑝𝑝̂ > k) = normalcdf(k, 10^99, p, � )
𝑛𝑛

𝑝𝑝𝑝𝑝

P(k1 < 𝑝𝑝̂ < k2) = normalcdf(k, k2, p, � )
𝑛𝑛

𝑝𝑝𝑞𝑞

k = invnorm(probability to left, p, � )𝑛𝑛For ΣX ~ N(nµ, n σ).
P(Σx < k) = normalcdf(-10^99, k, nµ, n σ)P(Σx > k) = normalcdf(k, 10^99, nµ, n σ)P(k1 <Σx < k2) = normalcdf(-10^99, k, nµ, n σ)
k = invnorm(probability to left, nµ, n σ)

377

Exercises for Chapter 8
1. A J-shaped distribution has a mean of 450 and a standard deviation of 28. Samples of sizen = 40 are drawn randomly from the population.
a. Find the mean of the sampling distribution for sample meansb. Find the standard error of meanc. State if the sampling distribution is approximately normal or normal.
2. A normal distribution has a mean of 580 and a standard deviation of 40. Samples of size
25 are drawn randomly from the population.
a. Find the mean of the sampling distribution for sample meansb. Find the standard error of meanc. State if the sampling distribution is approximately normal or normal.
3. Darla is a personnel manager in a large corporation. Each month she must review 16 of
the employees. From past experience, she has found that the reviews take her approximately
four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the randomvariable representing the time it takes her to complete one review; assume that X is normally
distributed. Let X-bar be the random variable representing the mean time to complete 16
reviews. Assume that the 16 reviews represent a random set of reviews.
a. Find the standard error of mean.
b. Find the probability that the mean time for a month’s reviews will require between
3.5 and 4.25 hours, and sketch the graph.
4. Darla is a personnel manager in a large corporation. Each month she must review 20 of
the employees. From past experience, she has found that the reviews take her approximately
4.5 hours each to do with a population standard deviation of 1.1 hours. Let Χ be the randomvariable representing the time it takes her to complete one review; assume that X is normally
distributed. Let X-bar be the random variable representing the mean time to complete 20
reviews. Assume that the 20 reviews represent a random set of reviews.
a. Find the standard error of mean.
b. Find the probability that an individual for a month’s reviews will require less than
4 hours, and sketch the graph.
c. Find the probability that the mean time for a month’s reviews will require less than
4 hours, and sketch the graph378

5. A normal distribution has a mean of 460 and a standard deviation of 30. Samples of size
25 are drawn randomly from the population. Round to tenths.
a. Find 25th percentile of the normal distribution.
b. Find the 25th percentile of the sampling distribution of sample means.
6. A J-shaped distribution has a mean of 320 and a standard deviation of 28. Samples of sizen = 45 are drawn randomly from the population. Find the 95th percentile for the sample
mean. Sketch the graph of the sampling distribution of sample means.
Use the following information to answer the next three exercises: A manufacturer
produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is
26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample
of 100 weights is selected.
7.

a. What is the distribution for the weights of one 25-pound lifting weight? What is
the mean and standard deviation of this distribution?
b. What is the distribution for the mean weight of one hundred 25-pound lifting
weights? What is the mean and standard deviation of the sampling distribution.
c. Find the probability that the mean actual weight for the 100 weights is less than
24.9. Draw the graph.

8. Find the probability that the mean actual weight for the 100 weights is greater than
25.2. Draw the graph.
9. Find the 90th percentile for the mean weight for the 100 weights. Draw the graph
Use the following information to answer the next four exercises: The lifetime in monthsof a particular smartphone's battery follows an exponential distribution with a rate of decayof 10%. A sample of 64 of these smartphones is taken.
10. a. What is the standard deviation for the population distribution?
b. What is the parameter µ?
11. Suppose that a sample of 64 batteries is selected. What is the distribution for the mean
length of time that the batteries last?
12. Find the probability that the sample mean for a sample of 64 batteries will be between 7months and 11 months.
13. Find the IQR for the mean amount of time for a sample of 64 batteries.

379

14. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally
distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample
49 fly balls.
a. Describe the distribution for x , the average distance in feet, for 49 fly balls.
b. What is the probability that the 49 balls travel a mean distance of less than 240 feet?
c. Find the 80th percentile of the distribution of the average of 49 fly balls.
15. Previously, De Anza Statistics students estimated that the amount of change daytime
Statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we
randomly pick 25 daytime statistics students.
a. In words, Χ =b. X ~( ,)c. In words, X =d. X ~( ,)e. Find the probability that an individual had between $0.80 and $1.00. Graph thesituation, and shade in the area to be determined.
f. Find the probability that the average of the 25 students was between $0.80 and $1.00.
Graph the situation, and shade in the area to be determined.
g. Explain why there is a difference in part e and part f.
16. Suppose that a category of world-class runners are known to run a marathon (26 miles)
in an average of 145 minutes with a standard deviation of 14 minutes. We select a random
sample of 49 races, and let x-bar represent the average time for the 49 running times.
a. X ~(,)

b. Find the probability that the runner will average between 142 and 146 minutes inthese 49 marathons.
c. Find the 80th percentile for the average of these 49 marathons.
d. Find the median of the average running times.
17. According to the Internal Revenue Service, the average length of time for an individual
to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is10.53 hours (without any attached schedules). The distribution is unknown. Let us assume
that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.
a. In words, Χ =b. In words, X =c. X ~( ,)d. Would you be surprised if the 36 taxpayers finished their Form 1040s in an
average of more than 12 hours? Explain why or why not in complete sentences.

380

e. Would you be surprised if one taxpayer finished his or her Form 1040 in morethan 12 hours? In a complete sentence, explain why.
18. In 2019 the average size of a U.S. farm was 444 acres, with a standard deviation of 55
acres. Suppose we randomly survey 38 farms from 2019. What is the probability that the
average size was more than 500 acres?
19. The length of songs in a collector’s iTunes album collection is uniformly distributed fromtwo to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are atotal of 43 songs on the five albums.
a. In words, Χ =b. Χ ~
c. In words, X = _____
d. Describe the distribution: X ~( ,)e. Find the first quartile for the average song length.
f. The IQR(interquartile range) for the average song length is from–.

20. The percent of fat calories that a person in America consumes each day is normally
distributed with a mean of about 36 and a standard deviation 10.
Suppose that 16
individuals are randomly chosen.
a. Find the probability that the mean percent of fat calories for a group of 16 is greaterthan 40.
b. Find the first quartile for the average percent of fat calories.
21. Determine which of the following are true and which are false. Then, in complete
sentences, justify your answers.
a. When the sample size is large, µ X is approximately equal to µ.
b. When the sample size is large, X is approximately normally distributed.
c. When the sample size is large, the standard deviation of the distribution for X isapproximately the same as the standard deviation for X.
22. Which of the following is NOT TRUE about the distribution of sample means?
a.
b.
c.
d.

The mean, median, and mode are equal.
The area under the curve is one.
The curve never touches the x-axis.
The curve is skewed to the right.

23. The distribution of income in some summer paid internships is considered wedge shaped(many very low paid, very few middle paid, and even fewer high paid). Suppose we pick a

381

company with a wedge-shaped distribution. Let the average summer salary be $2,000 persummer with a standard deviation of $4,000. We randomly survey 100 interns.
a. In words, Χ =b. In words, X =c. X ~
d. How is it possible for the standard deviation to be greater than the average?
e. Why is it more likely that the average of the 100 interns will be from $2,000 to$2,200 than from $2,200 to $2,400?
24. The attention span of a two-year-old is exponentially distributed with a mean of about
eight minutes. Suppose we randomly survey 60 two-year-olds.
a. In words, Χ =b. Χ ~( , )c. In words, X =d. X ~( ,)e. Before doing any calculations, which do you think will be higher? Explain why.
i. The probability that an individual attention span is less than ten minutes.
ii. The probability that the average attention span for the 60 children is less thanten minutes?
f. Calculate the probabilities in part e.
25. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution
with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay
Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas
stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:a. X ~ N(4.59, 0.10)
b. X ~ N(4.59,

0.10
16

c. X ~ N(4.59,)

d. X ~ N(4.59,

0.10
)
16
16
0.10

)

26. The time to wait for a particular rural bus is distributed uniformly from zero to 75
minutes. One hundred riders are randomly sampled to learn how long they waited. Would
you be surprised, based upon numerical calculations, if the sample average wait time (inminutes) for 100 riders was less than 30 minutes?
27. The time to wait for a particular rural bus is distributed uniformly from zero to 75
minutes. One hundred riders are randomly sampled to learn how long they waited. Find the90th percentile sample average wait time (in minutes) for a sample of 100 riders.

382

28. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution
with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay
Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas
stations. Find the probability that the average price for 30 gas stations is less than $4.55.
29. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution
with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay
Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas
stations. What's the approximate probability that the average price for 16 gas stations is over
$4.69?
30. The average length of a maternity stay in a U.S. hospital is said to be normally distributedwith mean of 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women
who recently bore children in a U.S. hospital.
a. In words, X =b. In words, X =c. X ~
d. Is it likely that an individual stayed more than five days in the hospital? Why or
why not?
e. Is it likely that the average stay for the 80 women was more than five days? Whyor why not?
h. Which is more likely:i. An individual stayed more than five days.
ii. the average stay of 80 women was more than five days.
31. Salaries for teachers in a particular elementary school district are normally distributedwith a mean of $44,000 and a standard deviation of $6,500. We randomly survey ten
teachers from that district.
a. Find the 90th percentile for an individual teacher’s salary.
b. Find the 90th percentile for the average teacher’s salary.
32. Men have an average weight of 172 pounds with a standard deviation of 29 pounds.
a. Find the probability that 20 randomly selected men will have a total weightexceeding 3600 lbs.
b. If 20 men have a sum weight greater than 3500 lbs, then their total weightexceeds the safety limits for water taxis. Based on (a), is this a safety concern?
Explain.
33. Never Ready Batteries has engineered a newer, longer lasting AAA battery. The company
claims this battery has an average life span of 17 hours with a standard deviation of 0.8
hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries
and find that the sample mean life span is 16.7 hours. If the process is working properly,383

what is the probability of getting a random sample of 30 batteries in which the sample mean
lifetime is 16.7 hours or less? Is the company’s claim reasonable?
34. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20
randomly selected adults are given an IQ test, what is the probability that the sample meanscores will be between 85 and 125 points?
35. Your company has a contract to perform preventive maintenance on thousands of airconditioners in a large city. Based on service records from previous years, the time that atechnician spends servicing a unit averages one hour with a standard deviation of one hour.
In the coming week, your company will service a simple random sample of 70 units in thecity. You plan to budget an average of 1.1 hours per technician to complete the work. Will
this be enough time?
36. According to ClinicalTrials.gov, the rapid, onside COVID-19 detection had an average
result in 90 minutes. Let’s assume the standard deviation of detection time is 15 minutes.
Random samples of size 40 are made. Find the probability that sample mean of 60 tests isless than 1 hour.
37. Certain coins have an average weight of 5.201 grams with a standard deviation of 0.065
g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291
g, what is the expected number of rejected coins when 280 randomly selected coins areinserted into the machine?
CLT for proportions
38. Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get toschool. Each day the driver would be chosen by randomly selecting one of the four names.
They carpool to school for 96 days. Use the normal approximation to the binomial tocalculate the following probabilities.
a. Find the probability that Janice is the driver at most 20 days.
b. Find the probability that Roberta is the driver more than 16 days.
c. Find the probability that Barbara drives exactly 24 of those 96 days.
39. Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of
the population favor a charter school for grades K through five. A simple random sample of
300 is surveyed. Calculate following using the normal approximation to the binomial
distribution.
a.
b.
c.
d.

Find the probability that less than 100 favor a charter school for K through 5.
Find the probability that at least 170 favor a charter school for K through 5.
Find the probability that less than 141 favor a charter school for K through 5.
Find the probability that there are fewer than 130 that favor a charter school for384

K through 5.
e. Find the probability that exactly 150 favor a charter school for K through 5.
40. Suppose that 54% of students register two weeks before the start of the semester. If you
randomly survey 70 students, what is the probability your survey says at most 20% register
two weeks before the start of the semester?
41. According to USA Today article, 56% of 18 to 24 year olds voted for Hilary Clinton in
the 2016 election. If we randomly survey thirty 18 to 24 year olds, what is the probability
that those selected at least 60% voted for Hilary Clinton?
42. A truckload of watermelons arrive at a packing facility. A random sample of 200 is
selected and examined for bruises and other defects. The truckload will be rejected if morethan 5% of the sample has defects. Suppose that in fact 8% of the watermelons on the truck
have defects. What’s the probability that the shipment will not be rejected?
43. Based on past experience, a bank claims that 24% of the people who receive student
loans will not make the payments on time due to the pandemic. The bank recently
reviewed 400 student loans.
a. What is the mean of the proportion of students in this group who may not maketimely payments?
b. What is the standard deviation of the proportion of students in this group who may
not make timely payments?
c. What’s the probability that over 30% of these students will not make the payments
on time?
44. According to Nielsen media, 44% of Hispanics adult consumers are more likely to beconsidered what they call “recovery optimists” after the COVID-19 pandemic. If we
randomly survey 50 Hispanic adult consumers, what is the probability that those selectedless than 30% are considered “recovery optimists”?
CLT for sums
Use the following information to answer the next four exercises: An unknown distribution
has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly
from the population.
45. Find the probability that the sum of the 95 values is greater than 7,650.
46. Find the probability that the sum of the 95 values is less than 7,400.
47. Find the sum that is two standard deviations above the mean of the sums.
48. Find the sum that is 1.5 standard deviations below the mean of the sums.

385

Use the following information to answer the next five exercises: The distribution of resultsfrom a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40
is drawn randomly.
49. Find the probability that the sum of the 40 values is greater than 7,500.
50. Find the probability that the sum of the 40 values is less than 7,000.
51. Find the sum that is one standard deviation above the mean of the sums.
52. Find the sum that is 1.5 standard deviations below the mean of the sums.
53. Find the percentage of sums between 1.5 standard deviations below the mean of thesums and one standard deviation above the mean of the sums.
Use the following information to answer the next six exercises: A researcher measures theamount of sugar in several cans of the same soda. The mean is 39.01 with a standard
deviation of 0.5. The researcher randomly selects a sample of 100.
54. Find the probability that the sum of the 100 values is greater than 3,910.
55. Find the probability that the sum of the 100 values is less than 3,900.
56. Find the probability that the sum of the 100 values falls between 3900 and 3910.
57. Find the sum that has a z-score of –2.5.
58. Find the sum that has a z–score of 0.5.
59. Find the probability that the sums will fall between the z = -2 and z = 1.
Use the following information to answer the next four exercises: An unknown distribution has
a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object ofinterest.
60. What is the mean of ΣX ?
61. What is the standard deviation of ΣX ?
62. What is P(Σx = 290)?
63. What is P(Σx > 290)?
64. True or False: Only the sums of normal distributions are also normal distributions.
65. In order for the sums of a distribution to approach a normal distribution, what must be
true?
386

66. What three things must you know about a distribution to find the probability of sums?
67. An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one

object from this distribution. What is the sample size if the standard deviation of ΣX is 42?
68. An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = theobject of interest. What is the sample size if the mean of ΣX is 15,200?

69. A market researcher analyzes how many electronics devices customers buy in a single
purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples
400 customers. What is P(Σx < 1,186)?

Use the following information to answer the next three exercises: An unknown distribution
has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one objectof interest.
70. What is the mean of ΣX?
71. What is the standard deviation of ΣX?
72. What is P(Σx > 9,000)?
Use the following information to answer the next four exercises: A manufacturer produces
25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26
pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of
100 weights is selected
73. a. What is the distribution for the sum of the weights of 100 25-pound lifting weights?
b. Find P(Σx < 2,450).

74. Draw the graph for Exercise 8.73.
75. Find the 90th percentile for the total weight of the 100 weights.
76. Draw the graph for Exercise 8.75.
Use the following information to answer the next three exercises: The length of time aparticular smartphone's battery lasts follows an exponential distribution with a mean of ten
months. A sample of 64 of these smartphones is taken.
77. What is the distribution for the total length of time that 64 batteries last?
78. Find the 80th percentile for the total length of time that 64 batteries will last.
79. Find the middle 80% for the total amount of time that 64 batteries will last.
80. Find P(Σ x > 740)
387

Use the following information to answer the next six exercises:
A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 isselected.
81. Find P(Σx > 420).
82. Find the 90th percentile for the sums.
83. Find the 15th percentile for the sums.
84. Find the first quartile for the sums.
85. Find the third quartile for the sums.
86. Find the 80th percentile for the sums.

REFERENCES
8.1 The Central Limit Theorem for Sample Means (Averages) and Sample Proportions
Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available
online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email(accessed May 17, 2013).
Fisheries, NOAA. “Atlantic Salmon (Farmed).” NOAA, www.fisheries.noaa.gov/species/atlanticsalmon-farmed (accessed July 4, 2021).
8.2 Using the Central Limit Theorem
Data from the Wall Street Journal.
“National Health and Nutrition Examination Survey.” Center for Disease Control and Prevention.
Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013).
Castillo, Walbert, et al. “How We Voted - by Age, Education, Race and Sexual Orientation.” USA
Today, Gannett Satellite Information Network, 9 Nov. 2016, college.usatoday.com/2016/11/09/howwe-voted-by-age-education-race-and-sexual-orientation/ (accessed July 17, 2018).
“Hispanic Consumers Are Recovery Optimists; Black Consumers Are Cautious Optimists.” Nielsen, 4
May 2021, www.nielsen.com/us/en/insights/article/2021/hispanic-consumers-are-recoveryoptimists-black-consumers-are-cautious-optimists/.
8.3 The Central Limit Theorem for Sums
Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The
Flurry Blog, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May
17, 2013).

388

CHAPTER 8 SOLUTIONS:
1) a. = 450; σ = 28/√40 ; approximately normal.
3) a. 𝜎𝜎𝑋𝑋̄ = 1.2/√16 b. P(3.5 < 𝑥𝑥̅ < 4.25) = normalcdf(3.5, 4.25, 4, 1.2/√16)5) a. invnorm(0.25, 460, 30) = 439.77

b. invnorm(0.25, 460, 30/√25 = 455.95

7) a. X ~ U(24, 26); µ = 25 σ = 0.5774

b. X ~ N(25, 0.0577) c. 0.0416

9) 25.07

13) 1.69

11) 𝑋𝑋� ~ N(10, 10/√40)15) a. Χ = amount of change students carryb. Χ ~ E(0.88, 0.88)

c. x = mean amount of change for sample of 25 students. X ~ N(0.88, 0.176)
e. 0.0819
f. 0.1882
g. The distributions are different. Part a is exponential and part b is normal.
17) a. X = length of time for an individual to complete IRS form 1040, in hours.
b. X = mean length of time for a sample of 36 taxpayers to complete IRS form 1040
c. X ~ N(10.53, 0.333)
d. Yes, we would be surprised, since P( x > 12) is approximately 0.
e. No; would not be totally surprised because the probability is P(x > 12) = 0.2312.
So about 23% of individuals take more than 12 hours to complete the form.
19) a. X = the length of a song, in minutes, in the collection
b. X ~ U(2, 3.5)
c. X = the average length, in minutes, of the songs from a sample of five albums
d. X ~ N(2.75, 0.066)
e. Q1 = invNorm(.25, 2.75, 0.066) = 2.71 minutes
21)

a. True. By the CLT, the mean of a sampling distribution of the means isapproximately the mean of the data distribution for large n.
b. True. According to the Central Limit Theorem, the larger the sample, the closer
the sampling distribution of the means becomes normal.
c. False. The standard deviation of the sampling distribution of the means
decreases as the sample size increases; so σ x =23)

f. 0.09 minutes

σwill be much smaller than σ.
na. X = the summer income of an internb. X = the average income from samples of 100 interns of a company
c. X ~ N(2000, 400)
d. Very wide differences in data values can have averages smaller than standard
deviations.
e. The distribution of the sample mean will have higher probabilities closer to the
population mean.

389

25)

option b27) k = invnorm(.90, 37.5, 2.17) = 40.3 minutes. Use invnorm because the sampling
distribution of sample means is approximately normal.
29) normalcdf(4.69, 10^99, 4.59, .10/√16) = 0.0000317
31) a. $52,330

b. $46,634

33) We are given that μ = 17, σ = 0.8, x = 16.7, and n = 30. To calculate the probability, we
0.8) = 0.020. If the process is working
use P( x < 16.7) = normalcdf(-10^99, 16.7, 17,30
properly, then there is only about a 2% probability that a sample of 30 batteries would havean average lifetime of 16.7 hours or less. So the sample data appears to be incompatible
with the company’s claim. Therefore, the class was justified to question the claim.
35) We are given that μ = 1, σ = 1, x = 1.1, and n = 70. We calculate the probability P( x <1) = 0.7958. So there is an 80% chance that the average1.1) = normalcdf(-10^99, 1.1, 1,
70
service time will be less than 1.1 hours. It would be wise to schedule more time since there
is a 20% chance that the mean maintenance time will be greater than 1.1 hours.
0.065) ≈ 1, we can conclude37) Since P(5.111 < x < 5.291) = normalcdf(5.111, 5.291, 5.201,280that virtually all the coins are within the limits. We would not expect any coins to be
rejected in a sample of 280.
39) a. 0b. 0.101c. 0.0186

d. 0.000395 e. 0.0268

41) 0.329= .002543) a.)24%b.)2.14%c.) P(𝑝𝑝̂ > 10%) = normalcdf(.30, 10^99, .24, .0214)

45) 0.3345

47) 7,833.92 49) 0.0089

57) Σx = 3,888.5

59) 0.8186

65) The sample size n gets larger.
71) σΣx = 1,00061) σΣx = 563) 0.9772

67) 4969) 0.1587

73) a. ΣX ~ N(2,500, 5.7735)

75) 2,507.40 77) X~N(640, 80)
invnorm(.90, 640, 80) = 742.5
81) 0.0072

51) 7,326.49 53) 77.45%

83) 391.54

55) 0.4207

b. 079) P10 = invnorm(.10, 640, 80) = 537.5, P90 =85) 405.51

390

9 | CONFIDENCE INTERVALSThis Photo by Unknown Author is licensed under CC BY-SA-NC

Have you ever wondered the average number of downloaded APPs a tablet user gets a month? You can
use confidence intervals to answer this question.

Chapter Objectives
By the end of this chapter, the student should be able to:••••

Calculate and interpret confidence intervals for estimating a population mean and a
population proportion.
Interpret the Student's t probability distribution as the sample size changes.
Discriminate between problems applying the normal and the Student's t distributions.
Calculate the sample size required to estimate a population mean and a population
proportion given a desired confidence level and margin of error.

Suppose you were trying to determine the mean rent of a two-bedroom apartment in yourtown. You might look in the classified section of the newspaper, write down several rents
listed, and average them together. You would have obtained a point estimate of the true391

mean. If you are trying to determine the percentage of times you make a basket whenshooting a basketball, you might count the number of shots you make and divide that by the
number of shots you attempted. In this case, you would have obtained a point estimate forthe true proportion.

9.1 INTRODUCTION
We use sample data to generalize about an unknown population. This part of statistics iscalled inferential statistics. The sample data help us to make an estimate of a population
parameter. We realize that the point estimate is most likely not the exact value of the
population parameter, but close to it. Therefore, point estimate is a sample statistic which
is a starting measurement to estimate the population parameter.
Example 9.1
If you worked in the marketing department of an entertainment company, you might beinterested in the mean number of APPS a consumer downloads a month from their APP store.
If so, you could conduct a survey and calculate the sample mean, x = 4 apps, and the sample
standard deviation, s = 0.1. You would use to estimate the population mean and s to estimate
the population standard deviation. The sample mean, x is the point estimate for the
population mean, μ. The sample standard deviation, s, is the point estimate for the population
standard deviation, σ.
After calculating point estimates, we construct interval estimates, called confidence
intervals. Intervals in mathematics represent a range of values that the variable can be.
Recall in Algebra, a variable that is between two values can be written in two ways:
Inequality notation: 3 < x < 10, here we are saying x is between 3 and 10.
Interval notation: (3, 10)
A confidence interval is another type of estimate but, instead of being just one number, it isan interval of numbers. The interval of numbers is a range of values calculated from a given
set of sample data. The confidence interval is likely to include an unknown population
parameter between a lower bound value and upper bound value with a certain level ofconfidence.
Confidence Interval Types
Mean:

lower bound value < µ < upper bound value

Proportion:

lower bound value < p < upper bound valueStandard
Deviation:

lower bound value < σ < upper bound value

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NOTE: For mean and proportion lower bound value is the point estimate – margin of error.
The upper bound value is the point estimate + margin of error.
The interval notation of this form is (point estimate – margin of error, point estimate +margin of error)
The empirical rule, which applies to bell-shaped distributions, says that in approximately
95% of the samples, the sample mean, x , will be within two standard deviations of the
population mean µ. From Example 9.1, two standard deviations is (2)(0.1) = 0.2. The sample
mean x is likely to be within 0.2 units of µ. Therefore, 0.2 is considered the margin of error.
Because x is within 0.2 units of µ, which is unknown, then μ is likely to be within 0.2 unitsof x in 95% of the samples. The population mean µ is contained in an interval whose lower
bound value is calculated by taking the sample mean and subtracting two standard
deviations (2)(0.1) and whose upper bound value is calculated by taking the sample mean
and adding two standard deviations. In other words, µ is between x − 0.2 and x + 0.2 in
95% of all the samples.
For the Example 9.1, the unknown population mean μ is between x − 0.2 = 4 − 0.2 = 3.8(lower bound value) and x + 0.2 = 4 + 0.2 = 4.2 (upper bound value).
3.8 apps < µ < 4.2 apps
We say that we are 95% confident that the unknown population mean number of apps
downloaded from the app store per month is between 3.8 and 4.2. The 95% confidence intervalis (3.8, 4.2).
Recap of Terms•

Point estimate: starting sample statistic (measurement) to aid in estimating thelower and upper bounds of the confidence intervals. 𝑥𝑥̅ the point estimate for estimatingµ. 𝑝𝑝̂ is the point estimate for estimating p. s is the point estimate for estimating σ.

•

Margin of Error: maximum difference from point estimate to the lower or upperbound. E = upper bound – point estimate.

•

Confidence Interval: the range of values that likely contains the population
parameter within a certain level of confidence.

393

Example 9.2
Find the point estimate for the population mean, µ, for a 90% confidence interval. Assume
the distribution is normally distributed with a population standard deviation of 6 minutes.
A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery
time of 36 minutes.
Solution 9.2
Point estimate for the population mean is the sample mean, x . Therefore x =36 minutes isthe point estimate.
Example 9.3
Find the point estimate for the population mean, µ, for a 90% confidence interval. Assume
the distribution is normally distributed with a population standard deviation of 6 minutes.
A random sample of 10 pizza delivery restaurants is taken and has the following raw data inminutes:30, 28, 41, 34, 29, 32, 36, 39, 34, 37
Solution 9.3
Point estimate for the population mean is the sample mean, x . We can find the sample meanusing the calculator. First, we must enter the data into L1.

x = mean(L1) = 34 minutes is the point estimate.

Using the TI-83, 83+, 84, 84+ Calculator
To calculate the mean and standard deviation:
place the data set into L1 by pressing Stat, Edit
2nd Quit,
Press 2nd Stat,
Move cursor to MATH
Mean(L1)

394_

Example 9.4
In a Gallup poll that took place June 2021, 96 out of 1200 US adults mention “the coronavirus
pandemic” as the most important problem in the US. This proportion has gone down 7% fromthe previous month. (Adapted from News.Gallup.com) In constructing a 98% confidence
interval for the population proportion that mention the pandemic as the most important
problem in the US, find the point estimate.
Solution 9.4
Point estimate for the population proportion is the sample proportion, 𝑝𝑝̂ . Therefore 𝑝𝑝̂ =is the point estimate. 𝑝𝑝̂ =96
can be written as a decimal (0.08) or a percent (8%).
1200

96
1200

9.1 Find the point estimate for the population mean and population standard deviation,µ and σ. Assume the distribution is approximately normal. In a random sample of 8 people,
the mean commute to work was 35.5 minutes and the sample standard deviation was 7.2
minutes.

Point Estimates
Sample Mean: x
Sample Proportion: 𝑝𝑝̂Sample Standard deviation: sSample Variance: s2395

9.2 | Confidence Interval for a Single Population Mean when σ known
A confidence interval for a population mean with a known standard deviation is based on the𝜎𝜎fact that the sample means follow an approximately normal distribution; 𝑥𝑥̅ ∼N(µx,)√𝑛𝑛Remember that a confidence interval is created for an unknown population parameter likethe population mean, μ. Confidence intervals for this parameter has the form:(point estimate – margin of error, point estimate + margin of error)( x − E, x + E )The margin of error, E, depends on the confidence level, CL, or percentage of confidence
and the standard error of the mean, σ x =

σ

n

.

The value that represents the level ofconfidence for a certain distribution is known as a critical value. The critical value for theZ-distribution has the notation of Zc or Zα/2. Therefore, E = zα 2

σ

n

.

Critical Value, Zα/2.
The confidence level, CL, is the area in the middle of the Z-distribution (standard normal
distribution). The confidence level is often considered the probability that the calculated
confidence interval estimate will contain the true population parameter. However, it is moreaccurate to state that the confidence level is the percent of confidence intervals that containthe true population parameter when repeated samples are taken. Most often, it is the choiceof the person constructing the confidence interval to choose a confidence level of 90% or higher
because that person wants to be reasonably certain of his or her conclusions.
CL = 1 – α, so α is the area that is not in the center and is split equally between the two tails.
Therefore, α is the probability that the interval will not contain the true population
parameter. Each of the tails contains an area equal to α/2. The z-score that has an area tothe right of α/2 is denoted by Zα/2.
For example, when CL = 0.95, α = 0.05 and α/2 = 0.025; we write Zα/2 = Z0.025.

396

Recall in chapter 7 to find the value of Z when given the area to the left, we can use thecalculator to find this value.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate_

Press 2nd Vars, choose #3 invnorm

The left critical value is negative since it is on the left side of zero. The right critical value isthe symmetric opposite of the left critical value. Round critical values to two decimal places.
Left: - Zα/2 = -Z0.025 = -1.96
Right: Zα/2 = Z0.025 = 1.96

Example 9.5
Find the critical value for the Z-distribution for 98% confidence level.

Solution 9.5
C.L = .98, α = .02, α/2 = .01 we write Zα/2 = Z0.01.

Left: - Zα/2 = -Z0.01 = invnorm(.01) = -2.33
Right: Zα/2 = Z0.01 = 2.33

397

Calculating the Confidence Interval for mean when σ is known
Point estimate – margin of error < µ < point estimate + margin of error( x − E, x + E )where E = zα 2σn.

The steps to construct and interpret the confidence interval are:
•
•
•
•
•

Calculate the sample mean, x from the sample data. Remember, in this section wealready know the population standard deviation σ.
Find the critical value, Zα/2 that corresponds to the confidence level.
Calculate the margin of error, E.
Construct the confidence interval.
Write a sentence that interprets the estimate in the context of the situation in the
problem. (Explain what the confidence interval means, in the words of the problem.)

The interpretation should clearly state the confidence level (CL), explain what population
parameter is being estimated (here, a population mean), and state the confidence interval(both endpoints). "We estimate with % confidence that the true population mean (include
the context of the problem) is between and(include appropriate units)."

Example 9.6
Suppose scores on exams in statistics are normally distributed with an unknown population
mean and a population standard deviation of three points. A random sample of 36 scores istaken and gives a sample mean (sample mean score) of 68. Find a confidence interval
estimate for the population mean exam score (the mean score on all exams).
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Solution 9.6• You can use technology to calculate the confidence interval directly.
• The first solution is shown step-by-step (Solution A).
• The second solution uses the TI-83, 83+, and 84+ calculators (Solution B).

398

9.6 Solution A
To find the confidence interval, you need the sample mean, x , and the margin of error, E.

x = 68E = zα 2σnσ = 3; n = 36; the confidence level is 90% (CL = 0.90) so α = 1 – CL = 1 – 0.90 = 0.10
α = 0.05  Zα/2 = Z0.05 = the positive value of invnorm(.05) = 1.645

σ
 3 
=
E z=
1.645 =α 2
 .8225n 36 
x – E = 68 – 0.8225 = 67.1775
x + E = 68 + 0.83 = 68.8225
67.1775 < µ < 68.8225
The 90% confidence interval for the population mean is (67.1775, 68.8225).
9.6 Solution B
Press STAT, TESTS. #7:ZInterval.

The confidence interval for the population mean is (67.178, 68.822).
Interpretation:
We estimate with 90% confidence that the true population mean exam score for all statistics
students is between 67.18 and 68.82.
Explanation of 90% Confidence Level:
Ninety percent of all confidence intervals constructed in this way contain the true mean
statistics exam score. For example, if we constructed 100 of these confidence intervals, wewould expect 90 of them to contain the true population mean exam score.

399

9.2 Suppose average pizza delivery times are normally distributed with an unknown
population mean and a population standard deviation of six minutes. A random sample of 28
pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find
a 90% confidence interval estimate for the population mean delivery time.

Example 9.7
According to an article in moneysavingpro.com, Americans aged 25 to 64 pay more than thosein the UK for cell phone plans. The article stated that Americans pay an average of $114 permonth on the cell phone plan. We would like to see if $114 lies within a 98% confidence
interval for the population mean of the monthly post pay mobile phone bill. Here is a list of
randomly selected Americans aged 25 to 64.
$98, $131, $97, $89, $114, $145, $109, $141, $102, $110, $90, $92, $134, $144, $152,$140, $118, $98, $144, $87, $150, $118, $118, $131, $99, $120, $115, $105, $95, $96
Find a 98% confidence interval for the true (population) mean of the monthly post pay mobile
phone bill. Assume that the population standard deviation is σ = $18
Solution 9.7 Solution A
To find the confidence interval, start by finding the point estimate: the sample mean.

x can be found using 1-var Stat on the calculator (Chapter 3)
x = $116.07
Next, find the margin of error, E. Because you are creating a 98% confidence interval, CL =0.98.

Left: - Zα/2 = -Z0.01 = invnorm(.01) = -2.33

Right: Zα/2 = Z0.01 = 2.33

400

E = zα 2σnE = 2.33

18=$7.6630

NOTE: the units for E is the same units as 𝑥𝑥̅ and s.

To find the 98% confidence interval, find the lower and upper bounds of the interval.
Lower bound: x – E = $116.07 – $7.66 = $108.41
Upper bound: x + E = $116.07 + $7.66 = $123.73
Solution 9.7 Solution B
Press STAT, TESTS 7: ZInterval.

The confidence interval is $108.42 < µ < $123.72.
Interpretation: We estimate with 98% confidence that the true monthly post pay mobile
phone bill for Americans between 18 - 64 is between $108.42 and $123.72. Therefore, thearticle stating that the average American pays $114 is supported.

9.3 The data that follows, shows a different random sampling of 20 mobile phone bills. Use
this data to calculate a 93% confidence interval for the true (population) mean of the monthly
post pay mobile phone bill. Assume that the population standard deviation is σ = $18 asprevious stated.
$91, $137, $97, $103, $137, $127, $86, $118, $128, $138, $126, $90,
$144, $94, $104, $94, $130, $141, $93, $128

401

Changing the Confidence Level or Sample Size
We need to look at if the range of the confidence interval is affected by changing theconfidence level or by changing the sample size.

Example 9.8
Suppose we change the original problem in Example 9.6, which uses 90% confidence level,by using a 95% confidence level. Find a 95% confidence interval for the true (population)mean statistics exam score.
Solution 9.8 using Calculator
To find the confidence interval about the mean with the population standard deviation
known, we use Zinterval.

E = upper bound – point estimate
E = 68.98 – 68 = .98
Notice that the E is larger for a 95% confidence level than the 90% confidence of Example 9.6
90% Confidence Level
E = .8225
67.18 < µ < 68.82

95% Confidence Level
E = .98
67.02 < µ < 68.98

Comparing the results
The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98).
The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger
than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more
confident that the confidence interval actually does contain the true value of the population
mean for all statistics exam scores, the confidence interval necessarily needs to be wider.

402

Summary: Effect of Changing the Confidence Level••

Increasing the confidence level increases the margin of error, making the confidence
interval wider.
Decreasing the confidence level decreases the margin of error, making the confidence
interval narrower.

9.4 Refer back to the pizza-delivery 9.2 Try It exercise. The population standard deviation issix minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a
95% confidence interval estimate for the true mean pizza delivery time.

Example 9.9
Suppose we change the original problem in Example 9.6 to see what happens to the marginof error if the sample size is changed.
Leave everything the same except the sample size. Use the original 90% confidence level.
a. What happens to the margin of error and the confidence interval if we increase thesample size and use n = 100 instead of n = 36?
b. What happens if we decrease the sample size to n = 25 instead of n = 36?
Solution 9.9 by Calculatora.) Zintervalb.) Zinterval

403

Summary: Effect of Changing the Sample Size••

Increasing the sample size causes the margin of error to decrease, making theconfidence interval narrower.
Decreasing the sample size causes the margin of error to increase, making theconfidence interval wider.

9.5 Refer back to the pizza-delivery 9.2 Try It exercise. The mean delivery time is 36 minutes
and the population standard deviation is six minutes. Assume the sample size is changed to
50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the
population mean delivery time.
Calculating the Sample Size, n
If researchers desire a specific margin of error, then they can use the margin of error formula
to calculate the required sample size. The margin of error formula for a population mean
when the population standard deviation is known has the following formula:

E = zα 2σnWhen we solve for n, the formula for sample size is as follows: z ⋅σ n =  α /2 
 E 

2

NOTE: clue word for margin of error is the word “within”.

404

Example 9.10
The population standard deviation for the age of CLC students is 15 years. If we want to be
95% confident that the sample mean age is within two years of the true population mean ageof CLC students, how many randomly selected CLC students must be surveyed?
Solution 9.10
From the problem, we know that σ = 15 and E= 2.
Z = Z0.025 = -invnorm(0.025, 0, 1) = 1.96, because the confidence level is 95%.
22

 zα /2 ⋅ σ   1.96 ⋅15 =n =
 216.09 =
2 
 E  Use n = 217: Always round the answer UP to the next higher integer to ensure that the
sample size is large enough. Therefore, 217 CLC students should be surveyed in order to be
95% confident that we are within two years of the true population mean age of CLC students.

9.6 The population standard deviation for the height of high school basketball players is threeinches. If we want to be 95% confident that the sample mean height is within 1.5 inch of thetrue population mean height, how many randomly selected students must be surveyed?

405

9.3 | Confidence Interval for a Single Population Mean when σunknown

In practice, we rarely know the population standard deviation, σ. In the past, when thesample size was large, this did not present a problem to statisticians. They used the sample
standard deviation s as an estimate for σ and proceeded as before to calculate a confidence
interval with close enough results. However, statisticians ran into problems when the sample
size was small. A small sample size caused inaccuracies in the confidence interval.
William S. Gosset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into thisproblem. His experiments with hops and barley produced very few samples. Just replacing σwith s did not produce accurate results when he tried to calculate a confidence interval. He
realized that he could not use a normal distribution for the calculation; he found that theactual distribution depends on the sample size. This problem led him to "discover" what iscalled the Student's t-distribution. The name comes from the fact that Gosset wrote under
the pen name "Student."
Up until the mid-1970s, some statisticians used the normal distribution approximation forlarge sample sizes and used the Student's t-distribution only for sample sizes of at most 30.
With graphing calculators and computers, the practice now is to use the Student's tdistribution whenever s is used as an estimate for σ.

Characteristics/Properties of Student’s t-distribution:
•

•
•

•
•

•
•

The graph for the Student's t-distribution is similar to the standard normal curve;
however, it has more probability in its tails than the standard normal distribution
because the spread of the t-distribution is greater than the spread of the standard normal.
So the graph of the Student's t-distribution will be thicker in the tails and shorter in thecenter than the graph of the standard normal distribution.
The mean for the Student's t-distribution is 0 and the distribution is symmetric about 0.
The exact shape of the Student's t-distribution depends on the degrees of freedom. As thedegrees of freedom increases, the graph of Student's t-distribution becomes more like thegraph of the standard normal distribution.
t-score, 𝑡𝑡 =𝑥𝑥̅ −𝜇𝜇

�

𝑠𝑠
�
√𝑛𝑛

, has the same interpretation as the z-score which is the number ofstandard deviations is from the mean.
Degrees of Freedom, df = n – 1, come from the calculation of the sample standard
deviation s. We used n deviations (x – 𝑥𝑥̅ values) to calculate s. Because the sum of thedeviations is 0, we can find the last deviation once we know the other n – 1 deviations.
Notation for t-distribution where T is the random variable: T∼ tdf
Critical value notation for t-distribution, tα/2 or tc.

406

Critical Value, tα/2.
Recall, confidence level (CL) is the area in the middle of the distribution. The confidence
level is often considered the probability that the calculated confidence interval estimate willcontain the true population parameter.
For example, when CL = 0.95, α = 0.05 and α/2 = 0.025; we write tα/2 = t0.025.

There are three ways to find the critical value for the t-distribution. All three ways need thearea in the tail (α/2) and the degree of freedom (n – 1).
1.) For those who have, TI-83, 83+, 84 to find tα/2 you must use the t-distribution chart(located in appendix) or Excel (T.inv)

407

2.) For those who have the newer TI 84 + calculators
Press 2nd Vars, choose #4 invT(area in the tail, df)

Example 9.11
Find the critical value for the t-distribution for 98% confidence level where n = 12.
Solution 9.11 Solution A (Using TI 84+ calculators)
C.L = .98, α = .02, α/2 = 0.01 we write tα/2 = t0.01. d.f. = n – 1 = 11

Left: - tα/2 = -t0.01 = invT(.01, 11) = -2.718 Right: tα/2 = t0.01 = - invT(.01, 11) = 2.718
Recall that t-distribution is symmetric. Therefore, the left and right critical values are
symmetric opposites. This means that if you know the left critical value, then the right
critical value is the positive version of it.
Solution 9.11 Solution B (Excel or t-chart)
C.L = .98, α = .02, α/2 = 0.01 we write tα/2 = t0.01. d.f. = n – 1 = 11

Go to an empty cell and type equal sign,then start typing t.inv

408

After you press Enter, the answer
will appear.

All three methods of finding the left and right critical values for the t-distribution need thearea in the tail (1%) and degrees of freedom (df = n – 1)

Left: - tα/2 = -t0.01 = -2.718

Right: tα/2 = t0.01 = 2.718

Calculating the Confidence Interval for mean when σ is unknown
Point estimate – margin of error < µ < point estimate + margin of error( x − E, x + E )where E = tα 2s.
nThe steps to construct and interpret the confidence interval are:
•
•
•
•
•

Calculate the sample mean x and s from the sample data.
Find the critical value, tα/2 that corresponds to the confidence level.
Calculate the margin of error, E.
Construct the confidence interval.
Write a sentence that interprets the estimate in the context of the situation in the
problem. (Explain what the confidence interval means, in the words of the problem.)

409

The interpretation should clearly state the confidence level (CL), explain what population
parameter is being estimated (here, a population mean), and state the confidence interval(both endpoints). "We estimate with % confidence that the true population mean (include
the context of the problem) is between and(include appropriate units)."

Example 9.12
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You
measure sensory rates for 15 subjects with the results given. Use the sample data to construct
a 95% confidence interval for the mean sensory rate for the population (assumed normal)
from which you took the data.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9

Solution 9.12
• The first solution uses the TI-83+ and TI-84 calculators (Solution A).
• The second solution is step-by-step (Solution B).
Solution 9.12 Solution A
Press STAT, TESTS. #8: Tinterval.

Solution 9.12 Solution B
To find the confidence interval, you need the sample mean, x , sample standard deviation, s,
and the margin of error, E.

x and s can be found using 1-Var Stat on the calculator
x = 8.2267, s = 1.6722, n = 15 → d.f. = 14

E = tα 2sn410The confidence level is 95% (CL = 0.95)
CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05
α/2 = 0.025 t α/2 = t 0.025 = the positive value of invT(.025,14) = 2.14s
 1.6722 E tα==
2.14  =2
 .924n 15 
Lower bound: x - E = 8.2267 - 0.924 = 7.3
Upper bound: x + E = 8.2267 + 0.924 = 9.15
7.3 < µ < 9.15
The 95% confidence interval for the population mean is (7.3, 9.15).
Interpretation: We estimate with 95% confidence that the true population mean sensory rate
is between 7.30 and 9.15.

9.7 You do a study of hypnotherapy to determine how effective it is in increasing the numberof hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with thefollowing results. Construct a 95% confidence interval for the mean number of hours slept
for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

Example 9.13
Industrial chemicals may enter the body through pollution or as ingredients in consumer
products. In October 2008, the scientists at HTP (Human Toxome Project) tested cord blood
samples for 20 newborn infants in the United States. The cord blood of the "In utero/newborn"group was tested for 430 industrial compounds, pollutants, and other chemicals, including
chemicals linked to brain and nervous system toxicity, immune system toxicity, andreproductive toxicity, and fertility problems. There are health concerns about the effects ofsome chemicals on the brain and nervous system. The table below shows how many of thetargeted chemicals were found in each infant’s cord blood. Use this sample data to construct
a 90% confidence interval for the mean number of targeted industrial chemicals to be foundin an in infant’s blood.
79
137

145
83

147
156

160
94

116
121

100
144

411

159
123

151
114

156
139

126
99

Solution 9.13 Solution A
Press STAT, TESTS. #8: Tinterval.

We estimate with 90% confidence that the mean number of all targeted industrial chemicals
found in cord blood in the United States is between 117.412 and 137.488.

Solution 9.13 Solution B:
First, from the sample, you can calculate x = 127.45 and s = 25.965 using 1-varStats. We arealso given n = 20, which we can then find degrees of freedom, df = 20 – 1 = 19.
You are asked to calculate a 90% confidence interval, which gives us the following notation:
CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10 then α/2 = 0.05,
Second, we can find the critical value by using Calculator, Excel, or t-chart:t α/2 = t 0.05 = -invT(.05, 19) = 1.729.
Third, we find margin of error using the following formula:s 25.965 
E tα=
1.729  ==2
 10.038n 20 
Lastly, find lower and upper bounds of the interval:
Lower bound: x – E = 127.45 – 10.038 = 117.412
Upper bound: x + E = 127.45 + 10.038 = 137.488
117.412 < µ < 137.488

412

9.4 | Confidence Interval for a Single Population Proportion, pDuring an election year, we see articles in the newspaper that state confidence intervals interms of proportions or percentages. For example, a poll for a particular candidate runningfor president might show that the candidate has 40% of the vote within three percentage
points (if the sample is large enough). Often, election polls are calculated with 95%
confidence, so, the pollsters would be 95% confident that the true proportion of voters who
favored the candidate would be between 0.37 and 0.43:(0.40 – 0.03, 0.40 + 0.03).
How do you know you are dealing with a proportion problem? First, the underlying
distribution is a binomial distribution. (There is no mention of a mean or average.) If X is abinomial random variable, then X ~ B(n, p) where n is the number of trials and p is theprobability of a success. To form a proportion, take X, the random variable for the number ofsuccesses and divide it by n, the number of trials (or the sample size). The random variable𝑃𝑃� (read "P hat") is that proportion,
𝑋𝑋
𝑃𝑃� =
𝑛𝑛

When n is large and p is not close to zero or one, we can use the normal distribution toapproximate the binomial. When np < 10 and nq < 10, the normal approximation probabilities
for the binomial distribution are not close enough to the actual binomial probabilities.
Therefore, not best to use when less than 10.
𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛
𝑃𝑃�∼𝑁𝑁 � , √ 𝑛𝑛 � Recall that for Binomial µ = np and σ = �𝑛𝑛𝑛𝑛𝑛𝑛
𝑛𝑛𝑝𝑝𝑝𝑝
𝑃𝑃�∼𝑁𝑁 �𝑝𝑝, � � is the reduced version𝑛𝑛Calculating the Confidence Interval for proportion
Point estimate – margin of error < µ < point estimate + margin of error

𝑝𝑝�𝑞𝑞�

where 𝐸𝐸 = 𝑍𝑍𝛼𝛼/2 � 𝑛𝑛 and 𝑞𝑞� = 1 − 𝑝𝑝̂(𝑝𝑝̂ − 𝐸𝐸, 𝑝𝑝̂ + 𝐸𝐸)

NOTE: the confidence interval can be used only if the number of success 𝑛𝑛𝑝𝑝̂ and number of
failures 𝑛𝑛𝑞𝑞� are both greater than ten.

413

The steps to construct and interpret the confidence interval are:
•
•
•
•
•
•

𝑥𝑥
𝑛𝑛

Calculate the sample proportion, 𝑝𝑝̂ = , where x is the frequency, from the sample

data.
Determine if the number of success 𝑛𝑛𝑝𝑝̂ and number of failures 𝑛𝑛𝑞𝑞� are both greaterthan ten.
Find the critical value, Zα/2 that corresponds to the confidence level.
Calculate the margin of error, E.
Construct the confidence interval.
Write a sentence that interprets the estimate in the context of the situation in the
problem. (Explain what the confidence interval means, in the words of the problem.)Example 9.14
According a poll by Pharmaceutical Technology newsletter, they gathered information from4004 Verdict global readers about their health concern about the Covid-19 pandemic. The
newsletter stated in the article that “the Delta variant continued to spread to 96 countries by
the last week of June (2021), when health concern increased to 38%”. Using this informationfrom the newsletter, construct a 96% confidence interval on the true proportion of readersthat have a health concern about the pandemic.
Solution 9.14
• The first solution uses a function of the TI-83, 83+ or 84 calculators (Solution A).
• The second solution is step-by-step (Solution B).
Solution 9.14 Solution A
Using the TI-83, 83+, 84, 84+ Calculator
To calculate_

STAT, TESTS, A: 1-PropZint
NOTE: x represents the frequency of the sample proportion x = 𝑝𝑝̂ n rounded to a whole
number

36.4% < p < 39.6%

414

Solution 9.14 Solution B (step by step)
To calculate the confidence interval, you must first find p̂ (point estimate), which is given in
the problem.
𝑝𝑝̂ = 0.38 therefore 𝑞𝑞� = 1 − 𝑝𝑝̂ = 0.62

Second, determine if 𝑛𝑛𝑝𝑝̂ and 𝑛𝑛𝑞𝑞� are both greater than ten.

𝑛𝑛𝑝𝑝̂ = (4004)(0.38) = 1522 and 𝑛𝑛𝑞𝑞� = (4004)(0.62) = 2482

Third, find margin of error, E where the confidence level is 96% (CL = 0.96)
so α = 1 – CL = 1 – 0.96 = 0.04 and α/2 = 0.02
Zα/2 = Z0.025 = the positive value of invnorm(.02) = 2.05
𝑝𝑝�𝑞𝑞�

0.38*0.62=0.0164004

𝐸𝐸 = 𝑧𝑧𝛼𝛼/2 � 𝑛𝑛 = 2.05�

NOTE: the units of E are the same units as the point estimate; therefore in this section E isa percentage. E = 1.6%
Lastly, find the lower and upper bound of the interval.
Lower Bound: 𝑝𝑝̂ − 𝐸𝐸 = 0.38 – 0.016 = 0.364

Upper Bound: 𝑝𝑝̂ + 𝐸𝐸 = 0.38 + 0.016 = 0.396
0.364 < p < 0.396

36.4% < p < 39.6%
Interpretation:
We estimate with 96% confidence that between 36.4% and 39.6% of all Verdict global readers
have health concern about Covid-19 pandemic.
Explanation of 96% Confidence Level:
Ninety-six percent of the confidence intervals constructed in this way would contain the true
value for the population proportion of all Verdict global readers have health concern about
Covid-19 pandemic.

415

9.8 Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of
the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a
confidence interval estimate for the true proportion of people who own tablets.

Example 9.15
For a class project, a political science student at a large university wants to estimate thepercent of students who are registered voters. He surveys 500 students and finds that 59.7%
are registered voters. Compute a 90% confidence interval for the true percent of students who
are registered voters, and interpret the confidence interval.
Solution 9.15
• The first solution uses a function of the TI-83, 83+ or 84 calculators (Solution A).
• The second solution is step-by-step (Solution B).
Solution 9.15 Solution A
Using the TI-83, 83+, 84, 84+ Calculator
To calculate
STAT, TESTS, A: 1-PropZint

NOTE:
x on the calculator must be
in whole number form;
therefore, the calculator
solution will be slightly off
from Solution A.

0.562 < p < 0.634
56.2% < p < 63.4%

416_

Solution 9.15 Solution B
𝑝𝑝̂ = 0.597; 𝑞𝑞� = 1 − 𝑝𝑝̂ = 0.403

The confidence level is 90% (CL = 0.90) so α = 1 – CL = 1 – 0.90 = 0.10
α/2 = 0.05 z α/2 = z 0.05 = the positive value of invnorm(.05) = 1.645
𝑝𝑝�𝑞𝑞�

0.597*0.403=0.036500

𝐸𝐸 = 𝑧𝑧𝛼𝛼/2 � 𝑛𝑛 = 1.645�

Lower Bound: 𝑝𝑝̂ − 𝐸𝐸 = 0.597 – 0.036 = 0.81

Upper Bound: 𝑝𝑝̂ + 𝐸𝐸 = 0.597 + 0.036 = 0.874
0.561 < p < 0.633

56.1% < p < 63.3%
Interpretation:• We estimate with 90% confidence that the true percent of all students that are registeredvoters is between 56.2% and 63.4%.
• Alternate Wording: We estimate with 90% confidence that between 56.2% and 63.4% of
ALL students are registered voters.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true value forthe population percent of students that are registered voters.

9.9 A student polls his school to see if students in the school district are for or against the
new legislation regarding school uniforms. She surveys 600 students and finds that 79.8%
are against the new legislation. Compute a 90% confidence interval for the true percent ofstudents who are against the new legislation, and interpret the confidence interval.

417

“Plus Four” Confidence Interval for p
There is a certain amount of error introduced into the process of calculating a confidence
interval for a proportion. Because we do not know the true proportion for the population, we
are forced to use point estimates to calculate the appropriate standard deviation of thesampling distribution. Studies have shown that the resulting estimation of the standard
deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence
intervals. We simply pretend that we have four additional observations. Two of theseobservations are successes and two are failures. The new sample size, then, is n + 4, and the
new count of successes is x + 2.
Computer studies have demonstrated the effectiveness of this method. It should be used
when the confidence level desired is at least 90% and the sample size is at least ten.

Example 9.16
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in thepast week?” Ten students reported smoking within the past week. Use the “plus-four”
method to find a 95% confidence interval for the true proportion of statistics students who
smoke.
Solution 9.16
Ten students out of 25 reported smoking within the past week, so x = 10 and n = 25. Because
we are using the “plus-four” method, we will use x = 10 + 2 = 12 and n = 25 + 4 = 29.

We are 95% confident that the true proportion of all statistics students who smoke cigarettes
is between 23.5% and 59.3%.

9.10 Out of a random sample of 65 freshmen at State University, 31 students have declareda major. Use the “plus- four” method to find a 96% confidence interval for the true proportionof freshmen at State University who have declared a major.

418

Calculating the Sample Size n
If researchers desire a specific margin of error, then they can use the margin of error formula
to calculate the required sample size.
𝑝𝑝�𝑞𝑞�

The margin of error formula for a population proportion is 𝐸𝐸 = 𝑧𝑧𝛼𝛼/2 � 𝑛𝑛 where 𝑞𝑞� = 1 − 𝑝𝑝̂
Solving for n gives you an equation for the sample size:zn = pˆ (1 − pˆ ) α / 2 
 E 

2

Recall a clue word for E is “within”. NOTE: if the estimated sample proportion is not given,
then use p̂ =.5
Example 9.17
Suppose a mobile phone company wants to determine the current percentage of customers
aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should
the company survey in order to be 90% confident that the estimated (sample) proportion iswithin three percentage points of the true population proportion of customers aged 50+ who
use text messaging on their cell phones?
Solution 9.17
E = 0.03 Zα/2 = -invnorm(.05) = 1.645 because C.L. = 0.90p̂ =.5 since an estimated sample proportion is not given2z 
 1.645 n=pˆ (1 − pˆ )  α /2  =.5(.5)  .03  E n = 751.7

2

Round the answer to the next higher value. The sample size should be 752 cell phonecustomers aged 50+ in order to be 90% confident that the estimated (sample) proportion iswithin three percentage points of the true population proportion of all customers aged 50+
who use text messaging on their cell phones.

9.11 Suppose an internet marketing company wants to determine the current percentage ofcustomers who click on ads on their smartphones. How many customers should the company
survey in order to be 90% confident that the estimated proportion is within five percentage
points of the true population proportion of customers who click on ads on their smartphones?

419

9.5 | Confidence Interval for Standard Deviation, σ (Optional)
This section is repeated in chapter 12 (chi-square distribution) with more examples.
Let’s conduct the following statistical experiment. We select samples of size n from a normal
population, which has a standard deviation of σ. We find that the standard deviation in our
sample is equal to s. Given these data, we can define a statistic, called chi-square, using thefollowing formula:χ2 =(n − 1) s 2

σ2This new distribution has the following characteristics:
1.) All chi-square values are greater than or equal to 0
2.) There is a different chi-square curve for each degrees of freedom, n – 1 (figure below)
3.) The curve is nonsymmetrical and skewed to the right4.) The total area under the curve is 1, or 100%25.) The notation is X ∼ χ df
χ𝟐𝟐𝟐𝟐
χ𝟐𝟐𝟔𝟔

χ𝟐𝟐𝟏𝟏𝟏𝟏0
For the χ 2 distribution, the population mean is μ = df (degrees of freedom) and the population
standard deviation is σ = �2(𝑑𝑑𝑑𝑑). The random variable is shown as χ 2 , but may also be any
upper-case letter.

Critical values of the Chi-Square Distribution
Recall that critical values are values from the distribution that separate the confidence area
and the non-confidence area. We found Zα/2 by using ±invnorm(α/2) and tα/2 by using tdistribution chart or by ±invT(α/2, df). Since the Z-distribution and the t-distribution are
symmetrical, the left and right critical values are opposite values. However, in the chi-square
distribution, values are only positive.
••

χ𝟐𝟐𝐋𝐋 represents the left critical value

χ𝟐𝟐𝐑𝐑 represents the right critical value
420

The critical values for the χ 2 distribution are recorded in two tables (left critical value table
and right critical value table). To find these values, you need area in the tail and the degrees
of freedom.

Example 9.18
Find the left and right critical value for 95% confidence level for chi-square distribution witha degree of freedom of 11.
Solution 9.18
C.L. = 0.95, α = 0.05, α/2 = 0.025 and df = 11

Below you can see, that the chi-square distribution table is split into 2 tables. The first table
is the left critical values, χ𝟐𝟐𝐋𝐋 . The second table is the right critical values, χ𝟐𝟐𝐑𝐑 . The tail area
is 0.025 and df = 11 so χ𝟐𝟐𝐋𝐋 = 3.8157 using the table for the left critical values and χ𝟐𝟐𝐑𝐑 = 21.92

using the table for the right critical values. The complete two tables are found in the
appendix.

421

9.12 Find the left and right critical values for the chi-square distribution for C.L. = .98 and
df = 6

Critical values using Excel
Recall that critical values from the Z-distribution (standard normal) and the t-distribution(approximately normal) are symmetric opposites. The critical values from chi-square
distribution are both positive. Also, t-distribution and chi-square distribution are dependent
on degrees of freedom (df).
To use Excel functions, the first thing to type is an equal sign. Here are the Excel functions
for each critical value.
Left Z-distribution critical value, -Zc
Right Z-distribution critical value, Zc
Left t-distribution critical value, -tc
Right t-distribution critical value, tc
Left chi-square critical value, χ2L
Right chi-square critical value, χ2R=norm.s.inv(a/2)=-norm.s.inv(a/2)=t.inv(a/2, df)=-t.inv (a/2, df)=chisq.inv(a/2, df)=chisq.inv.rt(a/2, df)

In Example 9.18 C.L. = .95 and df = 11. If the confidence level is 95%, that leaves 5% in
both tails of the distribution. α = 0.05. Split the 5%, which gives us 2.5% = 0.025 as α/2.

Note: Chi-square distribution and t-distribution are dependent on degrees of freedom thatis why it is included in the input of the Excel functions.

422

Confidence Interval for Variance/Standard Deviation
Recall that an interval is a range of numbers. A confidence interval is a range of values,
calculated from sample data, that is used to estimate an unknown population parameter.
The interval gives us a lower bound value and upper bound for the parameter with a certain
level of confidence.
Here we present intervals for estimating σ2 and σ between a lower and upper bound.
Unlike other confidence intervals we have seen, these bounds can’t be found using the
calculator. They depend on the chi-square critical values.
Calculating the Confidence Interval for variance
Lower bound < σ2 < Upper bound(n − 1) s 2

χ R2

<σ2 <

(n − 1) s 2

χ L2

where χ L2 , χ R2 are critical values, where d.f. = n – 1, and s2 is the point estimate.
NOTE: χ L2 , χ R2 are not to be squared. The right critical value is on the left side and the left

on the right side. Also notice the numerators are the same. When you divide the same value
by a larger number, you get a smaller value.

Calculating the Confidence Interval for standard deviation
Lower bound < σ < Upper bound(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2

where s is the point estimate.

The sample variance, s2, can be found using the calculator or Excel function. Using the ticalculator, s2 can be found by: Pressing [STAT], Edit, enter values into L1
 2nd Quit, 2nd [STAT], move cursor to MATH
 Choose option 8: variance(, 2nd [1]423Using Excel:
 Type the data into a column
 Go to an empty cell, type =Var.s(highlight data)

Example 9.19
The weights (in pounds) of 15 dogs selected randomly from those adopted out by an animal
shelter last week are shown in the list below. Construct a 98% confidence interval for the
population variance.
25, 34, 29, 30, 31, 28, 27, 28, 33, 31, 28, 29, 32, 29, 29

Solution 9.19
First we need to find the sample variance (chapter 3) of the data set. Recall to find it bycalculator you enter the data into L1 (Stat, Edit). 2nd Stat, Math, #8 Variance (L1):s2 = 5.552
Second, we find the critical values using the chi-square distribution chart.
Let C.L. = 0.98, α = 0.02, α/2 = 0.01 and df = n – 1 = 15 – 1 = 14

α/2 = 0.01
α/2 = 0.01

C.L. = 0.98
0= 4.660= 29.141

Third, we plug in the values into the formula.

(n − 1) s 2

χ R2

<σ <
2

(n − 1) s 2

χ L2

14(5.552)
14(5.552)<σ2 <29.1414.660
2.667 < σ 2 < 16.680

424

Example 9.20
A random sample of 18 men have a mean height of 67.5 inches and a standard deviation of
1.8 inches. Construct a 99% confidence interval for the population standard deviation.
Solution 9.20 using chi-square distribution tables
Since s = 1.8, s2 = 3.24.
Let C.L. = 0.99, α = 0.01, α/2 = 0.005 and df = n – 1 = 18 – 1 = 17

χ L2 = 5.697 and χ R2 = 35.718(n − 1) s 2

χ

2
R

<σ <

(n − 1) s 2

χ

2
L

=>

17(3.24)
17(3.24)<σ <35.7185.697

1.242 < σ < 3.109
Solution 9.20 using Excel
Since s = 1.8, s2 = 3.24.
Let α/2 = 0.005 and df = 17

χ L2 = chisq.inv(.005, 17) = 5.697 and χ R2 = chisq.inv.rt(.005, 17) = 35.718(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2=> 1.242 < σ < 3.109

Lower Bound on Excel: =sqrt(17*3.24/highlight right critical value)
Upper Bound on Excel:=sqrt(17*3.24/highlight left critical value)425

KEY TERMS
Confidence Interval (CI) an interval estimate for an unknown population parameter. Thisdepends on:•••

the desired confidence level,
information that is known about the distribution (for example, known standard
deviation),
the sample and its size.

Confidence Level (CL) the percent expression for the probability that the confidence
interval contains the true population parameter; for example, if the CL = 90%, then in 90 out
of 100 samples the interval estimate will enclose the true population parameter.
Critical values are the values (±Za/2, ±ta/2, χ L2 , χ R2 ) from a distribution that separates theconfidence area from the non-confidence area.

Degrees of Freedom (df) the number of objects in a sample that are free to vary
Margin of error (E) depends on the confidence level, sample size, and known or estimatedpopulation standard deviation.
Inferential Statistics also called statistical inference or inductive statistics; this facet of
statistics deals with estimating a population parameter based on a sample statistic. For
example, if four out of the 100 calculators sampled are defective we might infer that four
percent of the production is defective.
Parameter a numerical characteristic of a population
Point Estimate a single number computed from a sample and used to estimate a population
parameter
Standard Deviation a number that is equal to the square root of the variance and measures
how far data values are from their mean; notation: s for sample standard deviation and σ for
population standard deviation
Student's t-Distribution investigated and reported by William S. Gossett in 1908 andpublished under the pseudonym Student; the major characteristics of the random variable(RV) are:•••

It is continuous and assumes any real values.
The pdf is symmetrical about its mean of zero. However, it is more spread out andflatter at the tails than the normal distribution.
It approaches the standard normal distribution as n get larger.

426

•

There is a "family of t–distributions: each representative of the family is completely
defined by the number of degrees of freedom, which is one less than the number ofdata.

Formula Review
Critical Values
Left Z-distribution critical value, -Zc
Right Z-distribution critical value, Zc
Left t-distribution critical value, -tc
Right t-distribution critical value, tc

Calculator Function
Invnorm(α/2, 0, 1)- Invnorm(α/2, 0, 1)
InvT(α/2,df)-InvT(α/2,df)

Excel Function=norm.s.inv(α/2)=-norm.s.inv(α/2)=t.inv(α/2, df)=-t.inv (α/2, df)

Calculating the Confidence Interval for mean when σ is knownσ( x − E, x + E ) where E = zα 2 n .
Calculator function: Zinterval

Calculating the Confidence Interval for mean when σ is unknown( x − E, x + E ) where E = t

α 2

s
.
n

Calculator function: Tinterval

Calculating the Confidence Interval for proportion
𝑝𝑝�𝑞𝑞�

(𝑝𝑝̂ − 𝐸𝐸, 𝑝𝑝̂ + 𝐸𝐸) where 𝐸𝐸 = 𝑧𝑧𝛼𝛼/2 � 𝑛𝑛 and 𝑞𝑞� = 1 − 𝑝𝑝̂
Calculator function: 1propZint

NOTE: the confidence interval can be used only if the number of success 𝑛𝑛𝑝𝑝̂ and numberof failures 𝑛𝑛𝑞𝑞� are both greater than ten.

427

Calculating the Sample Size for a population mean z ⋅σ n =  α /2 
 E 

2

Calculating the Sample Size for a population proportionzn = pˆ (1 − pˆ ) α / 2 
 E 

2

NOTE: if the estimated sample proportion is not given, then use p̂ =.5

Formulas (section 9.5)
Critical Values
Left chi-square critical value, χ2L
Right chi-square critical value, χ2R
Confidence Interval for Variance:

Calculator Function
None, use a chart
None, use a chart(n − 1) s 2

χ R2

<σ2 <

Confidence Interval for Standard Deviation:(n − 1) s 2

χ L2(n − 1) s 2

χ R2

428

Excel Function=chisq.inv(α/2, df)=chisq.inv.rt(α/2, df)<σ <(n − 1) s 2

χ L2

EXERCISES FOR CHAPTER 9
1. Find the critical value, Zα/2 , for the following different confidence levels
a. 92%
b. 94%
c. 97%
2. A random sample of 49 students has a grade point average with a population standard
deviation of 0.78. Find the margin of error if the confidence level is 98%.
3. A random sample of 19 students has a grade point average with a standard deviation of
0.78. Find the margin of error if the confidence level is 98%.
4. The standard deviation of the weights of elephants is known to be approximately 15
pounds. We wish to construct a 95% confidence interval for the mean weight of newborn
elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The
sample standard deviation is 11 pounds.
a. Identify the following:i. x =
ii. σ =
iii. n =iv. s =
v. Determine which of the above values the point estimate is.
b. Construct a 95% confidence interval for the population mean weight of newborn
elephants using σ.
i. State the confidence interval,
ii. Find the critical values and sketch the graph associated with the problem,iii. Calculate the margin of error.
c. What will happen to the confidence interval obtained, if 500 newborn elephants areweighed instead of 50? Why?
5. The U.S. Census Bureau conducts a study to determine the time needed to complete theshort form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. The population
standard deviation is 2.2 minutes. The population distribution is assumed to be normal.
a. Identify the following:i. x =
ii. σ =
iii. n =b. Construct a 90% confidence interval for the population mean time to complete theforms.
i. State the confidence interval,
ii. Find the critical values.

429

iii. Calculate the margin of error.
c. If the Census wants to increase its level of confidence and keep the margin of error
the same by taking another survey, what changes should it make?
d. If the Census did another survey, kept the margin of error the same, and surveyed
only 50 people instead of 200, what would happen to the level of confidence? Why?
e. Suppose the Census needed to be 98% confident of the population mean length of time.
Would the Census have to survey more people? Why or why not?
6. A sample of 20 heads of lettuce was selected. Assume that the population distribution ofhead weight is normal. The weight of each head of lettuce was then recorded. The mean
weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard
deviation is known to be 0.2 pounds.
a. Identify the following:

b.
c.
d.
e.
f.

i. x =
ii. σ =
iii. n =iv. s =
v. Determine which of the above values the point estimate for estimating
population mean is.
Construct a 90% confidence interval for the population mean weight of the heads oflettuce.
Construct a 95% confidence interval for the population mean weight of the heads oflettuce.
In complete sentences, explain why the confidence interval in part b is larger than inpart c.
What would happen if 40 heads of lettuce were sampled instead of 20, and the marginof error remained the same?
What would happen if 40 heads of lettuce were sampled instead of 20, and theconfidence level remained the same?

7. The mean age for all Foothill College students for a recent Fall term was 33.2. The
population standard deviation has been pretty consistent at 15. Suppose that twenty-five
Winter students were randomly selected. The mean age for the sample was 30.4. We areinterested in the true mean age for Winter Foothill College students. Let X = the age of a
Winter Foothill College student.
a. Identify the following:i. x =
ii. _____ = 15iii. n =b. Construct a 95% confidence interval for the true mean age of Winter Foothill College
students by working out then answering the next seven exercises.

430

What is x estimating?
How much area is in both tails (combined)? α
Identify the margin of error.
In one complete sentence, explain what the interval means.
Using the same mean, standard deviation, and level of confidence, suppose that n were
69 instead of 25. Would the margin of error become larger or smaller? How do youknow?
h. Using the same mean, standard deviation, and sample size, how would the margin oferror change if the confidence level were reduced to 90%? Why?

c.
d.
e.
f.
g.

i.

Fill in the blanks on the graph with the area, upper and lower bounds of the confidence
interval, and the sample mean

8. Find the critical value, tα/2 , for the following different confidence levels
a. 90% when n = 25
b. 95% when n = 10
c. 98% when n = 36
9. A hospital is trying to cut down on emergency room wait times. It is interested in theamount of time patients must wait before being called back to be examined. An investigationcommittee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample
standard deviation of 0.5 hours.
a. Construct a 95% confidence interval for the population mean time spent waiting.
i. State the confidence interval,
ii. Find the critical values and sketch the graph,iii. Calculate the margin of error.
b. Explain in complete sentences what the confidence interval means.
10. One hundred eight Americans were surveyed to determine the number of hours they
spend watching television each month. It was revealed that they watched an average of 151
hours each month with a standard deviation of 32 hours. Assume that the underlying
population distribution is normal.

431

a. Define the random variable X in words.
b. Define the random variable X in words.
c. Which distribution should you use for this problem? X ∼ ______________
d. Construct a 99% confidence interval for the population mean hours spent watching
television per month.
i. State the confidence interval,
ii. Find the critical valueiii. Calculate the margin of error.
e. Why would the margin of error change if the confidence level were lowered to 95%?
11. The data in the following table are the result of a random survey of 39 national flags (withreplacement between picks) from various countries. We are interested in finding a confidence
interval for the true mean number of colors on a national flag. Let X = the number of colors
on a national flag.
X
1
2
3
4
5

Freq
1
7
18
7
6

a. Calculate the following:
x=
i.
ii.
sx =
iii.
n=

x

b. Define the random variablein words.
c. Construct a 95% confidence interval for the true mean number of colors on nationalflags.
d. How much area is in both tails (combined)?
e. Fill in the blanks on the graph with the areas, the upper and lower limits of theConfidence Interval and the sample mean.

432

f. In one complete sentence, explain what the interval means.
g. Using the same x , sx, and level of confidence, suppose that n were 69 instead of 39.
Would the margin of error become larger or smaller? How do you know?
h. Using the same x , sx, and n = 39, how would the margin of error change if theconfidence level were reduced to 90%? Why?
12. Marketing companies are interested in knowing the population percent of women who
make the majority of household purchasing decisions.
a. When designing a study to determine this population proportion, what is theminimum number you would need to survey to be 90% confident that the population
proportion is estimated to within 0.05?
b. If it were later determined that it was important to be more than 90% confident anda new survey were commissioned, how would it affect the minimum number you needto survey? Why?
13. Suppose the marketing company did do a survey. They randomly surveyed 200
households and found that in 120 of them, the woman made the majority of the purchasing
decisions. We are interested in the population proportion of households where women makethe majority of the purchasing decisions.
a. Identify the following:
i.
x=
ii.
n=
iii.
p̂ =
b. Define the random variables X and p̂ in words.
c. Which distribution should you use for this problem? X ∼ _______________
d. Construct a 95% confidence interval for the population proportion of households where
the women make the majority of the purchasing decisions.
i.
State the confidence interval,
ii.
Sketch the graphiii.
Calculate the margin of error.
e. List two difficulties the company might have in obtaining random results, if thissurvey were done by email.
14. Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280
identified themselves as non-manual wage earners, 250 identified themselves as mid- level
managers, and 160 identified themselves as executives. In the survey, 82% of manuallaborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of midlevel managers preferred trucks, and 26% of executives preferred trucks.
a. Construct a 95% confidence interval for the percent of executives who prefer trucks.

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b. Calculate the margin of error.
c. Suppose we want to lower the sampling error. What is one way to accomplish that?
d. The sampling error given in the survey is ±2%. Explain what the ±2% means.
15. A poll of 1,200 voters asked what the most significant issue was in the upcoming election.
Sixty-five percent answered the economy. We are interested in the population proportion of
voters who feel the economy is the most important.
a. Find a 90% confidence interval, and state the confidence interval and the margin oferror.
b. What would happen to the confidence interval if the level of confidence were 95%?
16. According to a news release from the Bureau of Labor Statistics (US Department of Labor)
for June, 2021, a household survey states that 8.6% of employed persons that identify as
Hispanic/Latino ethnicity teleworked because of the pandemic. Let’s assume that the thenumber surveyed is 3400. Calculate the following:
i.
ii.
iii.

x=
n=

p̂ =

a. State the estimated distribution of X. X~ ___________
b. What is p̂ estimating?
c. Construct a 92% Confidence Interval for the true proportion of employed person
telework because of the pandemic.
d. How much area is in each tail?
e. Calculate margin of error. .
f. In one complete sentence, explain what the interval means.
g. Using the same p̂ and level of confidence, suppose that n were increased to 5000.
Would the margin of error become larger or smaller? How do you know?
h. Using the same p̂ and n = 3400, how would the margin of error change if thei.

confidence level were increased to 98%? Why?
If you decreased the allowable margin of error, why would the minimum sample size
increase (keeping the same level of confidence)?

17. Among various college baseball pitchers, the standard deviation of heights is known to beapproximately three inches. We wish to construct a 95% confidence interval for the meanheight of college baseball pitchers. Forty-eight baseball pitchers are surveyed. The sample
mean is 72.5 inches. The sample standard deviation is 2.8 inches.
a. Construct a 95% confidence interval for the population mean height of baseballpitchers using the sample standard deviation.
b. Find the critical values associated with this problem.

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c. What will happen to the level of confidence obtained if 200 baseball pitchers weresurveyed?
18. Announcements for 84 upcoming engineering conferences were randomly picked from astack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, witha standard deviation of 1.28 days. Assume the underlying population is normal. Construct a
95% confidence interval for the population mean length of engineering conferences.
19. Suppose that an accounting firm does a study to determine the time needed to complete
one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours.
There is a known standard deviation of 7.0 hours. The population distribution is assumed to
be normal.
a. Which distribution should you use for this problem? Explain your choice.
b. Construct a 90% confidence interval for the population mean time to complete the tax
forms.
c. Find the critical values.
d. Calculate the margin of error.
e. If the firm wished to increase its level of confidence and keep the margin of error thesame by taking another survey, what changes should it make?
f. If the firm did another survey, kept the margin of error the same, and only surveyed
49 people, what would happen to the level of confidence? Why?
g. Suppose that the firm decided that it needed to be at least 96% confident of the
population mean length of time to within one hour. How would the number of people
the firm surveys change? Why?
20. A sample of 16 small bags of the same brand of candies was selected. Assume that the
population distribution of bag weights is normal. The weight of each bag was then recorded.
The mean weight was two ounces with a standard deviation of 0.12 ounces. The population
standard deviation is known to be 0.1 ounce.
a.
b.
c.
d.
e.
f.

Which distribution should you use for this problem? Explain your choice.
Construct a 90% confidence interval for the population mean weight of the candies.
Calculate the margin of error.
Construct a 98% confidence interval for the population mean weight of the candies.
In complete sentences, explain why the confidence interval in part b is smaller thanthe confidence interval in part d.
In complete sentences, give an interpretation of what the interval in part d means.

21. A new coffee house owner is interested in the mean number of disposable cups used in amonth to budget total cost for the year. The owner took a look at 20 different similarcoffeehouses. The mean from the sample is 802 disposable cups used with a sample standard
deviation of 24.8.

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a. Construct a 90% confidence interval for the population mean number of disposable
cups used in a month by a coffeehouse.
b. What will happen to the margin of error and confidence interval if 100 coffee houses
were looked at? Why?
22. MULTIPLE CHOICE: What is meant by the term “90% confident” when constructing a
confidence interval for a mean?
a. If we took repeated samples, approximately 90% of the samples would produce thesame confidence interval.
b. If we took repeated samples, approximately 90% of the confidence intervals calculated
from those samples would contain the sample mean.
c. If we took repeated samples, approximately 90% of the confidence intervals calculated
from those samples would contain the true value of the population mean.
d. If we took repeated samples, the sample mean would equal the population mean inapproximately 90% of the samples.
23. The Federal Election Commission collects information about campaign contributions anddisbursements for candidates and political committees each election cycle. During the 2012
campaign season, there were 1,619 candidates for the House of Representatives across theUnited States who received contributions from individuals. The table below shows the total
receipts from individuals for a random selection of 40 House candidates rounded to thenearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200.
$2,309,200 $7,400$1,243,900 $2,900$10,900$400$391,000 $13,300$467,400 $56,800$632,500 $15,800$468,700 $353,900 $3,714,500 $5,800$733,200 $986,100 $1,109,300 $3,800$953,800 $88,600$1,113,500 $6,600$385,200$5,800$512,900 $75,200 $745,100 $378,200 $3,072,100 $9,500$581,500$3,600$405,200 $41,000 $202,400 $13,200$1,626,700 $8,000
a. Find the point estimate for the population mean.
b. Using 95% confidence, calculate the margin of error.
c. Create a 95% confidence interval for the mean total individual contributions.
d. Interpret the confidence interval in the context of the problem.
24. The American Community Survey (ACS), part of the United States Census Bureau,
conducts a yearly census similar to the one taken every ten years, but with a smaller
percentage of participants. The most recent survey estimates with 90% confidence that themean household income in the U.S. falls between $69,720 and $69,922. Find the point
estimate for mean U.S. household income and the margin of error for mean U.S. householdincome.

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25. The ASA (average speed of answer) is 82 seconds for a month at a call center that helps
employees with computer issues. Let’s say the standard deviation for all calls is 22 seconds.
The manager want to estimate the mean speed of answer within 5 seconds with 93%
confidence to report to the vice president. How many calls must the manager measure?
26. In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack
pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96%
confidence interval for the population proportion of Bam-Bam snack pieces.
a. Calculate the point estimate.
b. Construct a 96% confidence interval for the population proportion of Bam-Bam snack
pieces per bag.
c. Do you think that six packages of fruit snacks yield enough data to give accurate
results? Why or why not?
27. A random survey of enrollment at 35 community colleges across the United States yielded
the following figures:
6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750;
10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768;
7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population isnormal.
a. Construct a 95% confidence interval for the population mean enrollment at
community colleges in the United States.
b. Calculate the margin of error.
c. State the point estimate.
d. State the critical value.
e. What will happen to the margin of error and confidence interval if 500 community
colleges were surveyed? Why?
28. Suppose that a committee is studying whether or not there is waste of time in our judicial
system. It is interested in the mean amount of time individuals waste at the courthouse
waiting to be called for jury duty. The committee randomly surveyed 81 people who recently
served as jurors. The sample mean wait time was eight hours with a sample standard
deviation of four hours.
a. Construct a 95% confidence interval for the population mean time wasted.
b. Find the critical values.
c. Explain in a complete sentence what the confidence interval means.
29. A pharmaceutical company makes tranquilizers. It is assumed that the distribution forthe length of time they last is approximately normal. Researchers in a hospital used the drug
on a random sample of nine patients. The effective period of the tranquilizer for each patient(in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

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a.
b.
c.
d.

Which distribution should you use for this problem? Explain your choice.
Construct a 95% confidence interval for the population mean length of time.
Find the critical values.
What does it mean to be “95% confident” in this problem?

30. Suppose that 14 employees, who were trained to assemble electric switch boards, weresurveyed to determine how long they had to use training module. It was revealed that theyused them an average of 5.2 months with a sample standard deviation of 2.4 months. Assume
that the underlying population distribution is normal.
a. Construct a 99% confidence interval for the population mean length of time using
training modules.
b. Calculate the margin of error.
c. Why would the margin of error change if the confidence level were lowered to 90%?
31. The Teacher’s Retirement System of the State of Illinois (TRS) collects information aboutannual pensions of public employees. A public relations associate is writing an article
responding to critics that teachers have high pensions. This associate wants to state theaverage annual pension amount.
The following table shows the annual pensions of Illinois retired teachers for a random
selection of 30 teachers:$46,500.00$39,050.00$89,500.00$82,025.00$60,521.00$90,845.00$102,000.00$65,700.00$56,500.00$45,800.00$40,966.00$68,000.00$61,810.00$76,530.00$31,500.00$105,100.00$101,557.00$49,970.00$49,000.00$59,459.00$56,175.00$63,520.00$78,258.00$35,810.00$40,000.00$55,363.00$87,933.00$39,148.00$52,578.00$55,061.00

a. Find x = $___________
b. Find sample standard deviation.
c. Use this sample data to construct a 96% confidence interval for the mean amount ofannual pension for retired teachers in Illinois. Use the Student's t-distribution.
32. Forbes magazine published data on the best small firms in 2012. These were firms that
had been publicly traded for at least a year, have a stock price of at least $5 per share, andhave reported annual revenue between $5 million and $1 billion. The table shows the ages of
the corporate CEOs for a random sample of these firms. Use this sample data to construct a
90% confidence interval for the mean age of CEO’s for these top small firms.
48
74
60
47
49

58
63
57
55
67

51
53
46
57
67

61
50
55
43
55
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56
59
63
61
55

59
60
57
62
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33. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants
to estimate its mean number of unoccupied seats per flight over the past year. To accomplish
this, the records of 225 flights are randomly selected and the number of unoccupied seats isnoted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard
deviation is 4.1 seats.
a. Identify the following
i.
𝑥𝑥̅ =
ii.
s =
iii.
n=
b. Which distribution should you use for this problem? Explain your choice.
c. Construct a 92% confidence interval for the population mean number of unoccupiedseats per flight.
34. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard
deviation of $3,156. Assume the underlying distribution is approximately normal.
a. Construct a 95% confidence interval for the population mean cost of a used car.
b. Calculate the margin of error.
c. Explain what a “95% confidence interval” means for this study.
35. Six different national brands of chocolate chip cookies were randomly selected at thesupermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume theunderlying distribution is approximately normal.
a. Construct a 90% confidence interval for the population mean grams of fat per servingof chocolate chip cookies sold in supermarkets.
b. Find the critical value.
c. Calculate the margin of error.
d. If you wanted a smaller margin of error while keeping the same level of confidence,
what should have been changed in the study before it was done?
36. A survey of the mean number of cents off that coupons give was conducted by randomly
surveying one coupon per page from the coupon sections of a recent San Jose Mercury News.
The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢;65¢; 40¢. Assume the underlying distribution is approximately normal.
a. Find
i.
𝑥𝑥̅ =
ii.
s =
iii.
n=
b. Construct a 95% confidence interval for the population mean worth of coupons.
c. Find the critical values
d. Calculate the margin of error.

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e. If many random samples were taken of size 14, what percent of the confidence
intervals constructed should contain the population mean worth of coupons? Explain
why.
37. A quality control specialist for a restaurant chain takes a random sample of size 12 tocheck the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with asample standard deviation of 1.55. Assume the underlying population is normally
distributed.
a. Find the 95% Confidence Interval for the true population mean for the amount of soda
served.
b. What is the margin of error?
38. Insurance companies are interested in knowing the population percent of drivers whoalways buckle up before riding in a car.
a. When designing a study to determine this population proportion, what is theminimum number you would need to survey to be 95% confident that the population
proportion is estimated to within 0.03?
b. If it were later determined that it was important to be more than 95% confident anda new survey was commissioned, how would that affect the minimum number you
would need to survey? Why?
39. Suppose that the insurance companies did do a survey. They randomly surveyed 400
drivers and found that 320 claimed they always buckle up. We are interested in the
population proportion of drivers who claim they always buckle up.
a. Identify
i.
x=
ii.
n=
p̂ =
iii.
b. Construct a 95% confidence interval for the population proportion who claim they
always buckle up.
c. Calculate the margin of error.
d. If this survey were done by telephone, list three difficulties the companies might havein obtaining random results.

40. According to a recent survey of 1,200 people, 61% feel that the president is doing an
acceptable job. We are interested in the population proportion of people who feel the president
is doing an acceptable job.
a. Which distribution should you use for this problem? X ∼ __________________
b. Construct a 90% confidence interval for the population proportion of people who feelthe president is doing an acceptable job.
c. Calculate the margin of error.
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41. An article regarding racial gaps on advanced math courses focusing on 8th grade waspublished in Sage Journals. It was reported that 67% of Asian students (n = 80), 45% of white
students (n = 180), 38% of Latinx students (n = 130), and less than 20% of black students
take algebra I by 8th grade (n = 90). These gaps widen as students advance through highschool. Let’s assume n = 500 eighth graders. Construct a 90% confidence interval for the true
proportion of Latinx students who took algebra I by 8th grade.
a. Construct a 95% confidence interval.
b. Calculate the margin of error.
42. Refer to the information in Exercise 41
a. Construct 95% confidence intervals for the truei.
percentage of all black students who took algebra I by 8th grade.
ii.
percentage of all Asians students who took algebra I by 8th grade.
iii.
percentage of all white students who took algebra I by 8th grade.
b. Even though the three point estimates are different, do any of the confidence intervals
overlap? Which?
c. For any intervals that do overlap, in words, what does this imply about thesignificance of the differences in the true proportions?
d. For any intervals that do not overlap, in words, what does this imply about thesignificance of the differences in the true proportions?
43. Stanford University conducted a study of whether running is healthy for men and womenover age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus
Fitness Association died. We are interested in the proportion of people over 50 who ran anddied in the same eight-year period.
a. Define the random variables X and 𝑝𝑝̂ in words.
b. Construct a 97% confidence interval for the population proportion of people over 50
who ran and died in the same eight–year period.
c. Explain what a “97% confidence interval” means for this study.
44. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine.
One of the questions asked was “What is the main problem facing the country?” Twenty
percent answered “crime.” We are interested in the population proportion of adult Americans
who feel that crime is the main problem.
a. Define the random variables X and 𝑝𝑝̂ in words.
b. Which distribution should you use for this problem? Explain your choice.
c. Construct a 95% confidence interval for the population proportion of adult Americans
who feel that crime is the main problem.
d. Calculate the margin of error.
e. Suppose we want to lower the sampling error. What is one way to accomplish that?

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f.

The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is±3%. In one to three complete sentences, explain what the ±3% represents.

45. Refer to Exercise 44 Another question in the poll was “[How much are] you worried about
the quality of education in our schools?” Sixty-three percent responded “a lot”. We areinterested in the population proportion of adult Americans who are worried a lot about thequality of education in our schools.
a. Construct a 95% confidence interval for the population proportion of adult Americans
who are worried a lot about the quality of education in our schools.
b. Calculate the margin of error.
c. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is±3%. In one to three complete sentences, explain what the ±3% represents.
46. According to a Field Poll, 70% of Illinoisans (actual results are 420 out of 600 registeredvoters surveyed) feel that “property taxes are too high” in Illinois. We wish to construct a 90%
confidence interval for the true proportion of Illinois registered voters who feel that IL
property tax is too high.
a.
b.
c.
d.

Determine the point estimate for the true population proportion.
Find the 92% confidence interval for the population proportion. .
Find the critical value.
Find the margin of error.

47. Five hundred and eleven (511) homes in a certain southern California community are
randomly surveyed to determine if they meet minimal earthquake preparedness
recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum
recommendations for earthquake preparedness, and 338 did not.
a. Find the confidence interval at the 90% Confidence Level for the true population
proportion of southern California community homes meeting at least the minimum
recommendations for earthquake preparedness.
b. Find the point estimate for the population proportion of homes that do not meet theminimum recommendations for earthquake preparedness.
48. In July 2021, Gallup reported that of the 4843 adults surveyed, 18% of Americans saythat they completely or mostly isolate themselves from non-family members due to thepandemic. The confidence level for this study was reported at 95% with a ±2% margin oferror.
a.
b.
c.
d.

Determine the estimated proportion from the sample.
Determine the sample size.
Identify CL and α.
Calculate the margin of error based on the information provided.

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e. Compare the margin of error in part d to the margin of error reported by Gallup.
Explain any differences between the values.
f. Create a confidence interval for the results of this study.
g. A reporter is covering the release of this study for a local news station. How should
the reporter explain the confidence interval to their audience?
49. A national survey of 2175 likely U.S. voters was conducted on June 20 to 24, 2021 by
Rasmussen Reports. It concluded with 95% confidence that 37% to 41% of voters think thecountry is heading in the right direction.
a. Find the point estimate and the margin of error for this confidence interval.
b. Can we conclude with 95% confidence that less than half of all voters believe this?
c. Use the point estimate from part a and n = 2175 to calculate a 75% confidence interval
for the proportion of voters that believe the country is heading in the right direction.
d. Can we (with 75% confidence) conclude that less than half of all voters believe this?
50. Lending Tree recently conducted a survey asking adults across the U.S. about “borrowing”
someone else’s login for streaming services. When asked, 305 of the 586 participants
admitted that used someone else’s login info to use Netflix.
a. Create a 99% confidence interval for the true proportion of American adults who haveused someone else’s login for Netflix.
b. This survey was conducted through online surveying method feb. 11 – 16th, 2021. The
margin of error of the survey compensates for sampling error, or natural variability
among samples. List some factors that could affect the survey’s outcome that are notcovered by the margin of error.
c. Describe how the confidence interval would change if the CL changed from 99% to90%.
51. You plan to conduct a survey on your college campus to learn about the politicalawareness of students. You want to estimate the true proportion of college students on yourcampus who voted in the 2020 presidential election with 95% confidence and a margin oferror no greater than five percent. How many students must you interview?
52. According to College Pulse study where 2000 undergraduates were recorded in the surveyof 15 questions. 38% of students strongly agreed with the statement “Higher education has arole to play in racial justice and racial equality in the US” while 7% strongly disagree. Find
98% confidence intervals for true percentage of student who strongly agree with thestatement.
53. In a June 2021 survey by Insider, “52% Americans want all student-loan borrowers tohave their debt forgiven”. A random sample of 3600 adults through GoBankingRates –personal finance portal.

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a.) Find the 90% confidence interval for the true proportion of Americans that want
all student loan borrowers to have their debt forgiven.
b.) Let’s say the full sample has a margin of error of plus/minus 3.6 percentage points.
If we want a margin of error of 3 percentage points with 95% confidence, how large ofa sample should be used?
54. According to a survey conducted by Bankrate, which surveyed 2,194 adults, including
1,330 homeowners. It was seen that 29 percent of respondents with a mortgage either didn't
know their rate or wouldn't say.
a. State the point estimate as a fraction.
b. Find the 98% confidence of the true proportion of adults with a mortgage either
don’t know their rate or wouldn’t say.
c. interpret the confidence in sentence form
Section 9.5 Exercises
55. Find the right and left critical values for chi-square distribution for 98% confidence
interval and degrees of freedom is 20.
56. Find the right and left critical values for chi-square distribution for 90% confidence
interval and degrees of freedom is 26.
57. Find the right and left critical values for chi-square distribution for 95% confidence
interval and degrees of freedom is 12.
58. Find the right and left critical values for chi-square distribution for 99% confidence
interval and degrees of freedom is 9.
59. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants
to estimate its mean number of unoccupied seats per flight over the past year. To accomplish
this, the records of 25 flights are randomly selected and the number of unoccupied seats isnoted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard
deviation is 4.1 seats.
a. Identify the following
i. 𝑥𝑥̅ =ii. s =
iii. n =b. Construct a 90% confidence interval for the population standard deviation of thenumber of unoccupied seats per flight.

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60. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard
deviation of $3,156. Construct a 95% confidence interval for the population variance
61. Six different national brands of chocolate chip cookies were randomly selected at thesupermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Construct a 90%
confidence interval for the population variance
62. A survey of the mean number of cents off that coupons give was conducted by randomly
surveying one coupon per page from the coupon sections of a recent San Jose Mercury News.
The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢;65¢; 40¢. Assume the underlying distribution is approximately normal.
a. Find s =b. Construct a 95% confidence interval for the population standard deviation worth of
coupons.
63. A quality control specialist for a restaurant chain takes a random sample of size 12 tocheck the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with asample standard deviation of 1.55. Assume the underlying population is normally
distributed. Find the 95% Confidence Interval for the true population standard deviation for
the amount of soda served.

445

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“Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at
http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July2, 2013).

9.3 Confidence interval about a single population proportion
Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling.
Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013).
Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, andMeredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at
http://www.pewinternet.org/Reports/2013/Teens-Social- Media-And-Privacy.aspx (accessed July 2,2013).
Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.”
Pew Research Center: Internet and American Life Project. Available online at
http://www.pewinternet.org/~/media//Files/Questionnaire/2013/
Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013).
Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they
will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at
http://www.gallup.com/poll/162758/three-four- workers-plan-work-past-retirement-age.aspx(accessed July 2, 2013).
The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2,2013).
Martin, Emmie. “30% Of Homeowners Are Making a Mistake That Could Cost Them Thousands.”
CNBC, CNBC, 1 May 2018, www.cnbc.com/2018/05/01/a-third-of-homeowners-dont-know-theirmortgage-rate.html?__source=twitter%7Cmain. (accessed July 17, 2018)“Concern over COVID Spread Hits the Lowest in June.” Pharmaceutical Technology, Verdict Media
Limited, July 2021, www.pharmaceutical-technology.com/news/concern-over-covid-19-spreadhits-the-lowest-in-june-since-the-start-of-the-pandemic-poll/. (accessed July 9, 2021)
Irizarry, Yasmiyn. “On Track or Derailed? Race, Advanced Math, and the Transition to High School Yasmiyn Irizarry, 2021.” SAGE Journals, SAGE Journals, 11 Jan. 2021,
journals.sagepub.com/doi/full/10.1177/2378023120980293. (accessed July 9, 2021)
Saad, Lydia. “Strict Social Distancing in the U.S. Dwindles to 18%.” Gallup.com, Gallup, 7 July
2021, news.gallup.com/poll/352166/strict-social-distancing-dwindles.aspx. (accessed July 11,
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ack_jul5 (accessed July 11, 2021).
Papandrea, Dawn. “Nearly 4 in 10 Americans Are Mooching Off Someone Else's Streaming Account.”
LendingTree, 11 Mar. 2021, www.lendingtree.com/americans-mooching-streaming-accounts/(accessed July 11, 2021).
Sheffey, Ayelet. “52% Of Americans Want All Student-Loan Borrowers to Have Their Debt Forgiven,
New Survey Finds.” Business Insider, Business Insider, 21 June 2021,
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Ezarik, Maria. What Students Think about Racial Justice Efforts on Campus, 6 May 2021,
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448

CHAPTER 9 SOLUTIONS:
1) b. 1 – α = .94, α = .06, α/2 = .03; The critical values are ±Zα/2 = ±invnorm(.03) = ±1.88.
3) 1 – α = .98, α = .02, α/2 = .01; tα/2 = 2.55; Margin of error is: E = tα 2

s
= .46
n

5) a. 𝑥𝑥̅ = 8.2, σ =2.2, n = 200
b. Using Zinterval: (7.944, 8.456); ±zα/2 = ±invnorm(.05) = ±1.645; E = .26 minutesc. Increase nd. The level of confidence would decrease. The smaller the n, the larger the E but since
E is staying the same, the confidence level would decrease.
e. The larger the confidence level, the larger n should be.
7). a. 𝑥𝑥̅ = 30.4, σ =15, n = 25
b. Zinterval: (24.52, 36.28)
c. µ d. α = .05
e. E = 36.28 – 30.4 = 5.88 years
f. We are 95% confident that the true mean age for all Winger Foothill College
students is between 24.52 and 36.28
g. The margin of error becomes smaller because as n increases the denominator in theformula for E formula becomes larger.
h. The margin of error for the mean would decrease because as the CL decreases, youneed less area under the normal curve (which translates into a smaller interval) to
capture the true population mean
9) a. (1.38, 1.62); ±tc = ±1.995; E = 1.62 – 1.5 = .12
b. 95% confident that the population mean for wait time is between 1.38 hours and 1.62
hours.
11) a. 𝑥𝑥̅ = 3.26; s = 1.02, n = 39
c. Tinterval (2.93, 3.59)

b. average number of colors on national flags

d. 0.05e.

f. 95% confidence that the population mean is between 2.93 and 3.59 colors for nationalflags. g. Smaller, more accurate h. Smaller less confidence13.) a. x = 120, n = 200, 𝑝𝑝̂ =

120
= .6
200

b. X number of households where women makethe majority of the purchasing decisions. 𝑝𝑝̂ is the proportion of household where womenmake the majority of the purchasing decisions. c. X ~ N(.6, .035)
d. 1propZint:(.532, .668), E = .068 e. answers vary44915) a. 1propZint .627 < p < .673, E = .023

b. .623 < p < .677

17) a. Tinterval 71.69 < µ < 73.31 b. ±tc = ± 2.012
c. the level of confidence
would stay same but the interval would become closer together.
19) a. Z-distribution since σ is known b. Zinterval (22.45, 24.75)
d. E = 1.15 hours

e. Larger n21) a. Tinterval (792.41, 811.59)
narrower.

f. Decreasec. ±Zc = ±1.645

g. n = 206

b. Smaller margin of error, confidence interval is23) a. 𝑥𝑥̅ = $568,872.5 b. E = 281,758.5
c. ($287114, $850631)
d. The population
average of campaign contributions is between $287,114 and $850, 631
1.81∗22 2
� = 64 calls
525) 𝑛𝑛 = �27) a. Tinterval (6243.4, 11014)

b. E = 2385.3 c. 𝑥𝑥̅ = 8628.7 d. ±tc = ±2.03

e. margin of error is smaller and confidence interval is closer together.

29) a. T distribution because sigma is unknown b. Tinterval (2.27, 2.76) c. ±tc = ±2.31
d. 95% confidence that the population mean lies within 2.27 hours and 2.76 hours.
31) a. 𝑥𝑥̅ = $62,872.63

b. s = $21,050.45

33) a. 𝑥𝑥̅ = 11.6, s = 4.1, n = 225

c. Tinterval (54,608, 71,137)

b. T distribution because sigma is unknown

c. Tinterval (11.12, 12.08)35) a. Tinterval (7.64, 9.36) b. ±tc = ± 2.015

c. E = .86

37) a. Tinterval (12.32, 14.29)

b. E = .99 oz

39) a. x = 320, n = 400, 𝑝𝑝̂ = .80
variesb. 1propZint (.761, .839)

d. n is larger

c. E = 3.92% d. answer41) a. 1propZint (.294, .46) b. E = .08
43) a. X is the number of people over 50 who ran and died, 𝑝𝑝̂ is the sample proportion ofpeople over 50 who ran and died b. 1propZint (.00289, .0282) c. 97% chance the true
proportion of those over 50 who ran and die is between .2% and 2.8%.
45) a. 1propZint 60% < p < 66%
b. E = 3%
c. ±3% represents the margin oferror and that the true proportion of adult Americans who are worried a lot about thequality of education in our schools is between 60% and 66%.

45047) a. 1propZint (.304, .373)
49) a. 𝑝𝑝̂ = .39 E = .02

b. 1 – 𝑝𝑝̂ = .661

b. Yes, because the entire interval range is below 50%

c. 1propZint (.378, .402)

d. yes51) n = 385
53) a. 1propZint (.506, .534)
55) χ L2 = 8.2604

χ R2 = 37.5662

57) χ L2 = 4.4038

χ R2 = 23.3367

59) a. 𝑥𝑥̅ = 11.6, s = 4.1, n = 25

b. n = 1066b.

(24)4.12(24)4.12<σ < 3.33 < σ < 5.4
36.41513.84861) s = 1.1, n = 6(5)1.1(5)1.1<σ < .497 < σ2 < 4.801
11.0705
1.145863) s = 1.55, n = 12

(11)1.552(11)1.552<σ < 1.098 < σ < 2.632
21.92
3.816

451

This page is purposely left blank.

452

10 | HYPOTHESIS TESTING WITHONE SAMPLE

Figure 10.1 Scientist doing a PCR (Polymerase Chain Reaction) in electrophoresis gel. This technique isoften used for genetic and DNA testing. Photo by National Cancer Institute on Unsplash

Chapter Objectives
By the end of this chapter, the student should be able to:
•
•
•
•
•

Describe hypothesis testing in general and in practice.
Conduct and interpret hypothesis tests for a single population mean when the
population standard deviation is known.
Conduct and interpret hypothesis tests for a single population mean when the
population standard deviation is unknown.
Conduct and interpret hypothesis tests for a single population proportion.
Differentiate between Type I and Type II Errors

453

Introduction
One job of a statistician is to make statistical inferences about populations based on samples
taken from the population. In the last chapter, we used confidence intervals to estimate an
unknown population parameter. In this chapter we will use statistical inference to make adecision about a parameter; that is, given two competing claims about an unknown
parameter, we will decide which of them is more plausible. The method for doing this iscalled hypothesis testing. We already have all of the tools needed to implement this method– and we will shortly distill the process into a straightforward five-step method. We will
illustrate the basic ideas with an example:
The ACME chemical corporation makes a fabric cleaner; the company claims that thisproduct successfully removes 90% of all stains. A consumer organization questions this claim
and decides to test the product. They select a random sample of 100 stained garments andapply the product to each. They find that the stain remover successfully cleaned only 78 ofthem. Does this sample data provide evidence that the company is overstating theeffectiveness of their product? That is, does the sample data provide evidence that the successrate for the fabric cleaner is less than 90%?
We let p denote the success rate for the cleaner – that is, this is the proportion of all stains
in the population that will be removed by the product. Now we see that we have twocompeting claims:••

ACME claims that p = 0.90
The consumer organization claims that p < 0.90.

Which one of these claims is true? Of course, there is no way to determine p explicitly, but
we could use the methods of Chapter 8, along with the given sample data, to compute a
confidence interval for p. Doing this gives us a 95% confidence interval of (0.699, 0.861).
That is, we are 95% confident that 69.9% < p < 86.1%. Since the upper bound of the interval
is well under 90%, the sample data does indeed provide evidence that p is less than 90%.
But there is another way we could go about this. Note that if the company’s claim is true,
then we would expect to get about 90 successes in a sample of 100. But the number ofsuccesses in 100 trials will change from sample to sample, so the fact that we got fewer than
90 does not in itself provide evidence that the success rate is less than 90%. ACME could
claim that the discrepancy between their claimed success rate and the sample data is simply
due to sample variation. This is where probability comes in. Let us give the company thebenefit of the doubt. Let us assume that the success rate really is 90% and calculate theprobability of getting a sample of 100 garments with a success rate of 78% or less (a sample
proportion less than 78% would also support the consumer organization’s claim). In Chapter
7, we saw that if n = 100 and p = .90, then the sampling proportion for p̂ is approximately454

normal, with mean µ pˆ = p = .90 and standard deviation σ pˆ =Using the calculator, we see that the probability we want is:p(1 − p).90(.10)== 0.03.
n100

P(p̂ < 0.78) = normalcdf(-10^99, .78, .90, .03) = 0.000031.
So, if ACME’s claim really is true, then getting 78 or fewer successes in a sample of 100
garments would be highly unlikely (only about 3 times for every 100,000 samples).
However, the consumer group really did get a sample of 100 with only 78 successes! Thismeans that there are two possibilities: The success rate really is 90%, and a very rare event
occurred, or else the assumption that p = .90 is not correct. Given how small the probability
we calculated was, the second option seems to be more plausible. In other words, the sample
data is not compatible with ACME’s claim about the success rate of their product; so we would
reject the assumption that the success rate is 90%. Thus, this sample data provides evidence
that ACME was overstating the success rate of their product.
The line of reasoning we used here is known as hypothesis testing, and it is a widely usedmethod of statistical inference. In general, a hypothesis test will always consist of twocontradictory hypotheses or statements, a decision based on sample data, and a conclusion.
Note that in all of our problems, we will be given either the sample data or the summary
statistics. Reviewing our example, we see that a hypothesis test will involve the following
five steps:
1. Set up two contradictory hypotheses, which we call the null hypothesis and thealternative hypothesis.
2. Determine the correct sampling distribution to perform the calculations.
3. Assuming that the null hypothesis is true, we calculate the probability of getting
sample data like the data we have actually observed.
4. If this probability is sufficiently small, we reject the null hypothesis.
5. Interpret the decision to write a meaningful conclusion; i.e. interpret the decision toanswer the original question.
These five steps comprise what is called the Five Step Method of hypothesis testing.
In this chapter, we develop hypothesis tests on single means and single proportions. We will
also learn about the errors associated with these tests. In subsequent chapters, we learn
many other tests to apply these ideas to other situations – but every test we learn will follow
the Five Step Method.

455

10.1 | Elements of a Hypothesis Test
Null and Alternative Hypotheses
Every hypothesis test begins by considering two hypotheses. These are called the null
hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints, andalmost always are stated in terms of one or more unknown population parameters.
Population Parameters
Mean μ
Proportion ρStandard Deviation σ

The null hypothesis is denoted by H0. It is a statement about the parameter that either isbelieved to be true or is used to put forth an argument unless it can be shown to be incorrect
beyond a reasonable doubt. Thus, all of our calculations will be done using the assumptionthat the null hypothesis is true.
The alternative hypothesis is denoted by Ha. It is a claim about the parameter that iscontradictory to H0. The alternative Ha is also sometimes called the “research hypothesis”.
Since the hypotheses are contradictory, only one of them can be true; thus, if our sample data
leads us to reject H0, then we will have statistical evidence that Ha is true. And we only will
reject H0 when the sample data provides compelling evidence that it is false (this will be
discussed in greater detail in the next section).
Based on calculations using sample data, we will make a decision about H0. There are only
two possible options for a decision: They are:••

Reject H0
Do not reject H0

If the decision is to reject H0, then this means that the sample data is incompatible with theassumption that H0 is true. When the decision is to reject H0 then the sample data provides
evidence that Ha is true. On the other hand, if the decision is “do reject H0", then the sample
data simply does not provide enough evidence to reject the null hypothesis.
By its very nature, H0 asserts that the value of the parameter is known; thus, H0 will always
include an equal sign; so the null hypothesis will always be a statement involving either “=”,“<” or “>”.

456

On the other hand, Ha will never have a symbol with an equal in it; so Ha will be a statement
involving either “≠”, “>” or “<”. The choice of symbol depends on the claim being tested.
(Note that some practitioners write H0 as a simple statement of equality, even with > or < asthe symbol in the alternative hypothesis. This practice is also acceptable.)

Basic Rules for Setting up the Hypotheses
1. The null hypothesis must include an equal sign.
2. One of the hypotheses must represent the claim being tested.
3. The hypotheses should be contradictory statements.

Example 10.1
We wish to test the claim that more than 30% of registered voters in a certain county voted
in the primary election. Find the hypotheses for this test.
Solution 10.1
The claim here concerns a population proportion p, the proportion of all registered voters in
the county that voted in the primary. The claim is that more than 30% voted, or p > 0.30.
This claim must be represented by either H0 or Ha; since the claim does not include an equal
sign, it will be Ha. Remember that H0 and Ha are opposite of each other. Thus we get the
hypotheses:
H0: p ≤ 0.30 ,Ha: p > 0.30

10.1 A medical trial is conducted to test whether or not a new medicine reduces cholesterolby at least 25%. State the null and alternative hypotheses for the test.

457

Example 10.2
We want to test whether the mean GPA of students in American colleges is different from2.0 (out of 4.0). State the hypotheses for this test.

Solution 10.2
Here the claim being tested involves a population mean, μ; and the claim states that themean GPA is different from 2.0. So, the claim is that μ ≠ 2.0. Since this is a “not equal”
statement, it must be represented by Ha, and H0 will be the opposing statement. So, thehypotheses are:H0: μ = 2.0

vs. Ha: μ ≠ 2.0

10.2 We want to test whether the mean height of eighth graders is 66 inches. State the nulland alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null andalternative hypotheses:
a. H0: μ ____ 66

b. Ha: μ _____ 66

Example 10.3
Suppose that we want to test whether the average time needed to complete a bachelor’s
degree in the U.S. is less than five years.
a. State the hypotheses for the test.
b. Suppose that sample data has been collected and analyzed, and the decision is toreject H0. Interpret this decision to write a meaningful conclusion for this test.
Solution 10.3
a. The claim is that the population mean μ is less than five years; i.e. the claim is μ < 5.
This must be represented by the alternative hypothesis, and so the null andalternative hypotheses are:H0: μ ≥ 5 vs. Ha: μ < 5.
b. Since the decision is to reject H0, there is evidence that H0 is false. That is, there isevidence that Ha is true. Thus, there is significant evidence that the average timeneeded to complete a bachelor’s degree in the U.S. is less than five years.

458

10.3 We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the nulland alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null andalternative hypotheses.
a. H0: μ ___ 45

b. Ha: μ ____ 45

Example 10.4
In an issue of U.S. News and World Report, an article on school standards stated that about
half of all students in France, Germany, and Israel take advanced placement exams and of
those, a third pass. The same article stated that 6.6% of U.S. students take advanced
placement exams and 4.4% pass. Suppose that a test is conducted to test whether thepercentage of U.S. students who take advanced placement exams is more than 6.6%.
a. State the null and alternative hypotheses for the test.
b. Suppose that sample data has been collected and analyzed, and the decision is to not
reject H0. Interpret this decision to write a meaningful conclusion for this test.
Solution 10.4
a. The claim is that more than 6.6% pass; i.e. that p > 0.066. This is a strict inequality,
so it will be represented by Ha, and we get the hypotheses: H0: p ≤ 0.066 vs. Ha:
p > 0.066
b. Since the decision is to not reject H0, this means that there is not enough evidence tosupport Ha. Thus, there is not enough evidence to conclude that the percentage of
U.S. students who take advanced placement exams is more than 6.6%.

10.4 On a state driver’s test, about 40% pass the test on the first try. We want to test theclaim that more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for
the null and alternative hypotheses:
a. H0: p ____ 0.40

b. Ha: p _____ 0.40

459

The p-value and Significance Level for a Test
Recall the Five Step Method for Hypothesis Testing:
1. Set up the null hypothesis and the alternative hypothesis.
2. Select the correct test; that is, determine the correct sampling distribution to
perform the calculations.
3. Assume that the null hypothesis is true and calculate the probability of getting asample statistic that differs at least as much from the hypothesized mean as thestatistic computed from our sample data.
4. If this probability is sufficiently small, we reject the null hypothesis.
5. Interpret the decision to write a meaningful conclusion, i.e. interpret the decision toanswer the original question.
We have discussed at some length Steps 1 and 5 of the process; now we focus on Steps 3 and4. The probability calculated in Step 3 is called the p-value for the test. To compute thisvalue, we assume that the null hypothesis is true, and then find the probability of getting asample statistic that is at least as extreme as the statistic from our sample. Thus, the smaller
the p-value is, the stronger the evidence against H0.
For any test, the p-value will be calculated in an appropriate sampling distribution. For tests
about a single population mean or population proportion, we can use the techniques learnedin Chapter 8. For example, suppose that we want to test the claim that the average timespent to complete a bachelor’s degree is less than five years. As we saw in Example 10.3, thehypotheses for this test are:H0: μ ≥ 5 vs. Ha: μ < 5Further suppose that we collected a sample and found that x = 3.5. Then the p-value would
be the probability p-value = P( x < 5). Graphically, this probability is the area under thenormal curve that is to the left of x = 3.5. For this reason, a test in which the alternative
hypothesis is a “less than” statement is called a left-tailed test.

460

Similarly, if we were testing the hypotheses H0: μ < 65 vs. Ha: μ > 65, and the sample data
gave us x = 67, then the p-value would be the probability p = P( x > 67). Graphically, thisprobability is the area under the normal curve that is to the right of x = 67. So, a test inwhich the alternative hypothesis is a “greater than” statement is called a right-tailed test.

Finally, if the alternative hypothesis is a “not equal” statement, then the p-value will be thearea in both tails. And if the sampling distribution is symmetric, then the p-value will be
split evenly between the two tails. E.g. suppose we test the hypotheses:
H0: μ = 50

Ha: μ ≠ 50

Then the graphical representation of the p-value would be as follows:

Tests in which the alternative hypothesis is a “not equal” statement are called two-tailed
tests.
Once we have calculated the p-value, we still need to use it to decide whether or not to reject
the null hypothesis. According to Step 4 of the Five Step Method, we will reject the null
hypothesis H0 only if p is “sufficiently small”; and the criterion for this is provided by anumerical value called the significance level of the test.
Before conducting a hypothesis test, the researcher will decide on a maximum allowableprobability for a Type I error, which we denote as α. A Type 1 error occurs when the decisionat the end of the hypothesis test is to reject H0 when in fact the H0 is true. This value is a
threshold value: if the p-value is less than α, then we will reject H0.
The significance level chosen will depend on the consequences of making a Type I error. For
example, in the social sciences, it is not uncommon to use a significance level of α = .10,
whereas in the pharmaceutical industry the standard level of significance for drug testing is
α = .01. The most widely used level is α = .05.

461

Rejection Rule Using the p-value•

We reject H0 at significance level α if p < α.

•

If p > α, then we do not reject H0.

NOTE: We do not use the phrase “accept H0” if the conclusion is “do not reject H0”.
Example 10.5
It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born.
We let p be the proportion of male babies born to It’s A Boy clients. Suppose that we test the
hypotheses using a significance level of α = 0.01.
H0: p = 0.50,

Ha: p > 0.50

a. Is this test left-tailed, right-tailed, or two-tailed? Explain.
b. Suppose further that the sample data yields a p-value = 0.025. What would be thedecision for H0?
c. Interpret this decision to state a conclusion in simple, non-technical terms.
Solution 10.5
a. Since the alternative hypothesis is a “>” statement, this is a right-tailed test.
b. Since p – value = .025 > .01, we do not reject H0.
c. The alternative hypothesis represents the claim that the company improves a
couple’s chances of having a boy. Since we do not reject H0, there is not enoughevidence to conclude that Ha is true. Thus, there is not enough evidence tosupport the company’s claim that their procedures improve the chances ofa boy being born.
Non-technical terms (interpretation) of the conclusion
If claim is H0 and Reject H0, then “there is enough evidence at ___ % level of significance toreject the claim that …
If claim is Ha and Reject H0, then “there is enough evidence at ____% level of significance tosupport the claim that …

462

10.5 A certain computer component is known to have an average time of 520 hours between
failures. A modification is made to the component that is supposed to extend the time
between failures. To test that the modification is successful, a researcher samples 100 of
the components and uses the data to test the hypotheses H0: μ < 520, Ha: μ > 520 using a
0.05 significance level.
a. Is this test left-tailed, right-tailed or two-tailed? Explain.
b. Suppose further that the sample data yields a p-value of p =0.0141. What would be
the decision for H0?
c. Interpret this decision to state a conclusion in simple, non-technical terms.

10.2 | The Z-Test for a Single Population Mean using p-value
Suppose that we wish to test a claim about an unknown population mean, μ. The appropriate
point estimate for μ is the sample mean, x . From the Central Limit Theorem, we know thatthe sampling distribution for x will be approximately normal provided either the population
distribution is normal, or if the sample size is n > 30. Moreover, if we assume that the
population standard deviation σ is known, then the mean and standard deviation of thesampling distribution are:µx = µand σ x =σ.
nWhen these conditions are met, we can use the normal distribution to find the p-value forthe test. And such a test is referred to as a Z-Test. We will illustrate this method with an
example:Example 10.6
A baker claims that his bread height is more than 15 cm, on average. Several of his
customers do not believe him. To persuade his customers that he is right, the baker decides
to do a hypothesis test. The baker knows from baking hundreds of loaves of bread that thestandard deviation for the height is 1.5 cm. and the distribution of heights is normal. He
makes 30 loaves of bread and finds that the mean height for the sample is 17 cm. Using a.05 significance level, does this sample data provide evidence to that the mean bread heightis more than 15 cm?
Solution 10.6
The claim being tested is that the mean is more than 15 cm, or μ > 15. Since this is a strict
inequality, this claim will be represented by Ha. The null hypothesis must contradict thealternate hypothesis, so we get the hypotheses:
H0: μ ≤ 15 and Ha: μ > 15463

We are sampling from a normal distribution and the population SD is known (σ = 1.5 cm.),1.5= 0.27386.
so the sampling distribution for x is normal with mean µ x = 15 and σ x =30
Next, we calculate the p-value for the test. Suppose the null hypothesis is true: that is,
assume that the mean height of the loaves is no more than 15 cm. Using this assumption,
we find the probability of observing a sample mean that is greater than or equal to 17; since
the sampling distribution is normal, we can use any of the methods from Chapter 7 to do thecalculation. For example, we can use the normalcdf function in the TI-84 calculator:
p = P( x > 17) = normalcdf(17, 10^99, 15,

1.530) = 1.42 x 10-13

In other words, if the population mean really is μ = 15, then the probability of selecting asample of 10 with x > 17 is p = 0.00000000000014, which is virtually zero.

Since p < .05 we reject H0. In fact, with such a small p-value, we would have rejected thenull hypothesis at any level of significance. A p-value of approximately zero tells us that, if
the population mean height really been 15 cm, it would be virtually impossible to get a sample
mean of 17 cm. purely by chance. Since the outcome of 17 cm. is so unlikely, its occurrence
provides strong evidence against the null hypothesis. There is sufficient evidence that thetrue mean height for the population of the baker's loaves of bread is greater than 15 cm.

10.6 A machine at a bottling plant is programmed to fill bottles to an average volume of 12.2
oz. The volumes are normally distributed with a standard deviation of σ = 0.6 oz. The line
manager suspects that the machine may be overfilling the bottles, and so he will test theclaim that the mean is greater than 12.2. A sample of 36 is selected; the sample mean is 12.5
oz, and the p-value is calculated to be p = 0.0013. What can we conclude from this test?
Note that in our example, we will follow the Five Step Method using p-value. The steps
are listed in the box below. This method will be used for the different types of hypothesis
tests in the chapter.

464

Five Step Method for Hypothesis Testing using p-value
Set up the hypotheses H0 and Ha.
Select the correct test and identify the significance level α.
Calculate the p-value.
Use these results to make the decision about H0.
In particular, if p < α, then we reject H0.
5. Interpret the decision about H0 in the context of the given problem to state aconclusion.

1.
2.
3.
4.

The Five Step Method streamlines and organizes our work by providing a template for
performing a hypothesis test. Note that of the five steps involved, only Step 3 requires any
calculation; and these calculations are programmed into the TI-83 and TI-84 calculators:
Using the TI-83, 83+, 84, 84+ Calculator_

To calculate the p-value (and test statistic) for a Z-Test, go to the STAT menu, and thenscroll over to the TESTS menu.
•

If we are given summary statistics σ, x and n, then we select the Stats option.
Enter the mean hypothesized in H0 for μ0; enter the values for σ, x and n.
In the row marked as “μ: ”, scroll over to the option that matches the alternative Ha.
Put the cursor on Calculate, and press Enter. The p-value and the test statistic z
will be displayed.
If you instead put the cursor on Draw, the calculator will show a graph of thenormal distribution, with the area corresponding to the p-value shaded; this option
also shows the test statistic z and p-value.

•

If we are given raw data, go to the EDIT menu and enter the data into a list.
Enter the hypothesized mean in H0 for μ0, the value for σ, and the list name(e.g. press 2nd 1 for list L1). In the row marked as “μ: ”, scroll over to the optionthat matches the alternative Ha. Press Calculate and Enter to see the p-value andthe test statistic.

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Example 10.7
Jeffrey, as an eight-year-old, established a mean time of 16.43 seconds for swimming the 25yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey
could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of
expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims,
Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim
faster than the 16.43 seconds. Conduct a hypothesis test using a significance level of α = 0.05.
Assume that the swim times for the 25-yard freestyle are normal.
Solution 10.7
We will follow the Five Step Method:(Step 1) Set up the hypotheses: This is a test of a single population mean; and claim isthat Jeffrey swims faster with the new goggles. For Jeffrey to swim faster, his time will beless than 16.43 seconds; the claim is that μ < 16.43. This claim will be the alternative Ha,
and we have:
H0: μ ≥ 16.43

Ha: μ < 16.43 (claim)

The "<" in the alternative means that this is a left-tailed test.
(Step 2) Select the correct test and identify the significance level α. Since the swim times
are normally distributed, and the population standard deviation is known (σ = 0.8), we use
a Z-Test. The significance level α is 0.05.
(Step 3) Calculate the p-value. In the TI-84, go to the STAT menu and then the TESTS
menu; select Z-Test (item 1). Enter the following:
μ0: 16.43(this is the mean hypothesized in H0)
σ: 0.8x : 16n : 15
μ : < μ0
Put the cursor on Calculate and press Enter to get the p-value = 0.0187.
The graph looks like:

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(Step 4) Use these results to make the decision about H0. Since p-value < α, we reject H0.
Note that if H0 is true, there is a 0.0187 probability that Jeffrey's mean time to swim the
25-yard freestyle will 16 seconds or less. Because a 1.87% chance is small, the mean time of
16 seconds or less is unlikely to have happened randomly. It is a rare event.
(Step 5) Interpret the decision about H0 in the context of the given problem to state aconclusion. At the 5% significance level, we conclude that Jeffrey swims faster using the
new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time toswim the 25-yard freestyle is less than 16.43 seconds.

10.7 The mean throwing distance of a football for Marco, a high school freshman quarterback,is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his
grip to more distance. The coach records the distances for 20 throws. For the 20 throws,
Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco getthrow farther than 40 yards. Test the coach’s claim using a significance level of α = 0.05.
Assume that the throwing distances for footballs are normally distributed, with a standard
deviation of σ = 10 yards.

10.3 | T-Test for a Single Population Mean using p-value
In the preceding section, we tested claims involving a population mean μ. Throughout that
section we assumed that we knew the population SD, σ. However, in actual applications, it
is often the case that σ is unknown; and in these cases, it is necessary to estimate σ by the
sample standard deviation, s.
If we are sampling from a normal distribution, the sampling distribution for x will beapproximately normal. But for a small sample, the sample standard deviation may be a poor
estimate for σ. To compensate for this potential error, we use the t-distribution introducedin Chapter 9 to calculate our p-values and critical values for our hypothesis tests. Recall that
this distribution is bell-shaped like the normal distribution, but with a larger variance.
Moreover, there is a different t-distribution for every sample size, and the smaller the sample
size the larger the variance. Finally, as the sample size gets very large, the t-distribution
approaches the normal distribution.
A test for a population mean μ that is performed under these conditions (i.e. in which σ isunknown) is called a T-Test. We start with an example:

467

Example 10.8
Statistics students believe that the mean score on the first statistics test is 65. A statistics
instructor thinks the mean score is higher than 65. To test his claim, he selects a random
sample of ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71.
He uses this data to conduct a hypothesis test using a 5% level of significance. The data are
assumed to be from a normal distribution. Does this sample data provide evidence that themean score for all students is more than 65?
Solution 10.8
(Step 1) This is a test of a single population mean. Since the instructor thinks the averagescore is more than 65, his claim is represented by the alternative hypothesis, and we have:
H0: μ ≤ 65

Ha: μ > 65

(Step 2) A 5% level of significance means that α = 0.05. If you read the problem carefully,
you will notice that there is no population standard deviation given. You are only given n =10 sample data values. Notice also that the data come from a normal distribution. This meansthat we must use a t-test.
(Step 3) Therefore, the distribution for the test is t9 where n = 10 and so df = 10 - 1 = 9. The
mean of the distribution will be µ x = µ = 65. We do not have an explicit formula for the
standard deviation of the sampling distribution as we did for normal distributions. But thep-value can easily be calculated using the TI-84:

Using the TI-83, 83+, 84, 84+ Calculator_

To calculate the p-value (and test statistic) for a T-Test, follow the same basic instructions
as for a Z-Test, but in the TESTS menu we will select T-Test.
•

Go the STAT menu, and then to the TESTS menu; select T-Test (option 2). In thisproblem, we are given raw data, so we first go to the STAT menu, select the Editmenu, and press ENTER. Next, we enter the 10 data values into a list, say L1.

•

Press STAT, arrow over to the TESTS menu, and select T-Test. For Inpt:, select theData option, and press ENTER. Then, enter the following:
μ0 : 65 (This is the value of μ that is hypothesized in H0)
List: L1 (The name of the list where the data is entered)
Freq: 1 (This will be 1 unless we are given the data as a frequency distribution)
μ : > μ0
(This is the alternative hypothesis)

468

•

Arrow down to Calculate and press ENTER.
The display screen will show the p-value, p = 0.0396.
It is useful to have a graphical representation of this p-value; this is a right-tailed
test, so the p-value is P( x > 67), the area to the right of the observed sample mean:(Step 4) Next we make a decision about H0:
Since p –value = 0.0396 < 0.05 = α, the decision is to reject H0.
(Step 5) And since we reject μ = 65, we have sufficient evidence to conclude that Ha is true.
At a 5% level of significance, the sample data provide sufficient evidence that the mean testscore is more than 65.
NOTES:•

If we select Draw instead of Calculate, the calculator will graph the area
corresponding to the p-value; the p-value and test statistic and will still appear on
this screen.

•

The output screen also shows the alternative hypothesis, μ > 65 at the top. It isalways a good idea to double check that the correct alternative hypothesis was
selected.

•

The output screen also displays the sample mean, sample standard deviation andsample size: x = 67, s = 3.1972, and n = 10, respectively. One of the advantages ofusing the data option is that the calculator will automatically calculate the key
sample statistics for us. We just need to enter the data!

Before starting another example, we will take this opportunity to show another way to
calculate the p-value. First, we can quickly use 1-VarStats to calculate the sample mean andsample standard deviation. Then we calculate the t-statistic using the formula:t=x − µ0s/ n=

67 − 65
3.1972 / 10

= 1.978 .

This is a right-tailed test, so the p-value = P( x > 67); and this probability can be rewritten
as p-value = P( x > 67) = P(t > 1.978).

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The latter probability can be calculated using the tcdf function in the TI-84 calculator:
Press 2nd VARS to access the DISTR menu and select the tcdf function; this function issimila1r to the normalcdf function that we have used many times. However, the tcdf functionneeds only three inputs – the left endpoint, right endpoint, and degrees of freedom. For thisproblem, we havep = P( x > 67) = P(t > 1.98) = tcdf(1.978, 10^99, 9) ≈ 0.0396.

10.8 It is believed that a stock price for a particular company will grow, on average, at a
rate of $5 per week with a standard deviation of $1. An investor claims that the stock won’t
grow as quickly. The changes in stock price are recorded for ten weeks and are as follows:$4, $3, $2, $3, $1, $7, $2, $1, $1, $2.
Use this data, along with a 5% level of significance, to test the claim that the mean.

Example 10.9
The National Institute of Standards and Technology provides exact data on conductivity
properties of materials. Following are conductivity measurements for 11 randomly selected
pieces of a particular type of glass.
1.11;

1.07;

1.11;

1.07;

1.12;

1.08;

.98; .98

1.02;

.95;

.95

Using α = .05, does this data provide evidence that the average conductivity of this type ofglass is greater than one? Assume the population is normal.
Solution to 10.9
We follow the Five Step Method:(Step 1) The claim being tested is that the average conductivity of the selected glass isgreater than one; i.e. the claim is that μ > 1, which will be the alternative hypothesis. So,
the hypotheses will be:
H0: μ ≤ 1 vs. Ha: μ > 1(Step 2) We are testing a sample mean and the population standard deviation is unknown.
Therefore, we need to use a Student's-t distribution. (Assume the underlying population isnormal)(Step 3) Do the calculations: In the TI-84, go to the STAT menu, then to the Edit menu;
enter the data into one of the lists. Go to the STAT menu, then to the TESTS menu, andselect T-Test.
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Select the data option and enter the following:
μ0 : 1 (This is the value of μ that is hypothesized in H0)
List: (Enter the name of the list where the data is entered)
Freq: 1
μ : > μ0 (this is a right-tailed test)
Arrow down to Calculate, and press Enter to get t = 2.014 and p-value = 0.0359.
Alternatively, arrow down to Draw, and press enter to see a graphical representation of thearea corresponding to the p-value; it is the area in the right tail:(Step 4) Since p-value = .0359 < .05, we reject H0 at the .05 level of significance.
(Step 5) Thus the results are statistically significant; there is sufficient evidence to conclude
that the mean conductivity for this type of glass is greater than 1.

10.4 | Hypothesis Test for a Single Population Proportion using
p-value
In this section we will discuss tests for a single population proportion p. If we are testing aclaim about a population proportion, then the appropriate random variable will be p̂ , thesample proportion. From Chapter 9, we know that if we select random samples of size n,
where bothnp > 10 and n(1 – p) > 10,
then the sampling distribution for p̂ will be approximately normal. Moreover, the mean andp(1 − p),nrespectively. Using this information and the Five Step Method, we can test claims like theone in the following example:

standard deviation for this sampling distribution are µ pˆ = p and σ pˆ =471

Example 10.10
A consumer group asserts that the proportion of households that have three cell phones is30%. A cell phone company has reason to believe that the proportion more than 30%. Before
they start a big advertising campaign, they conduct a hypothesis test. Their marketing people
survey 150 households with the result that 60 of the households have three cell phones; thisdata is then used to test the claim that more than 30% of households have three cell phones.
At the 5% significance level, is there sufficient evidence to support the company’s claim?
Solution to 10.10(Step 1) The phone company’s claim is that p > 0.30, which will be represented by thealternative hypothesis Ha. The null hypothesis will be contradictory statement, giving us
the hypotheses:
H0: p < 0.30

Ha: p > 0.30

(Step 2) The significance level is 5% or 0.05. Since the claim involves a population
proportion, the appropriate sample statistic will be a sample proportion, p̂ .
(Step 3) Note that np =150(.30) = 45 and n(1 – p) = 150(.70) = 105, so the sampling
distribution for p̂ will be approximately normal with mean μ = 0.30 and standard deviation

σ pˆ =

p(1 − p).30(.70)=≈ 0.03742 .
n150Note that the sample proportion is p̂ = 60/150 = 0.40.
And since this is a right-tailed test, the p-value is:p –value = P( p̂ > 0.40) = normalcdf(0.40, 10^99, .30, .03742) = 0.00377.
(Step 4) This p-value is less than the significance level, so our decision is to reject Ho.
Moreover, this p-value tells us that if the null hypothesis is true, there is only about a.377% chance that the sample proportion p̂ for a randomly selected sample will either be atleast as far off from the expected proportion of .30.
(Step 5) Thus, at the α = .05 level of significance, the sample data provides sufficientevidence that the percentage of households with three cell phones is more than 30%. In
other words, we have significant statistical evidence that the company’s claim is correct.
As was the case with the Z-test and T-Test, the p-value and test statistic for a hypothesis
test involving a proportion can be easily calculated using the TI-83 or TI-84 calculator:

472

Using the TI-83, 83+, 84, 84+ Calculator_

To calculate the p-value and the test statistic for a test involving a proportion, we go to the
STAT menu, and then to the TESTS menu. Select 1-PropZTest (item 5).
••

Enter the proportion hypothesized in H0 as p0; this must be a decimal.
Enter the number of successes in the sample as x; if a sample proportion p̂ isgiven, then use x = n p̂ . If necessary, round this value to the nearest whole number.

••Enter the sample size for n and select the appropriate alternative on the last line.
Put the cursor on Calculate and press Enter – the p-value and the test statistic will
be displayed.

Caution: The output screen will show two values labeled as p. The first one is the p-valueor the test. The second is the sample proportion p̂ (but the ^ is very small and easy to miss)10.9 Marketers believe that 92% of adults in the United States own a cell phone. A cell phone
manufacturer believes that percentage is actually lower. A random sample of 200 American
adults are surveyed, of which 174 report owning cell phones. Test the manufacturer’s claim
using a 5% level of significance. State the null and alternative hypothesis, find the p-value,and state your conclusion.

Example 10.11
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test theclaim that cell phone users developed brain cancer at a greater rate than that for non-cell
phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a
critical issue, use a 0.005 significance level.

Solution 10.11
We will follow the Five Step Method.
(Step 1) Let p be the proportion of cell phone users that develop brain cancer; then theclaim is that p > 0.00034. This claim will be represented by the alternative hypothesis, so
we get:H0: p ≤ 0.00034,

Ha: p > 0.00034

473

(Step 2) We are testing a claim about a proportion, so we use a one-proportion z-test. Notethat the sample is sufficiently large because np = 420,019(0.00034) = 142.8, and n(1 – p) =420,019(0.99966) = 419,876.2.
(Step 3) In the TI-84, we go to STAT >> TESTS and select 1-PropZTest. Enter p0 = 0.00034,
x = 172 and n = 420,019. Select the “>” option for the alternative, place the cursor either
on Calculate or Draw and press Enter to get the following output:(Step 4) Since the p-value = 0.0073 is greater than our alpha value = 0.005, we do not reject
H0.
(Step 5) We conclude that there is not enough evidence to support the claim of higher brain
cancer rates for cell phone users.
Remember that the claim was assigned to the alternative hypothesis (Ha) since it does nothave equality. This example, the null hypothesis is the opposite statement of the claim.
We assume the null hypothesis to be true and we are not rejecting the null hypothesis. We
cannot say that cell phone users developed brain cancer at a greater rate than that for noncell phone users.
Example 10.12
According to the US Census there are approximately 268,608,618 residents aged 12 andolder. Statistics from the Rape, Abuse, and Incest National Network indicate that 207,754
rapes occur each year (male and female), on average, for persons aged 12 and older. Thistranslates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there werereported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to
determine if there is a statistically significant difference between the local sexual assault
percentage and the national sexual assault percentage. Use a significance level of 0.01.
Solution 10.12
We again follow the Five Step Method.

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(Step 1) The claim is that the proportion of sexual assaults in Daviess County, KY issignificantly different from the national average. The proportion for the nation is207,754/268,608,618, which is about 0.00078. This means that the claim is p ≠ 0.00078, andso the hypotheses are:H0: p = 0.00078, Ha: p ≠ 0.00078(Step 2). Since we are working with proportions, we will use a one-proportion z-test. The
level of significance is 0.01.
(Step 3) Go to STAT >> TESTS and select 1-PropZTest. Enter 0.00078 as the hypothesized
proportion p0, enter 11 for x and 37,937 for n. Select the ≠ option for the alternative. Scroll
down to Calculate and press Enter to get the following output screen:

The figure on the right shows the output obtained by using the DRAW option.
(Step 4) Since the p-value, p = 0.00063, is less than the alpha level of 0.01, the sample data
indicates that we should reject the null hypothesis.
(Step 5) The sample data support the claim that the proportion of sexual assaults inDaviess County, Kentucky is different from the national average proportion.
Here is a summary of the types of tests we have looked at so far using p-values.
Parameter Being Tested

Parameter

Test Type

Mean - with known
population standard
deviation
Mean - with unknown
population standard
deviation
Proportion - ratio orpercentμZ-testμT-Testρ

1-PropZtest

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10.5 | Critical Values and the Critical Value Method
Before statistical software and graphing calculators became widely available, probabilities
had to be calculated using the tables. This is time consuming, so statistical practitioners
instead used the concept of a critical value to make the decision whether or not to reject H0.
Critical values can be calculated for distributions we have looked at previously in the chapter,the normal distribution and t-distribution as well as other probability distributions. The
critical values are the values separating unlikely values from those that are likely to occur.
The critical value method compares the critical value with the test statistic from the
hypothesis test.

Finding Critical Values for the Normal Distribution
Let’s look at the normal distribution. We already encountered critical values in Chapter 9;recall that the value z0 is the z-value so that the area to the right is exactly equal to α. For
example, suppose that we are performing a right-tailed z-test for a population mean, using
a .05 significance level. Note that we can find z.05 using the invNorm function in thecalculator:
Z.05 = invNorm(.95) ≈ 1.645
We can restate this by saying P(z > 1.645) = 0.05. Also note that as the z value increases,
the area to the right of z decreases; this is easy to see using graphs:

Distribution Plot

Normal, Mean=0, StDev=1
0.4

Density

0.3

0.2

0.1
0.05
0.0

0
X

1.645

The z value is on the horizontal axis (1.645). The area in red is 0.05.

476Distribution Plot

Normal, Mean=0, StDev=1
0.4

Density

0.3

0.2

0.1
0.04457
0.0

0
X

1.7

The z value is on the horizonal axis (1.7). The red area is greater than the previous graph.
Distribution Plot

Normal, Mean=0, StDev=1
0.4

Density

0.3

0.2

0.1
0.04006
0.0

0
X

1.75

The z-value is on the horizonal axis (1.75). The red area is larger than previous graphs.
It follows that that the area to the right of z is less than 0.05 if and only if z > 1.645. Thus:
the p-value for the test is less than .05 if and only if the test statistic z is greaterthan .05.
We would then refer to the area to the right of z.05 = 1.645 as the rejection region for thetest since we would reject H0 if and only if the test statistic z is in this region.
Of course, we could find the critical value and rejection region for any type of test, and forany significance level. But the most tests use the significance levels α = .10, .05 or .01.
Z-value critical values would be used for the population mean where the standard deviation
is known (z-test) and population proportion (1-propztest). The rejection region for z-tests
using these significance levels is summarized in the table below:

477

Rejection Region for Z-Test
SignificanceLevel
α = .10

Left-Tailed

Right Tailed

Two-Tailedz < -1.28

z > 1.28

z < -1.645 or z > 1.645

α = .05z < -1.645

z > 1.645

z < -1.96 or z > 1.96

α = .01z < -2.33

z > 2.33

z < -2.576 or z > 2.576

Finding Critical Values for the Student’s t-Distribution
Critical values for the Student’s t-distribution were also discussed. Recall that theStudent’s t-distribution is different for different sample sizes. The degrees of freedom, n -1,
changes the distribution, which will change the critical values. Therefore, the critical values
for the t-test are dependent on the level of significance, the tail of the test, and the degrees
of freedom.
Using a sample size of 10 and a level of significance of 0.05. If we are performing a righttailed t-test, we can find t.05 , the critical value, using the invT function in the calculator
where df = 9 or n - 1:
t.05 = invT(0.95, 9) = 1.833
We can restate this by saying P(t > 1.833) = 0.05 when n = 10 and it is right-tailed test. The
rejection region would be when t > 1.833.
Recall that the Student’s t-distribution is symmetric around zero like the normal
distribution so that in a left-tailed test where n = 10, the t critical value would be negative,
t.05 = -1.833 and the rejection region would be when t < - 1.833.
If the test was two-tailed with n = 10 and the level of significance of 0.05, we can find thet.025 using the invT function in the calculator:
t.025 = invT(0.975, 9) = 2.26
This means that the critical values for a two tailed t-test with a level of significance of 0.05
and n = 10 are -2.26 and 2.26. The rejection regions would be t < -2.26 and t > 2.26.
Unfortunately, there is a not concise table for rejection regions for t-tests like the z-tests
because critical values change depending on the degrees of freedom, tails of test, and levelof significance,

478

P-value method versus critical value method
The first few steps are the same for both methods in hypothesis testing. Instead of using
the p-value we will use the test statistic. After finding the test statistic, the difference
comes between the two methods. P-value is the probability of the test statistic happening.
Critical value is the value from the distribution that separates the rejection region from thenon-rejection region. Here is a recap of the rejection rules for both methods:
P-value method: Reject Ho if p-value < αCritical Value method: Reject Ho if test statistic is in rejection region.
Below are examples of hypothesis tests using the five-step method using critical value
method. Each type of test previously done in the chapter can done using critical values
instead of comparing p-values.
Example 10.13
According to a recent study, the mean cost of a heart-bypass operation is slightly less than
$26,100, and approximately 230,000 operations are performed annually. A sample of 36
bypass operations showed a mean cost of $25,000. Assume the population standard
deviation is $2400. Develop and test an appropriate hypothesis to see whether or not themean bypass operation is less than $26,100 using α = 0.05.
Solution 10.13
This is a test for mean when the population standard deviation is known. We will follow the
five-step method using the critical values.
(Step 1) The claim being tested is μ < 26,100, which will be the alternative hypothesis, Ha.
So the hypotheses are: H0: μ > 26,100 vs. Ha: μ < 26,100.
(Step 2) Since the alternative hypothesis is less than, “<”, it is a left tailed test. The level ofsignificance is 0.05.
(Step 3) We are testing a claim about a mean μ, and the population SD σ is known, so weuse a Z-Test.
The test statistic is z =

x−µ

σ

n

=

25000 − 26100
2400

36

= −2.75 .

This is a left-tailed test, so the critical value will be the z-score that has an area of .05 to theleft; i.e. the critical value is z.05 = -1.645.
(Step 4) The test statistic, -2.75, is to the left of the critical value, -1.645. Since z = -2.75 < 1.645, we reject H0.

479

(Step 5) There is sufficient evidence to conclude that μ < 26,100. That is, there issignificant evidence that the mean cost of a bypass operation is less than $26,100.
Example 10.14
According to an article in the Wall Street Journal, they claim that the mean time it takes ataxi from Manhattan to LaGuardia is now more than 35 minutes. Suppose new study using
40 taxis has an average of 42 minutes travel time from Manhattan to LaGuardia. Assume
the population standard deviation is 5.1 minutes. Develop and test an appropriate
hypothesis to see whether or not the mean travel time is more than 35 minutes using α =0.05.
Solution 10.14
This is a test for mean when the population standard deviation is known. We will follow the
five-step method using the critical values.
(Step 1) Set up the hypothesis:

H0: μ ≤ 35
Ha: μ > 35 (claim)(Step 2) Set up the correct test and identify the level of significance:
Right-tailed test and α = 0.05
(Step 3) Calculate the test statistic: Since we have the population mean perform a Z-test
μ0: 35σ: 5.1x : 42n : 40μ : > μ0

(this is the mean hypothesized in H0)

Z = 8.68

p-value = 1.99E -18 which is
0.00000000000000000199

Z.05 = -invnorm(.05,0,1) = 1.645 (criticalvalue)

Reject Ho because p-value < .05

Reject Ho because 8.68 is in the rejection
region480

(Step 4) Use these results to make the decision about H0. The z critical value is 1.645 for aright tailed test when the level of significance is 0.05. The test statistic for the z-test is 8.68.
8.68 > 1.645, so we reject H0(Step 5) There is enough evidence at 5% level of significance to support the claim that themean travel time from Manhattan to LaGuardia is now more than 35 minutes.

Example 10.15
An engineer at Duracell has designed a new battery for industrial use. The old-style battery
had an average life of 29.2 hours. The engineer claims the new battery is an improvement
over the old one since the new design has a longer life. A sample of 20 new batteries finds amean life of 32.3 hours with a standard deviation of 3.9 hours. Test the engineer's claim
using α = 0.01.
Solution 10.15
This is a test for mean when the population standard deviation is unknown. We will follow
the five-step method using the critical values.
(Step 1) The engineer’s claim is that the new battery is an improvement, meaning that it
will have a longer average life. So, the claim is that μ > 29.2 hours, and the hypotheses are:H0: μ < 29.2 vs. Ha: μ > 29.2
(Step 2) The alternative hypothesis is “greater than” so we have right tailed test. The levelof significance, which is given in the problem, is 0.01.
(Step 3) We are testing a claim about a population mean μ and the population standard
deviation is unknown, so we use a t-test.
x − µ 32.3 − 29.2=
= 3.55 .
Using the given sample statistics, the test value is t =3.9 20s n(Step 4) Before deciding about the null hypothesis we must find the critical value; thecritical value will be t.01, the t value so that the area to the right of it is exactly 0.01. Thisvalue can be found using the invT function; this function works just like the invNorm
function that we have used many times. The inputs will be the cumulative probability andthe degrees of freedom. In this problem, n = 20, so df = 19. And since the area in the righttail will be .01, the area to the left – the cumulative probability – will be .99.

481

On the calculator press 2nd VARS to access the DISTR menu and select the invT function.
The critical value is:t.01 = invT(.99, 19) = 2.539.
So we will reject H0 if and only if t > 2.539; the rejection region is shown below:
Distribution PlotT, df=19

0.4

Density

0.3

0.2

0.1

0.0

0.01
0
X

2.539

Since t = 3.55 > 2.539, we reject H0 at the 0.01 significance level.
(Step 5) Thus, the data provides sufficient evidence that μ > 29.2 hours. There is sufficient
statistical evidence to support the engineer’s claim.
Example 10.16
When Mendel conducted his famous genetic experiments with peas, one sample of offspring
consisted of 428 green peas and 152 yellow peas. Use a 0.01 level of significance to test
Mendel’s claim that under the same circumstances, 25% of the offspring will be yellow.
Solution to 10.16
This is a test for population proportion. We will follow the five-step method using thecritical values.
(Step 1) Set up the hypothesis H0 and Ha:
H0: p = 0.25
Ha: p ≠ 0.25
(Step 2) Select the correct test and identify the significance level α:
Since Ha is “not equal to” it is a two-tailed test. The significance level is 0.01.

482

(Step 3) Calculate the test statistic.
Since this test is for proportion, using the percent, on the calculator use 1-PropZTest.
P0: 0.25 (the percent changed to a decimal)
x: 152n: 428+152 = 580
prop: ≠ p0
Calculate: The test statistic, z = 0.671pˆ − p 0

Using the formula to calculate the test statistic: z =p 0 (1 − p 0 ) / n

=

�

152
�−0.25580

�0.25(1−0.25)/580= 0.671(Step 4) Use these results to make a decision about Ho. First, we must find the criticalvalues for the two-tailed test at a 0.01 significance level. Since it is two-tailed we divide
0.01/2.
Z0.005 = invNORM(0.995) = 2.576
So, the rejection regions are when z < -2.576 or z > 2.576. Since the test value 0.671 < 2.576,
we fail to reject H0.
(Step 5) Interpret the decision about H0 in the context of the given problem to state aconclusion.
There is not enough evidence to reject the claim that 25% of the offspring peas will beyellow.
Z-TestT-TestOne Proportion Z-TestUsed to test a claim about apopulation mean μ.
Assumptions:
Data comes from a random
sampleSampling from a normal
distribution, or sample sizen > 30The population SD, σ, isknown.
x − µ0
Test statistic: z =σ/ nUsed to test a claim about apopulation mean μ.
Assumptions:
Data comes from a random
sampleSampling from a normal
distribution.
The population SD, σ, is notknown.

Used to test a claim about apopulation proportion p.
Assumptions:
Data comes from a random
sampleThe sample size an
hypothesized proportion
satisfy: np0 > 10 and n(1 –p0) > 10.

Calculator function:

Calculator function:

Z-TestT-test

Test statistic:t=

x − µ0s/ nTest statistic:pˆ − p 0z=
p 0 (1 − p 0 ) / n

Calculator function:
1-PropZTest

483

10.6 | Type I and Type II Errors
We have looked briefly at Type I errors, but now we will spend more time investigating
Type I and Type II errors. When we perform a hypothesis test, the decision to reject H0depends on randomly selected sample data – so there is always the possibility of an error.
No matter how careful we are, there is always a small probability that we will just get an
unusual sample which leads us to the wrong decision. Since there are two possible
decisions, and each of these can be either correct or incorrect, there are a total of four
possible outcomes for a test, which are summarized in the following table:
H0 is true

H0 is false

Reject H0

Type I Error

Correct Decision

Fail to Reject HoCorrect Decision

Type II Error

As we see, there are two outcomes where the correct decision is made, and two outcomes
where an error occurs.
A Type I error occurs when the decision is to reject H0 when in fact H0 is true. The
probability of a Type I error is denoted by the Greek letter α.
A Type II error occurs when the decision is to not reject H0 when in fact H0 is false. The
probability of a Type II error is denoted by the Greek letter β.
Obviously, both α and β should be as small as possible because they are probabilities oferrors.
There is one more probability related to this chart; the Power of the Test is the probabilityof rejecting the null hypothesis when H0 is false. Thus, the Power of the test is 1 – β.
Ideally, we want a high power that is as close to one as possible. Increasing the sample size
can increase the Power of the Test.
The following are examples of Type I and Type II errors.
Example 10.17
Suppose the null hypothesis, H0, is: Frank's rock-climbing equipment is safe. Find the TypeI and Type II error. Also explain α and β.

484

Solution 10.17
Type I error: Frank thinks that his rock-climbing equipment may not be safe when, in fact,
it really is safe.
Type II error: Frank thinks that his rock-climbing equipment may be safe when, in fact,
it is not safe.
α = probability that Frank thinks his rock-climbing equipment may not be safe when, in
fact, it really is safe.
β = probability that Frank thinks his rock-climbing equipment may be safe when, in fact, it
is not safe.
Notice that, in this case, the error with the greater consequence is the Type II error; if
Frank thinks his rock-climbing equipment is safe, he will go ahead and use it.

10.10 Suppose the null hypothesis, H0, is: The blood cultures contain no traces of pathogen
X. State the Type I and Type II errors.

Example 10.18
Suppose the null hypothesis, H0, is: The victim of an automobile accident arriving at the
emergency room of a hospital is infected with HIV.
a. What type of error has a greater consequence, a Type I or Type II error?
b. Describe the probability α in words.
Solution 10.18
Type I error: The emergency crew thinks that the victim does not have HIV when, in
reality the victim is infected.
Type II error: The emergency crew thinks that the victim does have HIV, when in reality
the patient is not infected.
a. The error with the greater consequence is the Type I error. If the emergency crew
mistakenly believes that the victim is uninfected with HIV, then they may not take proper
precautions to avoid infection while treating him.
b. Recall that α is the probability of a Type I error; that is, the probability that the
emergency room staff thinks the patient does not have HIV when in fact he/she does havethe virus.

48510.11 Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has thegreater consequence, Type I or Type II?
Example 10.19
A certain experimental drug claims a cure rate of at least 75% for males with prostatecancer. Describe both the Type I and Type II errors in context. Which error is the more
serious, a Type I or Type II error?

Solution 10.19
The claim is that the cure rate is at least 75%. So the claim is p > 0.75; this statement
includes an equal sign, and so it will be the null hypothesis H0. Thus, the errors are:
Type I Error: A cancer patient believes the cure rate for the drug is less than 75% when it
actually is at least 75%.
Type II Error: A cancer patient believes the experimental drug has at least a 75% cure rate
when it has a cure rate that is less than 75%.
In this scenario, the Type II error contains the more severe consequence. If a patient
believes the drug works at least 75% of the time, this most likely will influence the patient’s(and doctor’s) choice about whether to use the drug as a treatment option.

10.12 Determine the Type I and Type II errors for the following scenario:
Assume a null hypothesis, H0, states that the percentage of adults with jobs is at least 88%.
Identify the Type I and Type II errors from these four statements.
a. Not to reject the null hypothesis that the percentage of adults who have jobs is atleast 88% when that percentage is actually less than 88%.
b. Not to reject the null hypothesis that the percentage of adults who have jobs is atleast 88% when the percentage is actually at least 88%.
c. Reject the null hypothesis that the percentage of adults who have jobs is at least
88% when the percentage is actually at least 88%.
d. Reject the null hypothesis that the percentage of adults who have jobs is at least
88% when that percentage is actually less than 88%.

486

KEY TERMS and FORMULA REVIEW
Hypothesis Test: A procedure for determining whether the hypothesis stated is areasonable statement and should not be rejected or is unreasonable and should be rejected.
Hypothesis: A statement about the value of a population parameter, in case of two
hypotheses, the statement assumed to be true is called the null hypothesis (notation H0)
and the contradictory statement is called the alternative hypothesis (notation Ha).
Level of Significance: In hypothesis testing, the Level of Significance is a thresholdvalue – it is the maximum value allowed for a Type I error. And when the p-value is below
the level of significance, the null hypothesis must be rejected.
p-value: The probability that an event will happen purely by chance assuming the null
hypothesis is true. The smaller the p-value, the stronger the evidence is against the null
hypothesis.
Student's t-Distribution: Investigated and reported by William S. Gossett in 1908 andpublished under the pseudonym Student. The major characteristics of this random variablet are:•••••

It is continuous, bell shaped curve.
The pdf is symmetrical about its mean of zero. The variance is more than that of
the normal distribution.
There is a "family" of t distributions: every representative of the family iscompletely defined by the number of degrees of freedom, which is one less than thesample size.
The smaller the sample size, the greater the variation
As the sample size n gets larger, the t-distribution approaches the standard normal
distribution.

Type 1 Error: The decision is to reject the null hypothesis when, in fact, the null
hypothesis is true.
Type 2 Error: The decision is not to reject the null hypothesis when, in fact, the null
hypothesis is false.
T-Test: A test for a population mean, μ that is used when the population SD, σ, isunknown.
Z-Test: A test for a population mean μ that is used when the population SD, σ, is known.

487

CHAPTER REVIEWAll hypothesis tests use the same basic steps, which we have called the Five Step Method:
1.
2.
3.
4.
5.

Set up the hypotheses H0 and Ha.
Select the correct test and identify the significance level σ.
Calculate the test statistic and p-value.
Use these results to make the decision about H0.
Interpret the decision about H0 in the context of the given problem to state aconclusion.

(Step 1) The first step is to set up the hypotheses. The null hypothesis H0 and thealternative hypothesis Ha are contradictory claims about one or more unknown population
parameters. There are a few basic rules for setting these up:•••

The null hypothesis must include an equal sign.
One of the hypotheses must represent the claim being tested.
The hypotheses should be contradictory statements.

(Step 2) The second step is to select the correct test:•••

When testing a population mean μ, with a known population SD σ we use a Z-Test.
When testing a population mean μ, with unknown population SD σ we use a T-Test
When testing a population proportion, p, we use a one-proportion z-test.

(Step 3) The third step is to calculate the p-value and the test statistic. We can get both very
easily using the TI-84 family of calculators.
The p-value is the probability that, if the null hypothesis is true, the results from anotherrandomly selected sample will be as extreme or more extreme as the results obtained fromthe given sample.
The test statistic is a measure of relative position, and the formula depends on the type oftest being used (see below).
(Step 4) The fourth step is to make the decision about H0. In a test, there are only two possible
decisions: Either we reject H0 or we do not reject H0. This decision is made by comparing thep-value found in Step 3 to a predetermined significance level α according to the following rule:••

We reject H0 at significance level α if p < α.
If p > α, then we do not reject H0.

(Step 5) The fifth step is to write a conclusion. After we make our decision about H0, weinterpret the decision in the context of the given problem to write a meaningful conclusion
for the test in terms of the given problem. That is, we should write a statement, in plain
English, that explains the result of the test. A few things to keep in mind:
488

•
•
•

If we reject H0, then there is sufficient evidence to conclude that H0 is incorrect. Thus,
rejecting H0 provides evidence that the alternative hypothesis Ha is true.
If we fail to reject H0, then there is not sufficient evidence to conclude that thealternative hypothesis Ha is true.
Failing to reject H0 does not mean that we have evidence that H0 is true. It simply
means that the sample data have failed to provide sufficient evidence to cast serious
doubt about the truthfulness of H0.

Summary of Tests:
Z-Test:•Used to test a claim about a population mean μ when σ (the population SD) is
known.

•

Data comes from a random sample.

•

Sampling from a normal distribution, or sample size n > 30•

Test statistic:•

Calculator function: Z-Test

Significance
Level

z=

x − µ0σ/ nRejection Regions
for Z-Test
Left-Tailed

Right Tailed

α = .10z < -1.28

z > 1.28

z < -1.645 or z > 1.645

α = .05z < -1.645

z > 1.645

z < -1.96 or z > 1.96

α = .01z < -2.33

z > 2.33

z < -2.576 or z > 2.576

489

Two-Tailed

T-Test:•Used to test a claim about a population mean μ when σ (the population SD) is notknown.

•

Data comes from a random sample.

•

Sampling from a normal distribution•

Test statistic:•

Calculator function: T-testt=x − µ0s/ nOne-Proportion Z-test:•

Used to test a claim about a population proportion p.

•

Data comes from a random sample.

•

The sample size and hypothesized proportion satisfy: np0 > 10 and n(1 – p0) > 10.

•

Test statistic:•

Calculator function: 1-PropZTestz=pˆ − p 0
p 0 (1 − p 0 ) / n

490

Chapter 10 Exercises
1. A test is being conducted to determine whether the mean speed of a cable internet
connection is more than three Megabits per second.
a. What is the random variable? Describe in words.
b. State the null and alternative hypotheses for the test.
2. The mean entry level annual salary of an employee at a large company is $58,000. An
economist wishes to test the claim that the mean entry level salary is higher for IT
professionals in the company. State the null and alternative hypotheses.
3. A test will be conducted to test the claim that the mean number of children for American
families is 2.
a. What is the random variable? Describe in words.
b. State the null and alternative hypotheses for the test.
4. A sociologist claims that the probability that a person picked at random in Times Squarein New York City is visiting the area is 0.83. A tourism expert thinks that the proportion is
actually less. Assume that the tourism expert’s claim is tested using an appropriate
hypothesis test.
a. What is the random variable? Describe in words.
b. State the null and alternative hypotheses for the test.
5. In a population of fish, it is widely believed that approximately 42% are female. A biologist
will conduct a hypothesis test to determine if the proportion of female fish is less than 42%.
State the null and alternative hypotheses for the test.
6. An article in the 1990’s stated that the mean time spent in jail by a first–time convictedburglar was 2.5 years. A study was then done to see if the mean time in jail has increased in
the new century. A random sample of 26 first-time convicted burglars in a recent year was
picked. The mean length of time in jail from the survey was 3 years. Assume that the
distribution of the population is normal, and that the population standard deviation is 1.5. If
you were conducting a hypothesis test to determine if the mean length of jail time has
increased, what would the null and alternative hypotheses be?
7. A random survey of 75 death row inmates revealed that the mean length of time on deathrow is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis
test to determine if the mean time spent on death is equal to 15 years, what would the nulland alternative hypotheses be?
a. H0:

b. Ha:

491

8. The National Institute of Mental Health published an article stating that in any one-year
period, approximately 9.5 percent of American adults suffer from depression or a depressive
illness. Suppose that we are conducting a hypothesis test to determine if the true proportionof people in a given town suffering from depression or a depressive illness is lower than thepercent in the general adult American population, what would the null and alternative
hypotheses be?
a. H0:

b. Ha:

9. A market analyst claims that the mean price of mid-sized cars in the Midwest region is$32,000. A suitable hypothesis test will be conducted to determine if the claim is true. State
the Type I and Type II errors in complete sentences.
10. A group of doctors is deciding whether or not to perform an operation. Suppose the null
hypothesis H0, in words, is: The surgical procedure will go well.
a. State the Type I and Type II errors in complete sentences.
b. Which is the error with the greater consequence?
11. The power of a test is 0.981. What is the probability of a Type II error?
12. A group of divers is exploring an old sunken ship. Suppose the null hypothesis, H0, is:the sunken ship does not contain buried treasure. State the Type I and Type II errors.
13. A microbiologist is testing a water sample for E-coli. Suppose the null hypothesis, H0, is:the sample does not contain E-coli. The probability that the sample does not contain E-coli,
but the microbiologist thinks it does is 0.012. The probability that the sample does containE-coli, but the microbiologist thinks it does not is 0.002.
a. What is the power of this test?
b. Which is the error with the greater consequence?
14. A population mean is hypothesized to be μ = 13. A random sample of 20 measurements isselected and the sample mean is 12.8, and the sample standard deviation is two. What
distribution should you use to perform the hypothesis test? Assume the underlying
population is normal.
15. A population has a standard deviation of σ = 5. A test is to be conducted to test the claimthat the mean is μ = 25. A sample of 108 individuals yields a sample mean of 24. What
distribution should you use to perform a hypothesis test?
16. It is thought that 42% of respondents in a taste test would prefer Brand A. Suppose wewant to test the claim that the true population proportion is less than 42%. We select arandom sample of 100 people, and find that 39% preferred Brand A. What distribution should
we use to perform a hypothesis test?

492

17. Suppose we are conducting a hypothesis test of a single population mean using aStudent’s t-distribution. What must we assume about the distribution of the data?
18. Suppose we are conducting a hypothesis test for a single population proportion. What
must be true about the quantities of np and nq= n(1 – p)?
19. It is believed that the mean height of high school students who play basketball in a large
urban district is 73 inches with a standard deviation of σ = 1.8 inches. A random sample of
40 players is chosen. The sample mean was 71 inches, and the sample standard deviation
was 1.5 years. Do the data support the claim that the mean height is less than 73 inches?
a. State the null and alternative hypotheses for the test.
b. Suppose that the p-value is almost zero; state an appropriate conclusion.
20. It is conjectured that the mean age of graduate students at a large state university is at
most 31 years. To test this claim, a random sample of 15 graduate students is taken. The
sample mean is 32 years, and the sample standard deviation is two years. Assume that theages are normally distributed and conduct an appropriate test.
a. State the null and alternative hypotheses for the test.
b. Assume that the p-value is 0.0366. Is the data significant at the .05 level? Explain.
c. Is the data significant at the .01 level? Explain.
21. Consider the following graph:a. Does the shaded region represent a low or a high p-value compared to a level ofsignificance of 1%?
b. Is this a right-tailed, left-tailed or two-tailed test? Explain.
22. Consider the statement, “If you do not reject the null hypothesis, then H0 must betrue.” Is this statement correct? Explain why or why not.
23. Suppose that a recent article stated that the mean time spent in jail by a first-time
convicted burglar is 2.5 years. A study was then done to see if the mean time has increasedin the new century. A random sample of 26 first-time convicted burglars in a recent year isselected. The mean length of time in jail from the survey was three years with a standard
deviation of 1.8 years. Suppose that it is somehow known that the population standard
deviation is 1.5.
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Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume
the distribution of the jail times is approximately normal.
a.
b.
c.
d.
e.

Is this a test for a mean or a proportion?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
Is the population standard deviation known and, if so, what is it?
Find the following:a. x = _______
b. σ =
c. s =
d. n = _______
f. Since both σ and s are given, which should be used? Explaing. Which test should be used?
h. Find the p-value.
i. Using a significance level of σ = 0.05, what would be the decision about H0?
j. Interpret this decision to state a conclusion for the test.

24. A random survey of 75 death row inmates revealed that the mean length of time on deathrow is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to
determine if the mean time on death row for all inmates is at most 15 years.
a. Is this a test of one mean or proportion?
b. State the null and alternative hypotheses: H0:
Ha:
c. Is this a right-tailed, left-tailed, or two-tailed test?
d. What symbol represents the random variable for this test?
e. In words, define the random variable for this test.
f. Is the population standard deviation known and, if so, what is it?
g. Find the following:(i) x =
(ii) s =(iii) n =h. Which test should be used?
i. State the distribution to use for the hypothesis test.
j. Find the p-value.
k. Using a significance level of  = 0.05, what would be the decision about H0?
l. Interpret this decision to state a conclusion for the test.
25. Suppose that we conduct at test with hypotheses H0: μ = 9 and Ha: μ < 9.
Is this a left-tailed, right-tailed, or two-tailed test?
26. Suppose that we conduct at test with hypotheses H0: μ ≤ 6 and Ha: μ > 6.
Is this a left-tailed, right-tailed, or two-tailed test?
27. Suppose that we conduct at test with hypotheses H0: p = 0.25 and Ha: p ≠ 0.25.
Is this a left-tailed, right-tailed, or two-tailed test?
28. A certain brand bottle of water is labeled as containing 16 fluid ounces of water. You
believe that the bottles contain less than 16 oz. Suppose that you will conduct a hypothesis
test for your claim. Would this be a left-tailed, right-tailed, or two-tailed test?
494

29. A golf pro claims that his mean golf score is 63. A sportswriter wants to test the claimthat his mean score is higher 63. Would this be a left-tailed, right-tailed, or two-tailed test?
30. A bathroom scale claims to be able to correctly identify any weight within a pound. You
suspect that it cannot be that accurate and will conduct an appropriate test to test yourclaim. Would this be a left-tailed, right-tailed, or two-tailed test?
31. You flip a coin and record whether it shows heads or tails. The coin is supposed to bebalanced, but you suspect that the probability of getting for this particular coin is less than
50%. If you were to test your claim, would this be a left-tailed, right-tailed, or two-tailedtest?
32. If the alternative hypothesis has a not equals ( ≠ ) symbol, what type of test does thissignify?
33. Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed,right-tailed, or two-tailed test?
34. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed,right-tailed, or two-tailed test?
35. Each of the following statements refers to a claim for a hypothesis test; so, each could
be either the null hypothesis or the alternative hypothesis. In each case, state the null
hypothesis H0 and the alternative hypothesis Ha, in terms of the appropriate parameter(either μ or p).
a. The mean number of years Americans work before retiring is 34.
b. At most 60% of Americans vote in presidential elections.
c. The mean starting salary for San Jose State University graduates is at least$100,000 per year.
d. Twenty-nine percent of high school seniors get drunk each month.
e. Fewer than 5% of adults in Los Angeles ride the bus to work.
f. The mean number of cars a person owns in her lifetime is at most ten.
g. About half of Americans prefer to live away from cities, given the choice.
h. Europeans have a mean paid vacation of six weeks each year.
i. Fewer than 11% of all American women will develop breast cancer.
j. The mean tuition cost for private American universities is more than $20,000 per
year.
36. Refer to problem 35; classify each test as either left-tailed, right-tailed, or two-tailed.
37. Refer to problem 35; for each test, state the Type I and Type II errors in complete
sentences.

495

38. Over the past few decades, public health officials have examined the link between weight
concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected
teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls
were surveyed again. Sixty-three said they smoked to stay thin. This data is used to test theclaim that more than thirty percent of the teen girls smoke to stay thin. The alternative
hypothesis is:a. p < 0.30

b. p ≤ 0.30

c. p ≥ 0.30

d. p > 0.30

39. A statistics instructor wishes to test the claim that fewer than 20% of students at hercollege attended the opening night midnight showing of the latest Harry Potter movie. She
surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate
alternative hypothesis would be:a. p = 0.20

b. p > 0.20

c. p < 0.20

d. p ≤ 0.20

40. Some years ago, an organization reported that on average, teenagers spent 4.5 hours per
week on the phone. The organization thinks that the current mean is higher. A test will be
conducted to test whether the current mean is higher than the previously reported value of
4.5 hours per week. Fifteen randomly chosen teenagers were asked how many hours per week
they spend on the phone. The sample mean was 4.75 hours with a sample standard deviationof 2.0. The null and alternative hypotheses are:a. Ho: x = 4.5,

Ha : x > 4.5

c. Ho: μ = 4.75, Ha: μ > 4.75

b. Ho: μ ≥ 4.5,

Ha: μ < 4.5

d. Ho: μ = 4.5,

Ha: μ > 4.5

41. When a new drug is created, the pharmaceutical company must subject it to testing before
receiving the necessary permission from the Food and Drug Administration (FDA) to market
the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?
a.
b.
c.
d.

To conclude the drug is safe when in, fact, it is unsafe.
Not to conclude the drug is safe when, in fact, it is safe.
To conclude the drug is safe when, in fact, it is safe.
Not to conclude the drug is unsafe when, in fact, it is unsafe.

42. A statistics instructor believes that fewer than 20% of McHenry Community College(MCC) students attended the opening midnight showing of the latest Harry Potter movie. To
test this claim she surveys 84 of her students and finds that 11 of them attended the midnightshowing. What is the Type I error for the test?
a. To conclude that the percent of MCC students who attended is at least 20%, when infact, it is less than 20%.
b. To conclude that the percent of MCC students who attended is 20%, when in fact, it
is 20%.

496

c. To conclude that the percent of MCC students who attended is less than 20%, when
in fact, it is 20% or more.
d. To conclude that the percent of MCC students who attended is less than 20%, when
in fact, it is less than 20%.
43. It is believed that College of Lake County (CLC) Intermediate Algebra students get lessthan seven hours of sleep per night, on average. A survey of 22 CLC Intermediate Algebra
students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. A test is
conducted to test the claim that that CLC Intermediate Algebra students get less thanseven hours of sleep per night, on average. Which of the following statements is true?
a. The Type II error is to not reject that the mean number of hours of sleep CLC
students get per night is at least seven when, in fact, the mean number of hours ismore than seven hours.
b. The Type II error is to not reject that the mean number of hours of sleep CLC
students get per night is at least seven when, in fact, the mean number of hours is at
most seven hours.
c. The Type II error is to not reject that the mean number of hours of sleep CLC
students get per night is at least seven when, in fact, the mean number of hours is atleast seven hours.
d. The Type II error is to not reject that the mean number of hours of sleep CLC
students get per night is at least seven when, in fact, the mean number of hours is
less than seven hours.
44. Previously, an organization reported that teenagers spent 4.5 hours per week, on average,on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly
chosen teenagers were asked how many hours per week they spend on the phone. The sample
mean was 4.75 hours with a sample standard deviation of 2.0. Suppose that a hypothesis test
is conducted to test the claim that the mean hours is more than 4.5 hours per week. What is
the Type I error?
a. To conclude that the current mean is higher than 4.5 hrs/week, when in fact, it ishigher.
b. To conclude that the current mean is more than 4.5 hrs/week, when in fact, it still
4.5 hrs.
c. To conclude that the current mean hours per week is 4.5 hrs/week, when in fact, it ishigherd. To conclude that the current mean is no higher than 4.5 hrs/week, when in fact,
it is not higher.

497

45. It is believed that College of Lake County (CLC) Intermediate Algebra students get lessthan seven hours of sleep per night, on average. A survey of 22 CLC Intermediate Algebra
students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. A test is
conducted to test the claim that that CLC Intermediate Algebra students get less thanseven hours of sleep per night, on average.
The distribution to be used for this test is: X ~a.


1.93 
N  7.24,
22 


b.

N(7.24, 1.93)

c. t22

d. t21

46. The National Institute of Mental Health published an article stating that in any one-year
period, approximately 9.5 percent of American adults suffer from depression or a depressive
illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered fromdepression or a depressive illness. Conduct a hypothesis test to determine if the true
proportion of people in that town suffering from depression or a depressive illness is lower
than the percent in the general adult American population.
a. Is this a test of one mean or proportion?
b. State the null and alternative hypotheses: H0:
Ha:
c. Is this a right-tailed, left-tailed, or two-tailed test?
d. What symbol represents the random variable for this test?
e. In words, define the random variable for this test.
f. Find the following:(i) x = ______ (ii) n =__(iii) p̂ = _____
g. What distribution is used for this test?
h. Find the p-value.
j. If the significance level is a = 0.05, what is the decision about H0?
k. Interpret the decision to write a conclusion for the test.
For the remaining exercises, use the FIVE STEP METHOD and a suitable hypothesis test to
answer each question. For problems involving a Student's-t distribution, you may assume
that the underlying population is normally distributed.
47. A particular brand of tires claims that its deluxe tire averages at least 50,000 miles
before it needs to be replaced. From past studies of this tire, the standard deviation is
known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires
surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles.
Using α = 0.05, is the data highly inconsistent with the claim?

498

48. From generation to generation, the mean age when smokers first start to smoke varies.
However, the standard deviation of that age remains constant of around 2.1 years. A surveyof 40 smokers of this generation was done to see if the mean starting age is at least 19. The
sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support theclaim at the 5% level?
49. The cost of a daily newspaper varies from city to city. However, the variation among prices
remains steady with a standard deviation of 20¢. A study was done to test the claim that themean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard
deviation of 18¢. Do the data support the claim at the 1% level?
50. An article in the San Jose Mercury News stated that students in the California state
university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose
you believe that the mean time is longer. You conduct a survey of 49 students and obtain asample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim
at the 1% level?
51. The mean number of sick days an employee takes per year is believed to be about ten.
Members of a personnel department do not believe this figure. They randomly survey eight
employees. The number of sick days they took for the past year are as follows:
12;

4;

15;

3;

11;

8;

6;

8.

Let x = the number of sick days they took for the past year. Does this data provide evidence
to support the personnel department’s claim the mean number of sick days differs from ten?
Conduct an appropriate test using a significance level of α = .05.
52. In 1955, Life Magazine reported that the 25-year-old mother of three worked, on
average, an 80-hour week. Recently, many groups have been studying whether or not the
women's movement has, in fact, resulted in an increase in the average work week for
women (combining employment and at-home work). Suppose a study was done to determine
if the mean work week has increased. 81 women were surveyed with the following results.
The sample mean was 83; the sample standard deviation was 10. Using a 5% significancelevel, does this data provide evidence that the mean number of hours worked by womeneach week has increased?
53. Your statistics instructor claims that at least 60 percent of the students who take her
Elementary Statistics class go through life feeling more enriched. For some reason that she
can't quite figure out, most people don't believe her. You decide to check this out on your own.
You randomly survey 64 of her past Elementary Statistics students and find that 34 feel moreenriched as a result of her class. Now, what do you think?

499

54. A Nissan Motor Corporation advertisement read, “The average man’s IQ is 107. The
average brown trout’s IQ is 4. So why can’t man catch brown trout?” Suppose you believethat the brown trout’s mean IQ is greater than four. You catch 12 brown trout. A fish
psychologist determines the IQ’s for the fish as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5.
Use this data to conduct a hypothesis test for the claim that the mean IQ for brown trout isgreater than 4.
55. Refer to the previous exercise. Conduct a hypothesis test to see if your decision andconclusion would change if your belief were that the brown trout’s mean IQ is different from4.
56. According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China,
the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of thepercent of girls born in China. You conduct a study. In this study, you count the number ofgirls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys bornof the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?
57. A poll done for Newsweek found that 13% of Americans have seen or sensed the presenceof an angel. A contingent doubt that the percent is really that high. It conducts its own survey.
Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As aresult of the contingent’s survey, would you agree with the Newsweek poll? In complete
sentences, also give three reasons why the two polls might give different results.
58. The mean work time each week for engineers in a start-up company is believed to beabout 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineeringfriends in start-ups for the lengths of their mean work weeks. The data is as follows:
70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
Use this data to test the claim that the mean number of hours worked each week is lessthan 60.
59. Toastmasters International cites a report by Gallop Poll that 40% of Americans fear
public speaking. A student believes that less than 40% of students at her school fear public
speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public
speaking. Conduct a hypothesis test to determine if the percent at her school is less than40%.
60. A recent report states that 68% of online courses taught at community colleges nationwide
were taught by full-time faculty. To test if this proportion is the same for College of Lake
County (CLC) in Illinois, was randomly selected for comparison. In that same year, 34 of the
44 online courses CLC offered were taught by full-time faculty. Does this data provide
evidence that the percentage of online courses at community colleges that are taught by fulltime faculty is different from 68% in Illinois?

500

61. According to an article in Bloomberg Businessweek, New York City's most recent adultsmoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nineout of 70 randomly chosen N.Y. City residents reply that they smoke. Use this data to test
the claim that the proportion of NYC adults who smoke has decreased.
62. The mean age of College of Lake County students in a previous term was 26.6 years old.
An instructor thinks the mean age for online students is older than 26.6. She randomly
surveys 56 online students and finds that the sample mean is 29.4 with a standard deviationof 2.1. Conduct a hypothesis test to test the instructor’s claim.
63. Nationwide, registered nurses earn an average annual salary of $69,110. A survey was
conducted of 41 California registered nurses to determine if the annual salary is higher than$69,110 for California nurses. The sample average was $71,121 with a sample standard
deviation of $7,489. Conduct an appropriate test to test the claim that the mean salary fornurses in California is higher than the national average.
64. La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children
much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned
their children. The mean weaning age was nine months (3/4 year) with a standard deviationof 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. isless than four years old.
65. Over the past few decades, public health officials have examined the link between weight
concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected
teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls
were surveyed again. Sixty-three said they smoked to stay thin. Does this data provide
sufficient evidence to conclude that more than thirty percent of teen girls smoke to stay thin?
After conducting the test, your decision and conclusion are
a. Reject H0: There is sufficient evidence to conclude that more than 30% of teen girls
smoke to stay thin.
b. Do not reject H0: There is not sufficient evidence to conclude that less than 30% ofteen girls smoke to stay thin.
c. Do not reject H0: There is not sufficient evidence to conclude that more than 30% ofteen girls smoke to stay thin.
d. Reject H0: There is sufficient evidence to conclude that less than 30% of teen girls
smoke to stay thin.
66. A statistics instructor believes that fewer than 20% of students at her community college
attended the opening night midnight showing of the latest Harry Potter movie. She surveys
84 of her students and finds that 11 of them attended the midnight showing. At a 1% levelof significance, an appropriate conclusion is:501

a. There is insufficient evidence to conclude that the percent of EVC students who
attended the midnight showing of Harry Potter is less than 20%.
b. There is sufficient evidence to conclude that the percent of EVC students who
attended the midnight showing of Harry Potter is more than 20%.
c. There is sufficient evidence to conclude that the percent of EVC students who
attended the midnight showing of Harry Potter is less than 20%.
d. There is insufficient evidence to conclude that the percent of EVC students who
attended the midnight showing of Harry Potter is at least 20%.
67. Previously, an organization reported that teenagers spent an average of 4.5 hours per
week on the phone. The organization thinks that the mean is now higher. To test the claim,
15 randomly chosen teenagers were asked how many hours per week they spend on thephone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Use thisdata to test the claim that the current mean is greater than 4.5 hours. At a significancelevel of a = 0.05, what is the correct conclusion?
a. There is enough evidence to conclude that the mean number of hours is more than
4.75.
b. There is enough evidence to conclude that the mean number of hours is more than
4.5.
c. There is not enough evidence to conclude that the mean number of hours is morethan 4.5.
d. There is not enough evidence to conclude that the mean number of hours is morethan 4.75.
68. According to the Center for Disease Control website, in 2011 at least 18% of high school
students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY
conducted a hypothesis test at the local high school (a medium sized–approximately 1,200
students–small city demographic) to determine if the local high school’s percentage waslower. One hundred fifty students were chosen at random and surveyed. Of the 150 students
surveyed, 82 have smoked. Use this data with a significance level of 0.05 to test whether the
local high school’s percentage is less than 18%.
69. A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A
broker wanted to determine if this survey could be valid. He surveyed a random sample of
250 families and found that 142 owned some type of stock. At the 0.05 significance level, canthe survey be considered to be accurate? In other words, at the .05 level, does this data
provide evidence that the true proportion of families who own stock differs from 48.8%?
70. Driver error can be listed as the cause of approximately 54% of all fatal auto accidents,
according to the American Automobile Association. Thirty randomly selected fatal accidents
are examined, and it is determined that 14 were caused by driver error. Using  = 0.05, is
the AAA proportion accurate?

502

71. The US Department of Energy reported that 51.7% of homes nationwide were heated bynatural gas. A random sample of 221 homes in Kentucky found that 115 were heated bynatural gas. Does this data provide evidence to support the claim that the percentage ofhomes heated by natural gas differs from the national average? Use a significance level of α
= 0.05 to test.
72. For Americans using library services, the American Library Association claims that at
most 67% of patrons borrow books. The library director in Gurnee, Illinois feels this is nottrue, so she asked a local college statistic class to conduct a survey. The class randomly
selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that thepercentage was higher in Gurnee, IL? Use an α = 0.01 level of significance. What is thepossible proportion of patrons that do borrow books from the Warren Newport Library in
Gurnee?
73. The Weather Underground reported that the mean amount of summer rainfall for thenortheastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected
and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3
inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported
average? What if α = 0.01? Assume the amount of summer rainfall follows a normal
distribution.
74. A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4
minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’scommute time is less and wants to publicize this fact. The mean for 25 randomly selected
commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is
the Austin, TX commute significantly less than the mean commute time for the 15 largest
US cities?
75. A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8
times each year. A hospital administrator believes that the average is actually higher thanreported. A random sample of 20 women results in these yearly visit totals:
3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1
At the α = 0.05 level, can it be concluded that the mean number of visits for women is higher
than 5.8 visits per year?
76. According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A professor
claims that mean family size is greater than the reported number above. A sample of a college
math class at a large university resulted in the following family sizes:
5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2At α = 0.05 level, is there evidence to support the professor’s claim? Does this mean that the
Almanac result is incorrect? Why or why not?
503

77. According to a Harvard Health Publishing article, the normal body temperature may belower than 98.6 degrees according to data samples from the past 160 years. A sample of 12
random adults from a biology class had the following temperatures in Fahrenheit:
96.6;

97.5;

98.9;

96.4;

98.1;

98.9;

97.6;

97.8;

98.1;

97.9;

96.7;

98.6

At α = 0.05, is there evidence to support the traditional belief that the average bodytemperature is 98.6 degrees?
78. Normal resting heart rates for women are between 60 – 100 beats per minute. A group of
13 women who exercise three days a week at the local YMCA recorded their resting heartrate, and the data is below:
67;
72

71;

82;

56;

94;

67;

78;

83;

63;

62;

74;

90;

At α = 0.05, is there evidence to support the claim that the resting heart rate of women who
work three days a week at the YMCA is less than 80 beats per minute?
79. According to the CDC adults 18-60 years old need 7 or more hours of sleep per night. It
is claimed that the average sleep time is greater than 7 hours. The sleep times of a random
sample from a CLC statistics class who are adults between 18-60 are listed below:
8;

8.5;

5.5;

7.5;

6;

9;

8.5

7.5;

6.5;

9;

8;

7.5

At α = 0.05 can it be concluded that the average is more than 7 hours of sleep per night?
80. According to the CDC adults 61 – 64 years old need 7 – 8 hours of sleep per night. The
sleep time of a random sample of adults in a sleep study of adults between 61 – 64 are given
below:
6.5;
7

7;

9.5;

7.5;

6.5;

6;

10;

7;

6.5;

8.5;

9;

7.5;

At the α = 0.05 level, is there evidence to support the claim that the mean of the sleep studyis 7.5 hours per sleep at night?
81. IQ, or intelligence quotient, is one way to measure a person’s ability to reason andproblem solve. The IQ population standard deviation 15 points. A statistics teacher claims
that students in his classes have an IQ greater than 100. He samples 20 students, and the
IQ scores are listed below:
86;
105;

90;
95;

110;
103;

115;
114;

100;
102;

95;
99;

504

91;
120;

103;
120;

98;
93;

87;
101

At the α = 0.05 level, is there evidence to support the teacher’s claim?
82. Another statistics teacher claims that the students in her classes have an IQ of 100, whichis the average. The IQ population standard deviation is 15 points. She samples 24 students,
and the IQ scores are listed below:
100;
88;

87;
99;

90;
82;

93;
109;

110;
97;

99;
85;

89;
99;

102;
98;

101;
101;

87;
91;

101;
92;

112
96

Test the teacher’s claim using a 5% level of significance.
83. The student academic group on a college campus claims that freshman students study atleast 2.5 hours per day, on average. One Introduction to Statistics class was skeptical; theyclaim that freshmen study less than 2.5 hours per day on average. To test the claim the classselected a random sample of 30 freshman students and found a mean study time of 137
minutes with a standard deviation of 45 minutes. At α = 0.01 level, is there evidence tosupport the Stats class’ claim?
84. A billing company that collects bills for medical offices in the surrounding area claims
that there is a change in the percent of bills being paid by Medicare. In the past, thepercentage that is paid by Medicare is 30%. A study of 7500 recent bills shows that 33% of
these bills are being paid by Medicare. Test the claim using a 5% level of significance.
Show all 5 steps.

505

REFERENCES
Data from the National Institute of Mental Health. Available online at
http://www.nimh.nih.gov/publicat/depression.cfm.
Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
Data from Bloomberg Businessweek. Available online at http://www.businessweek.com/news/201109-15/nyc-smoking- rate-falls-to-record-low-of-14-bloomberg-says.html.
Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013). Data from
Gallup®. Available online at www.gallup.com (accessed June 27, 2013). Data from Growing byDegrees by Allen and Seaman.
Data from La Leche League International. Available online at
http://www.lalecheleague.org/Law/BAFeb01.html.
Data from the American Automobile Association. Available online at www.aaa.com (accessed June27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed
June 27, 2013).
Data from the Bureau of Labor Statistics. Available online at
http://www.bls.gov/oes/current/oes291111.htm.
Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov(accessed June 27, 2013)
Data from the U.S. Census Bureau, available online at
http://quickfacts.census.gov/qfd/states/00000.html (accessed June27, 2013).
Data from the United States Census Bureau. Available online at
http://www.census.gov/hhes/socdemo/language/.
Data from Toastmasters International. Available online at http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1.
Data from Weather Underground. Available online at www.wunderground.com (accessed June 27,2013).
Harvard Health Publishing. “Time to redefine normal body temperature?” Shmerling, Robert H.
March 13, 2020. Available online at http://www.health.harvard.edu/blog/time-to-redefine-normalbody-temperature-2020031319173 (Accessed July 14, 2021)
Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in theState of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at
http://www.disastercenter.com/kentucky/crime/ 3868.htm (accessed June 27, 2013).

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“Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available
online at http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf.
Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a
Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer
Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed
June 27, 2013).
Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009.
Available online at http://www.rainn.org/get-information/statistics/frequency-of-sexual-assault(accessed June 27, 2013).

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CHAPTER 10 SOLUTIONS:
1) a. The RV is x = the sample mean Internet speed in Megabits per second.
b. H0: μ < 3, Ha: μ > 33) a. The RV is x = mean number of children from a random sample of American families.
b. H0: μ = 2, Ha: μ ≠ 25) a. H0: p ≥ 0.42
b. Ha: p < 0.42
7) a. H0: μ = 15 b. Ha: μ ≠ 159) Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not$32,000.
Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is$32,000.
11) β = 1 – Power = 0.019
13) a. Power = 1 – β = 0.998 b. The Type II error15) a normal distribution for a single population mean17) It must be approximately normally distributed.
19) a. H0: μ > 73 Ha: μ < 73b. The p-value is almost zero, which means there issufficient data to conclude that the mean height of high school students who play basketballon the school team is less than 73 inches. The data do support the claim.
21) a. The shaded region shows a low p-value.
b. The p-value is the area in the right tail, so it is a right-tailed test.
23) a. It is a test for a population mean. b. x
c. the mean time spent in jail for 26 first time convicted burglars
d. σ = 1.5 e. x = 3, σ = 1.5, s = 1.8, n = 26.
f. Since we know σ = 1.5, we should use it; using s instead would cause errors in thecalculations.
g. This will be a Z-test; h. p-value = 0.0446
i. Reject Ho
j. There is enough evidence at 5% level of significance to support the claim that the mean
length of jail time has increased.
25) This is a left-tailed test. 27) This is a two-tailed test.
29) a right-tailed test 31) a left-tailed test 33) This is a left-tailed test.
35) a. H0: μ = 34; Ha: μ ≠ 34
b. H0: p ≤ 0.60; Ha: p > 0.60
c. H0: μ ≥ 100,000; Ha: μ < 100,000
d. H0: p = 0.29; Ha: p ≠ 0.29
e. H0: p = 0.05; Ha: p < 0.05
f. H0: μ ≤ 10; Ha: μ > 10
g. H0: p = 0.50; Ha: p ≠ 0.50
h. H0: μ = 6; Ha: μ ≠ 6
j. H0: μ ≤ 20,000; Ha: μ > 20,000
i. H0: p ≥ 0.11; Ha: p < 0.1137) a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years.
Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
b. Type I error: We conclude that more than 60% of Americans vote in presidential
elections, when the actual percentage is at most 60%. Type II error: We conclude that at
most 60% of Americans vote in presidential elections when, in fact, more than 60% do.

508

c. Type I error: We conclude that the mean starting salary is less than $100,000, when it
really is at least $100,000. Type II error: We conclude that the mean starting salary is atleast $100,000 when, in fact, it is less than $100,000.
d. Type I error: We conclude that the proportion of high school seniors who get drunk eachmonth is not 29%, when it really is 29%. Type II error: We conclude that the proportion ofhigh school seniors who get drunk each month is 29% when, in fact, it is not 29%.
e. Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los
Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that
5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
f. Type I error: We conclude that the mean number of cars a person owns in his or her
lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude
that the mean number of cars a person owns in his or her lifetime is not more than 10
when, in fact, it is more than 10.
g. Type I error: We conclude that the proportion of Americans who prefer to live away from
cities is not about half, though the actual proportion is about half. Type II error: We
conclude that the proportion of Americans who prefer to live away from cities is half when,
in fact, it is not half.
h. Type I error: We conclude that the duration of paid vacations each year for Europeans is
not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of
paid vacations each year for Europeans is six weeks when, in fact, it is not.
i. Type I error: We conclude that the proportion is less than 11%, when it is really at least
11%. Type II error: We conclude that the proportion of women who develop breast cancer is
at least 11%, when in fact it is less than 11%.
j. Type I error: We conclude that the average tuition cost at private universities is morethan $20,000, though in reality it is at most $20,000. Type II error: We conclude that the
average tuition cost at private universities is at most $20,000 when, in fact, it is more than
$20,000.
39) c
41) b
43) d
45) d
47) Hypotheses: H0: μ ≥ 50,000 , Ha : μ < 50,000
Use a Z-test since σ = 8000 is known. From the TI-84, z = -2.315 and p = 0.0103.
Since p < .05, we reject the null hypothesis. There is sufficient evidence to conclude that themean lifespan of the tires is less than 50,000 miles.
49) Hypotheses: H0: μ = $1.00
b. Ha: μ ≠ $1.00.
Use a Z-test since σ = .20 is known. From the TI-84, z = -0.866 and p = 0.3865.
Since p > .01, we do not reject the null hypothesis. There is not enough evidence to conclude
that the mean cost of daily papers is different from $1. The mean cost could reasonably be
$1.

509

51) Hypotheses: H0: μ = 10
Ha: μ ≠ 10
Use T-test because population SD is unknown. From TI-84, t = –1.12 and p = 0.300
Since p > 0.05, do not reject the null hypothesis. At the 5% significance level, there isinsufficient evidence to conclude that the mean number of sick days is different from 10.
53) Hypotheses: H0: p ≥ 0.6
Ha: p < 0.6
Use 1-PropZTest since testing a claim about a proportion.
From the TI-84, we have z = -1.12 and p-value = 0.1308.
Since p > .05, we do not reject Ho. There is not enough evidence to conclude that less than
60% of people who take Statistics feel more enriched.
55) Hypotheses: H0: μ = 4
Ha: μ ≠ 4
Use T-test because population SD is unknown. From TI-84, t = 1.95 and p-value = 0.076.
Since p > 0.05, we do not reject the null hypothesis. There is insufficient evidence toconclude that the average IQ of brown trout is different from four.
57) Hypotheses: H0: p ≥ 0.13 Ha: p < 0.13
Use 1-PropZTest since testing a claim about a proportion.
From the TI-84, we have z = -2.688 and p-value = 0.0036. Since p < 0.05, reject Ho.
There is sufficient evidence to conclude that the percentage of Americans who have seen or
sensed an angel is less than 13%.
59) Hypotheses: H0: p = 0.40 Ha: p < 0.40
Use 1-PropZTest since testing a claim about a proportion. From the TI-84, z = -1.01 and pvalue = 0.1563. Since p > 0.05, do not reject Ho. There is insufficient evidence to support
the claim that less than 40% of students at the school fear public speaking.
61) Hypotheses: H0: p = 0.14 Ha: p < 0.14
Use 1-PropZTest since testing a claim about a proportion. From the TI-84, z = -0.2756 andp-value = 0.3914. Since p > 0.05, do not reject Ho. There is insufficient evidence toconclude that the proportion of NYC residents who smoke is less than 0.14.
63) Hypotheses: H0: μ = 69,110
Ha: μ > 69,110.
Use T-test because population SD is unknown. From TI-84, t = 1.719 and p-value = 0.0466.
Since p < 0.05, we reject the null hypothesis. There is sufficient evidence to conclude thatthe mean salary of California registered nurses exceeds $69,110.
65) c
67) c
69) Hypotheses: H0: p = 0.488, Ha: p ≠ 0.488
Use 1-PropZTest since testing a claim about a proportion.
From the TI-84, z = 2.53 p-value = 0.0114. Since p < 0.05, reject the null hypothesis.
At the 5% level of significance, there is enough evidence to conclude that the percentage offamilies who own stocks is not 48.8%. So, the survey does not appear to be accurate.
71) Hypotheses: H0: p = 0.517, Ha: p ≠ 0.517
Use 1-PropZTest since testing a claim about a proportion.
From the TI-84, p-value = 0.9203. Since p > 0.05, do not reject the null hypothesis.
At the 5% significance level, there is not enough evidence to conclude that the proportion ofhomes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s510population is only the state of Kentucky. Second, it is reasonable to assume that homes inthe extreme north and south will have extreme high usage and low usage, respectively.

73) Hypotheses: H0: µ ≥ 11.52, Ha: µ < 11.52
Use T-test because population SD is unknown. From TI-84, p-value = 0.000002.
Since p < .05, Reject the null hypothesis. At the 5% significance level, there is enoughevidence to conclude that the mean amount of summer rain in the northeaster US is lessthan 11.52 inches, on average. Note that since the p-value is almost 0, we would reject Ho
at any level of significance.
75) Hypotheses: H0: µ ≤ 5.8 Ha: µ > 5.8
Use T-test because population SD is unknown. From TI-84, p-value = 0.9987.
Since p is much larger than .05, we do not reject the null hypothesis. At the 5% level ofsignificance, there is no evidence whatsoever to conclude that women visit their doctors anaverage of more than 5.8 times a year.
77) Hypotheses: H0: μ = 98.6, Ha ≠ 98.6
Use T-test because population SD is unknown. Two-tailed test with α = 0.05. From TI-84, pvalue = 0.0057.
Since p is smaller than 0.05, we reject the null hypothesis. At the 5% level of significance,
there is not enough evidence to support the claim that the class’s normal body temperature
is 98.6 degrees.
79) Hypotheses: H0 ≤ 7, Ha > 7
Use T-test because population SD is unknown. Right tailed test with α = 0.05. From TI-84,
p-value = 0.0409.
Since p is smaller than 0.05, we reject the null hypothesis. At the 5% level of significance,
there is enough evidence to support the claim that class’s mean is greater than 7 hours of
sleep.
81) Hypothesis: H0 ≤ 100, Ha > 100
Use Z-test because the population SD is known, 15. Right tailed test with α = 0.05. From
TI-84 p-value = 0.3437
Since p is greater than 0.05, we do not reject the null hypothesis. At the 5% level ofsignificance there is not enough evidence to support the teacher’s claim that the class mean
IQ is greater than 100.
83) Hypotheses: H0: µ ≥ 150 Ha: µ < 150 (measuring in minutes)
Use T-test because population SD is unknown. From TI-84, p-value = 0.0622.
Since p > 0.01, do not reject the null hypothesis. At the 1% significance level, there is notenough evidence to conclude that freshmen students study less than 2.5 hours per day, on
average. The student academic group’s claim appears to be correct.

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11 | HYPOTHESIS TESTING WITHTWO SAMPLESThis Photo by Unknown Author is licensed under CC BY-SA-NC

Figure 11.1 Student compares independent Marvel fans and DC Comic fans

Chapter Objectives
By the end of this chapter, the student should be able to:• Classify hypothesis tests by type.
• Conduct and interpret hypothesis tests for two population means when the population
standard deviations are known.
• Conduct and interpret hypothesis tests for two population means when the population
standard deviations are unknown.
• Conduct and interpret hypothesis tests for two population proportions.
• Conduct and interpret hypothesis tests for matched or paired samples.
• Construct and interpret two-sample confidence intervals.

513

Introduction
Studies often require the comparison of two unknown population means or two unknown
population proportions. For example, medical researchers have done several studies tomeasure the effect aspirin has in preventing heart attacks. Typically, the treatment groupis given aspirin and the control group is given a placebo. Then the heart attack rate for eachgroup is monitored and compared over several years.
There are other situations that deal with the comparison of two groups. For example, studies
compare various diet and exercise programs. Politicians compare the proportion ofindividuals from different income brackets who might vote for them. Students are interested
in whether SAT or GRE preparatory courses really help raise their scores.
Now that we have learned how to conduct hypothesis tests on single means and single
proportions, we will expand our methods to test claims involving two unknown means or two
unknown proportions. The basic framework – the Five Step Method – will be the same, but
the sampling distributions in which we do the calculations will be slightly different.
Here we will be comparing parameters from two different populations, using random samples
chosen from each. In the first three sections we will assume that the samples selected are
independent of one another. That is, the sample values selected from one population are
not related in any way to sample values selected from the other population. In Section 11.4
we will compare two population means using “matched pairs”; for this test the data consistof two samples that are dependent on one another.
NOTE
This chapter relies heavily on either a calculator or a computer to calculate thedegrees of freedom, the test statistics, and p-values. The TI-83+ and TI-84 calculators have
utilities for doing tests involving two independent samples. Instructions for the TI-83+ and
TI-84 calculators are included, as well as formulas for the test statistic of each test. We will
also be able to use these calculators to do the calculations related to matched pairs tests.

This chapter will develop the following hypothesis tests:• Test of two population means using independent samples (either 2SampTTest or
2SampZTest)• Test of two population proportions using independent samples (2PropZTest)• Test of two population means using a Matched Pairs design (samples are dependent).
• Test of the two population proportions by testing one population mean of differences.

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11.1 | Two Population Means with Known σ1 and σ2
Suppose we want to test a claim about two unknown population means, µ1 and µ2, where µ1
is the mean for population 1 and µ2 is the mean for population 2. Note that any inequality
relating these two parameters can be rewritten in terms of µ1 – µ2. For example, thestatement µ1 < µ2 can be rewritten as µ1 – µ2 < 0. Similarly, µ1 > µ2 can be rewritten as µ1– µ2 > 0. Thus, we can really think of claims comparing two different population means, µ1
and µ2, as claims involving a single new parameter µ1 – µ2. An obvious point estimate forthis parameter is the difference of sample means, x1 − x 2 , where x1 and x 2 are calculated

from random samples selected from populations 1 and 2, respectively.
Assumptions:

1. We are using independent, random samples selected from the two populations.
2. For the two populations from which we are sampling:ooIf the sample sizes are small, the distributions from which we are sampling should
be approximately normal.
If the sample sizes are both large (n > 30), then the distributions are not important;they need not be normal.

Since the parameter µ1 – µ2 is estimated by the difference of sample means, x1 − x 2 , we will
be working with the sampling distribution for the random variable X 1 − X 2 . This randomvariable represents the differences of all possible sample means when random samples of size
n1 and n2 are selected from the respective populations. Assuming that the population SD’sfor the two populations are σ1 and σ2, respectively the basic properties of this random variable
are as follows:

Sampling Distribution for Difference of Sample Means
Random Variable: X 1 − X 2
Mean:µ x1 − x2 = µ1 − µ 2Standard Deviation:

σ x1 − x2 =σ 12 σ 2 2+n1n2When we are sampling from normal distributions, or the sample size is large, then thesampling distribution is approximately normal: X 1 − X 2 ~ N(µ1 – µ2,515

σ 12 σ 2 2
+
)
n1
n2

Some important notes:•••µ1 and µ2 are the unknown population means.
σ1 and σ2 are the population standard deviations.
n1 and n2 are the sample sizes.

•

The variance of the distribution for X 1 − X 2 is the sum of the variances of thesampling distributions for X 1 and X 2 .

When testing claims about a single population mean, we used one of two sampling
distributions: either a normal distribution (z-test) or a t-distribution, depending on whether
or not the population standard deviation is known. This will be the case for two-sample
tests as well. When both of the population standard deviations are known, we will use a( x − x 2 ) − ( µ1 − µ 2 )Two-Sample Z-Test; the test statistic for this test is: z = 1. Note that forσ x1 − x2

most tests, we will start with a null hypothesis that includes an equal sign; that is, our null
hypothesis will generally include the case µ1 – µ2 = 0. So the numerator will be simplified,
and we have the test statistic:z=( x1 − x 2 )σ 12 σ 2 2+n1n2Once we calculate our test statistic, we could use the fact that the sampling distribution for

X 1 − X 2 is approximately normal and either compute the p-value or use an appropriate

critical value for our test. However, the TI-83 and TI-84 calculators have a built-in utility
that will give us both the test statistic and p-value for our test.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the test statistic z and p-value for a Two-Sample Z-test:_

Go to the STAT, menu and then to TESTS. Select option 3, which is 2-SampZTest.
Enter the population SD’s, the sample means and the sample sizes; select the appropriate
alternative hypothesis. Arrow down to Calculate and press ENTER. The test statistic andp-value will appear on the output screen. If we select the Draw option instead of Calculate,
we can view a graph showing the region corresponding to the p-value.
NOTE: The first inputs we are prompted for are the population standard deviations, σ1 andσ2. If these are not known, we should not be using 2SampZTest.

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Example 11.1
The mean lasting time of two competing floor waxes is to be compared. The amount of timethat each brand lasts is normally distributed. Each brand of wax is applied to 20 randomly
chosen floors for testing tested for how the finish lasts under normal wear; the data arerecorded in the table below:
Wax
Sample Mean
Brand A
3 months
Brand B
2.8 months

Population SD
0.33 months
0.36 months

Using a 5% significance level, does this data suggest that Brand A wax lasts longer, on
average, than Brand B?
Solution 11.1:
This is a test of two means, using independent samples. The random variable we will beworking with is X 1 − X 2 = difference in the mean number of months the competing floor

waxes last; and since the population standard deviations known, we can use 2SampZTest.

The claim is that Brand A wax lasts longer Brand B, on average. I.e. the claim is μA > μB.
So this is a right tail test, and the hypotheses can be written as either:
H0: μA ≤ μB
Ha: μA > μBor

H0: μA – μB ≤ 0
Ha: μA – μB > 0

However, the first is preferred since it more naturally shows the claim and also matches the
inputs for the calculator. Go to 2SampZTest, and enter the following:
σ1: 0.33
σ1: 0.36
x1 : 3

n1: 20x 2 : 2.8

n2: 20µ1: > µ2

Put the cursor on Calculate and press enter to get Z = 1.83 and p-value = 0.0335
Since p-value < .05, we reject H0. At the 5% significance level, there is sufficient evidence
to conclude that Brand A wax lasts longer, on average, than Brand B wax.

517

11.1 The means of the number of revolutions per minute of two competing engines are to becompared. Thirty engines are randomly assigned to be tested. Both populations havenormal distributions. The table below shows the result. Using a 5% level of significance, do
the data indicate that Engine 2 has higher RPM than Engine 1?
Engine12

Sample
Mean
1500
1600

Population
SD
50
60

Example 11.2
A field researcher is gathering data on the trunk diameters of mature pine and spruce
trees in a certain area. The following are the results of his random sampling.
Pine Trees Spruce Trees
Sample mean (cm)3530
Sample size (n)4080Population Variance160
160
Using a .10 significance level, can he conclude that the average trunk diameter of a pinetree is greater than the average diameter of a spruce tree?
Solution 11.2 using p-value:
This is a test of two means, using independent samples. The random variable we will beworking with is X 1 − X 2 , where X 1 represents a random sample mean diameter of pines, andX 2 is the random sample mean diameter of spruces.
(Step 1) Let µ1 be the population mean of diameters of mature pine trees, and µ2 thepopulation mean of diameters of mature spruce trees. The claim is that pines have larger
mean diameter, or μ1 > μ2.
H0: μ1 ≤ μ2
Ha: μ1 > μ2
(Step 2) So this is a right tail test because Ha contains a greater than symbol, >(Step 3) Since the population standard deviations are known, we can use 2SampZTest.
CALCULATOR: 2SampZTest, and enter the following:

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NOTE: The calculator asks for the standard
deviation, which is the square root of thevariance.

Put the cursor on Calculate and press enter to get Z = 2.04 and p-value = 0.0206.
(Step 4) Since p-value < .10, we will reject H0.
(Step 5) At the .10 significance level, there is sufficient evidence to conclude that the mean
diameter for pine trees is larger than the mean diameter for spruce trees.
Solution 11.2 using critical value:(Step 1) Let µ1 be the population mean of diameters of mature pine trees, and µ2 thepopulation mean of diameters of mature spruce trees. The claim is that pines have larger
mean diameter, or μ1 > μ2.
H0: μ1 ≤ μ2
Ha: μ1 > μ2
(Step 2) So this is a right tail test because Ha contains a greater than symbol, >(Step 3) Since the population standard deviations known, we can use 2SampZTest.
CALCULATOR: 2SampZTest
Put the cursor on Calculate and press enter to get Z = 2.04
(Step 4) a right tailed Z-test with significance level α = .10. The critical value for the test
will be z-value that has an area of .10 to the right of it; so the area to the left of this value is.90. This is easily found using the calculator:
Z.10 = invNorm(.90) = 1.282.
Thus, we would reject Ho if and only if the test statistic Z > 1.282. Since our test statistic
is Z = 2.04, our decision again is to reject Ho.
(Step 5) At the .10 significance level, there is sufficient evidence to conclude that the mean
diameter for pine trees is larger than the mean diameter for spruce trees.

519

For convenience we again show the rejection regions for the three most commonly usedsignificance levels:

SignificanceLevel

Rejection Regions
for Z-Test
Left-Tailed

Right Tailed

Two-Tailed

α = .10z < -1.28

z > 1.28

z < -1.645 or z > 1.645

α = .05z < -1.645

z > 1.645

z < -1.96 or z > 1.96

α = .01z < -2.33

z > 2.33

z < -2.576 or z > 2.576

Summary of Two-Sample Z-Test•Used for testing a claim about two unknown means µ1, µ2 with data fromindependent samples.

•Use only when the population standard deviations σ1, σ2 are known.

X 1 − X 2 ~ N (µ1 – µ2,•

Random variable and distribution:•

Test statistic: z =•Calculator function: 2-SampZTest( x1 − x 2 )σ 12 σ 2 2
+
n1
n2

520

σ 12 σ 2 2
+
)
n1
n2

11.2 | Two Population Means when σ1 and σ2 are Unknown
When we want to compare two unknown population means, it is often the case that the
population standard deviations are also unknown. The test comparing two independent
population means with unknown and possibly unequal population standard deviations iscalled the Two Sample T-Test, or Aspin-Welch t-test. (The formula for the degrees of freedom
developed by Aspin and Welch)
Again we are using independent samples, and working with the random variable X 1 − X 2 .

And again the mean of this sampling distribution is µ x1 − x2 = µ1 − µ 2 . But when we do not know

the population standard deviations, we must estimate them using the two sample standard
deviations (s1 and s2) from our independent samples. This in turn provides an estimate forthe standard deviation of the sampling distribution:

σ x1 − x2 ≈ s x1 − x2 =

2

2

s1
s
+ 2
n1
n2

The expression is called the standard error of the difference of sample means. The test
( x − x 2 ) − ( µ1 − µ 2 ), where µ1 – µ2 is the difference in means
statistic for the test will be t = 1
standard error
hypothesized in H0. As with the Two Sample Z-Test, we almost start with a null hypothesisthat includes an equal sign, so for our calculations we will be assuming that µ1 – µ2 = 0, giving
the simplified test statistic:t=( x1 − x 2 )
2

2

s1
s
+ 2
n1
n2

The number of degrees of freedom (df) requires a somewhat complicated calculation.
However, a computer or calculator will find it for us very easily. The df are not always a whole
number. The test statistic calculated previously is approximated by the Student's tdistribution with df as follows:2 s1 2 s 2 2 


+

 n1
n
2


df =
4
4
s1
s2
+ 2
2
n1 (n1 − 1) n 2 (n 2 − 1)521

It is not necessary to compute this by hand. The TI-83 and TI-84 calculator willcompute it for us automatically (thank goodness!)
The calculator instructions for a Two-Sample T-Test are essentially the same as for a TwoSample Z-test, except that there is one addition input. When using 2SampTTest, we will
be asked whether we want to “Pool the Data”. If we know that the population variances
are equal, then we can combine the two samples to get a better estimate of the common
population variance. This is called “pooling the data”; we only pool the data if we know
the population variances are equal.

Using the TI-83, 83+, 84, 84+ Calculator
To calculate the test statistic and p-value for a Two Sample T-Test:_

Go to the STAT, menu and then to TESTS. Select option 4, which is 2-SampTTest.
Enter the sample means, sample SD’s and sample sizes; select the appropriate alternative
hypothesis. For 2-SampTTest, we are also asked if we want to Pool the data, and use the
rule:
•
•

If we know that the population variances are equal, we say YES (pool the data)
If we do not know, or if the population variances are not equal, we say NO.

Arrow down to Calculate and press ENTER. The test statistic and p-value will appear on the
output screen. If we select the Draw option instead of Calculate, we can view a graph showing
the region corresponding to the p-value.
NOTE: Any time we are unsure of whether to pool the data, we should select NO.
Example 11.3
The average amount of time students in a hybrid Stats class spend on a unit exam iscompared to students in a traditional Stats class. A study is done and data are collected,
resulting in the data in the table below. Each population has a normal distribution.
Hybrid Stats Class
Traditional Stats Class

Sample Size
916

Sample Mean Sample SD
52 minutes
5.2
44 minutes
6.3

Using a there a 5% level of significance, is there a difference in the mean amount of time
hybrid stats students spend on a unit exam than traditional stats student?

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Solution 11.3 using p-value
The population standard deviations are not known. Let h be the subscript for hybrid and t
be the subscript for traditional. Then, μh is the population mean for girls and μt is the
population mean for boys. This is a test for two population means using independent
samples, and the population standard deviations are not known – so we use 2-SampTTest.
(Step 1) This will use the random variable𝑋𝑋̄ℎ − 𝑋𝑋̄𝑡𝑡 , which is the difference in the sample
means between hybrid and traditional. We are testing whether there is a difference, so theclaim is that μh ≠ μt and the hypotheses are:H0: μh = μt
Ha: μh ≠ μt
(Step 2) α = .05; two-tailed test.
(Step 3) Go to 2SampTTest, and enter the following:

Since we are unsure if the population standard deviations are equal, we state NO for
pooling.
Put the cursor on Calculate and press enter to get t = 3.42 and p-value = 0.0028. Recall,p-value is the probability of the test value and its extreme happening (both tails in thiscase)

(Step 4) Since p-value < .05, we will reject H0.

523

(Step 5) At the 5% significance level, there is sufficient evidence to conclude that there is adifference in the mean time spent taking unit exams between hybrid section students andtraditional section students.
Solution 11.3 using critical value(Step 1) We are testing whether there is a difference, so the claim is that μh ≠ μt and the
hypotheses are:H0: μh = μt
Ha: μh ≠ μt
(Step 2) α = .05; two-tailed test.
(Step 3) Go to 2SampTTest, and enter the following:

Since we are unsure if the population standard deviations are equal, we state NO for
pooling.
Put the cursor on Calculate and press enter to get t = 3.42. Note also that the degrees of
freedom is shown on the output screen: df = 19.55.
(Step 4) Find 2 critical values since it is two-tailed we use α/2 = .025 and df = 19.55
By calculator: ± invT(.025, 19.55) = ±2.09
By Excel: ± t.inv(.025, 19.55) = ±2.09
Reject Ho because 3.42 (test value) is within the rejection regions

524

(Step 5) At the 5% significance level, there is sufficient evidence to conclude that there is adifference in the mean time spent taking unit exams between hybrid section students andtraditional section students.

11.2 Data from two independent samples are shown in the table below. Both have normal
distributions. The means for the two populations are thought to be the same. Does thissample data provide evidence of a difference in the population means? Test at the 5% level ofsignificance.
Population A
Population B

Sample Size Sample Mean Sample SD
25
5
1
16
4.7
1.2

Example 11.4
A study is done by a community group in two neighboring colleges to determine which school
graduates students with more math classes. College A samples 11 graduates; the average forthis sample is four math classes with a standard deviation of 1.5 math classes. College B
samples nine graduates. Their average is 3.5 math classes with a standard deviation of one
math class. The community group believes that, on average, students who graduate from
College A have taken more math classes than students from College B. Assume that both
populations have a normal distribution, and we are testing a 1% significance level. Answer
the following questions.
a. Is this a test of two means or two proportions?
b. Are the population standard deviations known or unknown?
c. Which distribution do you use to perform the test?
d. What is the random variable?
e. What are the hypotheses for the test?
f. Is this test right, left, or two-tailed?
g. What is the p-value?
h. Should H0 be rejected or not?
i. What conclusion can we draw from the test?
Solution 11.4
a. Test for two meansb. These are unknown

c. Student's t-distribution

d. X A − X Be.

H0 : μA ≤ μB

vs. Ha : μA > μB

f. Right-tailed

525

g. From TI-84, p = 0.1928

h. Do not reject H0 since p-value > .01
i. There is not sufficient evidence at the 1% level of significance to conclude that a student
who graduates from college A has taken more math classes, on the average, than a student
who graduates from college B.

11.3 A study is done to determine if Company A retains its workers longer than Company B.
Company A samples 15 workers, and their average time with the company is five years witha standard deviation of 1.2. Company B samples 20 workers, and their average time with thecompany is 4.5 years with a standard deviation of 0.8. The populations are normally
distributed.
a. Are the population standard deviations known?
b. Conduct an appropriate hypothesis test of the claim using a 5% significance level. Whatis your conclusion?

Example 11.5
A professor at a large community college wanted to determine whether there is a differencein the means of final exam scores between students who took his statistics course online andthe students who took his face-to-face statistics class. He believed that the mean of the final
exam scores for the online class would be lower than that of the face-to-face class. The
randomly selected 30 final exam scores from each group are listed below:
Online class:67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8
58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8

88.2 70.6
79.4 79.4

Face-to-face Class:77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6
85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4
Is the mean of the Final Exam scores of the online class lower than the mean of the Final
Exam scores of the face-to-face class? We will test this claim with a 5% significance level.
Solution 11.5 using p-value method
We let μ1 be the population mean for online classes, and μ2 be the population mean for faceto-face classes. The claim being tested is that scores for online classes are lower, on average,
than scores for face-to-face classes.
(Step 1) the claim is that μ1 < μ2, and so the hypotheses are:526

H0: μ1 ≥ μ2 and Ha: μ1 < μ2
(Step 2) α = .05; left-tailed test
(Step 3) This is a test for two unknown population means, using independent samples andunknown population SD’s, so we use a 2-SampTTest.
STAT menu and then to EDIT; enter the sample data into two lists (e.g. L1, L2).
Arrow over to TESTS and select option 4: 2SampTTest.
Choose the Data option, and press ENTER. Arrow down and enter L1 for the first listand L2 for the second list.
Select the “<” alternative; do not pool the data (i.e. select Pooled: No) Arrow down to
Calculate and press ENTER to get:t = -3.23 (test value)

andp-value = 0.0011(Step 4) Since p-value < .05, we reject H0.
(Step 5) At the 5% level of significance, there is sufficient evidence to conclude that themean of the final exam scores for the online class is less than the mean of final exam scoresof the face- to-face class. So it appears that the professor was correct.
Solution 11.5 using critical value method(Step 1) the claim is that μ1 < μ2, and so the hypotheses are:H0: μ1 ≥ μ2 and Ha: μ1 < μ2
(Step 2) α = .05; left-tailed test

527

(Step 3) This is a test for two unknown population means, using independent samples andunknown population SD’s, so we use a 2-SampTTest.
t = -3.23 (test value) and df = 51.6
(Step 4) Find left critical value since it is left-tailed and state conclusion. Recall that a left
critical value of the t-distribution is negative. Reject Ho if the test value is in the rejection
region.
-tc = invT(.05, 51.6) = -1.675 or by Excel t.inv(.05, 51.6) = -1.675
Reject Ho because -3.23 is in the rejection region.

(Step 5) At the 5% level of significance, there is sufficient evidence to conclude that themean of the final exam scores for the online class is less than the mean of final exam scoresof the face- to-face class. So it appears that the professor was correct.

Cohen's Standards for Small, Medium, and Large Effect Sizes (Optional)
Cohen's d statistic is a measure of effect size based on the differences between two means.
Named for American statistician Jacob Cohen, d measures the relative strength of thedifferences between the means of two populations based on sample data. This statistic isthe difference of the observed sample means, divided by spooled =22(n1 − 1) s1 + (n 2 − 1) s 2.
n1 + n 2 − 2This is the sample standard deviation obtained by “pooling” the data – that is, by combining
the two samples into a single, large sample. Note that we can get this value from thecalculator; if we use 2-SampTTest and select YES for Pooling, the last entry in the output
screen is Sxp, which is the pooled standard error.
Now, the Cohen’s d-statistic is defined as d =( x1 − x 2 ).
s pooled

528

The calculated value of effect size is then compared to Cohen’s standards of small, medium,and large effect sizes according to the following table:
Size of Effect

D

Small

0.2

Medium

0.5

Large

0.8

Cohen's Standard Effect Sizes

Example 11.6
Calculate Cohen’s d for Example 11.4. Is the size of the effect small, medium, or large?
Explain what the size of the effect means for this problem.
Solution 11.6

x1 = 4, s1 = 1.5, n1 = 11;

x 2 = 3.5, s2 = 1, n2 = 9

Go to 2-SampTTest, and enter the data; select YES for pooling, then scroll to Calculate andpress Enter to get spooled = Sxp = 1.302. Using the formula, we getd = ( x1 – x 2 )/spooled = (4 – 3.5)/1.302 = 0.384.

We would classify the effect as small because 0.384 is between Cohen’s value of 0.2 forsmall effect size and 0.5 for medium effect size. Since the effect size for the data is small,this indicates that there is not a significant difference between the means for the twocolleges.
Example 11.7
Calculate Cohen’s d for Example 11.3. Is the size of the effect small, medium or large?
Explain what the size of the effect means for this problem.

Solution 11.7
Go to 2-SampTTest, and enter the data; select YES for pooling, then scroll to Calculate andpress Enter to get spooled = Sxp .

529

spooled = Sxp = 5.94
Using the formula, we getd = ( x1 – x 2 )/spooled = (52 – 44)/5.94 = 1.347
This indicates a large effect size, because 1.347 is greater than Cohen’s cutoff of 0.8 for alarge effect size. The size of the differences between the means of hybrid sections andtraditional sections for unit exams indicating a significant difference in the means.

11.4 Weighted alpha is a measure of risk-adjusted performance of stocks over a period of ayear. A high positive weighted alpha signifies a stock whose price has risen while a small
positive weighted alpha indicates an unchanged stock price during the time period. Weightedalpha is used to identify companies with strong upward or downward trends. The weightedalpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq
on May 24, 2013 are listed in the tables below:.
94.2 75.2 69.6 52
48
41.9 36.4 33.4 31.5 27.6
77.3 71.9 67.5 50.6 46.2 38.4 35.2 33
28.7 26.5
76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.6 26
Northeast

126
70.6 65.2 51.4 45.5 37
33
29.6 23.7 22.6
116.1 70.6 58.2 51.2 43.2 36
31.4 28.7 23.5 21.6
78.2
68.2 55.6 50.3 39
34.1 3125.3 23.4 21.5
West

Is there a difference in the mean weighted alpha for banks in the northeast and in thewest? Test at a 5% significance level. Explain what the size of the effect means for thisproblem.

530

Summary of Two-Sample T-Test•Used for testing a claim about two unknown means µ1, µ2 with data fromindependent samples.

•

Use when sampling from two normal distributions and the population standard
deviations σ1, σ2 are not known.
2

 s1 2 s 2 2 


+

 n1
n
2


df =
4
4
s1
s2
+ 2
2
n1 (n1 − 1) n 2 (n 2 − 1)

( x1 − x 2 )•Test statistic: t =•Calculator function: 2-SampTTest (pooled or not pooled)•

Select the “Pooled” option only when the population variances are equal.
If in doubt, do not pool the data.

2

2

s1
s
+ 2
n1 n 2

11.3 | Comparing Two Population Proportions
In this section we develop a test for testing claims involving two unknown population
proportions, p1 and p2. As we did for comparing unknown means, we will look at theparameter p1 – p2, the difference of the two population proportions. The obvious point
estimate for this parameter is pˆ 1 − pˆ 2 , where p̂1 and p̂ 2 are sample proportions chosen fromthe respective populations. When conducting tests comparing two population proportions (p1 − p2 =hypothesized value), we will assume the following:
1. We are using two independent, random samples.
2. The number of successes is at least 10, and the number of failures is at least 10, for eachof the samples. That is, we should have n1 p1 ≥ 10 and n1 (1 − p1 ) ≥ 10 as well as n2 p2 ≥ 10 andn2 (1 − p2 ) ≥ 10 .
These assumptions imply that the sampling distribution for pˆ 1 − pˆ 2 will be approximately

normal. The mean of the sampling distribution will be p1 – p2, and the variance will be thesum of the variances of the sampling distributions for p̂1 and p̂ 2 . That is, we have:µ pˆ1 − pˆ 2 = p1 − p 2 and σ pˆ − pˆ = p1 q1 + p 2 q 2 .
1

2

531

n1

n2

Generally, the null hypothesis will state that the two population proportions are the same.
That is, H0: p1 = p2; and since the population proportions are equal, the variances of theindividual sampling variations will also be equal. So we can get a better approximation tothis common variance by using a pooled proportion, pc. This is the total number of successes,
divided by the total sample size: p c =x1 + x 2. Inserting pc in place of p1 and p2 in the formulan1 + n 2

above and simplifying, we get: σ pˆ1 − pˆ 2 =1 1
p c (1 − p c ) + 
 n1 n 2 

Now the test statistic for testing a claim about two proportions will be:z=( pˆ 1 − pˆ 2 ) − ( p1 − p 2 ).
σ pˆ1 − pˆ 2

Because the null hypothesis assumes that p1 and p2 are equal, we have p1 – p2= 0;
substituting the expression for σ pˆ1 − pˆ 2 above, we get the test statistic:z=pˆ 1 − pˆ 2
11 
p c (1 − p c ) + 
 n1 n 2 .

This may look complicated, but it is programmed into the TI-83 and TI-84 calculators. As
with other tests, the calculator will give us both the test statistic and the p-value.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the test statistic and p-value for a Two Proportion Z-Test:_

Go to the STAT, menu and then to TESTS. Select option 6, which is 2-PropZTest.
Enter the values for x1, n1, x2, and n2.
Note that both x1 and x2 must be whole numbers; so if you are given sample
proportions, use x1= n1 p̂1 and x2 = n 2 p̂ 2 , and round to the nearest whole number.
Select the appropriate alternative hypothesis.
Arrow down to Calculate and press ENTER.
The test statistic and p-value will appear on the output screen. Note that the pooled
proportion pc also shows on the output screen as p̂ (without a subscript). As withother tests, we can select the Draw option instead of Calculate to view a graph showing
the region corresponding to the p-value.

532

0 which also includes ≥ and ≤.
NOTE: the following summary is focusing on H 0 : p1 − p2 =Test value: z =pˆ 1 − pˆ 2
11 
p c (1 − p c ) + 
 n1 n 2 Assumptions: 1. The samples are independently chosen random samples
2. Both sample sizes are large:ˆ 2 ≥ 10 ; n 2 (1 − pˆ 2 ) ≥ 10 .
n1 pˆ 1 ≥ 10 ; n1 (1 − pˆ 1 ) ≥ 10 ; n 2 p

Example 11.8
Two types of medication for hives are being tested to determine if there is a difference in the
proportions of adult patient reactions. Twenty out of a random sample of 200 adults given
medication A still had hives 30 minutes after taking the medication. Twelve out of anotherrandom sample of 200 adults given medication B still had hives 30 minutes after taking the
medication. Test the claim that there is a difference in the proportions using a 1% level ofsignificance.
Solution 11.8 using p-value method
This is a test of two proportions. Let pA and pB be population proportions of individual withhives after 30 minutes for medications A and B respectively. The problem asks if there is adifference in proportions,(Step 1) the claim being tested is that pA ≠ pB and so the hypotheses are:H0: pA = pB and Ha: pA ≠ pB
(Step 2) α = .01; two-tailed test
(Step 3) we have data from independent random samples. From the given information, we
can see that the number of successes and failures in each sample is at least 10, so thesampling distribution for pˆ A − pˆ B will be approximately normal. Thus, we will use

2-PropZTest. In the calculator, go to the STAT menu, then to TESTS and select 2PropZTest.

533

Enter the following:
x1: 20n1: 200
x2: 12n2: 200
p : ≠ p0

Note: both x1 and x2 must be whole numbers; so if you are
given sample proportions, use x1= n1 p̂1 and x2 = n 2 p̂ 2 , andround to the nearest whole number.

Scroll down to Calculate and press Enter to get:z = 1.47 and p-value = 0.1404.
Using the Draw feature, we get a graph that looks like:(Step 4) Since p-value = 0.1404 > 0.01, we do not reject H0.
(Step 5) Thus, at a 1% level of significance, this sample data does not provide sufficientevidence to conclude that there is a difference in the proportions of adult patients who did
not react after 30 minutes to medication A and medication B.
Solution 11.8 using critical value method
difference in proportions,(Step 1) the claim being tested is that pA ≠ pB and so the hypotheses are:H0: pA = pB and Ha: pA ≠ pB
(Step 2) α = .01; two-tailed test
(Step 3) 2propZtest, z = 1.47
(Step 4) Find two-tailed critical value for the Z-distribution±Zc = ±invnorm(.005, 0, 1) = ±2.58
Do not reject Ho because 1.47 is not in the rejection region.

534

(Step 5) Thus, at a 1% level of significance, this sample data does not provide sufficientevidence to conclude that there is a difference in the proportions of adult patients who did
not react after 30 minutes to medication A and medication B.

11.5 Two types of valves are being tested to determine if there is a difference in pressure
tolerances. In a random sample of 100 Brand A valves, 14 cracked under 4,500 psi. In arandom sample of 100 of Brand B valves, 19 cracked under 4,500 psi. Using a 5% level ofsignificance, test the claim that there is a difference in the proportion of valves that crack at
4,500 psi.

Example 11.9
PC Magazine wrote an article that stated IOS is more popular in US than UK. IOS is theoperating system used by iPhone. Is there sufficient evidence to conclude that the
proportion of IOS users in USA is greater than the proportion of IOS users in the UK?
USA UK
IOS
206
161
Total surveyed 355
310
Test at a 1% level of significance.
Solution 11.9 using p-value method
This is a test of two population proportions. Let USA and UK be the subscripts for US and
UK. Let 𝑝𝑝𝑈𝑈𝑈𝑈𝑈𝑈 and 𝑝𝑝𝑈𝑈𝐾𝐾 be the proportions for USA and UK respectively. The claim is that the
proportion of IOS users in US is greater than the proportion of IOS users in UK:(Step 1) Claim is 𝑝𝑝𝑈𝑈𝑈𝑈𝑈𝑈 > 𝑝𝑝𝑈𝑈𝐾𝐾 . Thus we have the hypotheses:
H0: 𝑝𝑝𝑈𝑈𝑈𝑈𝑈𝑈 ≤ 𝑝𝑝𝑈𝑈𝐾𝐾

Ha: 𝑝𝑝𝑈𝑈𝑈𝑈𝑈𝑈 > 𝑝𝑝𝑈𝑈𝐾𝐾

535

(Step 2) α = .01 the test is right-tailed test
(Step 3) In the TI-84, go to STAT, then TESTS, and select 2-PropZTest.
Enter the following:
x1: 206 ,n1: 355 ,x2: 161,

n2: 310,

p : > p0

Scroll down to Calculate and press Enter to get:z = 1.57 and p-value = 0.0575.
(Step 4) Since p = .0575 > .01, we do not reject H0.
(Step 5) At the 1% level of significance, there is not sufficient evidence to conclude that the
proportion of USA IOS users is greater than the proportion of UK IOS users.

Example 11.10
US department of education did a study on undergraduate students and compared the
proportion of students who change their major. It is claimed that the proportion ofundergraduate students in a Bachelor’s degree program is the same as the proportion ofundergraduate students in an Associate’s degree program that change their major. Use 5%
level of significance to test the claim.
Bachelor’s Associate’s
Changed majors 205
196
Total surveyed
620
700
Solution 11.10 using p-value method
Let pB and pA are be the proportions of students in Bachelor’s degree programs andproportions of students in Associate degree programs, respectively.
(Step 1) claim pB = pA
H0: pB = pA
H a : p B ≠ pA

536

(Step 2) α = .05; two-tailed test.
(Step 3) 2-PropZTest and enter the following:
x1: 205
n1: 620x2: 196n2: 700
p : ≠ p0
Scroll down to Calculate and press Enter to get: z = 2 and p-value = 0.0458(Step 4) Since p-value < .05, we reject H0.
(Step 5) At the 5% level of significance, there is sufficient evidence to conclude the proportionof students in Bachelor’s degree is the same as the proportion of students in Associate’s
degree that switch majors.
Solution 11.10 using critical value method
proportions of students in Associate degree programs, respectively.
(Step 1) claim pB = pA
H0: pB = pA
H a : p B ≠ pA
(Step 2) α = .05; two-tailed test.
(Step 3) 2-PropZTest and enter the following:
x1: 205
n1: 620x2: 196n2: 700
p : ≠ p0
Scroll down to Calculate and press Enter to get: z = 2(Step 4) Find critical values which is both left and right since it is two-tailed test. ±Zc =±invnorm(.025, 0, 1) = ±1.96. Reject Ho because 2 (test value) is in the right rejection region.
(Step 5) At the 5% level of significance, there is sufficient evidence to conclude the proportionof students in Bachelor’s degree is the same as the proportion of students in Associate’s
degree that switch majors.

537

11.6 Among new graduates who have student loans, the department of education looks at the
number of students who default for the first time. The percentage of students in public
institutions that are in default is 10.8%. They looked at a sample of 2100 students. The
percentage of student in private for-profit institution that are in first time default is 15%.
They looked at 1500 students. At a 5% significance level, is there a difference in the
proportions?

Summary of Two-Proportion Z-Test•Used for testing a claim about two unknown proportions p1, p2 with data fromindependent samples.

•

Assume that n1 pˆ 1 ≥ 10 , n1 (1 − pˆ 1 ) ≥ 10 and that n 2 pˆ 2 ≥ 10 and n 2 (1 − pˆ 2 ) ≥ 10 .

That is, there are at least 10 successes and 10 failures in each sample.

pˆ 1 − pˆ 2•

Test statistic: z =•Calculator function: 2-PropZTest11 
p c (1 − p c ) + 
 n1 n 2 , where p c =538

x1 + x 2
n1 + n 2

11.4 | Matched Pairs Test
Two samples are dependent when the subjects are paired or matched in some way. For
example, to study whether or not a medication helps lower cholesterol, researchers mighttake a random sample of patients and measure their cholesterol. Then after a specified timetaking the medication, the patients would have their cholesterol measured again – this would
produce pairs of “before” and “after” data values, with one pair for each patient.
We could then analyze the actual differences, D = before – after, of the pairs using an ordinary
t-test. Thus the data is the gain or loss in cholesterol readings. Such a test is called amatched-pairs test. One of the key advantages of this experiment design is that it controls
for a lot of individual attributes of the patients, such as diet, exercise, overall health, etc.
When using a hypothesis test for matched or paired samples, the following conditions must
be met:1. Simple random sampling is used.
2. Two measurements are drawn from the same pair of individuals.
3. Differences are calculated for each matched pair.
4. The differences comprise the sample that is used for the hypothesis test.
5. Either the population of differences are normally distributed, or else the number ofdifferences in the sample is sufficiently large to ensure that the sampling distribution isapproximately normal.
Notation: We will use d to represent individual differences, so d is the sample mean ofthe differences, and µd is the population mean of all differences. From the Central Limit
Theorem, we get the mean and standard deviation of the sampling distribution for d :µ d = µ d and s d = s d .
nFrom this, we get a formula for the test statistic: t =d − µdsd / n.

Of course, we will rarely (if ever) use this formula, since we can get the test statistic, as wellas the corresponding p-value from the calculator.
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the test statistic and p-value for a dependent T test:
First place the individual differences into L1
2nd Quit, Press STAT, go to TESTS
T-test, choose DATA, enter the followingµ0: 0; LIST: L1; FREQ: 1; highlight correct Ha symbol (≠, <, >)539_

Example 11.11
A study was conducted to investigate the effectiveness of hypnotism in reducing pain.
Results for randomly selected subjects are shown in the table:
Subject:
Before
After

A
6.6
6.8

B
6.5
2.4

C
D
E
9.0 10.3 11.3
7.4 8.5
8.1

F
8.1
6.1

G
H
6.3 11.6
3.4
2.0

A lower score indicates less pain. The "before" value is matched to an "after" value and thedifferences are calculated. Assume that the population of differences has a normal
distribution. Using a 5% significance level, is there evidence that the sensory measurements
are lower, on average, after hypnotism?
Solution 11.11 using p-value method
The corresponding "before" and "after" values form matched pairs; we will let
d = "before" – "after"
If the sensory readings are lower after hypnotism, then “before” readings should, on
average, be larger than “after” readings and so we would expect the mean of all d values tobe positive.
(Step 1) Thus the claim we want to test is µd > 0. So the hypotheses are:H0: µd < 0, Ha: µd > 0(Step 2) α = .05; right-tailed test.
(Step 3) Now we calculate the differences; we can easily do this by adding another row to
our table:
Subject:
BeforeAfter
Difference:

A
6.6
6.8
-0.2

B
6.5
2.4
4.1

C
D
E
9.0 10.3 11.3
7.4 8.5
8.1
1.6 1.8
3.2

Go to t he STAT menu and then to EDIT. Enter thedifferences into L1.
Go to STAT again, then to the TESTS menu and selectT-test. Select the DATA option.
Set µ0 : 0, enter the list name (type 2nd 1 to get L1),and select the “greater than” alternative.
Scroll down to Calculate and hit Enter to get:t = 3.036 and p-value = 0.0095

540

F
8.1
6.1
2.0

G
H
6.3 11.6
3.4 2.0
2.9 9.6

(Step 4) Reject Ho because 0.0095 < 0.05
(Step 5) At the 5% level of significance, there is sufficient evidence to conclude that thesensory measurements are lower, on average, after hypnotism. Hypnotism appears to beeffective in reducing pain.
NOTE:
For the TI-84 calculator, you can either calculate the differences (before – after)
ahead of time and put the differences directly into a list, or you can put the before
data into one list L1 and the after data into a second list, L2. Then go to L3 andarrow up to the name; that is, put the cursor on L3 at the top of the screen.
Type L1 – L2 and press Enter; the calculator will do the subtractions and store thedifferences List 3.

Solution 11.11 using critical value method(Step 1) Thus the claim we want to test is µd > 0. So the hypotheses are:H0: µd < 0, Ha: µd > 0(Step 2) α = .05; right-tailed test.
(Step 3) Now we calculate the differences; we can easily do this by adding another row to
our table:
Subject:
BeforeAfter
Difference:t = 3.036

A
6.6
6.8
-0.2

B
6.5
2.4
4.1

C
D
E
9.0 10.3 11.3
7.4 8.5
8.1
1.6 1.8
3.2

F
8.1
6.1
2.0

G
H
6.3 11.6
3.4 2.0
2.9 9.6(test value)(Step 4) Find critical value. Since it is a t-test and right-tailed. We find the right tc.
tc = -inv.T(.05, 7) = 1.895 where d.f. = n – 1 = 7
Reject Ho because 3.036 is in the rejection region.
(Step 5) At the 5% level of significance, there is sufficient evidence to conclude that thesensory measurements are lower, on average, after hypnotism. Hypnotism appears to beeffective in reducing pain.

541

11.7 A study was conducted to investigate how effective a new diet was in loweringcholesterol. Results for the randomly selected subjects are shown in the table. The
differences have a normal distribution. Are the subjects’ cholesterol levels lower on averageafter the diet? Test at the 5% level.
Subject:
A
B
C
D
E
F
G
H
I
Before209 210 205 198 216 217 238 240 222
After
199 207 189 209 217 202 211 223 201

Possible claim scenarios for Matched Pair Tests:
1.) There is a decrease from before to after, which means after is “lower” than
before: this scenario will result in the average difference being positive, µd > 0.
2.) There is an increase from before to after, which means after is “higher” than
before: this scenario will result in the average difference being negative, µd < 0.
3.) There is a difference from before to after, which means after is not the same as
before: this scenario will result in the average difference being not equal to 0,µd ≠ 0.
4.) There is no difference from before to after, which means after is the same as
before: this scenario will result in the average difference being equal to 0, µd = 0.
Example 11.12
A pharmaceutical company wishes to test a new drug with the expectation of loweringcholesterol levels. Ten subjects are randomly selected and pretested. The subjects wereplaced on the drug for a period of 6 months, after which their cholesterol levels were tested
again.
The test results, before and after, are listed below. (All units are milligrams per deciliter.)
Subject
Before
After

1
195
180

2
225
220

3
202
210

4
195
175

5
175
170

6
250
243

7
235
205

8
268
250

9
190
183

10
240
225

Use this data, along with a 1% significance level, to test the claim that the drug is effective
in lowering cholesterol.

542

Solution 11.12 using p-value method
Here we are comparing two population means, using a matched pairs design. Thus, we will
use a t-test on the differences: 15, 5, -8, 20, 5, 7, 30, 18, 7, 15
Let d = cholesterol level before taking drug – cholesterol level after taking drug.
(Step 1) If the drug lowers cholesterol, then the cholesterol level should behigher before the drug so if the claim is true, we expect mean of the differences to bepositive, giving us the hypotheses:
Ho: μd < 0
Ha: μd > 0(claim)(Step 2) α = .05; right-tailed test.
(Step 3) Enter the differences into a list, and go to STAT >> TESTS >> TTest. Enter 0 forµ0, specify the list name, and select the > µ0 alternative.

This gives us a test statistic of t = 3.44, and a p-value = .0037.
(Step 4) Since p-value < 0.01, we reject Ho.
(Step 5) There is sufficient evidence to conclude that the mean of the differences is negative.
Thus, there is sufficient evidence to conclude that the drug helps lower cholesterol.

11.8 A new prep class was designed to improve SAT test scores. Nine students wereselected at random. Their scores on two practice exams were recorded, one before the classand one after. Are the scores, on average, higher after the class? Test at a 5% level. The
data recorded in the table below:
Student
1st Score
2nd Score

A
480
400

B
510
520

C
530
550

D
540
530

543

E
550
580

F
560
580

G
600
610

H
620
640

I
660
690

Example 11.13
Seven eighth graders at Kennedy Middle School measured how far they could push the shotput with their dominant (writing) hand and their weaker (non-writing) hand. They thoughtthat they could push equal distances with either hand. The data were collected and recordedin the table below; the distances shown are in feet.

Dominant
Hand
Weaker Hand

Student1

Student2

Student3

Student
4

Student
5

Student
6

Student
7

28

14

27

18

17

26

16

30

26

34

17

19

26

20

Conduct a hypothesis test to determine whether there is a significant difference in mean
distances between the children’s dominant and weaker hands. Use a significance level of α
= .05.
Solution 11.13 using p-value method
First, we find the differences for each pair. Here we let
d = distance for dominant hand – distance for weaker hand.
(Step 1) Since we asked whether there is a difference in the means:
H0: μd = 0,

Ha: μd ≠ 0(Step 2) α = .05; two-tailed test.
(Step 3)

We will assume that the differences have a normal distribution, and will use a T-test.

Enter the differences into a list, and go to STAT >> TESTS >> TTest. Enter 0 for µ0, specify
the list name, and select the ≠ µ0 alternative.

This gives us a test statistic of t = 2.18, and a p-value of p = .0716.

544

(Step 4) Since p > 0.05, we do not reject Ho.
(Step 5) Thus, there is not enough evidence to conclude that there is a significant differencein the means.
Note that if we had selected the DRAW option, we would see the following graph:

The graph shows that had this been a right-tailed test, our decision would have been to reject
Ho. However, our hypotheses are based on the question that has been asked, not on thesample data. And it would not be appropriate to change the hypotheses once we saw the datain order to get a different result.
Notice the other statistics that are given on the calculator screen (𝑥𝑥̅ and Sx). These values
are the sample average difference and sample standard deviation of the differences. The
symbols that should be used for the values are as follows:𝑑𝑑̅ = 3.71 and Sd = 4.5.

545

11.5 | Confidence Intervals from Two Samples
In this chapter, we have learned how to investigate claims involving two parameters; but
sometimes we actually need an estimate of the difference of two parameters. For example, if
we conducted a hypothesis test and conclude that µ1 > µ2, a natural follow-up question might
be: by how much? Using a confidence interval, we can actually give bounds for how large thedifference between these parameters really is. In this section we will extend the ideas of
Chapter 9 to develop confidence interval estimates for the difference of two means and thedifference of two proportions. Like the intervals developed in Chapter 9, these will have thebasic form:(point estimate – margin of error,

point estimate + margin of error)

And we now have all of the ingredients needed. For example, if we want to find a confidence
interval for µ1 – µ2 using data from two independent samples, we know that the appropriate
point estimate is the difference of sample means, x1 − x 2 ; and this random variable will either

follow a normal distribution or a t-distribution. When the population standard deviations
are known, we can use a z-distribution, and the margin of error will be
E = z α / 2σ x1 − x2 = z α / 2

σ 12
n1

+

σ 22
n2

( x1 − x 2 ) − zα / 2. Thus, the confidence interval will have the form:
σ 12
n1

+

σ 22
n2

< µ1 − µ 2 < ( x1 − x 2 ) + z α / 2

σ 12
n1

+

σ 22
n2

.

When the population standard deviations are not known, then we would estimate σ1 and σ2
by s1 and s2 and use a critical t values in place of the Z critical values:( x1 − x 2 ) − t α / 2

2

2

s1
s
+ 2 < µ1 − µ 2 < ( x1 − x 2 ) + t α / 2
n1
n2

2

2

s1
s
+ 2 .
n1
n2

While these formulas look complicated, we should remember that they are really the same
basic format as the intervals for a single mean µ that we developed in Chapter 9 (but the
point estimates and standard errors are different).
Moreover, we never need to actually use these formulas, because they are built into the TI83 and TI-84 calculators.

546

Using the TI-83, 83+, 84, 84+ Calculator
To calculate a confidence interval for the difference of two means, µ1 – µ2:_

Go to the STAT, menu and then to TESTS.
If the population SD’s are known, then Select option 9, which is 2-SampZInt.
If the population SD’s are not known, then Select option 10, which is 2-SampTInt.
Enter the SD’s, the sample means and the sample sizes just as we did for two-sample tests.
Specify the confidence level. Arrow down to Calculate and press ENTER. The confidence
interval will appear on the screen.
To illustrate this, we revisit the question from Example 10.2; in that example, we used ahypothesis test to compare the mean trunk diameters for pine spruce trees. In that test,
we concluded that the average trunk diameter for pine tree is in fact greater than
the average diameter of spruce trees.
Example 11.14
A field researcher is gathering data on the trunk diameters of mature pine and sprucetrees
in a certain area. The following are the results of his random sampling.
Sample mean (cm)Sample size
Population Variance

Pine Trees
35
40
160

Spruce Trees
30
80
160

Use this sample data to find a 90% confidence interval for the difference in mean trunk
diameter for pine and spruce trees. Interpret this interval.
Solution 11.14
We let µ1 be the mean diameter for pine trees, and µ2 be the mean diameter for spruce
trees.
We will calculate a confidence interval for µ1 – µ2; the population variances are known, so
we will use 2-SampZInt.
Go to 2SampZInt, and enter the following:547

Hit CALCULATE to get the interval. So we are 90% confident that, 0.971 < µ1 – µ2 < 9.03.
That is, the mean for pine trees is at least 0.971 inches more than the mean for spruce trees.
And, the mean for pine trees is at most 9.03 inches more than the mean for spruce trees.
Recall also Example 11.5; in that example, a professor at a large college wanted to compare
the mean final exam scores for two populations of students – online students and students intraditional, face to face courses. We conducted a test and found evidence that the mean score
for online classes was lower than the mean for traditional courses. Here we will calculate aconfidence interval for the difference of means.
Example 11.15
A professor at a large community college wanted to compare the mean final exam scores
between students who took his statistics course online and the students who took his face-toface statistics class. The randomly selected 30 final exam scores from each group are listed
below:
Online class:67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6
58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4
Face-to-face Class:77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6
85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4
Calculate a 95% confidence interval for the difference in mean scores; interpret thisinterval.
Solution 11.15
We let μ1 be the population mean for online classes, and μ2 be the population mean for faceto-face classes. We want a confidence interval for μ1 – μ2; the population SD’s are notknown, so will use 2-SampT-Int. Moreover, instead of sample statistics, we are given raw
data, so we will use the DATA option.
- First go the STAT menu and then to EDIT; enter the sample data into two lists (e.g. L1,
L2).
- Go to STAT again, then to TESTS; select option 10: 2SampTInt. Choose the DATA option.
- Arrow down and enter L1 for the first list and L2 for the second list.
- Set C-Level: .95- Set Pooled: No (since we have no reason to think the population variances are equal)- Arrow down to Calculate and press Enter to get the interval: (-19.67, -4.59).
Thus, we are 95% confident that -19.67 < µ1 – µ2 < -4.59.

548

The mean for online courses is at least 4.59 points lower than the mean for traditionalcourses. And, the mean for online courses is at most 19.67 points lower for traditionalcourses.

We can also find a confidence interval for p1 – p2, the difference of the two population
proportions. The point estimate for this parameter is pˆ 1 − pˆ 2 , where p̂1 and p̂ 2 are sample
proportions chosen from the respective populations. And we know that if the sample sizes
are sufficiently large (see Section 10.3), then the sampling distribution for pˆ 1 − pˆ 2 will beapproximately normal, so the margin of error will be E = zα / 2σ pˆ1 − pˆ 2 = zα / 2

p1 q1 p 2 q 2+. Thusn1
n2

we get the following formula for the confidence interval:( pˆ 1 − pˆ 2 ) − zα / 2

p1 q1 p 2 q 2+< p1 − p 2 < ( pˆ 1 − pˆ 2 ) + zα / 2n1
n2

p1 q1 p 2 q 2+
n1n2

Again, we would rarely (if ever) use this formula, as it is built into the TI-84 calculator:
Using the TI-83, 83+, 84, 84+ Calculator
To calculate a confidence interval for the difference of two means, p1 – p2:_

Go to the STAT, menu and then to TESTS. Select option B, which is 2-PropZInt.
Enter the values for x1, n1, x2, and n2. Note that both x1 and x2 must be whole numbers; so if
you are given sample proportions, use x1= n1 p̂1 and x2 = n 2 p̂ 2 , and round to the nearest whole
number.

Specify the Confidence level, as a decimal. Arrow down to Calculate and press ENTER.
The confidence interval will appear on the output screen.

Example 11.16
Researchers for a cell phone company conducted a study of smartphone use among adults.
They wish to compare the proportion of smartphone users among two different populations:adult whites (non-Hispanic) and African American adults. The results of the survey indicatethat of the 232 African American cell phone owners randomly sampled, 5% have an iPhone.
Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Use this datato construct a 95% confidence interval for the difference in population proportions.

549

Solution 11.16
Let pW and pA are be the proportions of people who own iPhones in the white and AfricanAmerican populations, respectively. Then we want a confidence interval for pW – pA.
The conditions for using a Two-Proportion Z-Interval are easily met, so we go STAT, then to
TESTS, and scroll down to 2-PropZInt. Enter the following:
x1: 134
n1: 1343
x2: 12n2: 232
C-Level: .95(this is 10% of 1343, rounded to a whole number)(this is 5% of 232, rounded to a whole number)

Arrow down to Calculate and press Enter to get the interval: (0.01536, 0.08075).
Thus, we are 95% confident that .01536 < pW – pA < .08075. The percentage difference
can range from 1.536% difference to 8.075% difference.

550

KEY TERMS and FORMULA REVIEW
Cohen’s effect size: This is a measure of the relative strength of the difference between
two means based on sample data. Cohen’s d-statistic is defined as d =( x1 − x 2 ), wheres pooled22(n1 − 1) s1 + (n 2 − 1) s 2.
n1 + n 2 − 2spooled is the pooled standard error, spooled =Degrees of freedom: The number of objects in a sample that are free to vary. For a onesample t-test, the degrees of freedom was simple: df = n – 1. However, for a two-sample ttest, the calculation is more complicated, given by Welch’s formula:2

 s1 2 s 2 2 


+

 n1
n
2 

df =
4
4
s1
s2
+
2
2
n1 (n1 − 1) n 2 (n 2 − 1)
Pooled proportion: Used in a hypothesis test for two proportions to estimate thecommon value of p1 and p2: p c =x1 + x 2.
n1 + n 2

Sampling Distribution for pˆ 1 − pˆ 2 : The mean and standard deviation are:µ pˆ1 − pˆ 2 = p1 − p 2 and σ pˆ − pˆ = p1 q1 + p 2 q 2 .
1

2

n1

n2

Provided np1 and np2 are both greater than 10, the distribution will be approximately
normal.
Sampling Distribution for the random variable X 1 − X 2 :The mean and standard deviation are:σ 12 σ 2 2+µ x1 −x2 = µ1 − µ 2 and σ x1 − x2 =n1
n2

If we are sampling from normal distributions, or the sample sizes are both large, then theσ 12 σ 2 2+)
sampling distribution is approximately normal: X 1 − X 2 ~ N(µ1 – µ2,
n1
n2

551

Standard Error: This is an estimate of σ x1 − x2 obtained by estimating the unknown
population variances by the respective sample variances; the standard error is:
2

2

s
s
s x1 − x2 = 1 + 2 .
n1 n 2
Two Sample Confidence Intervals:
Z-Interval for estimating difference of two means:
σ 12( x1 − x 2 ) − zα / 2

+

n1

σ 22
n2

< µ1 − µ 2 < ( x1 − x 2 ) + z α / 2

σ 12
n1

+

σ 22
n2

Calculator: STAT, TESTS, 2SampZint
T-Interval for estimating difference of two means:( x1 − x 2 ) − t α / 2

2

2

s1
s
+ 2 < µ1 − µ 2 < ( x1 − x 2 ) + t α / 2
n1
n2

2

2

s1
s
+ 2
n1
n2

Calculator: STAT, TESTS, 2SampTint

Z-Interval for estimating difference of two proportions:( pˆ 1 − pˆ 2 ) − zα / 2

p1 q1 p 2 q 2+< p1 − p 2 < ( pˆ 1 − pˆ 2 ) + zα / 2n1
n2

p1 q1 p 2 q 2+
n1
n2

Calculator: STAT, TESTS, 2PropZint

552

CHAPTER REVIEW
11.1 Two Population Means with known Standard Deviations
If we are testing a claim comparing two unknown population means,µ1 and µ2, where thepopulation standard deviations σ1 and σ2 are known, then we use a Two-Sample Z-Test.
The key features of the test are:••

The data must come from independent samples
Use only when the population standard deviations σ1, σ2 are known.

σ 12 σ 2 2+• Random variable and distribution: X 1 − X 2 ~ N (µ1 – µ2,)n1n2( x1 − x 2 )•Test statistic: z =•Calculator function: 2-SampZTest

σ 12 σ 2 2+n1n2

11.2 Two Population Means with unknown Standard Deviations
If we are testing a claim comparing two unknown population means,µ1 and µ2, where thepopulation standard deviations σ1 and σ2 are not known, then we use a Two-Sample TTest;
the key features of the test are:••Assume that the data comes from independent samples.
Assume that the populations from which we are sampling are approximately
normal, and the population standard deviations σ1, σ2 are not known.

•

Test statistic: t =( x1 − x 2 )
2

2

s1
s
+ 2
n1 n 22•••

 s1 2 s 2 2 


+
 n1

n
2 

df =
4
4
s1
s2
+
2
2
n1 (n1 − 1) n 2 (n 2 − 1)
Calculator function: 2-SampTTest
Select the “Pooled” option only when the population variances are equal. If in
doubt, do not pool the data.

553

11.3 Comparing Two Independent Population Proportions
If we are testing a claim comparing two unknown population proportions, p1 and p 2, we use
a Two-Proportion Z-Test; the key features of the test are••

Assume the data comes from from independent samples.
Assume that n1 pˆ 1 ≥ 10 , n1 (1 − pˆ 1 ) ≥ 10 and that n 2 pˆ 2 ≥ 10 and n 2 (1 − pˆ 2 ) ≥ 10 .
That is, there are at least 10 successes and 10 failures in each sample.

pˆ 1 − pˆ 2•

Test statistic: z =•Calculator function: 2-PropZTest11 
p c (1 − p c ) + 
 n1 n 2 , where p c =x1 + x 2
n1 + n 2

11.4 Matched or Paired Samples
If we are testing a claim about two unknown population means, using dependent samples
consisting of matched pairs of data values, then we use an ordinary t-test on the
differences.
The key features of the test are:• Test the differences d by subtracting one measurement from the other measurement ineach pair.
• Random Variable: d = sample mean of the differences• Distribution: Student’s t-distribution with n – 1 degrees of freedom, where n is thenumber of matched pairs.
• If n (the number of differences) is less than 30, then we must assume that thedifferences are normally distributed.
• Test statistic: t =d − µdsd / n.

• Calculator function: place difference (d) into L1, T-test on L1, Enter 0 for µ0554

Exercises for Chapter 11
The next seven exercises all describe a claim or scenario that will be investigatedusing a hypothesis test. In each case:a. State the hypotheses in terms of the appropriate parameter(s)b. Determine if the samples are independent or dependentc. Identify the random variable, and its distribution (either normal or t-distribution),along and explain this choice.
d. Identify the test to be used.
1. It is believed that 70% of males pass their drivers test in the first attempt, while 65% offemales pass the test in the first attempt. A test will be conducted to check whether the
proportions are in fact equal.
2. The manufacturer of a new windshield treatment claims that their product will repel water
more effectively. To test the claim, ten windshields are tested by simulating rain without the
new treatment. The same windshields are then treated, and the experiment is run again.
This data is then used to test the manufacturer’s claim.
3. The known standard deviation in salary for all mid-level professionals in the financial
industry is $11,000. Company A and Company B are in the financial industry. Suppose
samples are taken of mid-level professionals from Company A and from Company B. The
sample mean salary for mid-level professionals in Company A is $80,000. The sample meansalary for mid-level professionals in Company B is $96,000. This data is used to test theclaim that mid-level professionals at Company A and Company B are paid differently, on
average.
4. It is believed that the average grade on an English essay in a particular school system forfemales is higher than for males. A random sample of 31 females had a mean score of 82 witha standard deviation of three, and a random sample of 25 males had a mean score of 76 witha standard deviation of four.
5. It is thought that teenagers sleep more than adults on average. A study is done to verify
this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2.
A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.
6. A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000
with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs
has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions
different?
7. The manufacturer of a new medicine claims that the drug helps improve sleep. Eightsubjects are picked at random and given the medicine. The means hours slept for each person555were recorded before starting the medication and after. This data is then used to test the
manufacturer’s claim.
8. A study is done to determine which of two soft drinks has more sugar. The researchers
believe that Beverage B has more sugar than Beverage A, on average. A random sample of
13 cans of Beverage A is selected; the mean amount of sugar in the sample is 36 grams witha standard deviation of 0.6 grams. A random sample of 10 cans of Beverage B is selected; themean amount of sugar in this sample is 38 grams with a standard deviation of 0.8 grams.
Both populations have normal distributions.
a. What are the hypotheses for this test?
b. What test should be used? Why?
c. Is this a one-tailed or two-tailed test?
9. The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years
for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey
death records for people born in 1900 in a certain county. Of the 124 whites, the mean life
span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean
life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test
to see if the mean life spans in the county were the same for whites and nonwhites.
a. State the null and alternative hypotheses.
b. Which distribution (normal or Student's t) would we use for this hypothesis test?
Explain.
c. Calculate the test statistic and p-value.
d. Using a significance level of α = 0.05, what is the decision about Ho?
e. What can we conclude from this test?
10. The mean speeds of fastball pitches from two different baseball pitchers are to becompared. The populations (of fastball speeds) for each pitcher have normal distributions. A
sample of 14 fastball pitches is measured from each pitcher; the results are shown in thetable:
Pitcher
WesleyRodriguez

Sample mean (mph)86
91

Population SD (mph)3
7Scouters believe that Rodriguez pitches a speedier fastball than Wesley; use the data above
to test this claim.
a. State the null and alternative hypotheses in terms of the appropriate parameters.
b. What test should be used, and why?
c. Calculate the test statistic and the p-value.

556

d. At the 1% significance level, what is the decision about Ho?
e. What can we conclude from this test?
11. A researcher is testing the effects of plant food on plant growth; the researcher thinks
the food makes the plants grow taller. Nine plants have been given the plant food. Anothernine plants have not been given the plant food. The heights of the plants (in inches) arerecorded after eight weeks. The following table shows the results.
Plant Group Sample Mean
Population SD
Food
16
2.5
No Food14
1.5
Assume the populations have normal distributions and answer the following:
a.
b.
c.
d.
e.

State the null and alternative hypotheses.
What type of test should be used, and why?
Find the test statistic and the p-value.
At the 1% significance level, what is the decision about Ho?
What can we conclude from the test?

12. Two metal alloys are being considered as material for ball bearings. The mean melting
point of the two alloys is to be compared, and the melting points of each alloy have normal
distributions. 15 pieces of each metal are being tested; the results are shown in thefollowing table. This data will be used to test the claim that Alloy Zeta has a different
melting point than Alloy Gamma.
Alloy Gamma
Alloy Zeta

Sample Mean (°F)
800
900

Population SD (°F)
95105a. State the null and alternative hypotheses.
b. Which type of test should be used, and why?
c. Calculate the test statistic and the p-value.
d. Using a 1% significance level, what is our decision about Ho?
e. State a conclusion for the test.
13. Two types of phone operating system are being tested to determine if there is a differencein the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones
with OS1 had system failures within the first eight hours of operation. Nine out of anotherrandom sample of 150 phones with OS2 had system failures within the first eight hours ofoperation.
a. State the null and alternative hypotheses.
b. What type of test should be used, and why?
c. Find the test statistic and the p-value.

557

d. At the 5% significance level, what is the decision about Ho?
e. What can you conclude about the two operating systems?
14. In the recent Census, three percent of the U.S. population reported being of two or moreraces. However, this proportion varie from state to state. Suppose that two random surveys
are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people
reported being of two or more races. In the second random survey, out of 500 Nevadans, 17
people reported being of two or more races. Conduct a hypothesis test to determine if the
population percentages are the same for the two states or if the percent for Nevada is
statistically higher than for North Dakota.
a. State the null and alternative hypotheses.
b. Is this a right-tailed, left-tailed, or two-tailed test?
c. Which test would we use for this hypothesis test, and why?
d. Calculate the test statistic and p-value.
e. Using a significance level of α = 0.05, what is the decision?
f. Write a conclusion for the test.
15. A study was conducted to test the effectiveness of a software patch in reducing system
failures over a six-month period. Results for randomly selected installations are shown in thetable below. Each installation was tested before and after the patch, and the number ofsystem failures recorded in the table below. Using a 1% significance level, test the claim thatthe average number of system failures is reduced after installing the patch. Assume thedifferences have a normal distribution.
Installation
A
B
C
D
E
F
G
Before
3
6
4
2
5
8
2
After
1
5
2
0
1
0
2
a. What is the random variable?
b. State the null and alternative hypotheses.
c. What kind of test should be used and why?
d. What are the test statistic and p-value?
e. What is the decision for Ho?
f. What conclusion can we draw about the software patch?

H
62

16. A study was conducted to test the effectiveness of a juggling class. Before the classstarted, six subjects juggled as many balls as they could at once. After the class, the same
six subjects juggled as many balls as they could. The differences in the number of balls arecalculated. Assume that the differences have a normal distribution, and test the claim thatthe mean number of balls increased after the class; use a 5% significance level.
Subject
Before
After

A
3
4

B
4
5

C
3
6

D
2
4

558

E
4
5

F
5
7

a. State the null and alternative hypotheses.
b. What kind of test should be used and why?
c. What are the test statistic and p-value?
d. What is the decision for Ho?
e. What conclusion can we draw about the juggling class?
17. A doctor wants to know if a medication is effective in lowering blood pressure. Six subjects
have their blood pressures recorded. After twelve weeks on the medication, the same six
subjects have their blood pressure recorded again. For this test, only systolic pressure is ofconcern. Test the doctor’s claim at the 1% significance level.
Patient
BeforeAfter

AB
161 162
158 159

C
165
166

a. State the null and alternative hypotheses.
b. What is the test statistic?
d. What is the decision about Ho?

DEF
162 166 171
160 167 169
c. What is the p-value?
e. What is the conclusion?

NOTE
In the following problems, if you are using a Student's t-distribution, including for paired
data, you may assume that the underlying population is normally distributed. If nosignificance level is given, assume that α = .05.
18. The mean number of English courses taken in a two–year time period by male and female
college students is believed to be about the same. An experiment is conducted and data arecollected from 29 males and 16 females. The males took an average of three English courses
with a standard deviation of 0.8. The females took an average of four English courses with astandard deviation of 1.0. Is there a significant difference in the means?
19. A student at a four-year college claims that mean enrollment at four–year colleges ishigher than at two–year colleges in the United States. Two surveys are conducted. Of the 35
two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of
4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard
deviation of 8,191.
20. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they
could hold their breath in a relaxed position. After a two-minute rest, they timed themselves
while jumping. The girls thought that the mean difference between their jumping and relaxedtimes would be zero. Test their hypothesis.

559

Relaxed time
26
47
30
22
23
45
37
29

Jumping Time
21
40
28
21
25
43
35
32

21. Mean entry-level salaries for college graduates with mechanical engineering degrees andelectrical engineering degrees are believed to be approximately the same. A recruiting office
thinks that the mean mechanical engineering salary is actually lower than the meanelectrical engineering salary. The recruiting office randomly surveys 50 entry levelmechanical engineers and 60 entry level electrical engineers. Their mean salaries were$46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210,
respectively. Conduct a hypothesis test to determine if there is evidence that the mean entrylevel mechanical engineering salary is lower than the mean entry-level electrical engineeringsalary. (Use α = .05)
22. Marketing companies have collected data implying that teenage girls use more ring toneson their cellular phones than teenage boys do. In one particular study of 40 randomly chosen
teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones forthe girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with astandard deviation of 0.8. Conduct a hypothesis test to determine if the means areapproximately the same or if the girls’ mean is higher than the boys’ mean.
Use the following information to answer the next two exercises: The Eastern andWestern Major League Soccer conferences have a new Reserve Division that allows new
players to develop their skills. Data for a randomly picked date showed the following annual
goals:
Western
Los Angeles 9
FC Dallas 3
Chivas USA 4
Real Salt Lake 3
Colorado 4
San Jose 4

Eastern
D.C. United 9
Chicago 8
Columbus 7
New England 6
MetroStars 5
Kansas City 3

23. Suppose that a hypothesis test is conducted to test the claim that Western Division
teams score more goals, on average, than Eastern Division teams. The exact distribution
for this hypothesis test would be:560

a. the normal distributionc. the uniform distribution

b. the Student's t-distribution
d. the exponential distribution

24. If the level of significance for the test in #23 is α = 0.05, then the conclusion would be:a. There is sufficient evidence to conclude that the W Division teams score fewer
goals, on average, than the E teams
b. There is insufficient evidence to conclude that the W Division teams score moregoals, on average, than the E teams.
c. There is insufficient evidence to conclude that the W teams score fewer goals, on
average, than the E teams score.
d. Unable to determine
Use the following information to answer the next two exercises: A statistics instructor
believes that there is no significant difference between the mean class scores of statistics day
students on Exam 2 and statistics night students on Exam 2. She takes random samples fromeach of the populations. The mean and standard deviation for 35 statistics day students were
75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41
and 19.73. The “day” subscript refers to the day students and the “night” subscript refers to
the night students.
25. If this data is used to test the instructor’s claim, which of the following would be theappropriate conclusion?
a. There is insufficient evidence to conclude that the statistics day students' meanon Exam 2 is better than the statistics night students' mean on Exam 2.
b. There is insufficient evidence to conclude that there is a significant difference
between the means of the statistics day students and night students on Exam 2.
c. There is sufficient evidence to conclude that there is a significant difference
between the means of the statistics day students and night students on Exam 2.
26. An appropriate alternative hypothesis for the hypothesis test is:a. µday > µnightb. µday < µnightc. µday = µnightd. µday ≠ µnight

27. Researchers interviewed street prostitutes in Canada and the United States. The mean
age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard
deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution
was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada
lower than the mean age in the United States? Test at a 1% significance level.
28. A test is conducted to compare two types of diet. A powder diet is tested on 49 people,
and a liquid diet is tested on 36 different people. The powder diet group had a mean weight561

loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean
weight loss of 45 pounds with a standard deviation of 14 pounds. Use this data and a 5%
significance level to test the claim that the liquid diet yields a higher mean weight loss thanthe powder diet.
29. Parents of teenage boys often complain that auto insurance costs more, on average, forteenage boys than for teenage girls. A group of concerned parents examines a random sample
of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls,
it was $559. From past years, it is known that the population standard deviation for eachgroup is $180. Does this data provide evidence that the mean cost for auto insurance forteenage boys is greater than that for teenage girls?
30. A group of transfer bound students wondered if they will spend the same mean amounton texts and supplies each year at their four-year university as they have at their community
college. They conducted a random survey of 54 students at their community college and 66
students at their local four-year university. The sample means were $947 and $1,011,
respectively. From past studies, the population standard deviations are known to be $254
and $87, respectively. Conduct a hypothesis test to determine if the means are statisticallythe same.
31. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-pergallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a meanof 31 mpg with a standard deviation of 7 mpg. And 31 non-hybrid sedans get a mean of 22
mpg with a standard deviation of 4 mpg. Suppose that the population standard deviations
are known to be 6 and 3, respectively. Conduct a hypothesis test to evaluate the
manufacturers’ claim.
32. A study is done to determine if students in the California State University system takelonger to graduate, on average, than students enrolled in private universities. One hundred
students from both the California State University system and private universities are
surveyed. Suppose that from years of research, it is known that the population standard
deviations are 1.5811 years and 1 year, respectively. The following data are collected. The
California State University system students took on average 4.5 years with a standard
deviation of 0.8. The private university students took an average of 4.1 years with a standard
deviation of 0.3. Use this data to test the claim.
33. One of the questions in a study of marital satisfaction of dual-career couples was to rate
the statement “I’m pleased with the way we divide the responsibilities for childcare.” Theratings went from one (strongly agree) to five (strongly disagree). The table below contains
ten of the paired responses for husbands and wives. Use this data to conduct a hypothesis
test to see if the mean difference in the husband’s versus the wife’s satisfaction level isnegative (meaning that, within the partnership, the husband is happier than the wife).

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Husband’s score 2 2 1 3 2 1 1 1 2 4
Wife’s score
2 2 3 3 4 2 1 1 2 4
34. A recent drug survey showed an increase in the use of drugs and alcohol among local highschool seniors as compared to the national percent. Suppose that a survey of 100 local seniors
and 100 national seniors is conducted to test whether the proportion of drug and alcohol use
is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within
the past month, while 60 national seniors reported using them. Does this data provide
evidence to support the claim?
35. A test is conducted to determine whether the proportions of female suicide victims thatare ages 15 to 24 are the same for whites and for African-Americans in the United States.
We randomly pick one year, 1992, to compare the populations. The number of suicides
estimated in the United States in 1992 for white females is 4,930. Five hundred eighty wereaged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let
female suicide victims be our population.
Use the following information to answer the next three exercises. Neuro-invasive West
Nile virus is a severe disease that affects a person’s nervous system. It is spread by the Culex
species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion ofneuro-invasive West Nile virus cases more than the 2010 proportion of neuro-invasive West
Nile virus cases? Using a 1% level of significance, we conduct an appropriate hypothesis test.
36. This is:a. a test of two proportions
c. a test of a single meanb. a test of two independent meansd. a matched pairs test.

37. An appropriate null hypothesis is:a. p2011 ≤ p2010
b. p2011 ≥ p2010
c. µ2011 ≤ µ2010
d. p2011 > p2010
38. The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is:a. There is sufficient evidence to conclude that the proportion of people in the
United States in 2011 who contracted neuro-invasive West Nile disease is less thanthe proportion of people in the United States in 2010 who contracted neuro-invasive
West Nile disease.
b. There is insufficient evidence to conclude that the proportion of people in the
United States in 2011 who contracted neuro-invasive West Nile disease is more thanthe proportion of people in the United States in 2010 who contracted neuro-invasive
West Nile disease.

563

c. There is insufficient evidence to conclude that the proportion of people in the
United States in 2011 who contracted neuro-invasive West Nile disease is less thanthe proportion of people in the United States in 2010 who contracted neuro-invasive
West Nile disease.
d. There is sufficient evidence to conclude that the proportion of people in the
United States in 2011 who contracted neuro-invasive West Nile disease is more thanthe proportion of people in the United States in 2010 who contracted neuro-invasive
West Nile disease.
39. A recent year was randomly picked from 1985 to the present. In that year, there were2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe
College, there were 321 Hispanic students out of a total of 2,441 students. Does this data
provide evidence that the percent of Hispanic students at the two colleges is significantly
different? Explain.

40. Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults
are considered obese if their body mass index (BMI) is at least 30. The researchers wanted todetermine if the proportion of women who are obese in the south is less than the proportionof southern men who are obese. The results are shown in the table below. At the1% level ofsignificance, is there evidence to support the claim?
Men
Women

Number who are obese
42,76967,169Sample size
155,525
248,775

41. Researchers conducted a study to find out if there is a difference in the use of eReaders
by different age groups. Randomly selected participants were divided into two age groups. In
the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309
participants 30 years old and older use eReaders. Does this data indicate that there is asignificant difference in the proportions between the age groups?

42. A group of friends debated whether a higher percentage of men use smartphones than
women. They consulted a research study of smartphone use among adults. The results of thesurvey indicate that of the 973 men randomly sampled, 379 use smartphones. For women,
404 of the 1,304 who were randomly sampled use smartphones. Use this data to test thefriends’ conjecture at the 5% level of significance.

564

43. Two computer users were discussing tablet computers. They conjectured that a higher
proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older.
The table below shows the number of tablet owners for each age group. Test the claim using
a 5% level of significance.
Own a Tablet
Sample size

16-29 yr old
69
628

30 yrs and older
231
2309

44. A teacher is interested in whether children’s educational computer software costs less,on average, than children’s entertainment software. Thirty-six educational software titles
were randomly picked from a catalog. The mean cost was $31.14 with a standard deviationof $4.69. Thirty-five entertainment software titles were randomly picked from the samecatalog. The mean cost was $33.86 with a standard deviation of $10.87. Use this data to test
the claim that children’s educational software costs less, on average, than children’s
entertainment software.

45. A social scientist recently claimed that the proportion of college-age males with at least
one pierced ear is as high as the proportion of college-age females. She conducted a survey
at a college to test her claim. Out of 107 males, 20 had at least one pierced ear. Out of 92
females, 47 had at least one pierced ear. Does this data provide evidence to reject her
claim?

46. Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data
are recorded in the table blow. Does this data provide evidence that their cholesterol levels
were significantly lowered?
Starting cholesterol level
140
220
110
240
200
180
190
360
280
260

565

Ending cholesterol level
140
230
120
220
190
150
200
300
300
240

Use the following information to answer the next two exercises. A new AIDS prevention
drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS
after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four
years. We want to test whether the method of treatment reduces the proportion of patientsthat develop AIDS after four years or if the proportions of the treated group and the untreatedgroup stay the same. Let the subscript t = treated patient and ut = untreated patient.
47. The appropriate hypotheses are:a. H0: pt < put and Ha: pt ≥ put
c. H0: pt = put and Ha: pt ≠ put

b. H0: pt ≤ put and Ha: pt > put
d. H0: pt ≥ put and Ha: pt < put

48. If the p-value is 0.0062 and we use α = .05, what is the conclusion?
a. The method has no effect.
b. There is sufficient evidence to conclude that the method reduces the proportion of
HIV positive patients who develop AIDS after four years.
c. There is sufficient evidence to conclude that the method increases the proportionof HIV positive patients who develop AIDS after four years.
d. There is insufficient evidence to conclude that the method reduces the proportionof HIV positive patients who develop AIDS after four years.
49. A golf instructor is interested in determining if her new technique for improving players’
golf scores is effective. She takes four new students. She records their 18-hole scores before
learning the technique and then after having taken her class. She conducts a hypothesis test.
The data are as follows:
Score before class
Score after classPlayer 1
83
80

Player 2
78
80

Player 3
93
86

Player 4
87
86

Does this data provide evidence that the technique is effective?
50. A local cancer support group believes that the estimate for new female breast cancer
cases in the south is higher in 2013 than in 2012. The group compared the estimates of new
female breast cancer cases by southern state in 2012 and in 2013. The results are shown in
the table:
States AL
AR
FL
GA
KY
LA
MS
NC
OK
SC
TN
TX
2012
3450 2150 15540 6970 3160 3320 1990 7090 2630 3570 4680 15050
2013
3720 2280 15710 7310 3300 3630 2080 7430 2690 3580 5070 14980
Using a 5% significance level, does this data provide evidence that there were more cases in2013 than in 2012?

566

51. A traveler wanted to know if the prices of hotels are different in the ten cities that hevisits the most often. The list of the cities with the corresponding hotel prices (in dollars) for
his two favorite hotel chains is in the table below. At a 5% level of significance, is there
evidence that the prices are different, on average, between the two chains?
City
Atlanta
BostonChicagoDallasDenverIndianapolis
Los Angeles
New York City
Philadelphia
Washington, D.C.

Hyatt Regency Price
107
358
209
209
167
179
179
625
179
245

Hilton Price
169
289
299
198
169
214
169
459
159
239

52. Mean entry-level salaries for college graduates with mechanical engineering degrees andelectrical engineering degrees are believed to be approximately the same. A recruiting officer
wants to estimate the difference in the means. He randomly surveys 50 entry levelmechanical engineers and 60 entry level electrical engineers. Their mean salaries were$46,100 and $46,700, respectively, and their standard deviations were $3,450 and $4,210,
respectively. Construct a 95% confidence interval for the difference in the means. Does thisinterval suggest that the means could be the same? Explain.
53. Marketing companies have collected data implying that on average, teenage girls use
more ring tones on their cellular phones than teenage boys do. In one particular study of 40
randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean numberof ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was
1.7 with a standard deviation of 0.8. Use this data to construct a 95% confidence interval forthe difference in means between girls and boys. According to this interval, is there evidence
that on average, girls use more ring tones that boys? How big is this difference?
54. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-pergallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a meanof 31 mpg with a standard deviation of 7 mpg. And 31 non-hybrid sedans get a mean of 22
mpg with a standard deviation of 4 mpg. Suppose that the population standard deviations
are known to be 6 and 3, respectively. Use this data to construct a 95% confidence interval
for the difference in mean mpg between hybrid cars and non-hybrid cars.
55. A study is done to determine if students in the California State University system takelonger to graduate, on average, than students enrolled in private universities. One hundred
students from both the California State University system and private universities are
surveyed. Suppose that from years of research, it is known that the population standard

567

deviations are 1.5811 years and 1 year, respectively. The following data are collected. The
California State University system students took on average 4.5 years with a standard
deviation of 0.8. The private university students took an average of 4.1 years with a standard
deviation of 0.3. Use this data to find a 95% confidence interval for the difference in means
between the Cal State system and private universities.
56. A recent year was randomly picked from 1985 to the present. In that year, there were2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe
College, there were 321 Hispanic students out of a total of 2,441 students. Use this data toconstruct a 90% confidence interval for the difference in the proportions.
57. Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults
are considered obese if their body mass index (BMI) is at least 30. The researchers wanted todetermine if the proportion of women who are obese in the south is less than the proportionof southern men who are obese. The results are shown in the table below. Use this data toconstruct a 95% confidence interval for the difference in obesity rates between men andwomen.
Men
Women

Number who are obese
42,76967,169Sample size
155,525
248,775

58. An accounting firm is trying to decide between IT training that is conducted in-house or
conducted by consultants. The table below shows the average annual training cost per
employee. Are the mean costs significantly different using 5% level of significance? Assume
the population variances are equal. Show all hypothesis testing steps.
nIn-house
200
Consultants 185

mean Sample stdev
$485
$32
$502
$40

59. According to CNNMoney article, the millennials who are eligible for retirement planfrom workplace are saving at the same participation rate as older generations. Test thisclaim at a 1% level of significance. Show all hypothesis testing steps. Suppose thefollowing data is collected:
n
proportion
Millennials
300
94%
Other Generations 330
92%

568

REFERENCES
11.1 Two Population Means with Unknown Standard Deviations
Data from Graduating Engineer + Computer Careers. Available online at
http://www.graduatingengineer.com
Data from Microsoft Bookshelf.
Data from the United States Senate website, available online at www.Senate.gov (accessed June 17,2013).
“List of current United States Senators by Age.” Wikipedia. Available online at
http://en.wikipedia.org/wiki/ List_of_current_United_States_Senators_by_age (accessed June 17,2013).
“Sectoring by Industry Groups.” Nasdaq.Available online at
http://www.nasdaq.com/markets/barchart- sectors.aspx?page=sectors&base=industry (accessed June17, 2013).
“Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education,2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June17, 2013).
“World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseballalmanac.com/ws/wsmenu.shtml (accessed June 17, 2013).
11.2 Two Population Means with Known Standard Deviations
Data from the United States Census Bureau. Available online at
http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf
Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center,2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/(accessed June 17, 2013).
“Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smartphone-users-numbers (accessed June 17, 2013).
Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available
online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed
June 17, 2013).
“State-Specific Prevalence of Obesity Among Adults—Unites States, 2007.” MMWR, CDC.
Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17,2013).
“Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at:
http://www.disastercenter.com/ crime/txcrime.htm (accessed June 17, 2013).

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11.3 Comparing Two Independent Population Proportions
Data from Educational Resources, December catalog.
Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data
from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013).
Data from Statistics, United States Department of Health and Human Services. Data from Whitney
Exhibit on loan to San Jose Museum of Art.
Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed
June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994.
“State of the States.”Gallup,2013. Available online at http://www.gallup.com/poll/125066/StateStates.aspx?ref=interactive (accessed June 17, 2013).
“West Nile Virus.” Centers for Disease Control and Prevention. Available online at
http://www.cdc.gov/ncidod/dvbid/ westnile/index.htm (accessed June 17, 2013).
That's no surprise. After paying bills. “66% Of Millennials Have Nothing Saved for Retirement.”
CNNMoney, Cable News Network, money.cnn.com/2018/03/07/retirement/millennial-retirementsavings/index.html?sr=twCNN030718millennial-retirement-savings0103PMStory (accessed July 18,2018).
Cohen, Jason. “IOS More Popular in Japan and US, Android Dominates in China and India.”
PCMAG, PCMag, 4 Sept. 2020, www.pcmag.com/news/ios-more-popular-in-japan-and-usandroid-dominates-in-china-and-india.

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CHAPTER 11 SOLUTIONS:
1) a. H0: p1 = p2

Ha: p1 ≠ p2

b. independent samples

d. Two-proportion z-test.
Ha: μ1 ≠ μ23) a. H0: μ1 = μ2

c. RV = pˆ 1 − pˆ 2

b. independent samples

c. RV = x1 − x 2

d. Two-Sample z-test (because we know σ1 = σ2 = $11,000)
5) a. H0: μ1 < μ2
Ha: μ1 > μ2 (where μ1 represents mean for teens)
b. independent samples
c. RV = x1 − x 2
d. Two-Sample t-test7) a. H0: μd < 0

Ha: μd > 0 (where d = # hrs sleep before – #hrs of sleep after)b. dependent samples
c. RV = dd. Matched pairs t-test
Ha: μ1 ≠ μ2 b. Use a t-test, because population SD’s are unknown.
9) a. H0: μ1 = μ2
c. Using 2SampTTest, t = 5.42, p = 0.00000023
d. Reject Ho.
e. There is sufficient evidence to conclude that the mean life spans are different.
11) a. H0: μ1 < μ2
Ha: μ1 > μ2 (where μ1 represents mean for plants given the food)b. Use a z-test, because population SD’s are known.
c. Using 2SampZTest z = 2.05, p = 0.0198 d. Do not reject Ho since p > .01.
e. There is not enough evidence to conclude that food makes plants grow taller.
13) a. H0: p1 = p2
Ha: p1 ≠ p2
b. Two-Proportion z-test.
c. Using 2PropZTest z = 1.28, p = 0.2016 d. Do not reject Ho since p > .05.
e. There is not enough evidence to conclude that there is a difference in the crash rates.
15) a. The random variable is the sample mean of the differences between the number of
failures before the patch and the number of failures after the patch: RV = db. H0: μd < 0 Ha: μd > 0 (where d = # failures before – #failures after)c. This uses a t-test
d. t = 3.29, p = 0.0067
e. Reject Ho since p < .01.
f. There is sufficient evidence to conclude that the average number of system failures
is reduced after installing the patch.
17) a. H0: μd < 0 Ha: μd > 0 (where d = pressure before meds – pressure after meds)
b. t = 1.75
c. p = 0.0699
d. Do not reject Ho since p > .01.
f. There is not enough evidence to conclude that the medication helped lower systolicblood pressure.
19) The hypotheses are H0: μ1 < μ2 , Ha: μ1 > μ2 (where μ1 represents 4-yr schools)
Use 2-SampTTest, because population SD’s are unknown.
From the calculator, t = 0.248, p = 0.402 . Do not reject Ho since p > .05.
There is not enough evidence to conclude that the mean enrollment is higher for 4-yr
schools.
21) The hypotheses are H0: μ1 > μ2 , Ha: μ1 < μ2 (where μ1 represents mean salary formechanical engineers).
Use 2-SampTTest, because population SD’s are unknown.
From the calculator, t = -0.82, p = 0.2066 . Do not reject Ho since p > .05.
There is not enough evidence to conclude that the mean salary is lower for mechanicalengineers.
23) b.
25) b.

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27) The hypotheses are H0: μ1 > μ2 Ha: μ1 < μ2 ( μ1 is the mean age for Canada). Use 2SampTTest. From the calculator, t = -2.17 and p = 0.0157. Since p > .01, do not reject Ho.
There is not enough evidence at the .01 level of significance to conclude that the mean age
for entering prostitution is lower in Canada than in the U.S.
29) The hypothess are H0: µ1 ≤ µ2, Ha: µ1 > µ2 (where µ1 is the mean for boys) Use 2SampZTest, since the population SD’s are both known. From the calculator, z = 2.50 and
p-value: 0.0062. Since p < 0.05, Reject Ho. At the 5% significance level, there is sufficientevidence to conclude that the mean cost of auto insurance for teenage boys is greater thanthat for girls.
31) The hypotheses are H0: µ1 ≤ µ2, Ha: µ1 > µ2 (where µ2 is the mean for hybrid cars) Use
2-SampZTest, since the population SD’s are both known. From the calculator, z = 6.36 andp = 0. Since p < .05, we reject Ho. At the 5% significance level, there is sufficient evidence
to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid
sedans.
33) The hypotheses are H0: µd = 0, Ha: µd < 0 (d = husband’s score – wife’s score)This is a matched-pairs t-test. From the calculator, t = -1.86 and p = 0.0479. Since p < .05,
we reject the null hypothesis. There sufficient evidence (just barely!) to conclude that themean difference is negative. That is, at the .05 significance level, there is evidence that on
average, husbands are happier with the division of child-care responsibilities.
35) The hypotheses are: H0: pW = pB Ha: pW ≠ pB We use 2-PropZTest; from thecalculator we get z = - 0.1944 and p-value = 0.8458. Do not reject Ho, since p > .05.
There is not sufficient evidence to conclude that the proportions of white and black femalesuicide victims, aged 15 to 24, are different.
37) The hypotheses would be Ho: p2011 < p2010 , Ha: p2011 > p2010.

Answer a.

39) The hypotheses are: H0: p1 = p2, Ha: p1 ≠ p2.
Use 2-PropZTest; from the calculator, z = 4.29 and p-value = 0.00002. Since p < .05, we
reject Ho. There is sufficient evidence to conclude that the proportions of Hispanic students
at Cabrillo College and Lake Tahoe College are different.
41) The hypotheses are: H0: p1 = p2, Ha: p1 ≠ p2. Use 2-PropZTest; from the calculator,z = -2.94 and p-value = 0.0033. Since p < .05, we reject Ho. At the 5% level of significance,
there is sufficient evidence to conclude that the proportion of eReader users aged 16 to 29
years is different from the proportion of eReader users 30 and older.
43) The hypotheses are: H0: p1 < p2, Ha: p1 > p2 (where p1 is the proportion for 16-29 yrolds) Use 2-PropZTest; from the calculator, z = .721 and p-value = 0.2354. Since p > .05, we
do not reject Ho. At the 5% level of significance, there is not sufficient evidence to conclude572

that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years andolder.
45) The hypotheses are: H0: p1 > p2, Ha: p1 < p2 (where p1 is the proportion for malestudents) Use 2-PropZTest; from the calculator, z = -4.82 and p-value = 0.0000007.
Since p < .05, we do reject Ho. At the 5% significance level, there is sufficient evidence toconclude that the proportions of male students with at least one pierced ear is less than the
proportion for female students.
47) The hypotheses are H0: pt > put, Ha: pt < put (where pt is the proportion of untreatedpatients who develop AIDS) Answer d.
49) The hypotheses are H0: µd = 0, Ha: µd > 0 (d = score before class – score after) This is amatched-pairs t-test. From the calculator, t = 1.19 and p = 0.1595. Since p > .05, we do not
reject the null hypothesis. There is no evidence that the course helped improve scores.
51) The hypotheses are H0: µd = 0, Ha: µd ≠ 0 (d = Hyatt price – Hilton price)
This is a matched-pairs t-test. From the calculator, t = .41 and p = 0.6881
Since p > .05, we do not reject the null hypothesis. There is no evidence that there is adifference in prices between the chains.
53) Let µ1 = mean for girls, µ2 = mean for boys. Using 2SampTInt, we are 95% confident
that .7224 < µ1 – µ2 < 2.2775. So we are 95% confident that .7224 + µ2 < µ1. There isevidence that the mean number for girls is at least .7224 more than the mean for boys.
55) Let µ1 = mean for Cal State, µ2 = mean for private universities. Using 2SampZInt, weare 95% confident that -.1185 < µ1 – µ2 < .9185. Since the interval contains both positive
and negative values, this interval does not provide evidence that the mean time needed tograduate is higher for Cal State than it is for private schools.
57) Let p1 = proportion for men, p2 = women. Using 2PropZInt, we are 95% confident that
0.0022 < p1 – p2 < 0.0078. Thus, there is evidence that 0.0022 + p2 < p1. That is, we are
95% confident that the proportion for men is at least .0022 more than for women.
59)
Let p1 = proportion for millennials, p2 = older generation. Ho: p1 = p2 (claim) Ha: p1≠ p2 Using 2PropZTest, test value Z = .979; p-value = .327, Do not Reject Ho. There is notenough evidence to reject the claim that millennials save at the same proportion as older
generations.

573

This page is purposely left blank.

574

12 | THE CHI-SQUARE DISTRIBUTION
Figure: The chi-square distribution can be used
to determine if shoe styles are equally likely.
(Photo by Samuel Lopes on Unsplash)

Chapter Objectives
By the end of this chapter, the student should be able to:
•
•
•
•
•
•

Interpret the chi-square probability distribution as the sample size changes.
Conduct and interpret standard deviation and variance confidence intervals.
Conduct and interpret chi-square single variance hypothesis tests.
Conduct and interpret chi-square goodness-of-fit hypothesis tests.
Conduct and interpret chi-square test of independence hypothesis tests.
Conduct and interpret chi-square homogeneity hypothesis tests.

575

12.1 | Introduction to Chi-Square Distribution
Have you ever wondered if lottery numbers were evenly distributed or if some numbers
occurred with a greater frequency? How about if the types of movies people preferred weredifferent across different age groups? What about if a coffee machine was dispensing
approximately the same amount of coffee each time? You could answer these questions byconducting a hypothesis test.
You will now study a new distribution, one that is used to determine the answers to suchquestions. This distribution is called the chi-square distribution.
In this chapter, you will learn four major applications of the chi-square distribution:
1. the confidence interval for standard deviation or variance2. the test of a single variance, which tests variability, such as in the coffee example
3. the goodness-of-fit test, which determines if data fit a particular distribution, such as
in the lottery example
4. the test of independence, which determines if categorical variables are independent,such as in the movie example
NOTE: Though the chi-square distribution depends on calculators or computers for most of
the calculations, there are Chi-Square Distribution tables available (see Appendix). TI-83+
and TI-84 calculator instructions are included in the text.
The Chi-Square Distribution
Let’s conduct the following statistical experiment. We select samples of size n from a normal
population, which has a standard deviation of σ. We find that the standard deviation in our
sample is equal to s. Given these data, we can define a statistic, called chi-square, using thefollowing formula:χ2 =(n − 1) s 2

σ2This new distribution has the following characteristics:
1.) All chi-square values are greater than or equal to 0
2.) There is a different chi-square curve for each degrees of freedom, n – 1 (figure below)
3.) The curve is nonsymmetrical and skewed to the right4.) The total area under the curve is 1, or 100%5.) The notation is X ∼ χ df
2576

χ𝟐𝟐𝟐𝟐
χ𝟐𝟐𝟔𝟔

χ𝟐𝟐𝟏𝟏𝟏𝟏

0

NOTE: The degrees of freedom for the three major applications of Chi-square distribution
are each calculated differently.
For the χ 2 distribution, the population mean is μ = df (degrees of freedom) and the population
standard deviation is σ = �2(𝑑𝑑𝑑𝑑). The random variable is shown as χ 2 , but may also be any

upper-case letter.

Critical values of the Chi-Square Distribution
Recall that critical values are values from the distribution that separate the confidence area
and the non-confidence area. We found zα/2 by using ±invnorm(α/2) and tα/2 by using tdistribution chart or by ±invT(α/2, df). Since the z-distribution and the t-distribution are
symmetrical, the left and right critical values are opposite values. However, in the chi-square
distribution, values are only positive.
••

χ𝟐𝟐𝐋𝐋 represents the left critical value

χ𝟐𝟐𝐑𝐑 represents the right critical value

The critical values for the χ 2 distribution are recorded in two tables (left critical value table
and right critical value table). To find these values, you need area in the tail and the degrees
of freedom.
For example, let C.L. = 0.90, α = 0.10, α/2 = 0.05 and df = 14577

Below you can see, that the chi-square distribution table is split into 2 tables. The first table
is the left critical values, χ𝟐𝟐𝐋𝐋 . The second table is the right critical values, χ𝟐𝟐𝐑𝐑 . The tail area
is 0.05 and df = 14 so χ𝟐𝟐𝐋𝐋 = 6.571 using the table for the left critical values and χ𝟐𝟐𝐑𝐑 = 23.865

using the table for the right critical values. The complete two tables are found in the
appendix.

578

12.1 Find the left and right critical values for the chi-square distribution for C.L. = .98 and
df = 6

Critical values using Excel
Recall that critical values from the Z-distribution (standard normal) and the t-distribution(approximately normal) are symmetric opposites. The critical values from chi-square
distribution are both positive. Also, t-distribution and chi-square distribution are dependent
on degrees of freedom (df).
To use Excel functions, the first thing to type is an equal sign. Here are the Excel functions
for each critical value.
Left Z-distribution critical value, -Zc
Right Z-distribution critical value, Zc
Left t-distribution critical value, -tc
Right t-distribution critical value, tc
Left chi-square critical value, χ2L
Right chi-square critical value, χ2R=norm.s.inv(a/2)=-norm.s.inv(a/2)=t.inv(a/2, df)=-t.inv (a/2, df)=chisq.inv(a/2, df)=chisq.inv.rt(a/2, df)

The following example uses C.L. = .95 and df = 14. If the confidence level is 95%, that leaves
5% in both tails of the distribution. α = 0.05. Split the 5%, which gives us 2.5% = 0.025 as
α/2.

579

12.2 | Confidence Interval for Variance/Standard Deviation
Recall that an interval is a range of numbers. A confidence interval is a range of values,
calculated from sample data, that is used to estimate an unknown population parameter.
The interval gives us a lower bound value and upper bound for the parameter with a certain
level of confidence.
Here we present intervals for estimating σ2 and σ between a lower and upper bound.
Unlike other confidence intervals we have seen, these bounds can’t be found using the
calculator. They depend on the chi-square critical values.
Calculating the Confidence Interval for variance
Lower bound < σ2 < Upper bound(n − 1) s 2

χ R2

<σ2 <

(n − 1) s 2

χ L2

where χ L2 , χ R2 are critical values, where d.f. = n – 1, and s2 is the point estimate.
NOTE: χ L2 , χ R2 are not to be squared. The right critical value is on the left side and the left

on the right side. Also notice the numerators are the same. When you divide the same value
by a larger number, you get a smaller value.

Calculating the Confidence Interval for standard deviation
Lower bound < σ < Upper bound(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2

where s is the point estimate.

The sample variance, s2, can be found using the calculator or Excel function. Using the ticalculator, s2 can be found by: Pressing [STAT], Edit, enter values into L1
 2nd Quit, 2nd [STAT], move cursor to MATH
 Choose option 8: variance(, 2nd [1]580

Using Excel:
 Type the data into a column
 Go to an empty cell, type =Var.s(highlight data)

Example 12.1
The weights (in pounds) of 15 dogs selected randomly from those adopted out by an animal
shelter last week are shown in the list below. Construct a 98% confidence interval for the
population variance.
25, 34, 27, 27, 31, 28, 27, 28, 33, 31, 28, 29, 32, 29, 29

Solution 12.1
First we need to find the sample variance (chapter 2) of the data set. Recall to find it bycalculator you enter the data into L1 (Stat, Edit). 2nd Stat, Math, #8 Variance (L1):s2 = 6.314
Second, we find the critical values using the chi-square distribution chart.
Let C.L. = 0.98, α = 0.02, α/2 = 0.01 and df = n – 1 = 15 – 1 = 14

α/2 = 0.01
α/2 = 0.01

C.L. = 0.98
0= 4.660= 29.141

Third, we plug in the values into the formula.

(n − 1) s 2

χ R2

<σ <
2

(n − 1) s 2

χ L2

14(6.314)
14(6.314)<σ2 <29.1414.660
3.033 < σ 2 < 18.969
581

Example 12.2
A random sample of 18 men have a mean height of 67.5 inches and a standard deviation of
1.5 inches. Construct a 99% confidence interval for the population standard deviation.
Solution 12.2 using chi-square distribution tables
Since s = 1.5, s2 = 2.25.
Let C.L. = 0.99, α = 0.01, α/2 = 0.005 and df = n – 1 = 18 – 1 = 17

χ L2 = 5.697 and χ R2 = 35.718(n − 1) s 2

χ

2
R

<σ <

(n − 1) s 2

χ

2
L

=>

17(2.25)
17(2.25)<σ <35.7185.697

1.035 < σ < 2.591
Solution 12.2 using Excel
Since s = 1.5, s2 = 2.25.
Let α/2 = 0.005 and df = 17

χ L2 = chisq.inv(.005, 17) = 5.697 and χ R2 = chisq.inv.rt(.005, 17) = 35.718(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2=> 1.035 < σ < 2.591

Lower Bound on Excel: =sqrt(17*2.25/highlight right critical value)
Upper Bound on Excel:=sqrt(17*2.25/highlight left critical value)582

12.3 | Test of a Single Variance
A test of a single variance assumes that the underlying distribution is normal. The null andalternative hypotheses are stated in terms of the population variance, σ2 (or population
standard deviation, σ). The test statistic is:χ2 =(n − 1) s 2

σ2where:
n = the total number of data  df = n - 1
s2 = sample varianceσ2 = population variance
A test of a single variance may be right-tailed, left-tailed, or two-tailed.
Example 12.3
Math instructors are not only interested in how their students do on exams, on average, but
how the exam scores vary. To many instructors, the variance (or standard deviation) may be
more important than the average. Standard deviation tells us how consistent data is.
Suppose a math instructor believes that the standard deviation for his final exam is five
points. One of his best students thinks otherwise. The student claims that the standard
deviation is more than five points. If the student were to conduct a hypothesis test, what
would the null and alternative hypotheses be?
Solution 12.3
Even though we are given the population standard deviation, we can set up the test using
the population variance as follows. Recall variance is the square of standard deviation.
H0: σ2 ≤ 52
Ha: σ2 > 52 (claim)

 right-tailed test

12.2 A SCUBA instructor wants to record the collective depths each of his student’s divesduring their checkout. He is interested in how the depths vary, even though everyone shouldhave been at the same depth. He believes the standard deviation is three feet. His assistant
thinks the standard deviation is less than three feet. If the instructor were to conduct a test,
what would the null and alternative hypotheses be?
583

Example 12.4
With individual lines at its various windows, a post office finds that the standard deviation
for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The
post office experiments with a single, main waiting line and finds that for a random sample
of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.
With a significance level of 5%, test the claim that a single line causes lower variation among
waiting times (shorter waiting times) for customers.
Solution 12.4 using p-value
Since the claim is that a single line causes less variation, this is a test of a single variance.
The parameter is the population variance, σ2, or the population standard deviation, σ.
Random Variable: The sample standard deviation, s, is the random variable.
Let s = standard deviation for the waiting times.
Step 1:

Set up the null hypothesis and the alternative hypothesis

H0: σ ≥ 7.2
Ha: σ < 7.2 (claim)
The word "less" tells you this is a left-tailed test.
Step 2: Select the correct test, determine the correct sampling distribution2, where:Distribution for the test: χ 24••

n = the number of customers sampled
df = n – 1 = 25 – 1 = 24

Calculate the test statistic: Substituting n = 25, s = 3.5, and σ = 7.2 (the value of σ inHo), we get: χ 2 =(n − 1) s 2

σ2= 5.67

Step 3: Calculate the probability of getting a sample statistic that differ584

p-value = P ( χ 2 < 5.67) = χ2cdf(0,5.67, 24) = 0.000042
By Excel=CHISQ.DIST(5.67, 24, true) gives lefttailed p-value

2nd VARSNOTE: remember that chi-square distribution starts at 0
Step 4: State conclusion based on p-value. P-value is compared with alpha. Alpha is thelevel of significance.
Reject H0 because p-value < α, which α = 0.05
Step 5: Interpret the conclusion.
At a 5% level of significance, the data provides sufficient evidence to conclude that asingle line causes a lower variation among the waiting times. That is, with a single
line the standard deviation of customer waiting times is less than 7.2 minutes.

Solution 12.4 using critical value
Step 1:

Set up the null hypothesis and the alternative hypothesis

H0: σ ≥ 7.22
Ha: σ < 7.22 (claim)
Step 2: Select the correct test, determine the correct sampling distribution
Calculate the test statistic:

Substituting n = 25, s = 3.5, and σ = 7.2 (thehypothesized value of σ) we get: χ 2 =(n − 1) s 2

σ2= 5.67

Step 3: Calculate the critical value
We are only finding the left critical value since it is left-tailed. We do not split α = 0.05
since it is one-tailed. Therefore, using the chi-square ( χ 2 ) Left chart, we need the areaof the tail which is 0.05 and df = n – 1 = 24.

χ L2 = 13.848 or by Excel = chisq.inv(0.05,24)
Step 4: State the conclusion. In the critical value method, you reject Ho if the test statistic
is in the rejection region.

585

In this example, the rejection region is on the left side because it is a left-tailed test.
The left rejection region of a chi-square distribution starts at 0 and ends at the criticalvalue, χ L2 = 13.848 .

0

5.67

13.848The test statistic is 5.67 and is between 0 and 13.848; therefore, the conclusion is
Reject Ho.
Step 5: Interpret the conclusion.
Interpretation is the same as above.

12.3 The FCC conducts broadband speed tests to measure how much data per second passes
between a consumer’s computer and the internet. Let the standard deviation of Internet
speeds across Internet Service Providers (ISPs) was 10.8 percent. Suppose a sample of 20
ISPs is taken, and the standard deviation is 12.2. An analyst claims that the standard
deviation of speeds is more than what was reported. State the null and alternative
hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the pvalue, and draw a conclusion. Test at the 1% significance level.

586

12.4 | Goodness-of-Fit Test
In this type of hypothesis test, you determine whether the data "fits" a particular distribution
or not. For example, you may suspect your unknown data fits a binomial or a uniform
distribution. You use a chi-square test (meaning the distribution for the hypothesis test is
chi-square) to determine if there is a fit or not. The null and the alternative hypotheses forthis test may be written in sentences or may be stated as equations or inequalities.

χ =∑The test statistic for a goodness-of-fit test is:2(O − E ) ,
2

E

where
•
•
•

O = observed frequency
E = expected values (from theory)
Degrees of freedom = k – 1, where k = the number of different data cells or categories

The observed values are the data values and the expected values are the values you would
expect to get if the null hypothesis were true.
The goodness-of-fit test is almost always right-tailed. If the observed values and thecorresponding expected values are not close to each other, then the test statistic can get verylarge and will be way out in the right tail of the chi-square curve.
NOTE:
The expected value for each cell needs to be at least five in order for you to use this test.

Goodness of Fit P-value Steps
1.) Ho: p1 = %, p2 = %, … pk = %Ha: at least one percentage is different2.) Test value is χ =2∑(O − E )or

Ho: the frequency distribution fits ____ distribution
Ha: the frequency distribution does not fit ___distribution

2

E

Calculator: χ GOF Test
2

GOF p-value By Excel:

L1: Observed Values
L2: Expected Values=chisq.dist.rt(test value, df)

3.) P-value = χ 2 cdf(Test Value, 10^99, df) since it is always right-tailed
4.) Reject Ho if p-value < α5.) Interpret the decision to write a meaningful conclusion
587

Example 12.5
The Nike picture at the beginning of this chapter stated that chi-square distribution can beused to determine if shoe styles are equally likely. Suppose that a study from a Nike runningshoe order based on customer demand was done and found the following observed
frequencies.
Nike
Air Zoom
RunningAlphafly
Style
Frequency 33

React
Infinity

Zoom
Pegasus 38

React Miler

ZoomX
Invincible

41

54

47

40

a. Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test
to test the claim that Nike shoe styles are equally likely.
b. Find the degree of freedom.
c. Find the test valued. Find the p-valuee. State the conclusion
f. Interpret the decision
Solution 12.5a.) H0: p1 = p2 = p3 = p4 = p5 (claim)
Ha: at least 1 proportion is differentb.) df = number of categories – 1 = 4c.) To find the test value, first need to find the Expected values. To find the Expected values
when the Goodness of Fit test is about uniform distribution (equally likely), divide n by
the number of categories. Here, n = 215 and k = number of categories = 5. So the Expectedvalues are 215/5 = 43.
Now place the Observed values (33, 41, 54, 47, 40) into L1, place the Expected values (43,43, 43, 43, 43) into L2.
TI-83 (plus)
In L3, enter the formula byhighlighting L3 and then pressing
Enter. Your cursor will now appear
at the bottom of the screen. Type in
the formula (L1 – L2)^2/L2. Thenpress enter to fill in L3. Takesum(L3) to find test value.

TI-84 plus
In the newer calculators, you can 2nd Quit
after entering the Observed in L1 andExpected in L2. 2nd Quit, press Stat, Tests

588

Sum up L3 (2nd Stat, Math) to findthe Test Value, (2nd Vars, χ2CDF) tofind the p-value

χ 2 Test Value = 5.814
d.) Since Goodness of Fit tests are always right-tailed, the p-value is on the right side of
the chi-square distribution.
p-value By Excel:
χ 2 CDF(test value, 10^99, df) ==chisq.dist.rt(test value, df)
χ 2 CDF(5.814, 10^99, 4) = .213
=chisq.dist.rt(5.814,4)e.) Do not reject Ho because p-value > α.
f.) Since level of significance is not given, the standard level of significance is 5%.
Remember the claim is assigned to Ho in this example.
There is not enough evidence at 5% level of significance to reject the claimthat Nike shoe styles are equally likely.
Notice we are still following the 5 steps of hypothesis testing shown in chapter 9. The
alternative hypothesis uses words instead of inequality symbols (>, <, ≠). The type of test isalways right-tailed.

589

Example 12.6
Absenteeism of college students from math classes is a major concern to math instructors
because missing class appears to increase the drop rate. Suppose that a study was done todetermine if the actual student absenteeism rate follows faculty perception. The faculty
expected that a group of 100 students would miss class according to Table 12.1.
a. Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.
b. Can you use the information as it appears in the charts to conduct the goodness-of-fit
test? If not, create a new frequency distribution to continue.
c. Find the test valued. Find the p-valuee. State the conclusion
f. Interpret the decision
Number ofabsences per
term
0–2
3–5
6–8
9 – 11
12+

Table 12.1

Faculty
perception

Number ofabsences per
term
0–2
3-5
6–8
9 – 11
12+

50
30
12
6
2

Table 12.2

Observednumberstudents
35
40
20
1
4

of

A random survey across all mathematics courses was then done to determine the actual
number (observed) of absences in a course. The chart in Table 12.2 displays the results ofthat survey.
Solution 12.6a.) H0: Student absenteeism fits faculty perception.
Ha: Student absenteeism does not fit faculty perception.
b.) No, notice that the expected number of absences for the "12+" entry is less than five (it istwo). Combine that group with the "9–11" group to create new tables where the numberof students for each entry are at least five. The new results are in Table 12.3 and Table
12.4.
Number ofabsences per
term
0–2
3–5
6–8
9+

Table 12.3

Faculty
perception

Number ofabsences per
term
0–2
3-5
6–8
9+

50
30
12
8

Table 12.4

590

Observednumberstudents
35
40
20
5

of

c.) Degrees of freedom (df) = k – 1 = number of categories – 1 = 4 – 1 = 3d.) Test Value using calculator
Using the TI-83, 83+, 84, 84+ Calculator
To calculate_Enter Observed frequency values into L1. Enter Expected frequency values in to L2.
TI-83 (plus)
In L3, enter the formula byhighlighting L3 and then pressing
Enter. Your cursor will now appear
at the bottom of the screen. Type in
the formula (L1 – L2)^2/L2. Thenpress enter to fill in L3. Takesum(L3) to find test value.

TI-84 plus
In the newer calculators, you can 2nd Quit
after entering the Observed in L1 andExpected in L2. 2nd Quit, press Stat, Tests

Sum up L3 (2nd Stat, Math) to findthe Test Value, (2nd Vars, χ2CDF) tofind the p-value

p-value By Excel:=chisq.dist.rt(test value, df)=chisq.dist.rt(14.29,3)

χ 2 Test Value = 14.29e.) p-value = χ 2 CDF(14.29, 10^99, 3) = .00254

f.) Reject Ho (Student absenteeism fits faculty perception.) because p-value < αg.) Therefore, Student absenteeism does not fit faculty perception.

591

Example 12.7
Employers want to know which days of the week employees are absent in a five-day work
week. Most employers would like to believe that employees are absent uniformly during theweek. Suppose a random sample of 60 managers were asked on which day of the week they
had the highest number of employee absences. The results were distributed as in the tablebelow. For the population of employees, do the days for the highest number of absences occur
with equal frequencies during a five-day work week? Test at a 5% significance level.
Number of
Absencesn = 60

Monday15

Tuesday
12

Wednesday
9

Thursday9

Friday15

Solution 12.7
H0: The absent days occur with equal frequencies, that is, they fit a uniform distribution.
Ha: The absent days occur with unequal frequencies, that is, they do not fit a uniform
distribution.
If the absent days occur with equal frequencies, then, out of 60 absent days there would be
12 absences expected on each day. These numbers are the expected (E) values. The values
in the table are the observed (O) values or data.
L1: Observed (O) values (15, 12, 9, 9, 15)
L2: Expected (E) values (12, 12, 12, 12, 12)592

χ 2 Test Value = 3

p-value By Excel:

p-value = χ 2 CDF(3, 10^99, 4) = .5578=chisq.dist.rt(test value, df) =chisq.dist.rt(3, 4)

Conclusion: Do Not Reject Ho (The absent days occur with equal frequencies) because pvalue ≥ αInterpretation: At a 5% level of significance, the sample data does not provide sufficientevidence to conclude that the absences do not occur with equal frequencies.

12.4 Teachers want to know which night each week their students are doing most of their
homework. Most teachers think that students do homework equally throughout the week.
Suppose a random sample of 49 students were asked on which night of the week they did themost homework. The results were distributed as in the following table:
SundayMondayTuesday

Wednesday Thursday Friday

Saturday

Number
of
11
8
10
7
10
5
5
Students
From the population of students, do the nights for the highest number of students doing the
majority of their homework occur with equal frequencies during a week? What type of
hypothesis test should you use?

Example 12.8
One study claims that the number of televisions that American families have is as follows:
10% of families have 0 televisions, 16% have 1 television, 55% have 2 televisions, 11% have
3 televisions, and 8% have 4+ televisions. A random sample of 600 families in the far western
United States resulted in the following table.
Number of
Televisions
0
1
2
3
4+
Total = 600

Frequency
66
119
340
60
15

593

At the 1% significance level, does it appear that the distribution “number of televisions" of
far western United States families is different from the distribution for the American
population as a whole?
Solution 12.8
Ho: p1 = 10%, p2 = 16%, p3 = 55%, p4 = 11%, p5 = 8% (claim)
Ha: at least one percentage is different
To find the expected values, we need to take the percentages in H0 (since H0 is assumed true)and multiply each by n (n = 600):
Number ofPercent
Expected
Televisions
from Ho
Frequency
010%(0.10)(600) = 60
116%
(0.16)(600) = 96
255%
(0.55)(600) = 330
311%(0.11)(600) = 66
4+8%(0.08)(600) = 48
Remember that df for Goodness of Fit tests is k – 1 instead of n – 1.
df = 5 – 1 = 4; df ≠ 600 – 1
L1: 66, 119, 340, 60, 15 (Observed frequency)
L2: 60, 96, 330, 66, 48
Calculate the test statistic: χ 2 = χ 2 GOF-Test or Sum(L3) = 29.65

p-value By Excel:
p-value = P( χ 2 > 29.65) = χ 2 CDF(29.65, 10^99, 4) = 0.000006
Conclusion: Reject H0 because p-value < α=chisq.dist.rt(test value, df)=chisq.dist.rt(29.65,4)This means you reject the belief that the distribution for the far western states is the sameas that of the American population as a whole.

594

Interpretation: At the 1% significance level, from the data, there is sufficient evidence toconclude that the "number of televisions" distribution for the far western United States isdifferent from the "number of televisions" distribution for the American population as awhole.
Example 12.9
Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are
the coins fair? Test at a 5% significance level.
Solution 12.9
This problem can be set up as a goodness-of-fit problem. The sample space for flipping two
fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same assaying, "Does the distribution of the coins (20 HH, 27 HT, 30TH, 23 TT) fit the expecteddistribution?"
Random Variable: Let X = the number of heads in one flip of the two coins. X takes on thevalues 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number ofcells is three. Since X = the number of heads, the observed frequencies are 20 (for two heads),
57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (fortwo heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.
H0: The coins are fair (uniformly distributed)
Ha: The coins are not fair.
Distribution for the test: χ 2 where df = 3 – 1 = 2.
Calculate the test statistic: χ 2 = 2.14

p-value By Excel:

Graph:=chisq.dist.rt(test value, df)=chisq.dist.rt(2.14, 2)

p-value = P( χ 2 > 2.14) = χ 2 CDF(2.14, 10^99, 2) = 0.3430
Conclusion: Do Not Reject H0 because p-value ≥ αInterpretation: There is insufficient evidence to conclude that the coins are not fair.
595

12.5 | Test of Independence and Test for Homogeneity
Test of Independence
Tests of independence involve using a contingency table of observed (data) values. The test
statistic for a test of independence is similar to that of a goodness-of-fit test:χ =∑2(O − E )

2

E

i• j

where:••••

O = observed values
E = expected values
i = the number of rows in the tablej = the number of columns in the table(O − E )There are i ⋅ j terms of the form

2E

A test of independence determines whether two factors are independent or not.
NOTE:
The expected value for each cell needs to be at least five in order for you to use this test.
In a test of independence, we state the null and alternative hypotheses in words. Since thecontingency table consists of two factors, the null hypothesis states that the factors are
independent and the alternative hypothesis states that they are not independent(dependent). If we do a test of independence using the example, then the null hypothesis is:
H0: Factor 1 and Factor 2 are independent
Ha: Factor 1 and Factor 2 are dependent
The test of independence is always right-tailed because of the calculation of the test statistic.
If the expected and observed values are not close together, then the test statistic is very large
and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.
The number of degrees of freedom for the test of independence is:
df = (number of columns - 1)(number of rows - 1)

596

The following formula calculates the expected number (E):E=

(row total )(column total )total number surveyed

12.5 A sample of 300 students is taken. Of the students surveyed, 50 were music students,
while 250 were not. Ninety- seven were on the honor roll, while 203 were not. If we assume
being a music student and being on the honor roll are independent events, what is the
expected number of music students who are also on the honor roll?

Example 12.10
In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend
time with a disabled senior citizen. The program recruits among community college students,
four-year college students, and nonstudents. In the following table is a sample of the adultvolunteers and the number of hours they volunteer per week. Is the number of hours
volunteered independent of the type of volunteer?
Type of Volunteer
CommunityStudents
Four-Year
Students
Nonstudents

1 – 3 hours 4–
hours
College 111
96

6 7 – 9 hours
48

College 96

133

61

91

150

53

Table: Number of Hours Worked Per Week by Volunteer Type (Observed) The table contains
observed (O) values (data).

Solution 12.10
The observed table and the question at the end of the problem, "Is the number of hour’s
volunteered independent of the type of volunteer?" tell you this is a test of independence.
The two factors are number of hours volunteered and type of volunteer. This test is always
right-tailed.
H0: The number of hours volunteered is independent of the type of volunteer.
Ha: The number of hours volunteered is dependent on the type of volunteer.
Distribution for the test: χ 42 since df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4
Calculate the test statistic: χ 2 = 12.99 (calculator or computer)597

Using the TI-83, 83+, 84, 84+ Calculator
To calculate test value of independence test_

Enter the 3 by 3 Contingency Table into Matrix [A] by going to 2nd Matrix, Edit
Press 2nd Quit to save the matrix and exit.
Go to STAT, TESTS, and select χ2-Test; make sure Observed: [A].

Press Calculate.

NOTE: You do not have to enter anything into Matrix [B]. Matrix [B] will automatically befilled with Expected Values.

Graph:

p-value By Excel:
p-value = P( χ 2 > 12.99) = 0.0113
Since no α is given, assume α = 0.05. p-value = 0.0113.

=chisq.dist.rt(test value, df)=chisq.dist.rt(12.99,4)

Conclusion: Reject H0 because p-value < α.
This means that the factors are not independent.
Interpretation: At a 5% level of significance, there is sufficient evidence to conclude thatthe number of hours volunteered and the type of volunteer are dependent on one another.

598

Using the TI-83, 83+, 84, 84+ Calculator
To see the Expected Values_

Expected Values of the Independence test were placed into Matrix [B]. To view them, Press
2nd Matrix, Edit, [B]Type
CC students
4-YR
students
Nonstudents

1–3
hrs
90.572103

4–6
hrs
115.19131

7–9hrs
49.23755.995104.42132.8156.76812.6 The Bureau of Labor Statistics gathers data about employment in the United States. A
sample is taken to calculate the number of U.S. citizens working in one of several industry
sectors over time. Table below shows the results:
Industry Sector
Nonagricultural wage and salary
Goods-producing, excluding
agriculture
Services-providing
Agriculture, forestry, fishing, andhunting
Nonagricultural self-employed andunpaid family worker
Secondary wage and salary jobs inagriculture and private householdindustries
Secondary jobs as a self-employed or
unpaid family worker

2000
13,2432457

2010
13,0441771

2020
15,0181950

10,786240

11,273214

13,068
201

931

894

972

14

11

11

196

144

152

We want to know if the change in the number of jobs is independent of the change in years.
State the null and alternative hypotheses and the degrees of freedom.

599

Example 12.11
College of Lake County is interested in the relationship between anxiety level and the needto succeed in school. A random sample of 400 students took a test that measured anxiety leveland need to succeed in school. Following table shows the results. CLC wants to know ifanxiety level and need to succeed in school are independent events.

Needto
Succeedin
SchoolHigh Need
Med Need
Low Need
Column Total

High
Anxiety
35
18
4
57

MedHigh
Anxiety
42
48
5
95

MediumAnxiety

Med-Low
Anxiety

Low
Anxiety

Row Total

53
63
11
127

15
33
15
63

10
31
17
58

155
193
52
400

a. Test for independence using 1% level of significance using critical valuesb. How many high anxiety level students are expected to have a high need to succeed inschool?
Solution 12.11
a. Test for independence
H0: The need to succeed in school is independent of anxiety level
Ha: The need to succeed in school is dependent of anxiety level
Distribution for the test: χ 42 since df = (5 columns – 1)(3 rows – 1) = (4)(2) = 8
Calculate the test statistic: χ 2 = 48.42 (calculator or computer)Enter the 3 by 5 Contingency Table into Matrix [A] by going to 2nd Matrix, Edit
2nd Quit, STAT, TESTS, χ 2 -Test, Calculate
Find the Critical Value: α = 0.01 (do not split since it is right-tailed test), df = 8

χ R2 = 20.090 by chart or By Excel =chisq.inv.rt(.01, 8)
Recall that the critical value separates the rejection region from the non-rejection region.
Here the rejection region is on the right side since it is a right-tailed test. The rejection
region starts are 20.090 and ends at positive infinity. Reject Ho if the test value is inside
the rejection region, which means that the test value is greater than the critical value.

600

Conclusion: Reject H0 because 48.42 (test value) > 20.090 (critical value)
Rejecting Ho means we are rejecting that the variables are independent. If we arerejecting independence, we are saying that the variables are dependent. Another phrase
for variables being dependent is that the variables are related. Therefore, theinterpretation for this example is as follows:
Interpretation: The need to succeed in school is related to the level of anxiety.

b. How many high anxiety level students are expected to have a high need to succeed inschool?
Here we are looking for an expected value. So we need to look at Matrix [B].

Needto High
Succeedin Anxiety
SchoolHigh Need22.088
Med Need27.503
Low Need
7.41

MedHigh
Anxiety
36.81345.838
12.35

MediumAnxiety

Med-Low
Anxiety

Low
Anxiety49.21361.27816.51

24.41330.3988.19

22.47527.985
7.54

You can expect about 22 high anxiety level students to have high need to succeed inschool.
Test for Homogeneity
A special type of independence test, called the test for homogeneity, can be used to draw aconclusion about whether two populations have the same distribution. To calculate the test
statistic for a test for homogeneity, follow the same procedure as with the test of
independence. However, the hypotheses statements are stated differently.
Hypotheses
H0: The distributions of the two populations are the same.
Ha: The distributions of the two populations are not the same.

601

Test Statistic (value)
Use a χ 2 test statistic. It is computed in the same way as the test for independence, using thematrix feature of the calculator.
Common Uses:
Comparing two populations. For example: freshman vs sophomore, before vs. after, east vs.
west. The variable is categorical with more than two possible response values.

Example 12.12
It is claimed that online and in-class college students have the same grade distribution.
Suppose that 250 randomly selected online college students and 300 randomly selected inclass college students’ grades were looked at. The results are shown in the following table.
Do online and in-class college students have the same grade distribution? Use a level ofsignificance of 0.05.
Onlinestudents
In-class
students

A grade48

B grade84

C grade49

D grade45

F grade
24

60

86

88

35

31

Solution 12.12
H0: The distribution of grades for online college students is the same as the distribution ofgrades for in-class college students.
Ha: The distribution of grades for online college students is not the same as the distributionof grades for in-class college students
Degrees of Freedom, df = 4
Distribution for the test: χ 2
Calculate the test statistic: χ 2 -Test = 10.138 (calculator or computer)p-value = P( χ 2 > 10.138) = .0382
Since no α is given, assume α = 0.05.

p-value by Excel:=chisq.dist.rt(test value, df) =chisq.dist.rt(10.138,4)

602

Conclusion: Reject H0 because p-value < α. This means that the distributions are not thesame.
Interpretation: At a 5% level of significance, from the data, there is sufficient evidence toconclude that the distribution of grades for online college students is not the same as the
distribution of grades for in-class college students
Notice that the conclusion is only that the distributions are not the same. We cannot use thetest for homogeneity to draw any conclusions about how they differ.

12.7 Do families and singles have the same distribution of cars? Use a level of significance of
0.05. Suppose that 100 randomly selected families and 200 randomly selected singles wereasked what type of car they drove: sport, sedan, hatchback, truck, and van/SUV. The results
are shown in the table below. Do families and singles have the same distribution of cars? Test
at a level of significance of 0.05.
FamilySingle

Sport
5
45

Sedan
15
65

Hatchback35
37

This Photo by Unknown Author is licensed under CC BY-SA-NC

603

Truck
17
46

Van/SUV
28
7

Example 12.13
Both before and after a recent earthquake, surveys were conducted asking voters which of
the three candidates they planned on voting for in the upcoming city council election. Has
there been a change since the earthquake? Use a level of significance of 0.05. The following
table shows the results of the survey. Has there been a change in the distribution of voter
preferences since the earthquake?
Before
After

Perez
167
214

Chung
128
197

Stevens
135
225

Solution 12.13
H0: The distribution of voter preferences was the same before and after the earthquake.
Ha: The distribution of voter preferences was not the same before and after the earthquake.
df = (number of columns – 1)(number of rows – 1) = 2(1) = 2
Distribution for the test: χ 2
Calculate the test statistic: χ 2 = 3.2603 (calculator: χ 2 – Test)
p-value by Excel:

p-value = P( χ 2 > 3.2603) = 0.1959
α = 0.05 and the p-value = 0.1959. p-value ≥ α=chisq.dist.rt(test value, df)=chisq.dist.rt(3.2603,2)

Do not reject Ho.
Interpretation: At a 5% level of significance, from the data, there is insufficient evidence toconclude that the distribution of voter preferences was not the same before and after theearthquake.

Key Terms
Critical Value is a value from the distribution that separates the confidence area from thenon-confidence area.
Chi Square Distribution is a family of curves created by using variance and dependent on
degrees of freedom. The shape of the curves is skewed right. The values of the distribution
are greater than or equal to zero. µ = df, σ = �2(𝑑𝑑𝑑𝑑)604

Formula Review
Critical Values
Left chi-square critical value, χ2L
Right chi-square critical value, χ2R

Calculator Function
None, use a chart
None, use a chart(n − 1) s 2

Confidence Interval for Variance:

χ R2

<σ <

Confidence Interval for Standard Deviation:

2

Excel Function=chisq.inv(α/2, df)=chisq.inv.rt(α/2, df)(n − 1) s 2

χ L2(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2

Review of Tests
You have seen the χ 2 test statistic used in four different circumstances. The following
bulleted list is a summary that will help you decide which χ 2 -test is the appropriate one touse.
•

Test of a Single Variance/Standard Deviation: Use the test to determine variation.
Test statistic: χ =2(n − 1) s 2

σ2

Degrees of freedom: df = n – 1; Test may be left-, right-, or two-tailed•

Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an
unknown distribution "fits" a known distribution. In this case there will be a single
qualitative survey question or a single outcome of an experiment from a single population.
Goodness-of-Fit is typically used to see if the population is uniform (all outcomes occur
with equal frequency), the population is normal, or the population is the same as anotherpopulation with a known distribution. The null and alternative hypotheses are:H0: p1 = %, p2 = %, … pk = %Ha: at least 1 proportion is different
than stated

Test statistic: χ =2∑(O − E )E

or

H0: The population fits the given
distribution.
Ha: The population does not fit the given
distribution.

2(calculator: χ2GOF-Test)

Degrees of freedom: df = k – 1
Test is always right-tailed

605

•

Independence: Use the test for independence to decide whether two variables (factors)
are independent or dependent. In this case there will be two qualitative survey questions
or experiments and a contingency table will be constructed. The goal is to see if the twovariables are unrelated (independent) or related (dependent). The null and alternative
hypotheses are:
H0: The two variables (factors) are independent.
Ha: The two variables (factors) are dependent.
Test statistic: χ =
2

∑
i• j

E=

(O − E )2(calculator: χ 2 -Test)

E(row total )(column total )total number surveyed

Degrees of Freedom: df = (number of rows – 1)(number of columns – 1)
Test is always right-tailed•

Homogeneity: Use the test for homogeneity to decide if two populations with unknown
distributions have the same distribution as each other. In this case there will be a single
qualitative survey question or experiment given to two different populations. The nulland alternative hypotheses are:
H0: The two populations follow the same distribution.
Ha: The two populations have different distributions.
Test statistic: χ =
2

∑
i• j

E=

(O − E )E

2(calculator: χ 2 -Test)(row total )(column total )total number surveyed

Degrees of Freedom: df = (number of rows – 1)(number of columns – 1)
Test is always right-tailed

606

EXERCISES FOR CHAPTER 12
1. If the number of degrees of freedom for a chi-square distribution is 25, what is the
population mean and standard deviation?
2. If df > 90, the distribution is _______________. If df = 15, the distribution is______________.
3. When does the chi-square curve approximate a normal distribution?
4. Find the critical values, χ L2 and χ R2 for C.L. = 0.95 and df = 11.
5. Find the critical values, χ L2 and χ R2 for C.L. = 0.98 and df = 19.
6. Find the critical values, χ L2 and χ R2 for C.L. = 0.99 and df = 23.
7. Find the critical values, χ L2 and χ R2 for C.L. = 0.90 and df = 35.
8. The grade point averages for 10 randomly selected students are listed below. Construct a
90% confidence interval for the population standard deviation, σ.
2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8
9. The outputs of 12 selected Keurig coffee makers are listed below. The quality control
needs the standard deviation to be low. Construct a 98% confidence interval for the
population standard deviation, σ.
11.85, 11.96, 11.9, 11.92, 12.01, 11.95, 11.96, 11.89, 11.97, 12.03, 11.88, 11.94
10. A large health maintenance organization (HMO) uses control charts to monitor theprocess of directing patient calls to the proper department or doctor’s receptionist. The
standard deviation of 21 control charts is 6.8 minutes. Construct a 95% confidence
interval for the population variance, σ2.
11. An archer’s standard deviation for his hits is six (data is measured in distance from thecenter of the target). An observer claims the standard deviation is less.
a. What type of test should be used?
b. State the null and alternative hypotheses.
c. Is this a right-tailed, left-tailed, or two-tailed test?
12. The standard deviation of heights for students in a school is 0.81. A random sample of 50
students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the school
is greater than 0.81.
a. State the null and alternative hypotheses.
b. df =
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13. The average waiting time in a doctor’s office varies. The standard deviation of waitingtimes in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes thevariance of waiting times is greater than originally thought.
a. What type of test should be used?
b. What is the test statistic?
c. What is the p-value?
d. What can you conclude at the 5% significance level?
14. Suppose an airline claims that its flights are consistently on time with an average delayof at most 15 minutes. It claims that the average delay is so consistent that the varianceis no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled
traveler calculates the delays for his next 25 flights. The average delay for those 25 flights
is 22 minutes with a standard deviation of 15 minutes.
a. Is the traveler disputing the claim about the average or about the variance?
b. A sample standard deviation of 15 minutes is the same as a sample variance of ______
minutes.
c. Is this a right-tailed, left-tailed, or two-tailed test?
d. H0:
e. df =
f. chi-square test statistic =g. p-value =h. Graph the situation. Label and scale the horizontal axis. Mark the mean and test
statistic. Shade the p-value.
i. Let α = 0.05. What is the conclusion? Rewrite out the conclusion in complete
sentences.
j. How did you know to test the variance instead of the mean?
k. If an additional test were done on the claim of the average delay, which distribution
would you use?
15. A plant manager is concerned her equipment may need recalibrating. It seems that theactual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard
deviation should be at most 0.5 oz. In order to determine if the machine needs to berecalibrated, 84 randomly selected boxes of cereal from the next day’s production were
weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to berecalibrated? Use α = 0.01.

608

16. Consumers may be interested in whether the cost of a particular calculator varies fromstore to store. Based on surveying 43 stores, which yielded a sample mean of $84 and asample standard deviation of $12, test the claim that the standard deviation is greaterthan $15.
17. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for
her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder
looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes,
and 57 minutes. Use α = 0.10.
18. Airline companies are interested in the consistency of the number of babies on each flight,
so that they have adequate safety equipment. They are also interested in the variation of
the number of babies. Suppose that an airline executive believes the average number ofbabies on flights is six with a variance of nine at most. The airline conducts a survey. The
results of the 18 flights surveyed give a sample average of 6.4 with a sample standard
deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. Use α = 0.05.
19. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility
rate has been attributed to the law passed in 1979 restricting births to one per woman.
Suppose that a group of students studied whether or not the standard deviation of births
per woman was greater than 0.75. They asked 50 women across China the number ofbirths they had had. The results are shown in the following table. Does the students’survey indicate that the standard deviation is greater than 0.75? Use α = 0.01.
# of Births
Frequency
0
5
1
30
2
10
3
5
20. According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, witha standard deviation of two. His friend, also an aquarist, does not believe that thestandard deviation is two. She counts the number of fish in 15 other 20-gallon tanks.
Based on the results that follow, do you think that the standard deviation is different
from two?
Use α = 0.10.
Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11609

21. The manager of "Frenchies" is concerned that patrons are not consistently receiving thesame amount of French fries with each order. The chef claims that the standard deviation
for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may behigher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and astandard deviation of two oz.
a. Find 90% confidence interval for population standard deviation.
b. Test the claim at 5% level of significance.
22. You want to buy a specific computer. A sales representative of the manufacturer claims
that retail stores sell this computer at an average price of $1,249 with a very narrow
standard deviation of $25. You find a website that has a price comparison for the same
computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95;$1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation
than claimed by the manufacturer? Use the 5% significance level. As a potential buyer,what would be the practical conclusion from your analysis?
23. A company packages apples by weight. One of the weight grades is Class A apples. Class
A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance
of 5% above or below the mean for apples in the same consumer package. A batch of apples
is selected to be included in a Class A apple package. Given the following apple weightsof the batch, does the fruit comply with the Class A grade weight tolerance requirements?
Conduct an appropriate hypothesis test.
(a) at the 5% significance level(b) at the 1% significance level
Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157;
171; 152; 161; 141; 166; 172;
24. Determine the appropriate test to be used in the following exercises:a. An archeologist is calculating the distribution of the frequency of the number ofartifacts she finds in a dig site. Based on previous digs, the archeologist creates an
expected distribution broken down by grid sections in the dig site. Once the site has
been fully excavated, she compares the actual number of artifacts found in each gridsection to see if her expectation was accurate.
b. An economist is deriving a model to predict outcomes on the stock market. He createsa list of expected points on the stock market index for the next two weeks. At the closeof each day’s trading, he records the actual points on the index. He wants to see how
well his model matched what actually happened.

610

c. A personal trainer is putting together a weight-lifting program for her clients. For a
90-day program, she expects each client to lift a specific maximum weight each week.
As she goes along, she records the actual maximum weights her clients lifted. She
wants to know how well her expectations met with what was observed.
25. A teacher predicts that the distribution of grades on the final exam will be as describedby the table on the left and the actual distribution for a class of 20 is given on the right:
Predicted Distribution:
Actual Distribution:
Grade
Proportion
Grade
Frequency
A
0.25
A
7
B
0.30
B
7
C
0.35
C
5
D
0.10
D
1
a. df =
b. State the null and alternative hypotheses.
c. χ2 test statistic =d. p-value =e. At the 5% significance level, what can you conclude?
26. The following data are real. The cumulative number of COVID-19 vaccine doses reported
for Lake County, IL on May 27, 2021 is broken down by ethnicity as in the following table.
Ethnicity
White
Hispanic
Black/AfricanAmerican
Asian, Pacific Islander
Other race

Number of Doses
388,833
81,67629,80351,89826,200a. If the ethnicities of COVID-19 vaccine doses followed the ethnicities of the total county
population, fill in the expected number of cases per ethnic group.
The percentage of each ethnic group in Lake County is as in following table (Census
2019 projections).
Ethnicity
White, not Hispanic
Hispanic
Black/AfricanAmerican
Asian, Pacific Islander
Other race

Percent of total county
60.6%
22.4%
7.5%
8.5%
1%

611

b. Perform a goodness-of-fit test to determine whether the occurrence of vaccine doses
follows the ethnicities of the general population of Lake County, IL.
H0:Ha:
c. Is this a right-tailed, left-tailed, or two-tailed test?
d. degrees of freedom =e. χ2 test statistic =f. p-value =g. Graph the situation. Label and scale the horizontal axis. Mark the mean and test
statistic. Shade in the region corresponding to the p-value.
h. Let α = 0.05. Decision:i. Conclusion (write out in complete sentences):
j. Does it appear that the pattern of vaccine doses in Lake County corresponds to the
distribution of ethnic groups in this county? Why or why not?
27. A survey by Computerworld mobile data service, asked “how would you rate the overall
trustworthiness of your mobile data provider?” The results from 2016 are shown in thefollowing table:
Very trustworthy
31%
Somewhat
41%
trustworthy
Somewhat9%
untrustworthy
Very untrustworthy
4%
No opinion
15%
Let’s say there is a claim that the distribution has changed since then. You randomly
select 500 students enrolled here this semester and survey them. The results are below:
Very trustworthy
38%
Somewhat
35%
trustworthy
Somewhat10%
untrustworthy
Very untrustworthy
7%
No opinion
10%
Use a level of significance of 5% to test the claim. Show all steps.
28. Determine the appropriate test to be used in the following exercises.
a. A pharmaceutical company is interested in the relationship between age andpresentation of symptoms for a common viral infection. A random sample is taken of
500 people with the infection across different age groups.

612

b. The owner of a baseball team is interested in the relationship between player salaries
and team winning percentage. He takes a random sample of 100 players from different
organizations.
c. A marathon runner is interested in the relationship between the brand of shoes
runners wear and their run times. She takes a random sample of 50 runners andrecords their run times as well as the brand of shoes they were wearing.
29. Transit Railroads is interested in the relationship between travel distance and the ticketclass purchased. A random sample of 200 passengers is taken. The following table shows
the results. The railroad wants to know if a passenger’s choice in ticket class is
independent of the distance they must travel.
Traveling
Distance
1–100 miles
101–200 miles
201–300 miles
301–400 miles
401–500 miles
Total

Third class

Second classFirst class

Total

21
18
16
12
6
73

14
16
17
14
6
67

6
8
15
21
10
60

41
42
48
47
22
200

a. State the hypotheses, H0 and Ha:b. df =
c. How many passengers are expected to travel between 201 and 300 miles and purchasesecond-class tickets?
d. How many passengers are expected to travel between 401 and 500 miles and purchasefirst-class tickets?
e. What is the test statistic?
f. What is the p-value?
g. What can you conclude at the 5% level of significance?
30. An article in the New England Journal of Medicine, discussed a study on smokers in
California and Hawaii. In one part of the report, the self-reported ethnicity and smoking
levels per day were given. The table below summarizes the results.
Smoking
level perday
1 – 10
11 – 20
21 – 3031 +AfricanNativeLatinoAmerican Hawaiian

Japanese
White
Americans

9886
6514
1671
759

8378
10680
4715
2305

2745
3062
1419
788

12831
4932
1406
800

613

7650
9877
6062
3970

a. State the hypotheses.
H0:Ha:
b. Find the Expected Values. Round to two decimal places. Place them into a table.
c. df =d. χ2 test statistic =e. p-value =f. Is this a right-tailed, left-tailed, or two-tailed test? Explain why.
g. Graph the situation. Label and scale the horizontal axis. Mark the mean and test
statistic. Shade in the region corresponding to the p-value.
h. State the decision and conclusion (in a complete sentence) for α = 0.05
i. State the decision and conclusion (in a complete sentence) for α = 0.01
31. A math teacher wants to see if two of her classes have the same distribution of test scores.
a. What test should she use?
b. What are the null and alternative hypotheses?
32. A market researcher wants to see if two different stores have the same distribution ofsales throughout the year. What type of test should he use?
33. A meteorologist wants to know if East and West Australia have the same distribution ofstorms. What type of test should she use?

34. Do private practice doctors and hospital doctors have the same distribution of working
hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are
selected at random and asked about the number of hours a week they work. The results
are shown in following table:

a.
b.
c.
d.
e.

20 – 29
30 – 39 40 – 49
Private Practice16
40
38
Hospital
84459State the null and alternative hypotheses.
df =What is the test statistic?
What is the p-value?
What can you conclude at the 5% significance level?

50 - 59639

35. Which test would you use to decide whether two factors have a relationship?
36. Which test would you use to decide if two populations have the same distribution?
37. How are tests of independence similar to tests for homogeneity?
61438. How are tests of independence different from tests for homogeneity?
39. Decide whether the following statements are true or false.
a. As the number of degrees of freedom increases, the graph of the chi-square
distribution looks more and more symmetrical.
b. The standard deviation of the chi-square distribution is twice the mean.
c. The mean and the median of the chi-square distribution are the same if df = 24.
40. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct
a hypothesis test to determine if the die is fair. The data in the following table are theresult of the 120 rolls.
Face Value
1
2
3
4
5
6

Frequency Expected frequency
15
29
16
15
30
15

41. The marital status distribution of the U.S. male population, ages 15 and older, is as shownin the following table.
Marital Status
Never married
MarriedWidowedDivorced/separated

Percent
31.3
56.1
2.5
10.1

Expected frequency

Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded thefollowing frequency distribution. We are interested in whether this age group of males fits
the distribution of the U.S. adult population. Calculate the frequency one would expect whensurveying 400 people. Fill in the Expected frequency in the above table, rounding to two
decimal places.
Marital Status
Never married
MarriedWidowedDivorced/separated

Frequency
140
238
2
20

615

42. The columns in the following table contain the Race/Ethnicity of U.S. Public Schools fora recent year, the percentages for the Advanced Placement Examinee Population for thatclass, and the Overall Student Population. Suppose the right column contains the resultof a survey of 1,000 local students from that year who took an AP Exam.
Race/Ethnicity

APExaminee
10.2%

OverallStudent
5.4%

SurveyFrequency
113

Asian, Asian American, orPacific IslanderBlack or African-American
8.2%
14.5%
94
Hispanic or Latino
15.5%
15.9%
136
American Indian or Alaska
0.6%
1.2%
10
Native
White
59.4%
61.6%
604
Not reported/other
6.1%
1.4%
43
a. Perform a goodness-of-fit test to determine whether the local results follow the
distribution of the U.S. overall student population based on ethnicity.
b. Perform a goodness-of-fit test to determine whether the local results follow the
distribution of U.S. AP examinee population, based on ethnicity.
43. The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total
population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the
Manhattan, NY, area yielded the data in the following table. Conduct a goodness-of-fit
test to determine if the self-reported sub-groups of Asians in the Manhattan area fit thatof the Lake Tahoe area.
Race

Asian
Indian
131

Lake
Tahoe
Frequency
Manhattan 174
Frequency

Chinese

Filipino

Japanese Korean

Vietnamese Other

118

1045

80

12

9

24

557

518

54

29

21

66

44. Read the statement and decide whether it is true or false:a. In a goodness-of-fit test, the expected values are the values we would expect if the null
hypothesis were true.
b. In general, if the observed values and expected values of a goodness-of-fit test are notclose together, then the test statistic can get very large and on a graph will be way out
in the right tail.
c. Use a goodness-of-fit test to determine if high school principals believe that students
are absent equally during the week or not.
d. The test to use to determine if a six-sided die is fair is a goodness-of-fit test.

616

e. In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null
hypothesis.

45. The CLC math department expects the grade distribution of College Algebra to be as
follows: 15% A grade, 25% B grade, 30% C grade, 10% D grade, 5% F grade, and 15% W.
W is a withdrawal. The spring 2021 observed grade distribution of --- students is listed
in the table:
Grade distribution
A
B
C
D
F
W

Actual grade
distribution (Observed)
59
85
53
29
28
81

Expected grade
distribution

Conduct a goodness-of-fit test to determine if the actual grade distribution fits the
distribution of the department’s expected distribution. Use the level of significance of 1%.

46. A sample of 212 commercial businesses was surveyed for recycling one commodity; acommodity here means any one type of recyclable material such as plastic or aluminum.
The following table shows the business categories in the survey, the sample size of eachcategory, and the number of businesses in each category that recycle one commodity.
Based on the study, on average half of the businesses were expected to be recycling one
commodity. As a result, the last column shows the expected number of businesses in eachcategory that recycle one commodity. At the 5% significance level, perform a hypothesis
test to determine if the observed number of businesses that recycle one commodity follows
the uniform distribution of the expected values.
Business Type

Numberclass
Office35
Retail/Wholesale
48
Food/Restaurants
53
Manufacturing/Medical 52
Hotel/Mixed
24

in Observed # thatrecycle
19
27
35
21
9

617

Expected # thatrecycle
17.5
24
26.5
26
12

47. The following table contains information from a survey among 489 participants classified
according to their age groups. The second column shows the number of obese people per
age class among the study participants. The last column comes from a different study atthe national level that shows the corresponding percentages of obese people in the sameage classes in the USA. Perform a goodness of fit hypothesis test at the 5% significancelevel to determine whether the survey participants are a representative sample of the
USA obese population.
AgeClass Frequency
of Expected USA average(Years)Obese(%)
participants
21 – 3075
15%
31 – 40132
30%
41 - 5068
15%
51 – 60109
20%
61 – 70105
20%
48. A recent debate about where in the United States skiers believe the skiing is best
prompted the following survey. Test to see if the best ski area is independent of the levelof the skier.
U.S. Ski Area
Beginner
Intermediate Advanced
Tahoe
20
30
40
Utah
10
30
60
Colorado
1040
50
49. Car manufacturers are interested in whether there is a relationship between the size ofcar an individual drives and the number of people in the driver’s family (that is, whether
car size and family size are independent). To test this, suppose that 800 car owners wererandomly surveyed with the results in the following table. Conduct a test of
independence.
Family Size
1
2
3-4
5+

Sub &
Compact
20
20
20
20

Mid-size

Fulll-size

Van & Truck

35
50
50
30

40
70
100
70

35
80
90
70

618

50. College students may be interested in whether or not their majors have any effect onstarting salaries after graduation. Suppose that 300 recent graduates were surveyed as
to their majors in college and their starting salaries after graduation. The following tableshows the data. Conduct a test of independence using α = .05.
Major
EnglishEngineering
Nursing
Business
Psychology< $50,000
5
10
10
10
20

$50,000 - $68, 999
20
30
15
20
30

$69, 000 +
5
60
15
30
20

51. Some travel agents claim that honeymoon hot spots vary according to age of the bride.
Suppose that 280 recent brides were interviewed as to where they spent theirhoneymoons. The information is given in the following table. Conduct a test of
independence using α = .10.
Location
Niagara Falls
Poconos
EuropeVirgin Islands

20 – 29
15
15
10
20

30 – 39
25
25
25
25

40 – 49
25
25
15
15

50 and over
20
10
5
5

52. A manager of a sports club keeps information concerning the main sport in which
members participate and their ages. To test whether there is a relationship between theage of a member and his or her choice of sport, 643 members of the sports club are
randomly selected. Conduct a test of independence using α = .05.
Sport
RacquetballTennisSwimming

18 -2542
58
72

26 – 3058
76
60

31 – 4030
38
65

41 and over
46
65
33

53. A major food manufacturer is concerned that the sales for its skinny French fries havebeen decreasing. As a part of a feasibility study, the company conducts research into thetypes of fries sold across the country to determine if the type of fries sold is independentof the area of the country. The results of the study are shown in table below. Conduct a
test of independence using α = .05.
Type of Fries
Skinny Fries

Northeast
70

South
50
619

Central
20

West
25

Curly Fries
100
60
15
30
Steak Fries
20
40
10
10
54. According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, thefollowing is a breakdown of the amount of life insurance purchased by males in thefollowing age groups. He is interested in whether the age of the male and the amount oflife insurance purchased are independent events. Conduct a test for independence using
α = .01.

Age of
Males
20-29
30-39
40-49
50+

None
40
35
20
40

< $200,000 $200,000$400,000
15
40
5
20
0
30
30
15

$400,001 $1,000,000
0
20
0
15

$1,000,001+
5
10
30
10

55. Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is arelationship between the level of education an individual has and salary. Conduct a testof independence using α = .05
Annual Salary< $30,000$30,000 - $39, 999$40,000 - $49,999$50.000 - $59,999$60,000 +Not a HS HS
graduate
graduate
15
25
20
40
10
20
5
10
0
5

College
graduate
10
70
40
20
10

Master’s/Doctorate
5
30
55
60
150

56. Read the statement and decide whether it is true or false
a. The number of degrees of freedom for a test of independence is equal to the sample
size minus one.
b. The test for independence uses tables of observed and expected data values.
c. The test to use when determining if the college or university a student chooses toattend is related to his or her socioeconomic status is a test for independence.
d. In a test of independence, the expected number is equal to the row total multiplied by
the column total divided by the total surveyed.

620

57. An ice cream maker performs a nationwide survey about favorite flavors of ice cream indifferent geographic areas of the U.S. Based on the following table, do the numberssuggest that geographic location is independent of favorite ice cream flavors? Test at the
5% significance level.
US
RegionStrawberry Chocolate Vanilla Rocky
Road

West
Midwest
East
South

12
10
8
15

21
32
31
28

22
22
27
30

19
11
8
8

Mint
Chocolate
Chip
15
15
15
15

Pistachio
86
7
6

58. The following table provides a recent survey of the youngest online entrepreneurs whose
net worth is estimated at one million dollars or more. Their ages range from 17 to 30.
Each cell in the table illustrates the number of entrepreneurs who correspond to thespecific age group and their net worth. Are the ages and net worth independent? Performa test of independence at the 5% significance level.
AgeGroup/Net 1 – 5Worth Value (inmillions)
17 – 258
26 – 30
6

6 – 24

25 +

7
5

5
9

59. Based on Pew Research Center, below is the number of registered voters who identify as
one of the following generation and political party in 2018.
Conduct a test of
independence at the 5% significance level. What is the expected number of Republican
Generation X?
Millennial

Race/Generation
Democratic
124
Republican
59

Gen X

BoomerSilent

105
76

112
75

111
78

60. A psychologist is interested in testing whether there is a difference in the distribution of
personality types for business majors and social science majors. The results of the study
are shown in the following table. Conduct a test of homogeneity. Test at a 5% level ofsignificance.
Business
Social Science

Open
41
72

Conscientious Extrovert Agreeable Neurotic
52
46
61
58
75
63
80
65
621

61. Do people from the Midwest and West coast select different breakfasts? The breakfasts
ordered by randomly selected Midwesterners and West coasters at a popular breakfast
place is shown in the table below. Conduct a test for homogeneity at a 5% level ofsignificance.
West Coast
MidwestFrench Toast
47
65

Pancakes
25
59

Waffles
28
55

Omelets
63
60

62. A fisherman is interested in whether the distribution of fish caught in Green Valley Lakeis the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selectedfish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 werebass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115
were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform atest for homogeneity at a 5% level of significance.
63. In 2017 according to the U.S. National Center for Education Statistics, the following table
you can see the GPA groupings with respect to enrollment in STEM. According to thesurvey results shown in the table, is the distribution of GPA same for STEM and nonSTEM enrollment? Provide your assessment at the 5% significance level. Did you expect
the result you obtained?
Undergraduate GPA
0.1 – 2.1
2.2 – 2.8
2.9 – 3.4
3.5 – 4.0

Enrolled in STEM
courses
32
249
235
282
202

Not enrolled in STEM
courses
171
274
132
170
253

64. When looking at energy consumption, we are often interested in detecting trends overtime and how they correlate among different countries. The information in the following
table shows the average energy use (in units of kg of oil equivalent per capita) in the USA
and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do theenergy use values in these two areas come from the same distribution? Perform theanalysis at the 5% significance level.
Year
2010
2009
2008
2007

European
Union
3413
3302
3505
3537

United States
7164
7057
7488
7758
622

2006
3595
7697
2005
3613
7847
65. The Insurance Institute for Highway Safety collects safety information about all types ofcars every year, and publishes a report of Top Safety Picks among all cars, makes, andmodels. In the following table presents the number of Top Safety Picks in six car
categories for the two years 2013 and 2017. Analyze the table data to conclude whether
the distribution of cars that earned the Top Safety Picks safety award has remained thesame between 2013 and 2017. Derive your results at the 5% significance level.
Year/Car
Type
2013
2017

Small Midsize Large
12
31

22
30

10
19

Small
SUV
10
11

Mid-Size
SUV
27
29

Large
SUV
6
4

REFERENCES
12.1 The Chi-Square Distribution
Data from Parade Magazine.
“HIV/AIDS Epidemiology Santa Clara County.” Santa Clara County Public Health Department, May2011.
12.3 Test of a Single Variance“AppleInsider Price Guides.” Apple Insider, 2013. Available online at
http://appleinsider.com/mac_price_guide (accessed May 14, 2013).
Data from the World Bank, June 5, 2012.
12.4 Goodness-of-Fit Test
Data from the U.S. Census Bureau
Data from the College Board. Available online at http://www.collegeboard.com. Data from the U.S.
Census Bureau, Current Population Reports.
Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene,“Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American
Journal of Epidemiology volume 158, no. 1, pages 85-92.
Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in
the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at
http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013).
Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington
Count, VA. Available online at623

http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May
24,2013).

12.5 Test of Independence/Homogeneity
DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary
Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to
Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013.
Available online at http://field.com/fieldpollonline/subscribers/ Rls2436.pdf (accessed May 24, 2013).
HarrisInteractive,“Favorite
Flavorof
Ice
Cream.”
Available
http://www.statisticbrain.com/favorite-flavor-of-ice- cream (accessed May 24, 2013)

onlineat“Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngestonline-entrepreneur-list (accessed May 24, 2013).
Data from the Insurance Institute for Highway
www.iihs.org/iihs/ratings (accessed May 24, 2013).

Safety,2013.

Available

onlineat“Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at
http://data.worldbank.org/indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013).
“Parent and Family Involvement Survey of 2007 National Household Education Survey Program(NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at
http://nces.ed.gov/pubsearch/ pubsinfo.asp?pubid=2009030 (accessed May 24, 2013).
Gralla, Preston. “The Votes Are in: Which Mobile Data Provider Is Best?” Computerworld,
Computerworld, 21 Dec. 2016, www.computerworld.com/article/3150992/wireless-carriers/the-votesare-in-which-mobile-data-provider-is-best.html.
“White Millennial Voters Are More Democratic than White Voters in Older Generations.” Pew
Research Center - U.S. Politics & Policy, Pew Research Center, 20 Mar. 2018,www.pewresearch.org/politics/2018/03/20/1-trends-in-party-affiliation-among-demographicgroups/2_10-4/.
U.S. Department of Education, National Center for Education Statistics, 2012/17 Beginning
Postsecondary Students Longitudinal Study (BPS:12/17).

624

CHAPTER 12 SOLUTIONS:
1) µ = df = 25; s = √2(df) = √50 ≈ 7.071.
3) The chi-square distribution approximates the normal distribution when df > 90.
5) Χ2L = 7.633, Χ2R = 36.191. 7) Χ2L = 22.465, Χ2R = 49.802.
9) Χ2L = 3.054, Χ2R = 24.725; s2 = .00282(n − 1) s 2

χ R2

<σ <

(n − 1) s 2

χ L2= .0354 < σ < .101

11) a. Use a Χ2 test for a single variance. b. Ho: σ = 6; Ha: σ < 6.
c. This is a left-tailed test.
13)

a. Use a Χ2 test for a single variance. b. Χ2 = 29(4.1)2/(3.4)2 = 42.17
c. p = Χ2 cdf(42.17, 109, 29) = 0.0542. d. Since p > .05, do not reject Ho.
There is not quite enough evidence to conclude that the standard deviation of
waiting times is greater than 3.4 minutes.

15)

The hypotheses are Ho: σ < 0.5; Ha: σ > 0.5. Use a Χ2 test for a single variance.
The test statistic is Χ2 = 83(.54)2/(.5)2 = 96.81, and the p-value is p = Χ2 cdf(96.81, 109,83) = 0.1426. Since p > .01, we do not reject Ho. There is not enough evidence toconclude that the machine needs recalibrating.

17)

The hypotheses are Ho: σ < 3 (claim); Ha: σ > 3. Use a Χ2 test for a single
variance. The test statistic is Χ2 = 4(3.194)2/(3)2 = 4.534, and the p-value is p = Χ2
cdf(4.534, 109, 4) = 0.3385. Since p > .10, we do not reject Ho. There is not enoughevidence to refute her claim.

19)

The hypotheses are Ho: σ < 0.75; Ha: σ > 0.75. Use a Χ2 test for a single variance.
The test statistic is Χ2 = 49(.789)2/(.75)2 = 54.23, and the p-value is p = Χ2 cdf(54.23,
109, 49) = 0.2818. Since p > .01, we do not reject Ho. There is not enough evidence toconclude that the standard deviation is more than 0.75.

21)

a. Round degrees freedom down to df = 45 to get Χ2L = 30.6123 and Χ2R = 61.6562.
Using Excel Χ2L = CHISQ.INV(0.05,48)= 33.098 and Χ2R = CHISQ.INV.RT(0.05,48)= 65.171 Apply the formula to get the interval: 1.76 < σ < 2.50 or using Excelvalues 1.72 < σ < 2.41
b. The hypotheses are Ho: σ < 1.5; Ha: σ > 1.5. Use a Χ2 test for a single variance.
The test statistic is Χ2 = 48(2)2/(1.5)2 = 85.33, and the p-value is p = Χ2 cdf(85.33,
109,48) = 0.00073. Since p < .05, we reject Ho. There is sufficient evidence toconclude that the machine needs the standard deviation is greater than 1.5 oz.
625

23)

Note that 5% of 150 is 7.5, so the hypotheses are Ho: σ < 7.5; Ha: σ > 7.5. Use a Χ2
test for a single variance; the test statistic is Χ2 = 14(10.43)2/(7.5)2 = 27.07, and thep-value is p = Χ2 cdf(27.07, 109, 14) = 0.0189.
a. Since p < .05, we reject Ho. At the .05 level of significance, there is sufficientevidence to conclude that the standard deviation exceeds the weight tolerance.
b. Since p > .01, we do not reject Ho at the .01 level of significance. I.e. at the .01
level of significance, there is not enough evidence to conclude that the standard
deviation exceeds the weight tolerance.

25) a. df = 3 (the number of categories minus 1)b. Ho: p1 = .25, p2 = .30, p3 = .35, p4 = .10;
Ha: At least one of the proportions is different than stated.
c. According to Ho, the expected values are 5 A’s, 6 B’s, 7 C’s and 2 D’s, so Χ2 = 2.038.
d. p = Χ2 cdf(2.038, 109, 3) = 0.5645. e. Do not reject Ho. There is not a significant
difference between the observed distribution and the expected distribution.
27)

Ho: p1 = .31, p2 = .41, p3 = .09, p4 = .04, p5 = .15; Ha: at least 1 proportion isdifferent than stated. Test value: Χ2 = 32.43; p-value = .00000156; Reject Ho; Thereis enough evidence to support the claim that the distribution has changed since
2016.

29)

a. Ho: Distance and ticket class are independent. Ha: Distance and ticket class are
dependent
b. df = (4)(2) = 8

c. 16.1 passengers expected d. 6.6 passengers expectede. Χ2 =15.92 f. p-value = .0435
g. Reject Ho. There is enough evidence at 5%
level of significance to reject the claim that distance and travel class are
independent.
31)

a. Χ2 test, test for Homogeneityb. Ho: class 1 and class 2 have the same distribution of test scores. Ha: class 1 andclass 2 do not have the same distribution of test scores.

33) Χ2 test, test for Homogeneity35) Χ2 test (test for independence)37) Both use Χ2 test with a contingency table39) a. trueb. false

c. false (they are close in number but not exactly the same)41) Expected frequency: 125.2, 224.4, 10, 40.4
43) Lake Tahoe frequency is expected values, Manhattan is observed values.

626

Ho: p1 = 9.2%, p2 = 8.3%, p3 = 73.6%, p4 = 5.6%, p5 = .8%, p6 = .6%, p7 = 1.7%
Ha: at least 1 proportion is not the same as statedΧ2 = 2035; p-value = 0; Reject Ho; There is enough evidence to reject that the selfreported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe
area.
45)

Expected frequency: A 50.25, B 83.75, C 100.5, D 33.5, F 16.75, W 50.25; X2-GOF
test value = 50.97, p-value = 8.77E-10, reject Ho because p-value < α, There isenough evidence to reject the claim that the grade distribution for College Algebra is
the following 15% A grade, 25% B grade, 30% C grade, 10% D grade, 5% F grade,and 15% W.

47)

Ho: Surveyed obese fit the distribution of expected obese
Ha: Surveyed obese does not fit the distribution of expected obese
Χ2 = 3.71; p-value = 0.4462 Do not Reject Ho. Surveyed obese does fit the expected.

49)

Ho: Family size and size of car are independent; Χ2 = 15.83; p-value = .071. Do not
reject Ho. There is not enough evidence to reject the claim that they are
independent.

51)

Ho: Location of honeymoon and Age are independent; Χ2 = 15.7; p-value = .0734;
Reject Ho. Location of honeymoon and Age are dependent.

53)

Ho: type of fries and area of US sold are independent; Χ2 = 18.84; p-value = .00445;
Reject Ho. They are dependent.

55)

Ho: Annual Salary and level of education are independent. Χ2 = 255.8; p = 0; Reject
Ho. They are related.

57)

Ho: Geographic location and favorite ice cream flavor are independent. Χ2 =14.06; pvalue = .521. They are independent.

59)

Ho: Generation and political party are independent; Χ2 =4.7; p-value = .194; Do Not
Reject Ho. They are independent. Expected Gen X and Republican is 70 people.

61)

Ho: Midwesterners and West coasters have the same distribution for breakfast
selection; Χ2 = 11.56; p-value = .0091; We can reject that they have the same
distribution.

63)

Ho: Applicable reasons and Most important reasons have the same distribution; Χ2 =158.75; Reject Ho. They do not have the same distribution.

65)

Ho: 2009 and 2013 distribution of cars is the same; Χ2 =6.65; p-value = .248; Do not
reject Ho

627

This page is purposely left blank.

628

13 | LINEAR REGRESSION AND
CORRELATION

Figure 13.1 There is a correlation between the amount of cheese consumption and the number ofpeople who died by becoming tangled in their bedsheets between 2000 – 2009. For more correlations
go to www.tylervigen.com. Photo by Katrin Leinfellner on Unsplash

Chapter Objectives
By the end of this chapter, the student should be able to:
•
•
•
•
•
•

Discuss basic ideas of linear regression and correlation.
Create and interpret a line of best fit.
Calculate and interpret the correlation coefficient.
Determine if a correlation is significant
Use the regression model to make predictions.
Identify and interpret outliers.

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INTRODUCTION
Professionals often want to know how two or more numeric variables are related. For
example, is there a relationship between the grade on the second math exam a student takes
and the grade on the final exam? If there is a relationship, what is the relationship and how
strong is it?
In another example, your income may be influenced by your education, your profession, youryears of experience, and your ability. The amount you pay a repair person for labor is often
determined by an initial amount plus an hourly fee. The type of data described in thischapter is bivariate data – the prefix "bi" indicating there are two variables. Although wewill only study models involving two variables, in many real-world situations statisticians
use multivariate data, meaning many variables.
In this chapter, we will be studying the simplest form of regression, "linear regression" with
one independent variable. More specifically, given sample data measured on two variables,x and y, we wish to find a linear equation that best fits the observed sample data. We will
also develop statistical measures and tests that measure how strong the linear relationshipis between the variables.

13.1 | Linear Equations
We start by reviewing the basic facts about linear equations. A linear equation is anequation of the form:y = a + bx
where a and b are constant numbers. This is called a linear equation because the graph ofthe equation is a straight line.
••

Independent variable: the variable x written on the horizontal axis.
Dependent variable: the variable y written on the vertical axis where y depends
on x. Given any value, we can substitute it for the independent variable x to obtainthe corresponding value for the dependent variable.

Example 13.1
Graph the following equations: which of these is linear?
a. y = -1 + 2x

b. y = 2e x

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Solution 13.1
a. Linearb. Not linear

13.1 Is the following an example of a linear equation? If so, sketch the graph.
y = -0.125 – 3.5x

Example 13.2
Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32
per hour plus a $31.50 one-time charge. The total cost to a customer depends on the numberof hours it takes to complete the job. Find the equation that expresses the total cost in termsof the number of hours required to complete the job.
Solution 13.2
Let x = the number of hours it takes to get the job done. Let y = the total cost to thecustomer.
The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)(x) is the cost of
the word processing only. The total cost is: y = 31.50 + 32x

13.2 Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 perclass as well as $20 per student in the class. The total cost Emma pays depends on the numberof students in a class. Find the equation that expresses the total cost in terms of the numberof students in a class.

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Slope and Intercept of a Linear Equation
Given a linear equation y = a + bx, we call b the slope, and a is called the y-intercept. Frombasic algebra, recall that the slope is a number that describes the steepness of the line, andthe y-intercept is the y coordinate of the point (0, a) where the line crosses the y-axis. The
sign of the slope b determines whether the line slopes upward or downward, as shown in the
figure below:(a) If b > 0, the line slopes upward to the right.
(b) If b = 0, the line is horizontal.
(c) If b < 0, the line slopes downward to the right.

More specifically, when the slope b is positive, an increase in x results in an increase in y.
And if the slope b is negative, then an increase in x results in a decrease in y.

Example 13.3
Svetlana tutors to make extra money for college. For each tutoring session, she charges a
one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total
amount of money Svetlana earns for each session she tutors is y = 25 + 15x.
What are the independent and dependent variables? What is the y-intercept and what is
the slope? Interpret these using complete sentences.
Solution 13.3
The independent variable x is the number of hours Svetlana tutors each session. The
dependent variable y is the amount, in dollars, Svetlana earns for each session.
The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a onetime fee of $25 (this is when x = 0). The slope is 15 (b = 15). For each additional hour shetutors, Svetlana earns an additional $15.

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13.3 Ethan repairs household appliances like dishwashers and refrigerators. For each visit,
he charges $25 plus $20 per hour of work. A linear equation that expresses the total
amount of money Ethan earns per visit is y = 25 + 20x.
What are the independent and dependent variables? What is the y-intercept and what is
the slope? Interpret them using complete sentences.

13.2 | Scatter Plots
Before we begin our discussion of linear regression and correlation, we need to a way todisplay the relation between two variables x and y. The most commonly used graph is ascatter plot or scatterplot (also called a scatter diagram, scattergram or scatter graph).
To make a scatterplot, we just plot the points, for each data pair using the x-value as the xcoordinate and the y-value as the y-coordinate. For small examples, this is easy to do byhand; but most statistical software packages will make very nice scatterplots, and they canbe done on the TI-84 as well. We will illustrate this with an example.

Example 13.4
SCUBA divers have maximum dive times they cannot exceed when going to different
depths. The data in the table below show different depths with the maximum dive times inminutes.
X (depth in feet)
50
60
70
80
90
100

Y (maximum dive time)
80
55
45
35
25
22

Construct a scatter plot. Let x = depth in feet and let y = maximum dive time in minutes.

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Solution 13.4
Using the TI-83, 83+, 84, 84+ Calculator
To create a scatterplot:_

1. Enter your X data into list L1 and your Y data into list L2.
2. Press 2nd STATPLOT ENTER to use Plot 1. On the input screen for PLOT 1,
highlight On and press ENTER. (Keep the other plots set to OFF.)
3. For TYPE: highlight the very first icon, which is the scatter plot, and press ENTER.
4. For Xlist: enter L1 and for Ylist: enter L2.
5. For Mark: it does not matter which symbol you highlight, but the square is theeasiest to see. Press ENTER.
6. Make sure there are no other equations that could be plotted. Press Y = and clear
any equations out.
7. Press the ZOOM key and then select option 9, "ZoomStat"; the calculator will fit thewindow to the data. You can press WINDOW to see the scaling of the axes.
Following these instructions, we get a graph like:
Scatterplot of Maximum Dive Time vs Depth

Maximum dive time (in minutes)

80
70
60
50
40
30
20
50

60

70
80
Depth (in feet)

90

100

The points in the scatterplot exhibit a linear pattern; this suggests that a linear equation
would be a good fit to this data. We also can see that as the depth increases, the maximum
dive time decreases.

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A scatter plot shows the direction of the relationship between the variables.
••

If the points in the graph slope upward as we move left to right, then we say that
there is a positive relationship between the variables x and y.
If the points in the graph slope downward as we move left to right, then we say that
there is a negative relationship between the variables x and y.

We can also roughly determine the strength of the linear relationship by looking at the
scatter plot; the more closely the points conform to a linear pattern, the better a linear
equation will describe the relationship. If the points in the graph appear to follow a nonlinear pattern (i.e. there is curvature) then perhaps a linear model is not appropriate.
When we look at a scatterplot, we want to notice the overall pattern and any deviations
from the pattern. The following scatterplot examples illustrate these concepts.

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Again, in this chapter, we are interested in scatter plots that show a linear pattern. The
linear relationship is strong if the points are close to a straight line, except in the case of ahorizontal line where there is no relationship. If we think that the points show a linear
relationship, we would like to fit a line to the points in the scatter plot. This line can becalculated through a process called linear regression, which we will learn in the next
section. We will also develop a numerical measure that provides a more objective measureof how well the line fits the data. We will only use the fitted line when we have determinedthat there is a statistically significant relationship between the variables x and y.

13.4 Amelia plays basketball for her high school. She wants to improve to play at the college
level. She notices that the number of points she scores in a game goes up in response to thenumber of hours she practices her jump shot each week. She records the following data:
X (hours practicing jump shot)
5
7
9
10
11
12

Y (points scored in a game)
15
22
28
31
33
36

Construct a scatter plot and state whether Amelia’s conjecture appears to be true.

Example 13.5
Two different tests are designed to measure employee productivity and dexterity. Several
employees of a company are randomly selected and asked to complete the tests. The results
are below. Draw a scatterplot using Excel, describe the scatterplot.
Dexterity

23 25 28 21 21 25 26 30 34 36

Productivity 49 53 59 42 47 53 55 63 67 75

636

Solution 13.5
First, place Dexterity values in column A and place Productivity values in column B.
Second, highlight both columns, Insert Scatter. Choose the Scatter that is not connected.

Description: Strong positive linear pattern637

13.3 | The Regression Equation
Data rarely fit a straight line exactly; usually we must be satisfied with estimates. In thissection we discuss how to calculate the Line of Best Fit. This equation is also known as
the Least-Squares Line.
This section covers the following parts of the regression line:
•
•
•
•
•
•

The line of best fit (regression line, or least squares line)
The residual, εoSlope and y-intercept
Interpolation and Extrapolation
The correlation coefficient, r
The coefficient of determination, r2

Example 13.6
A random sample of 11 statistics students produced the following data, where x is the third
exam score out of 80, and y is the final exam score out of 200. Using the scatterplot, does it
appear that we can predict the final exam score of a random student if we know the third
exam score?
X (third exam score) 65
67
71
71
66
75
67
70
71
69
69
Y (final exam score) 175 133 185 163 126 198 153 163 159 151 159
Solution 13.6
The scatterplot below exhibits a tight, linear pattern. This indicates a linear model could
be used to predict the final exam score from the score on the third exam.

638

Next, we wish to find a line that best "fits" the data. To do this, we use what is called a leastsquares regression line to obtain the line of best fit.
Consider the following diagram. Each point of data is of the form (x, y) and each point of theline of best fit using least- squares linear regression has the form (x, ŷ). The ŷ is read "y-hat"and is the estimated value of y; that is, it is the value of y obtained by plugging x into theregression line. It is not generally equal to y from data.

The term y0 – ŷ0 = ε0 is called the "error" or residual (ε is the Greek letter epsilon). It is not
an error in the sense of a mistake; instead, it is the difference between the estimate and thetrue value of y.
••

If the observed data point lies above the line, the residual is positive, and the line
underestimates the actual data value for y.
If the observed data point lies below the line, the residual is negative, and the line
overestimates that actual data value for y.

In the diagram above, y0 – ŷ0 = ε0 is the residual for the point shown; since the point lies above
the line, the residual is positive.
To get the line of best fit, our first impulse might be to minimize the sum of all the residuals– that is, minimize the sum of all vertical distances y – ŷ. However, this sum would end up
awfully close to zero. The reason is that if the line really passes through the points, then thepositive errors would cancel out the negative residuals, resulting in a sum of zero. To avoid
this cancellation effect, we might instead minimize the sum of the actual vertical distancesfrom the data points to the points on the line; these distances are given by the absolute valueof the residuals.
However, to minimize the sum, we need to use Calculus; and in Calculus the absolute valuefunction is difficult to work with – so instead, we will take the square of each residual, andminimize the sum of these squares. This is where the name least-squares line comes from.

639

For each data point, we calculate the residuals or errors, yi - ŷi = εi ; then we square theseresiduals and add them up. The regression line is the line for which the sum of theresiduals,

Σ (yi - ŷi)2 = Σ εi2

is minimized. This expression is called the Sum of Squared Errors and denoted as SSE.
Using Calculus, we can determine the values of a and b that make the SSE a minimum; theresulting line, ŷ = a + bx, is called the least squares regression line. The coefficientsare:a = y − bx and b =∑ ( x − x )( y − y )∑ (x − x)2Here, x and y are the mean of the observed x- and y-values, respectively. Note that the
regression line always passes through the point ( x , y ).
••Linear regression - the line of best fit or the least-squares line is the straight linethat “best” fits the scatterplot of the data.
Residual – the difference between the observed value of y and the y value predictedby the regression equation.

The process of fitting a line to a set of points in the plane is called linear regression. The
least-squares approach described above is widely used, and so it is no surprise that thesecalculations are programmed into most statistical software packages, as well as into many
calculators. So, there is no need to memorize the formulas above; instead, we will use the
TI-83/TI-84 calculators:
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the regression line yˆ = a + bx :_1. Go to STAT and then to the EDIT menu. Enter the x-values into list L1 and the yvalues into list L2. Make sure that the order is preserved – i.e. these must remain
paired so that the corresponding (x,y) values are next to each other in the lists.
2. Go to STAT and then to the TESTS menu; scroll down select LinRegTTest.
3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1
4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER
5. Leave the line for "RegEq:" blank
6. Highlight Calculate and press ENTER.
The output screen contains a lot of information. For now, we will focus on a few items fromthe output, and will return later to the other items. In particular, at the top of the screen
you will see y = a + bx which tells us that the intercept is a and the slope is b.
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Example 13.7
To illustrate, recall the situation from Example 13.6, the “Third Exam/Final Exam”
example. Find the regression line (or line of best fit).
There we had the data:
X (third exam score) 65
67
71
71
66
75
67
70
71
69
69
Y (final exam score) 175 133 185 163 126 198 153 163 159 151 159
Solution 13.7
Following the instructions above, we have the following input and output screens:

Scroll down to find the values a = -173.513, and b = 4.8273; the equation of best fit line isŷ = –173.51 + 4.83x
We have already seen how to make a scatterplot for the data; we can graph the regression
line on the same screen. To graph the best-fit line, press the "Y=" key and type theequation –173.5 + 4.83X into equation Y1. Press ZOOM 9 again to graph it.
Another way to graph the line after you create a scatter plot is to use LinRegTTest:
1. Make sure you have already made the scatter plot. Check it on your screen.
2. Go to LinRegTTest and enter the lists.
3. At RegEq: press VARS and arrow over to Y-VARS. Press 1 for 1:Function. Press 1
for 1:Y1.
Then arrow down to Calculate and do the calculation for the line of best fit.
4. Press Y = (you will see the regression equation).
5. Press GRAPH. The line will be drawn.
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Following either of these procedures, the graph of the line of best fit for the third
exam/final-exam example is as follows:

Now that we have the regression line, we want to give a practical interpretation of the slope
and intercept.
When y = a + bx:••

a is the value of the y-intercept, that is the value of y that corresponds to x = 0.
b is the expected change in y that corresponds to one-unit increase in x.

But in this case, the value a (-173.5) does not have any practical meaning – of course, itwould not make sense for a final exam score to be negative! Moreover, 0 is very far outside
the range of observed x-values, so we would not want to use the model to make such a
prediction in any case.
For this problem, we have b = 4.83, so for each one-point increase in the third exam score,
we would expect the final exam to increase by about 4.83 points, on average.
Example 13.8
In an area of the Midwest, records were kept on the relationship between the annual rainfall(in inches) and the yield of wheat (bushels per acre).
Rain (inches)
Yield (bushels/acre)

10.5 8.8
50.5 46.2

13.4 12.5
58.8 59.0

18.8 10.3 7.0
82.4 49.2 31.9

15.6 16.0
76.0 78.8

a. Calculate the regression equation.
b. Interpret the slope of the equation in the context of the given question.

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Solution 13.8
Go to the STAT menu, and then the EDIT menu. Enter the Temp values into L1, and theGrowth values into L2. Go to the STAT menu again, and then to the TESTS menu; scroll
down to LinRegTTest; set the alternative to the “≠”, scroll to Calculate and press Enter.

a. Scroll down in the output screen to find a = 4.26681 and b = 4.37908. Thus, theregression equation is: ŷ = 4.267 + 4.379x.
b. Slope is the coefficient of x. Slope = 4.379 this mean that for each additional inch ofrain each year, we expect the yield of wheat to increase by about 4.4 bushels/acre on
average.

13.5 The data below are ages and systolic blood pressures of 9 randomly selected adults.
Age
Pressure

38
116

41
120

45
123

48
131

51 53
142 145

57
148

61 65
150 152

a. Calculate the regression equation.
b. Interpret the slope of the equation in the context of the given question.

Using the Line of Best Fit for Prediction
If the scatter plot indicates that there is a linear relationship between the variables, we
may want to use the line of best fit to make predictions. There are two different types ofpredictions.
••Interpolation – predicted values that fall within the range of data points that weretaken.
Extrapolation – predicted values for points outside the range of data that weretaken.
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Interpolation is considered more accurate than extrapolation. Interpolation is reasonable touse the line to make predictions for y given x within the domain of x-values in the sample
data, but not necessarily for x-values outside that domain.
Extrapolation is less accurate because we don’t know how the variables are related outside
this range; it may be that for other data values the points in the scatterplot no longer havea linear pattern at all, and sharply curve away from the line. So, for values outside therange of observed x-values, the equation will be unreliable at best.

The Correlation Coefficient r
A scatterplot is a valuable visual tool; however, scatterplots can also be misleading. For
example, by changing the horizontal or vertical scale, we can make the linear pattern lookstronger or weaker than it really is. Other than examining the scatter plot and seeing that a
linear model seems reasonable, how else can we tell if there is a linear relationship? We needa numerical measure of the strength of the relationship between x and y, and that numericalmeasure is the correlation coefficient.
The correlation coefficient, r, developed by Karl Pearson in the early 1900s, is numericaland provides a measure of strength and direction of the linear association between the
independent variable x and the dependent variable y.
The correlation coefficient is calculated as

n

r=
[n

∑ ( xy) − (∑ x )(∑ y )∑ x − (∑ x ) ] [ n ∑ y − (∑ y ) ]
2

2

2

2

where n = the number of data pairs.
The formula for r looks complex, but it is easily calculated using the TI-83 and TI-84
calculators. In fact, we already know how to do it. It appears at the bottom of the output
screens for LinRegTTest.

Example 13.9
For example, if we use the data from the Third Exam/Final Exam used in Example13.6
and 13.7, we had the data:
X (third exam score) 65
67
71
71
66
75
67
70
71
69
69
Y (final exam score) 175 133 185 163 126 198 153 163 159 151 159

644

Solution 13.9
We return to the output screen for the Third Exam/Final Exam example and find r at thebottom of the LinRegTTest output table. We see that r = 0.663. Thus, we would say that
there is a moderately strong, positive correlation between scores on the third exam and scoreson the final exam.

Recall that the sign of r shows the direction of the relationship between x and y and can beseen in the figures below.

a) A scatter plot showing data with a positive correlation. 0 < r < 1b) A scatter plot showing data with a negative correlation. –1 < r < 0c) A scatter plot showing data with zero correlation. r = 0

NOTE
Even if we have a very strong correlation, this does not suggest that x causes y or y causes
x. Remember, "correlation does not imply causation."

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Properties of the Correlation Coefficient, r•-1 ≤ r ≤ 1. That is, the value of r is always between -1 and +1.

•

A positive value of r means that when x increases, y tends to increase andwhen x decreases, y tends to decrease. So, there is a positive correlation.

•

A negative value of r means that when x increases, y tends to decrease
and when x decreases, y tends to increase. So, there is a negative
correlation.

•

The sign of r is the same as the sign of the slope, b, of the best-fit line.

•

The sign of the correlation r indicates the strength of the linear
relationship between x and y. Values of r close to -1 or to +1 indicate astrong linear relationship between x and y.

•

If r = 0, there is absolutely no linear relationship between x and y (nolinear correlation).

•

If r = 1, there is a perfect positive correlation. If r = -1, there is a perfectnegative correlation. In both cases, all of the original data points lie on astraight line (this is very rare in the real world)•r is the same no matter what units are used to measure x and y; we caneasily see that whatever units are used. They will all cancel out of thenumerator and denominator.

•r is the same, no matter which variable is the independent or dependent.
If we switch x and y, r will be the same.

The Coefficient of Determination
The statistic r2 is called the coefficient of determination; as the notation suggests, thisis just the square of the correlation coefficient. This statistic is usually stated as a percent,rather than in decimal form, and it measures something very specific in the in the context
of the data.
When we calculate a regression, some of the variation in the y-values is due to variation in
the x-values. This variation is called explained variation. But unless the linear
relationship between x and y is a perfect correlation (very rare!) there will also be variationin the y-values that has nothing to do with x. These variations could be simple sampling
variation, or could be from other factors that influence the variable y. This type ofvariation is called unexplained variation.
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The coefficient of variation r2 measures the percentage of variation in thedependent variable y that is explained by variation in the independent variable x.
Therefore, r2 is usually written as a percentage. Note also that this interpretation of r2aligns with the properties of the correlation coefficient r. If the regression equation is agood fit to the observed data, then we would expect that a large percentage of the variationin y would be due to variation in x. That is, when there is a strong correlation, we would
expect r2 to be close to 100%. And r2 is close to 1 exactly when r is close to either -1 or 1.
Once we have calculated r, finding r2 is very easy; but the coefficient of determination also
appears on the output for LinRegTTest. Again, referring to the Third Exam/Final Exam
example, we can read the value of r2 from the output screen: r2 = 0.66312 = 0.4397. Since.4397 ≈ 44%, this means that approximately 44% of the variation in the final exam grades
can be explained by the variation in the grades on the third exam.

Example 13.10
The paired data below consist of the temperatures on randomly chosen days and theamount a certain kind of plant grew (in millimeters):
Temp
Growth
a.
b.
c.
d.

62
36

76
39

50
50

51
13

71
33

46
33

51
17

44
6

Draw a scatterplot.
Find linear regression equation.
Find the correlation coefficient r and interpret.
Find the coefficient of determination r2 and interpret.

Solution 13.10
a. Scatterplot

647

79
16

Go to the STAT menu, and then the EDIT menu. Enter the Temp values into L1, and theGrowth values into L2. Go to the STAT menu again, and then to the TESTS menu; scroll
down to LinRegTTest; set the alternative to the “≠”, scroll to Calculate and press Enter.

b. Linear regression equation: 𝑦𝑦� = 14.57 + .211𝑥𝑥
c. The correlation coefficient is at the bottom of the output screens: r = 0.1955. Thereis a weak positive correlation between temperature and growth.
d. The coefficient of determination is the second to last value in the output screens: r2= 0.038. Only about 3.8% of the total variation in growth of the plant is explained byvariation in temperature.

13.6 The data below are ages and systolic blood pressures of 9 randomly selected adults.
Age
Pressure

38
116

41
120

45
123

48
131

51
53
142 145

57
148

61
65
150 152

a. Find the correlation coefficient r and interpret.
b. Find the coefficient of determination r2 and interpret.
Finally, we will point out that the correlation coefficient is part of an alternative formula for
calculating the slope of the regression line. Let x and y be the mean of the observed x- andy-values, respectively, and let sx and sy be the standard deviations of the observed x- andy-values. Then we can calculate the slope using the formula:
b=r

sy
sx

.

Note that this formula clearly shows one of the properties listed earlier; namely, that the signof r is the same as the sign of the slope, b, of the best-fit line.

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13.4 | Testing the Significance of the Correlation Coefficient
The correlation coefficient, r, tells us about the strength and direction of the linear
relationship between x and y. However, our interpretation of this statistic is inherently
subjective. For example, we know that when r is “close to 1”, this indicates a strong
correlation. But this begs the question, how close? That is, just how close must r be to 1 inorder for the correlation to be considered “strong”? Similarly, how close does r need to be to
0 to call the correlation “weak”? More importantly, when is the correlation strong enough towarrant using the regression equation for predictions? What we need is an objective
criterion; in short, we need a hypothesis test for whether the correlation is statisticallysignificant.
This hypothesis test is called the test of the "significance of the correlation
coefficient” and will allow us to quickly and efficiently decide whether the linear
relationship in the sample data is strong enough to use to model the relationship in the
population. We can do the five-step method like we have used previously.
The sample data are used to compute r, the correlation coefficient for the sample. If we haddata for the entire population, we could find the population correlation coefficient; thisunknown parameter is denoted by ρ, the Greek letter "rho." The sample statistic r is apoint estimate for ρ.
The hypotheses for the test are:Null and Alternative Hypotheses Testing for Linear Relationships
Significant Linear
Relationship
Two-tailed test
Positive Linear
Relationships
Right-tailed test
Negative Linear
Relationship
Left-tailed test

The null hypothesis isthat there is no linear
correlation.

H0: ρ = 0

Ha: ρ ≠ 0

The null hypothesis isthat there is negative
correlation or nocorrelation.
The null hypothesis isthat there is apositive correlation or
no correlation.

H0: ρ ≤ 0

Ha: ρ > 0

H0: ρ ≥ 0

Ha: ρ < 0

The most common test involves testing for a significant linear correlation. Then the null
hypothesis states that the population correlation coefficient ρ is 0. This means that if we wereto calculate the regression line using all data pairs in the population, then the points in the
scatterplot would be randomly scattered about a horizontal line. That is, knowing the valueof x would not be of any use in predicting y. So, the null hypothesis essentially states that
there really is no linear relationship between x and y.
649

On the other hand, the alternative hypothesis states that the population correlation
coefficient is significantly different from zero. I.e., Ha states that there is a significant
linear relationship between x and y.
Thus, if we reject H0, we have evidence that there is a significant correlation
between x and y.
We sometimes shorten this by simply saying that r is significant.
As described above, the test of significance is two-tailed (because the alternative hypothesis
is a “≠” statement). And it is a t-test with df = n – 2, where n is the number of data pairs.
The test statistic is:

t=r

n−2
1− r2

To decide the test, we can use either critical values or p-values. Once the test statistic isfound, we can get the p-values can be found with the TI-83 and TI-84 calculators using thetcdf function (in the DISTR menu). Even better, we can get both the test stat and p-value
using LinRegTTest, and the following instructions:
Using the TI-83, 83+, 84, 84+ Calculator
To calculate the regression line yˆ = a + bx :_1. Go to STAT and then to the EDIT menu. Enter the x-values into list L1 and the yvalues into list L2.
2. Go to STAT and then to the TESTS menu; scroll down select LinRegTTest.
3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1
4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER.
5. Leave the line for "RegEq:" blank.
6. Highlight Calculate and press ENTER.
The test statistic t and p-value will appear on the output screen.

Hypothesis Tests for Linear Regression using p-value
Example 13.11
Recall the Third Exam/Final Exam case. We were given data on exam scores:
X (third exam score) 65
67
71
71
66
75
67
70
71
69
69
Y (final exam score) 175 133 185 163 126 198 153 163 159 151 159
We have already calculated the regression line for the data: ŷ = –173.51 + 4.83x, as well as
the correlation coefficient r = 0.663. Use this information to determine whether thecorrelation is significant at the 5% significance level.
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Solution 13.11
This is a hypothesis test for correlation, and we will use the five-step method using p-value.
(Step 1) Set up the hypothesis Ho and Ha. The hypotheses are H0: ρ = 0 and Ha: ρ ≠ 0, andthe significance level is α = 0.05.
(Step 2) Select the correct test and identify the significance level α. This is a two-tailed testand α = 0.05.
(Step 3) Calculate the p-value. Following the instructions above, we have the following
input and output screens:

From the calculator output, we have t = 2.66 and p = 0.0262.
(Step 4) Use these results to make the decision about Ho. Since p < .05, we reject H0.
(Step 5) Interpret the decision about Ho in the context of the given problem to state aconclusion. Thus, there is a significant correlation between the Exam 3 scores and the Final
Exam scores. And because r is significant, the regression line can be used to predict
final exam scores.

Example 13.12
Consider the following cereal grams of sugar and calories per serving data that was
sampled from big box store. A concerned shopper claims that there is a positive linear
correlation between sugar per serving and calories. Perform a test to determine if there is apositive linear correlation with a 5% level of significance.
Grams of sugar per serving 8
20
12
12
11
9
28
10
5
Calories per serving
130 190 130 210 160 170 130 160 150

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Solution 13.12
This is a hypothesis test for positive linear regression using p-values.
(Step 1) The hypotheses are Ho: ρ ≤ 0 and Ha > 0.
(Step 2) This is a right-tailed test, and the level of significance is 0.05.
(Step 3) Calculate the p-value using LinRegTTest on the TI 83/84.
Put the data into the L1 and L2 using STAT 1:Edit, finishing with quit (2nd mode). Use
STAT  TESTS F:LinRegTTest. Select lists, Freq 1, β & ρ > 0, Calculate.
t = - 0.1216 and the p-value = .5467(Step 4) Since the p-value is greater than α, (0.5467 > 0.05) do not reject the null
hypothesis.
(Step 5) There is not enough evidence to support the claim that there is a positive linear
correlation between calories and grams of sugar in the cereals at the store.
In Summary•••

If r is not significant, the regression equation should not be used for prediction.
If r is significant then the regression equation can be used to predict the value of yfor values of x that are within the domain of observed x values.
If r is significant, the line may NOT be appropriate or reliable for prediction
OUTSIDE the domain of observed x values in the data.

Hypothesis testing for linear regression using critical value (s)The p-value is quite easy to obtain using technology such as the TI-84 calculator. However,the calculator requires us to have the raw data available. But there are some cases when wejust have the correlation coefficient and want to use it to determine whether the correlation
is significant. In these instances, we could calculate the test statistic t and then compare it
to an appropriate critical value; but there are also tables of critical values for r. That is, we
can compare the correlation coefficient r directly to the critical value to make the decision
about H0.
There is a table for critical values for the Pearson correlation coefficient at the end of thischapter and in the appendix. We reproduce an excerpt here for convenience:652

Because this is a two-tailed t-test, we will reject H0 at the α = .05 level if either:r < -CV or r > CV
More concisely, we reject H0 at the α = .05 level if | r | > CV.
The table again shows how the sample size is also a factor in whether or not a correlation issignificant or not.
Consider once more the third exam/final exam example. The line of best fit is ŷ = –173.51+4.83x, r = 0.6631 and there are n = 11 data points. Can the regression line be used
for prediction?
We must determine if the correlation is significant; the hypotheses are:H0: ρ = 0 vs. Ha: ρ ≠ 0.
But instead of using the p-value as before, we will instead use the critical value. We knowthat df = n – 2 = 11 – 2 = 9.
From the table, the critical values are + 0.602; since 0.6631 > 0.602, we reject H0 and again
conclude that r is significant.
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There is sufficient evidence to conclude that there is a significant linear relationship between
the third exam score and the final exam score. Because r is significant, the regression line
can be used to predict final exam scores.
Example 13.13
Suppose you computed the following correlation coefficients using samples of the given size.
For each of the following cases, use the table at the end of the chapter to locate the criticalvalue and use it to determine if r is significant and the line of best fit associated with each r
can be used to predict a y value. If it helps, draw a number line. Assume Ha: ρ ≠ 0
a. r = 0.801, calculated using n = 10 data points and α = .05.
b. r = -0.624, calculated using n = 14 data points and α = .01.
c. r = 0.776, calculated using n = 16 data points and α = .02.
Solution 13.13
a. We computed r = 0.801 using n = 10 data points. Then df = n – 2 = 10 – 2 = 8, and thecritical values associated with df = 8 and α = .05 are + 0.632.
So we would reject H0 if either r < -0.632 or r > 0.632. Since r = 0.801 and 0.801 > 0.632, weconclude that r is significant, and the regression line may be used for prediction. It is
helpful to view this on a number line:r = 0.801 > +0.632. Therefore, r is significant.

b. We computed r = - 0.624 with 14 data points. So df =12 and α = .01; the critical values
are +0.661. Since –0.624 is not in the tails seen below r is not significant and the line
cannot be used for prediction.

r = -0.624 > -0.661. Therefore, r is not significant.

c. We have r = 0.776, calculated using n = 16 data points. So df = 14 and α = .02.
The critical values are +0.574. Since 0.776 > 0.574, r is significant, and the line should be
used for prediction.

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r = 0.776 > +0.574. Therefore, r is significant.

13.7 For each of the following, use critical values to determine whether the correlation issignificant. Then state whether or not the regression line can be used for prediction.
a. r = 0.5204 using n = 9 data points, α = .1, two-tailed
b. r = –0.7204 using n = 8 data points, α = .05, left-tailedc. r = 0.708 and n = 9, α = .02, two-tailed
d. r = 0.434 and n = 14, α = .05 , right-tailed
E
Example 13.14
In an area of the Midwest, records were kept on the relationship between the rainfall (ininches) and the yield of wheat (bushels per acre). Use α = .01
Rain (inches)
Yield (bushels/acre)

10.5 8.8
50.5 46.2

13.4 12.5
58.8 59.0

18.8 10.3 7.0
82.4 49.2 31.9

15.6 16.0
76.0 78.8

a. Test the significance of the correlation between Rain and Yield using the p-value.
b. Test the significance of the correlation using the critical value.
Solution 13.14
Go to the STAT menu, and then the EDIT menu. Enter the Temp values into L1, and theGrowth values into L2. Go to the STAT menu again, and then to the TESTS menu; scroll
down to LinRegTTest; set the alternative to the “≠”, scroll to Calculate and press Enter.
a. The test statistic and p-value are: t = 13.31 and p = 3.16 x 10-6 = 0.00000316. Since
p < .01, we reject H0. There is a significant relationship between Rain and Yield.
b. The correlation coefficient is r = .9808. Using df = 9 – 2 = 7, the critical values are+0.798. Since r > 0.798, we again reject H0 and conclude that there is a significant
relationship between Rain and Yield.

655

NOTE
The table provided in this textbook shows critical only for the significance level of 5%, α =0.05. But there are other tables available that show critical values for other significancelevels as well. Of course, if we are using the p-value method, we could test the significanceof the correlation using any desired significance level; we would not be limited to using α =0.05.

Assumptions in Testing the Significance of the Correlation Coefficient
Testing the significance of the correlation coefficient requires certain assumptions about
the population data. Remember, the purpose of this test is to determine whether the linear
relationship that we see between x and y in the sample data provides strong enoughevidence to conclude that there is in fact a linear relationship between x and y in the
population.
The regression equation that we calculate from the sample data gives the best-fit line forour particular sample. We want to use this best-fit line for the sample as an estimate of thebest-fit line for the population. Examining the scatterplot and testing the significance of thecorrelation coefficient helps us determine if it is appropriate to do this.
Assumptions underlying the test of significance are:1) There is a linear relationship in the population that models the average value of yfor varying values of x. In other words, the expected value of y for each particular xvalue lies on a straight line in the population. (We do not know the equation for theline for the population. Our regression line from the sample is our best estimate ofthis line in the population.)2)

The y values for any particular, fixed x-value are normally distributed about theline. This implies that there are more y values scattered closer to the line than are
scattered farther away. Assumption (1) states that these normal distributions are
centered on the line; so, for a fixed x-value, the mean of the normal distribution ofcorresponding y-values is the y-value predicted by the regression equation.

3) The standard deviations of the normal distributions described in (2) are all equal.
That is, the population of y-values corresponding to a fixed x will have the same
standard deviation σ, regardless of the value of x.
4) The residual errors are mutually independent (no pattern).
5)

The data are produced from a well-designed, random sample or randomized
experiment.
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Note that in the literature, Assumptions (2) and (3) are usually called the Normality
Assumption and the Constant Variance assumption, respectively. Combining the first
three assumptions, we see that no matter what the value of x, the distribution ofcorresponding y-values will have the same shape and spread about the regression line.
This is illustrated in the following figure:The y values for each x value are normally distributed about the line with the same standard
deviation. For each x value, the mean of the corresponding y values lies on the regression line.

Finally, the point estimate for the common standard deviation s in Assumption (3) is called
the standard error of the estimate. Note that this is just the standard deviation of theresiduals. This statistic provides a measure of the spread of y-values about the regressionline, and can be calculated using the formula:
s=

SSE
=
n−2

∑ ( y − yˆ ) 2
n−2

We rarely need to use this formula, however. Like other key statistics related to regression,
the standard error s appears on the output screens for LinRegTTest. For example, in the
Third Exam/Final Exam example, the standard error is s = 16.42.

Test of Significance of Correlation
Hypotheses: H0: ρ = 0 vs. Ha: ρ ≠ 0
H0: ρ ≤ 0 vs. Ha: ρ > 0
H0: ρ ≥ 0 vs. Ha: ρ < 0
Test statistic: t = r

n−2
1− r2

Calculator Function: LinRegTTest

If we reject H0, then we have evidence that there is a significant correlation
between x and y.
657

13.5 | Prediction
Recall the Third Exam/Final Exam example. We examined the scatterplot and showedthat there is a significant correlation between the Exam 3 scores and the Final Exam scores(i.e. the correlation coefficient is significant). We also found the equation of the best-fit line
for the final exam grade as a function of the grade on the third exam: ŷ = –173.51 + 4.83x.
We can now use the least-squares regression line for prediction. For example, suppose we
wanted to predict the mean final exam score of statistics students who received 73 on thethird exam. Note that the sample exam scores (x-values) range from 65 to 75. Since x = 73 isin the range of observed x-values 65 and 75, and the correlation is significant, we can use
the regression equation. So, we substitute x = 73 into the equation to get:
ŷ = –173.51 + 4.83(73) = 179.08
Of course, this does not mean that any student who scores 73 on the third exam isguaranteed to score 179.08 on final exam. There will be other factors that influence a
student’s score on the final exam (e.g. the number of hours spent studying for the exam).
What we know is that in the population, among all students who scored 73 on the thirdexam, the average score on the final exam for is the y-value on the regression line. So when
we say that 179.08 is the predicted score, this means:
We predict that statistics students who earn a grade of 73 on the third exam will earn a
grade of 179.08 on the final exam, on average.

Example 13.15
Recall the Third Exam/Final Exam example from Example 13.6 We have shown thatthe correlation is significant, and the regression equation is ŷ = –173.51 + 4.83x.
a. What would you predict the final exam score to be for a student who scored a 66 on
the third exam?
b. What would you predict the final exam score to be for a student who scored a 90 on
the third exam?
Solution 13.15
a. ŷ = –173.51 + 4.83(66) = 145.27.
b. The observed x-values in the sample data are between 65 and 75. Ninety is welloutside of the domain of these observed x-values, so we cannot reliably predict the
final exam score for this student.

658

Of course, we can easily enter 90 into the equation for x and calculate the corresponding yvalue, but the predicted value will not be reliable. To illustrate how unreliable the prediction
can be for x-values outside the range of observed data, make the substitution x = 90 into theequation:y = –173.51 + 4.84(90) = 261.19
The final-exam score is predicted to be 261.19, which is not even possible - the largest the
final-exam score can be is 200.
Recall that the process of predicting inside of the observed x values observed in the data iscalled interpolation. The process of predicting outside of the observed x values observed in
the data is called extrapolation. In general, extrapolation should be done with extreme
caution, if at all.

13.8 Data are collected on the relationship between the number of hours per weekpracticing a musical instrument and scores on a math test. The line of best fit is as follows:ŷ = 72.5 + 2.8x
What would you predict the score on a math test would be for a student who practices amusical instrument for five hours a week?

Prediction and r•

If r is not significant, the regression equation should not be used for prediction.
Use the mean of the observed y values for prediction.

•

If r is significant then the regression equation can be used to predict the value of yfor values of x that are within the domain of observed x values.

•

If r is significant, the line may NOT be appropriate or reliable for prediction
OUTSIDE the domain of observed x values in the data.

659

13.6 | Outliers
In some data sets, there are values (observed data points) called outliers. Outliers areobserved data points that are far from the least squares line. They have large "errors", wherethe "error" or residual is the vertical distance from the line to the point.
Outliers need to be examined closely. Sometimes, for some reason or another, they should not
be included in the analysis of the data. It is possible that an outlier is a result of erroneous
data. Other times, an outlier may hold valuable information about the population understudy and should remain included in the data. The key is to examine carefully what causes
a data point to be an outlier.
Besides outliers, a sample may contain one or a few points that are called influential points.
Influential points are observed data points that are far from the other observed data points
in the horizontal direction. These points may have a big effect on the slope of the regression
line. To begin to identify an influential point, you can remove it from the data set and see if
the slope of the regression line is changed significantly.
Computers and many calculators can be used to identify outliers from the data. Computer
output for regression analysis will often identify both outliers and influential points so thatyou can examine them.

Identifying Outliers
We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However,
we would like some guideline as to how far away a point needs to be in order to be considered
an outlier.
As a rough rule of thumb, we can flag any point that is located further than two
standard errors above or below the best-fit line as an outlier.
We can do this visually in the scatter plot by drawing an extra pair of lines that are twostandard deviations above and below the best-fit line. Any data points that are outside thisextra pair of lines are flagged as potential outliers. Or we can do this numerically by
calculating each residual and comparing it to twice the standard deviation. On the TI-83,
83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followedby the numerical calculations. You would generally need to use only one of these methods.
Example 13.16
In the Third Exam/Final Exam example, use the graphical method to determine if there
is an outlier or not. If there is an outlier, we will delete it and fit the remaining data to a newline; the new line ought to fit the remaining data better. This means the SSE and standard
error should both be smaller, and the correlation coefficient ought to be closer to +1.
660

Solution 13.16
There are multiple ways to determine if a value is an outlier.
Visually with Graph
With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphicallyand visually. If we were to measure the vertical distance from any data point to thecorresponding point on the line of best fit and that distance were equal to 2s or more, thenwe would consider the data point to be "too far" from the line of best fit. We need to find andgraph the lines that are two standard deviations below and above the regression line. Any
points that are outside these two lines are considered outliers. These lines are Y2 and Y3:
Recall that the regression equation for this example is ŷ = –173.51 + 4.83x, and the standard
error is s = 16.4. (This statistic appears on the output screen for LinRegTTest)
We want to draw two lines that are parallel to the regression line, and such that the verticaldistance to the regression line is 2s at each point; to do this, we add and subtract 2s = 2(16.4)
to the equation for the regression line:
andY2 = –173.5 + 4.83x –2(16.4)
Y3 = –173.5 + 4.83x + 2(16.4)

Graph the scatterplot with the best fit line in equation Y1, as before.
Then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9.

We see that the only data point that is not between lines Y2 and Y3 is the point x = 65, y =175. The outlier is the student who had a grade of 65 on the third exam and 175 on the final
exam; this point is further than two standard deviations away from the best-fit line.
Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult
to tell if the point is between or outside the lines; for this example, the point is just barely
outside the parallel lines. On a computer, enlarging the graph may help; on the calculator
screen it can help to zoom in.
661

Numerical with Line of Best Fit
If we are in doubt, we just need to numerically verify that the point is an outlier.
1. Calculate the predicted final exam score for a student who scores 65 on the thirdexam:ŷ = –173.51 + 4.83(65) = 140.44.
2. Calculate the difference between the observed and predicted values (i.e. theresidual):y – ŷ = 175 – 140.44 = 34.56.
3. Determine if the residual is more than two standard errors. Since s = 16.4. Then 2s= 32.8. Since 34.56 > 32.8, the observation is an outlier.
Using the Residual Table
We can also use a numerical method to identify the outlier.
In the table below, the first two columns are the third-exam and final-exam data. The thirdcolumn shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x.
The residuals, or errors, have been calculated in the fourth column of the table:
Residual = observed y value − predicted y value = y − ŷ.
Recall that the standard error s is the standard deviation of the residuals, which is s = 16.4.
x
65
67
71
71
66
75
67
70
71
69
69

y
175
133
185
163
126
198
153
163
159
151
159

ŷ
140
150
169
169
145
189
150
164
164
160
160

y−ŷ
35
-17
16
-6
-19
9
3
-1
-10
-9
-1

We are looking for all data points for which the residual is greater than 2s = 2(16.4) = 32.8 orless than -32.8. Compare these values to the residuals in column four of the table. The only
such data point is the student who had a grade of 65 on the third exam and 175 on the final
exam; the residual for this student is 35.
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NOTE: This method involves a lot of calculation, but it is easily automated. In fact, moststatistical packages (such as Minitab or SPSS) will automatically calculate the residuals andflag data points that appear to be outliers.
Now that we have identified the point (65, 175) as an outlier, we would re-examine the data
for this point to see if there are any problems with the data. If there is an error, we should
fix the error if possible, or if not delete the data. If the data is correct, we would leave it in
the data set. For this example, let us suppose that after reviewing the data, we found that
this data pair data really was an error and should be removed.
We remove the data point and compute a new best-fit line and correlation coefficient using
the ten remaining points; using the calculator as before, we obtain the new regression
equation:ŷ = –355.19 + 7.39xMoreover, the new model has correlation coefficient r = 0.9121. Recall that for the original
data, we had r = 0.6631, so the new line exhibits a stronger correlation. This means thatthe new line is a better fit to the ten remaining data values. Moreover, the new standard
error is s = 9.275, a significant reduction. Both of these tell us that the new regression line
can better predict the final exam score given the third exam score.

NOTE
When outliers are deleted, the researcher should either record that data was deleted, andwhy, or should provide results both with and without the deleted data. If data is erroneousand the correct values are known (e.g., student one actually scored a 70 instead of a 65), thenthis correction can be made to the data.

13.9 Identify the potential outlier in the scatter plot. The standard deviation of theresiduals is approximately 8.6.

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Example 13.17
The Consumer Price Index (CPI) measures the average change over time in the prices paidby urban consumers for consumer goods and services. The CPI affects nearly all Americans
because of the many ways it is used. One of its biggest uses is as a measure of inflation. By
providing information about price changes in the Nation's economy to government,
business, and labor, the CPI helps them to make economic decisions. The President,
Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary andfiscal policies. In the following table, x is the year and y is the CPI.
x
y
x
y
1915
10.1 1969
36.7
1926
17.7 1975
49.3
1935
13.7 1979
72.6
1940
14.7
1980
82.4
1947
24.1 1986
109.6
1952
26.5 1991
130.7
1964
31.0 1999
166.6
a.
b.
c.
d.
e.
f.

Draw a scatterplot of the data.
Calculate the least squares line; write the equation in the form ŷ = a + bx.
Graph the line on the same axes as the scatterplot.
Find the correlation coefficient. Is it significant?
What was the average CPI for the year 1990?
List the residuals and find any outliers.

Solution 13.17
a. Parts a and c:

Go to STAT, and then the EDIT menu. Enter the x-values into L1 and the y-values
into L2.
Go to STAT, and then the TESTS menu; scroll down to LinRegTTest. Specify thelists, select the “≠” alternative, place the cursor on Calculate and press Enter to get
the output screens.

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NOTE: To enter Y1 in RegEQ, press VARS, choose Y-Vars, choose Function Enter, Y1 enterb. The least-squares regression line is: ŷ = –3204 + 1.662x.
c. See the graph in a.
d. The correlation coefficient is r = 0.8694. To determine if it significant we run ahypothesis test.
(Step 1) The hypotheses are H0 = 0 and Ha ≠ 0.
(Step 2) Two-test test and the level of significance is 0.05.
(Step 3) We can get the test statistic and p-value from the output screen: t = 6.09
and p = 0.000054.
(Step 4 and 5) Since p < .05, the correlation is significant. (using p-value)
Alternatively, there are n = 14 data points, so df = 12; the corresponding criticalvalue is 0.532.
Since 0.8694 > 0.532, r is significant. (using critical values)e. ŷ = –3204 + 1.662(1990) = 103.4 CPI
f. To list residuals using your calculator, you must first place Y1 in RegEQ as seen in
the picture above AND make sure you Calculate. Then go back into your List,
highlight L3, Press Enter, Press Vars, Function, Y-vars, Y1. Then Press ( 2nd L1 ),Press Enter. L3 will be filled with the predicted values. Highlight L4, Press Enter,
L2 – L3, Press Enter. L4 will be filled with the residuals.

665

NOTE
In the example, notice the pattern of the points compared to the line. Although thecorrelation coefficient is significant, the pattern in the scatterplot indicates that a curve
might be a more appropriate model to use than a line. If we were to try and fit a non-linear
function to the data, we might obtain a more accurate model.

13.10 The following table shows economic development measured in per capita income
PCINC.
Year
1870
1880
1890
1900
1910
a.
b.
c.
d.
e.
f.

PCINC
340
499
592
757
927

Year
1920
1930
1940
1950
1960

PCINC
1050
1170
1364
1836
2132

What are the independent and dependent variables?
Draw a scatter plot.
Use regression to find the line of best fit and the correlation coefficient.
Is there a significant linear relationship between the variables? Explain.
Interpret the slope of the regression equation.
Use the line of best fit to estimate PCINC for the years 1925 and for 2000.

13.7 Summary of Concepts using TI-83/84 and Excel
A lot of concepts have been introduced in this chapter. The following is a comprehensive
example bringing all the concepts together.
Example 13.18
The following data represents the thread count of various bed sheets and theircorresponding price. Use the following data to solve the following use TI-83/84 and Excel
when possible.
Thread Count300 800 1600 600 500 400 1000 700 1500 1800
Price in Dollars 53
54
77
67
55
40
97
70
80
96

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a. Draw a scatterplot using TI-83/84 and Excel.
b. Determine the line of best fit. Interpret the slope, the y-intercept and the correlation
coefficient, r.
c. Calculate the residuals.
d. Determine if there are any outliers.
e. Calculate the coefficient of determination, r2 .
f. Test the claim that there a linear correlation between thread counts and cost in
dollars at a 5% level of significance using p-value.
Solution 13.18
a. Using TI-83/84 put Thread Count into a list and Price in dollars into another list. To
edit lists go to STAT, 1:Edit.
Go to Stat Plot (2nd y =). Pick a plot. Select ON, Type: Scatter plot, Xlist: ThreadCount, Ylist: Cost in Dollars. You must change the window to fit the data. Window: x
max = 2000, y max = 150. Then press graph.

Using Excel place the data into cells and select the scatter graph.

667

NOTE
Lists can be cleared by going to the top and pushing clear. This clears the whole list. To
clear one item in the list, use the del button. Do not use del at the top of the list as it will
delete the whole list.
b. Using the TI-83/84 and the LinRegTTest on the calculator.
X list: Thread Count list
Y list: Cost in dollars list
Freq: 1
β & α: ≠ 0
RegEQ:
Calculate

LinRegTTesty = a + bx
β ≠ 0 and ρ ≠ 0
t = 3.509240232
p = .0079704493
df = 8a = 43.57142857
b = .0275310559
s = 12.59165158
r2 = .606197797
r = .7785870517

From this information the line of best fit is ŷ = 43.57 + 0.0275x
Using Excel and the LINEST function: =LINEST(y values, x values, True – to solve for b,
True – for additional regression statistics)

See the first row in the right table is the slope (m = 0.027531 and y-intercept = 43.57143)

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The slope is 0.027531 which meant that the as the thread count goes up by 1, the cost goesup by $0.0275310.
The y – intercept is 43.57143. This means that when x = 0 (That is a thread count of zero),the cost is $43.57. Notice that this extrapolation is inaccurate.
The correlation coefficient, r, is 0.778587 which is positive indicating a positive slope andpositive correlation.
c. The residuals, εo, can be calculated with both the TI-83/84 and Excel as long as you
use the calculations are done correctly in each column. Both tables would have thesevalues. Excel was used to make the table below.
Column 1: x values (from problem)
Column 2: y values (from problem)
Column 3: ŷ values (predicted values using y = mx + b, using column 1 x values)
Column 4: y - ŷ (the residuals, εo)d. Determine if there are any outliers. Looking at the LinRegTTest results chart inpart b. note that s = 12.59. Outliers are values that are greater or less than 2s =2(12.59) = 25.18
Look at the residual value when the thread count is 1000 and see 28.89757 isgreater than 25.18 so this point (1000, 97) is an outlier.

669

e. Calculate the coefficient of coefficient of determination, r2.
Looking at b for both the TI-83/84 and Excel, r2 = 0.606198. This means that 60.6%of the variation in the cost in dollars can be explained in the variation in the threadcount.
f.

Test the claim that there is a correlation between thread count in sheets and cost in
dollars using p-value.
(Step 1) Ho: ρ = 0 and Ha: ρ ≠ 0, The claim is the null hypothesis.
(Step 2) It is two-tailed test, and the level of significance is 0.05.
(Step 3) See b for the test statistic, t = 3.5092, and the p-value, p = 0.00797(Step 4) p-value < α: 0.00797 < 0.05. Do not reject the null hypothesis.
(Step 5) At a 5% level of significance there is enough evidence to support the claimthat there is a linear correlation between thread count and cost in dollars.

KEY TERMS and FORMULA REVIEWCorrelation Coefficient: A measure developed by Karl Pearson (early 1900s) thatmeasures the strength and direction of linear relationship between two variables x and y.
It is given by the formula:
n

r=
[n

∑ ( xy) − (∑ x )(∑ y )∑ x − (∑ x ) ] [ n ∑ y − (∑ y ) ]
2

2

2

2

Coefficient of determination: The square of the correlation coefficient, which is r2. The
coefficient of determination measures the percentage of variation in the dependent variable
that is explained by variation in the dependent variable.
Constant Variance Assumption: This assumption states that for any fixed x-value, thepopulation of corresponding y-values is normally has the same variance.
Least Squares Regression Line: This is the equation for the line of best fit to a set of ndata points (x, y). The equation is ŷ = a + bx, where:a = y − bx and b =670∑ ( x − x )( y − y )∑ (x − x)2Here, x and y are the mean of the observed x- and y-values, respectively; the regression
line always passes through the point ( x , y ).
Normality Assumption: This assumption states that for any fixed x-value, the populationof corresponding y-values is normally distributed.
Outlier: An observation that does not fit the rest of the data. For regression, this is anobserved data point that is located at a vertical distance more than 2s from the regressionline, where s is the standard error of the estimate.
Residual. The difference y – ŷ between an observed y-value and the y-value predicted by
the regression line.
Scatterplot: A graph of the ordered pairs (x, y) of observed data, where x is theindependent variable and y is the dependent variable.
Standard error of the estimate: This is the standard deviation of the residuals; it is alsoa point estimate for the common standard deviation from the Constant Variance assumption.
This statistic provides a measure of the spread of y-values about the regression line, and is
calculated using:
s=

SSE
=
n−2

∑ ( y − yˆ ) 2
n−2

CHAPTER REVIEW
13.2 Scatter Plots
If we have measurements on two quantitative variables, x and y for a sample of individuals
in a population, then we can use a scatterplot to make an initial determination as to whether
there is a linear relationship between the variables. In particular, a scatterplot can tell us
the direction of the relationship between the x variables and the y variables and provide arough idea of the strength of the relationship.
13.3 The Regression Equation
The procedure of fitting a linear equation to a collection of observed data points is called
linear regression.
The regression line is the “line of best fit” to the data. To understand what this means, we
calculate the residuals, the differences y – ŷ between the observed y-values and the y-values
predicted by line. The residuals measure the vertical distances between the observed data
points and the corresponding points on the regression line.
The sum of the squares of the residuals, Σ(y – ŷ)2, is called the SSE, or Sum of Squared
Errors. The regression line is found by minimizing the SSE and so is also called the least671

squares line. This line, usually denoted as ŷ = a + bx, is easily calculated on the TI-84 using
LinRegTTest.
13.4 Testing the Significance of the Correlation Coefficient
The correlation coefficient r is a numerical measure of the strength and direction of thelinear association between x and y:•

The correlation coefficient r is always between –1 and +1.

•

When r is positive, the slope of the regression line is positive.

•

When r is negative, the slope of the regression line is also negative.

•

Values of r that are close to 0 indicate a weak correlation.

•

Values of r that are close to + 1 indicate a strong correlation.

The value of r also shows on the output screens for LinRegTTest.

These properties are intuitive and easy to understand but are somewhat subjective. So, we
also introduced a test to determine whether or not a correlation between two variables issignificant. This test involves the correlation parameter ρ, and the hypotheses are:H0: ρ = 0 vs. Ha: ρ ≠ 0
H0: ρ ≥ 0 vs. Ha: ρ < 0
H0: ρ ≤ 0 vs. Ha: ρ > 0n−2. We can get both the test statistic
1− r2
and p-value from the calculator using LinRegTTest. We also can make the decision about
Ho by comparing r directly to an appropriate critical value. More importantly
This is a two-tailed t-test with test statistic t = r

Rejecting H0 at the .05 significance level provides evidence that the correlation issignificant.
There are several important assumptions that we make when calculating a regression
equation, and judging how well it fits the data:
1. There is a linear relationship in the population that models the average value of y forvarying values of x. In other words, the expected value of y for each particular x-value
lies on a straight line in the population.

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2. The y-values corresponding to a fixed x-value are normally distributed about the line.
Assumption (1) states that these normal distributions are centered on the regressionline, so for a fixed x-value, the mean of the normal distribution of corresponding yvalues is the y-value predicted by the regression equation.
3. The standard deviations of the normal distributions described in (2) are all equal.
That is, the population of y-values corresponding to a fixed x will have the same
standard deviation σ, regardless of the value of x. The point estimate for this common
standard deviation is called the standard error of the estimate; it is the standard
deviation of the residuals, and so provides a measure of the spread of y-values about
the regression line. The standard error of the estimate also appears on the output
screens for LinRegTTest.
4. The residual errors are mutually independent (no pattern).
5. The data are obtained from a well-designed, random sample or randomized
experiment.

13.5 Prediction
Once we determine that there is a significant correlation between x and y, we can use theleast squares regression line to make predictions; that is, given a value of the independent
variable x, we can use the equation ŷ = a + bx to find the corresponding y-value. Some things
to remember:•••

If r is not significant, the regression equation should not be used for prediction.
If r is significant then the regression equation can be used to predict the value of yfor values of x that are within the domain of observed x values.
If r is significant, the line may NOT be appropriate or reliable for prediction
OUTSIDE the domain of observed x values in the data.

13.6 Outliers
An outlier is a point that lies well outside the pattern of the observed data; specifically, it isan observed data point that is located at a vertical distance more than 2s from theregression line, where s is the standard error of the estimate. We showed both a graphicalmethod and a numerical method for identifying outliers; see Example 13.16.

673

Critical Values of r

674

675

Exercises for Chapter 13
1. A specialty cleaning company charges an equipment fee and an hourly labor fee. The
total amount of the fee the company charges for each session is given by the equation y = 50
+ 100x.
a. What are the independent and dependent variables?
b. What is the y-intercept? Interpret in the context of the problem.
c. What is the slope? Interpret in the context of the problem.
2. The price of a single issue of stock can fluctuate throughout the day. The price of stock
for Shipment Express is y = 15 – 1.5x where x is the number of hours passed in an eighthour day of trading.
a. What are the slope and y-intercept? Interpret their meaning.
b. If you owned this stock, would you want a positive or negative slope? Why?
3. For each of the scatterplots, answer the following:i. Does there appear to be a linear relationship between the variables?
ii. Is the linear relationship strong or weak?
iii. Is the relationdship positive or negative?
a.

b.

c.

d.

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4. What does an r value of zero mean?
5. A random sample of ten professional athletes produced the following data where x is thenumber of endorsements the player has and y is the amount of money made (in millions ofdollars).
x
Y
a.
b.
c.
d.
e.

0
2

3
8

2
7

1
3

5
13

5
12

4
9

3
9

0
3

4
10

Draw a scatter plot of the data.
Use regression to find the equation for the line of best fit.
Draw the line of best fit on the scatter plot.
What is the slope of the line of best fit? Interpret.
What is the y-intercept of the line of best fit? What does it represent?

6. When testing the significance of the correlation coefficient,a. What is the null hypothesis?
b. What is the alternative hypothesis?
7. When n = 100 and r = -0.89, is there a significant correlation? Explain.
8. A landscaping company is hired to mow the grass for several large properties. The total
area of the properties combined is 1,345 acres. The rate at which one person can mow is asfollows: ŷ = 1350 – 1.2x where x is the number of hours and ŷ represents the number ofacres left to mow.
a. How many acres will be left to mow after 20 hours of work.
b. How many acres will be left to mow after 100 hours of work?
c. How many hours will it take to mow all of the lawns? (When is ŷ = 0?)
9. An electronics retailer used regression to find a simple model to predict sales growth in
the first quarter of the new year (January through March). The model is good for 90 days,
where x is the day. The model can be written as follows ŷ = 101.32 + 2.48x where ŷ is in
thousands of dollars.
a. What would you predict the sales to be on day 60?
b. What would you predict the sales to be on day 90?

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10. The following table shows cell phone usage in the U.S. for the years 2003-2009:
Year Cellular Usage (%)
2003
54.67
2005
74.19
2007
84.86
2009
90.82
a.
b.
c.
d.

Construct a scatterplot.
Find the line of best-fit.
Find the correlation coefficient.
Would the regression line provide a reliable estimate for cell phone usage in 2016?
Explain.

11. Find the regression equation and correlation coefficient for the following data:
x
y

16.3
7

39.3
10

42.6
6

10.8
7

30.6
4

12. Find the regression equation and correlation coefficient for the following data:
x
y

57
156

53
59
164 163

61
177

53
159

56
175

60
151

13. The following data shows the number of years spent studying a foreign language andthe score on a test of fluency.
Number of years (x)
3
4
4
2
5
3
4
5
3
2

Score (y)
57
78
72
58
89
63
73
84
75
48

Find the regression equation and correlation coefficient for the data.

678

14. In an area of the Midwest, records were kept on the relationship between the rainfall(in inches) and the yield of wheat (bushels per acre). The data is shown below:
Rain (inches)
Yield (bushels/acre)

10.5 8.8
50.5 46.2

13.4 12.5
58.8 59.0

18.8 10.3 7.0
82.4 49.2 31.9

15.6 16.0
76.0 78.8

Use this data to determine whether there is a significant correlation between the amount of
rainfall and yield of wheat.
15. The data below are the final exam scores of 6 randomly selected statistics students andthe number of hours they studied for the exam.
Hours
Score

5
64

10 4
86 69

6
86

10
59

9
87

Use this data to determine whether there is a significant correlation between the number ofhours studied and the exam score.
The following data shows data for the first two decades of AIDS reporting. Use this data toanswer questions #16-23.
Year
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002

#AIDS Cases Diagnosed
319
1170
3076
6240
11,776
19,032
28,56435,44742,674
48,634
59,66078,53078,83471,87468,50559,347
47,14938,39325,17425,52225,643
26,464679

16. Create a scatterplot for the data, using YEAR as the independent variable.
17. Find the regression equation. Round the coefficients to whole numbers.
18. Identify each of the following from the regression output:
r = _______

r2 = _______

s = _______

19. Is there a significant correlation between the year and the number of AIDS cases
diagnosed? Explain.
20. Interpret the slope of the regression line.
21. Calculate the following point predictions:a. When x = 1985, ŷ =b. When x = 1990, ŷ =22. Would it be appropriate to use the regression line to estimate the number of diagnosed
AIDS cases for the current year?
23. Does the line seem to fit the data? Why or why not? Does your answer affect how you
would use the regression equation for estimation?
24. Solve the following.
a. The SSE (Sum of Squared Errors) for a data set of 18 numbers is 49. What is thestandard error of the estimate?
b. The standard error for the estimate a data set is 9.8. What is the cutoff for thevertical distance that a point can be from the line of best fit to be considered an
outlier?
25. The following scatter plot shows the relationship between hours spent studying andexam scores. The line shown is the calculated line of best fit. The correlation coefficient is
0.69.

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a. Do there appear to be any outliers? If so, describe the point.
b. A point is removed, and the line of best fit is recalculated. The new correlation
coefficient is r = 0.98. Does the point appear to have been an outlier? Why or why
not?
c. What effect did the potential outlier have on the line of best fit?
d. Are you more or less confident in the predictive ability of the new line of best fit?
26. The Gross Domestic Product Purchasing Power Parity is an indication of a country’s
currency value compared to another country. The following table shows the GDP PPP of
Cuba as compared to US dollars.
Year
1999
2000
2001
2002
2003
2004
2005
2006
2008
2009
2010

Cuba’s PPP
1700
1700
2300
2900
3000
3500
4000
11,000
9500
9700
9900

a. Construct a scatter plot of the data.
b. Find the least-squares regression line.
c. Find the correlation coefficient.
27. Does the higher cost of tuition translate into higher-paying jobs? The table lists the top
ten colleges based on mid-career salary and the associated yearly tuition costs.
SchoolPrinceton

Mid-Career Salary (in
h

137

d )

YearlyT28,540i i

Harvey Mudd

135

40,133

CalTech127

39,900US Naval

1220West Point

120

0

MIT

118

42,050

Lehigh University

118

43,220

NYU-Poly

117

39,565Babson College

117

40,400Stanford

114

54,506681

a. Construct a scatter plot of the data, using tuition as the independent variable.
b. Find the correlation coefficient r. Is there a significant correlation?
c. From the scatterplot, does there appear to be any outliers?
d. After removing the outlier(s), is there a significant correlation?
28. a. Explain what it means when a correlation has an r2 of 0.72.
b. Can a coefficient of determination be negative? Why or why not?
29. Recently, the annual number of driver deaths per 100,000 for the selected age groups
was as follows:
Age

Midpt . Number of Driver Deaths (per 100,000)

16-19

17.5

38

20-24

22

36

25-34

29.5

24

35-54

44.5

20

55-74 . 64.5

18

75+

28

80

For each age group, pick the midpoint of the interval for the x value; for the 75+ group, use
80.
a. Using “ages” as the independent variable, calculate the least squares (best–fit) line.
b. Find the correlation coefficient. Is it significant?
c. Based on this data, is there a linear relationship between age of a driver and driver
fatality rate?
30. The table below shows the life expectancy for an individual born in the United States incertain years.
Year of
Bi1930hLifeE 59.7

1940

62.9

1950

70.2

1965

69.7

1973

71.4

1982

74.5

1987

75

1992

75.7

2010

78.7

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a. Which variable should be the independent and which should be the dependent
variable?
b. Draw a scatter plot of the ordered pairs.
c. Calculate the least squares line; put the equation in the form ŷ = a + bx.
d. Find the correlation coefficient. Is the correlation significant?
e. Find the estimated life expectancy for an individual born in 1950 and for one born in
1982.
f. Why aren’t the answers to part e the same as the values in the table?
g. Should we use the least squares line to find the estimated life expectancy for anindividual born in 1850? Why or why not?
h. Interpret the slope of the regression line.
31. The maximum discount value of the Entertainment® card for the “Fine Dining” section,
Edition ten, for various pages is given below:
Page
4 b

a.
b.
c.
d.
e.

Maximum value
($)
16

14

19

25

15

32

17

43

19

57

15

72

16

85

15

90

17

Decide which variable should be the dependent variable.
Draw a scatter plot of the ordered pairs.
Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Should the regression equation be used to find the maximum value for therestaurants on p.70?

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32. The table below gives the gold medal times for every other Summer Olympics for the
women’s 100-meter freestyle (swimming).
Year Time (seconds)
1912

82.2

1924

72.4

1932

66.8

1952

66.8

1960

61.2

1968

60.0

1976 55.65
1984 55.92
1992 54.64
2000 53.8
2008 53.1
a. Decide which variable should be the dependent variable.
b. Draw a scatter plot of the data.
c. Does it appear from inspection that there is a relationship between the variables?
Why or why not?
d. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx.
e. Find the correlation coefficient. Is the decrease in times significant?
f. Find the estimated gold medal time for 1932. Find the estimated time for 1984.
g. Why are the answers from part f different from the chart values?
h. Does it appear that a line is the best way to fit the data? Why or why not?
i. Use the least-squares line to estimate the gold medal time for the next Summer
Olympics. Is this estimate reliable? Why or why not?
33. Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies tostudy their population. They gather data for the percent of new sparrow hawks in eachcolony and the percent of those that have returned from migration.
Percent return: 74 66 81 52 73 62 52 45 62 46 60 46 38
Percent new:5
6
8
11 12 15 16 17 18 18 19 20
20
a. Enter the data into your calculator and make a scatter plot.
b. Use your calculator’s regression function to find the equation of the least-squares
regression line. Add this to your scatter plot from part a.
c. Interpret the slope and intercept of the regression line in the context of the problem.
d. How well does the regression line fit the data? Explain your answer.
e. Which point has the largest residual? Is this point an outlier? An influential point?
Explain.

684

f.

An ecologist wants to predict how many birds will join another colony of sparrowhawks to which 70% of the adults from the previous year have returned. What is theprediction?

34. The height (sidewalk to roof) of notable tall buildings in America is compared to thenumber of stories of the building (beginning at street level). The data is shown below:
Height (in feet) Stories
1,050

57

428

28

362

26

529

40

790

60

401

22

380

38

1,454

110

1,127

100

700

46

a. Using “stories” as the independent variable and “height” as the dependent variable,
make a scatter plot of the data.
b. Does it appear from inspection that there is a relationship between the variables?
c. Calculate the least squares line. Put the equation in the form: ŷ = a + bx
d. Find the correlation coefficient. Is it significant?
e. Find the estimated heights for a 32-story building.
f. Find the height for a 94-story building.
g. Based on this data, is there a linear relationship between the number of stories intall buildings and the height of the buildings?
h. Suppose we wanted to estimate the height of a building with six stories? Would theleast squares line give an accurate estimate of height? Explain why or why not.
i. Based on the least squares line, adding an extra story is predicted to add about how
many feet to a building

685

35. The following table consists of one student athlete’s time (in minutes) to swim 2000
yards and the student’s heart rate (beats per minute) after swimming on a random sample
of 10 days:
Swim Time Heart
R144
34.12
35.72

152

34.72

124

34.05

140

34.13

152

35.73

146

36.17

128

35.57

136

35.37

144

35.57

148

a. Enter the data into your calculator and make a scatter plot.
b. Find the equation of the least-squares regression line. Add this to your scatter plot
from part a.
c. Interpret the slope and y-intercept of the regression line.
d. How well does the regression line fit the data? Explain your response.
e. Which point has the largest residual? Explain what the residual means in context. Is
this point an outlier? An influential point? Explain.
36. The following table shows data on average per capita wine consumption (in liters) anddeaths from heart disease (per 100,000 residents) in a random sample of 10 countries:
Consumption
Deaths
a.
b.
c.
d.
e.

2.5
221

3.9
167

2.9
131

2.4
191

2.9
220

0.8
297

9.1
71

2.7
172

0.8
211

0.7
300

Enter the data into your calculator and make a scatter plot.
Find the regression equation and add to your scatter plot from part a.
Explain in words what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Do the data provide convincing evidence that there is a linear relationship between
the amount of wine consumed and the heart disease death rate? Carry out anappropriate test at the .05 significance level to justify your answer.

686

37. The percentage of female wage and salary workers who are paid hourly rates for theyears 1979 to 1992 are given in the following table:Year% of workers paid hourly rates

1979

61.2

1980

60.7

1981

61.3

1982

61.3

1983

61.8

1984

61.7

1985

61.8

1986

62.0

1987

62.7

1990

62.8

1992

62.9

a. Using “year” as the independent variable and “percent” as the dependent variable,
draw a scatter plot of the data.
b. Does it appear from inspection that there is a relationship between the variables?
Explain.
c. Calculate the least-squares line. Put the equation in the form: ŷ = a + bx
d. Find the correlation coefficient. Is it significant?
e. Find the estimated percentages for 1991 and 1988.
f. Based on the data, is there a linear relationship between the year and the percent offemale wage and salary earners who are paid hourly rates?
g. Are there any outliers in the data?
h. What is the estimated percent for the year 2050? Does the least-squares line give an
accurate estimate for that year? Explain why or why not.
i. What is the slope of the least-squares (best-fit) line? Interpret the slope.
38. The average number of people in a family that received welfare for various years isgiven in the following table:
Year Welfare family
1969 i 4.0
1973

3.6

1975

3.2

1979

3.0

1983

3.0

1988

3.0

1991

2.9
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a.
b.
c.
d.

Using “year” as the independent variable, draw a scatter plot of the data.
Calculate the least-squares line. Put the equation in the form: ŷ = a + bx
Find the correlation coefficient. Is the correlation significant?
Based on this data, is there a linear relationship between the year and the average
number of people in a welfare family?
e. Using the least-squares line, estimate the welfare family size for 1960 and 1995.
Does the least-squares line give an accurate estimate for those years? Explain why
or why not.

39. The following are advertised sale prices of color televisions at a major appliance store.
Size (inches)
9
20
27
31
35
40
60

Sale Price ($)
147
197
297
447
1177
2177
2497

Suppose that we wish to use the size of the television to predict the selling price.
a. Calculate the least-squares line. Put the equation in the form of: ŷ = a + b
b. Find the correlation coefficient. Is there a significant correlation between size andprice?
c. Find the estimated sale price for a 32-inch television. Find the cost for a 50-inch
television.
d. Interpret the slope of the regression line.
40. According to a flyer by a Prudential Insurance Company representative, the costs ofapproximate probate fees and taxes for selected net taxable estates are as follows:
Net Taxable Estate ($)
600,000
750,000
1,000,000
1,500,000
2,000,000
2,500,000
3,000,000

Approx. Probate Fees and Taxes ($)
30,00092,500203,000
438,000
688,000
1,037,000
1,350,000

Suppose that we wish to investigate the relationship between these two variables.

688

a.
b.
c.
d.

Which would be the independent variable? Which would be the dependent variable?
Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Find the estimated total cost for a next taxable estate of $1,000,000; why does thisvalue differ from the value shown in the table?
e. Interpret the slope of the least-squares regression line.
f. List the residuals for each x value.
41. The following table shows data for the average heights of American boys of different
ages. Suppose that we want to use a boy’s age to predict his height.
Age (years)
Birth
2
3
5
7
10
14

Height (cm)
50.8
83.8
91.4
106.6
119.3
137.1
157.5

a. Which variable should be the independent variable? Which should be the dependent
variable?
b. Calculate the least-squares line. Put the equation in the form: ŷ = a + bx
c. Find the correlation coefficient. Is the correlation significant? Explain.
d. Find the estimated average height for a one-year-old. Find the estimated averageheight for an eleven-year-old.
e. Interpret the slope of the least-squares line.
f. List the residuals for each x value.

689

REFERENCES
13.1 Linear Equations
Data from the Centers for Disease Control and Prevention.
Data from the National Center for HIV, STD, and TB Prevention.

13.5 Prediction
Data from the Centers for Disease Control and Prevention.
Data from the National Center for HIV, STD, and TB Prevention.
Data from the United States Census Bureau. Available online at
http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html
Data from the National Center for Health Statistics.

13.6 Outliers
Data from the House Ways and Means Committee, the Health and Human Services Department.
Data from Microsoft Bookshelf.
Data from the United States Department of Labor, the Bureau of Labor Statistics. Data from the
Physician’s Handbook, 1990.
Data from the United States Department of Labor, the Bureau of Labor Statistics.

690

CHAPTER 13 SOLUTIONS:
1) a. The independent variable is x = number of hours worked;
the dependent variable is y = total charge.
b. y-intercept is a = 50. This is the equipment fee (a fixed cost)c. The slope is b= 100. The labor fee is $100/hr.
3) a. Strong positive correlationb. Fairly strong negative correlationc. No linear relationshipd. Non-linear relationship.
5) a, cScatterplot of Y vs X

14
14

12

12

10

10

8
Y

8
Y

6
4

6
4

2

2

0

0

0

1

2

X

3

4

5

0

1

2

X

3

4

5

b. The regression equation is: y = 2.234 + 1.99 x.
d. The slope is b = 1.99. We expect athletes to earn about $1.99 million per endorsement.
e. The intercept is a = 2.234 An athlete with no endorsements is expected to earn about$2.234 million.
7) The critical value for n = 100 is CV = .205. Since | r | = .89 > CV, the correlation issignificant.
9) Given ŷ = 101.32 + 2.48x.
a. When x = 60, y = 250.12
b. When x = 90, y = 324.52
11) ŷ = 6.39 + 0.015 x; r = 0.094
13) ŷ = 31.53 + 10.90x; r = 0.911
15) Hypotheses are: Ho: ρ = 0 Ha: ρ ≠ 0. From the calculator, t = 0.46 and p = 0.6693.
Since p > .05, we do not reject Ho. The correlation is not significant.
17) ŷ = -3,448,225 + 1750 x.
19) Hypotheses are Ho: ρ = 0 Ha: ρ ≠ 0. From the calculator, t = 2.27 and p = 0.0344.
Since p < .05, we reject Ho. There is a significant correlation.
Alternatively, the correlation coefficient is r = 0.4526, and the critical value is CV = 0.423.
Since r > 0.423, the correlation is significant.
21) a. When x = 1985, ŷ = 25,525
b. When x = 1990, ŷ = 34,275
23) Visually, the line does not appear to be a good fit to the data; however, the hypothesis
test indicates a significant correlation, so using the regression line for point predictions isjustified.
However, some caution should be exercised, and the equation should be used only in therange of observed x-values.
25) a. Yes, there appears to be an outlier at (6, 58).
b. The fact that the correlation coefficient improved significantly when the point isremoved suggests that this really was an outlier.
c. The potential outlier flattened the slope of the line of best fit because it was below thedata set. It made the line of best fit less accurate is a predictor for the data.
d. We would be more confident in the predictive ability of the new regression line.
69127) a.
Scatterplot of Mid-Career Salary vs Tuition
140

Mid-Career Salary

135
130
125
120
115
0

10000

20000

30000
Tuition

40000

50000

60000

b. r = -0.148. The critical value is CV = .632. Since | r | < .632, there is not a significant
correlation between tuition and mid-career salary.
c. The two military institutions appear to be outliers (these have zero tuition)
d. Removing these two points, the new correlation coefficient is r = -0.715, and the new
critical value is CV = .707. Since we now have | r | > .707, the correlation is significant.
29) a. ŷ = 35.58 – 0.192 x
b. r = –0.579; LinRegTTest gives p-value = 0.2288 so we would not reject the null
hypothesis. I.e. the correlation is not significant.
c. There is not a significant linear relationship between deaths and age of driver.
31) a. Independent variable is x = page number; dependent variable is y = max value of
discount.
b.
Scatterplot of Max Value vs Page Number
19

Max Value

18

17

16

15
0

10

20

30

40
50
Page Number

60

70

80

90

c. ŷ = 17.218 – 0.014 x d. r = – 0.275; from the table, CV = 0.666. Since |r| < CV, there
is not a significant correlation between page number and maximum value of discount.
e. Since there is not a significant correlation, we should not use the regression equation for
estimation.

69233) a. in the calculator b. ŷ = 31.93 – 0.304 x.
c. The slope of the regression line is -0.304 with a y-intercept of 31.93. The y-intercept
indicates that when there are no returning sparrow hawks, there will be almost 32% new
sparrow hawks; this doesn’t make sense since because if there are no returning birds, then
the new percentage would have to be 100% (this shows the dangers of extrapolation). The
slope tells us that for each percentage increase in returning birds, the percentage of new
birds in the colony decreases by 0.304%.
d. The coefficient of determination is r2 = .560 which means that about 56% of the total
variation in the percent of new birds is explained by the model and the correlation
coefficient, r = 0.748, indicates a moderately strong correlation between returning and new
percentages.
e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when theobserved return percentage is 66%, our observed new percentage, 6%, is almost 6% lessthan the predicted new value of 11.87%. However, the standard error is s = 3.66, and so
this residual is less than two standard errors from the predicted value; hence the point is
not an outlier.
f. If there are 70% returning birds, we would expect to see y = -0.304(70) + 31.93 = 10.65%
new birds in the colony.
35) a. in the calculator b. ŷ = 193.88 – 1.495 x.
c. We have a slope of –1.495 with a y-intercept of 193.88. The slope, in context, indicatesthat for each additional minute added to the swim time, the heart rate will decrease by 1.5
beats per minute. If the student is not swimming at all, the y-intercept indicates that hisheart rate will be 193.88 beats per minute. While the slope has meaning (the longer ittakes to swim 2,000 meters, the less effort the heart puts out), the y-intercept does notmake sense. If the athlete is not swimming (resting), then his heart rate should be very low.
d. The coefficient of determination is r2 = .015 which means that only 1.5% of the variationin heart rate is explained by this regression equation. So, the model is a very poor fit to theobserved data, and we conclude that there is not a linear relationship between swim timeand heart rate.
e. The point (34.72, 124) generates the largest residual of –11.82. This means that ourobserved heart rate is almost 12 beats less than our predicted rate of 136 beats per minute.
The standard error is s = 10.02 – since the residual is less than two standard errors, thispoint is not considered an outlier. When this point is removed, the coefficient ofdetermination increases to .098, which is better, but still indicates a poor fit to the data.
So, the point is not influential either.

69337) aScatterplot of Percent paid hourly rates vs Year

Percent paid hourly rates

63.0

.

62.5

62.0

61.5

61.0

1980

1982

1984

1986
Year

1988

1990

1992

b. Yes, the scatterplot exhibits a fairly strong linear pattern. c. ŷ = −266.9 + 0.166 x
d. r = 0.944; the correlation is significant.
e. When x = 1991, y = 63.6; when x = 1988, y = 63.1.
f. Yes, the scatterplot, correlation coefficient and hypotheses test all indicate a linear
relationship.
g. Yes, the point (1987, 62.7) gives a residual of -.242. The standard error is s = .247, so
the observe y-value is more than three standard errors from the predicted y-value.
h. When x = 2050, we would get y = 73.4. But this would not be an accurate prediction,
since x = 2050 is too far outside the range of observed data for a reliable prediction.
i. The slope is b = 0.1656. The percent of workers paid hourly rates tends to increase by
0.1656 (1/6 of 1%) each year.
39) a. ŷ = –745.25 + 54.76 x b. r = 0.8944; yes, it is significant.
c. When x = 32, y= $1006.93; when x = 50, y = $1992.53.
d. For each additional inch, the price increases by $54.76.
41) a. The independent variable is x = age; the dependent variable is y = height.
b. ŷ = 65.09 + 7.09 x
c. The correlation coefficient is r = 0.976. Test the hypotheses Ho: ρ = 0 Ha: ρ ≠ 0.
From the calculator, t = 10.05 and p = 0.00017. Since p < .05, we reject Ho.
The correlation is significant.
d. When x = 1, y= 72.18 cm; when x = 11, y = 143.08 cm.
e. The slope is b = 7.09. On average, boys grow by about 7.09 cm per year.
f. Residuals: -14.29, 4.52, 5.03, 6.04, 4.55, 1.06, -6.92

694

14 | F-DISTRIBUTION AND 1-WAY
ANOVA

This Photo by Unknown Author is licensed under CC BY

Figure 13.1 One-way ANOVA can be used to compare average roasting time of 3 or more blends.

Introduction
Chapter Objectives
By the end of this chapter, the student should be able to:••••Interpret the F probability distribution as the number of groups and the sample sizechange.
Discuss two uses for the F distribution: one-way ANOVA and the test of two variances.
Conduct and interpret one-way ANOVA.
Conduct and interpret hypothesis tests of two variances.

Many statistical applications in psychology, social science, business administration, and thenatural sciences involve several groups. For example, an environmentalist is interested in
knowing if the average amount of pollution varies in several bodies of water. A sociologist isinterested in knowing if the amount of income a person earns varies according to his or her
upbringing. A consumer looking for a new car might compare the average gas mileage ofseveral models.
695

14.1 | Test of Two Variances/Standard Deviations
One use of the F distribution is testing two variances. It is often desirable to compare twovariances rather than two averages. For instance, college administrators would like twocollege professors grading exams to have the same variation in their grading. In order for alid to fit a container, the variation in the lid and the container should be the same. A
supermarket might be interested in the variability of check-out times for two checkers.
In order to perform an F test of two variances, it is important that the following are true:
1. The populations from which the two samples are drawn are normally distributed.
2. The two populations are independent of each other.
Unlike most other tests in this book, the F test for equality of two variances is very sensitive
to deviations from normality. If the two distributions are not normal, the test can give higher
p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggestthat students not use this test at all, but in the interest of completeness we include it here.
We will again be using the 5 steps of hypothesis testing.
Suppose we sample randomly from two independent normal populations. Let σ 12 and σ 22 be
the population variances and s12 and s22 be the sample variances. Let the sample sizes be n1
and n2. Since we are interested in comparing the two sample variances, we use the F ratio:

s12

F=

σ 12
s22

σ 22
F has the distribution similar to chi-square distribution but dependent on two degrees of
freedom.

The F distribution has the following characteristics:
1.) All F values are greater than or equal to 0
2.) There is a different F curve for each pair of degrees of freedom n1 – 1, n2 – 1(Figure below).
3.) The curve is nonsymmetrical and skewed to the right4.) There is 100% under the curve5.) The notation is F ~ F(n1 – 1, n2 – 1) where n1 – 1, are the degrees of freedom for thenumerator and n2 – 1 are the degrees of freedom for the denominator.
6.) µ =

d . f .N .
d . f .D. − 1
696

F2,5
F8, 26
F16, 7F3, 11Since the null hypothesis will have equality σ 12 = σ 22 (or σ 12 ≥ σ 22 or σ 12 ≤ σ 22 ) then the F
Ratio becomes

s12
F= 2
s2
Using the TI-83, 83+, 84, 84+ Calculator
To calculate F test value_

Press STAT, TESTS, 2SampFTestNOTE:When using the 2-SampFtest, standard
deviation is requested instead ofvariance. Make sure you know which is
given in the problem.

Critical Values of the F-distribution
Recall, critical value is a value of the distribution that separates the confidence area from thenon-confidence area. We will be using the F- distribution table from Statistics Online
Computational Resource site (http://www.socr.ucla.edu/Applets.dir/F_Table.html) or Excel tofind the critical values. The same rules apply in the previous distributions (Z, t, and χ 2 )when it applies to alpha (α): if the test is two-tailed you will split α.
To find FR, the right critical value:
1. Scroll to the correct α page
2. Look up the degrees of freedom of numerator (d.f.1) at the top of the chart and look up
the degrees of freedom of the denominator (d.f.2) on the side of the chart.
3. Round to the nearest degrees of freedom if necessary.
697

To find FL, the left critical value:
1. Switch the degrees of freedom: d.f.N. will now be d.f.D. and vice versa
2. Look up the new d.f.N. at the top of the chart and the new d.f.D on the side of thechart
3. FL =

1chart value

NotationFR
FL

Calculator Function
none
none

Excel Function=F.inv(α or α/2, dfn, dfd)
=F.inv.rt(α or α/2, dfn, dfd)

Example 14.1
Find the right and left critical value of the F-distribution when given the degrees of
freedom of numerator is 5 and the degrees of freedom of the denominator is 10. Use 90%
confidence level for a two-tailed test.
Solution 14.1 using F-distribution table
First, find probability in the tails. CL = 0.90 which gives α/2 = .05 (area in 1 tail). Look for0.05 page in the tables.
Second, for FR keep the dfn = 5 and dfd = 10 the same.
FR = 3.3258
To find FL, dfn = 10 and dfd = 5
FL =

1
1
=
= 0.2112
4.7351chart value

698

Solution 14.1 using Excel
Here dfn = 5 and dfd = 10 stays the same for both FR and FL.
FL = F.inv(.05, 5, 10) = 0.2112
FR = F.inv.rt(.05, 5, 10) = 3.3258

Example 14.2
Two college instructors are interested in whether or not there is any variation in the way
they grade math exams. They each grade the same set of 30 exams. The first instructor's
grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test
the claim that the first instructor's variance is smaller. (In most colleges, it is desirable forthe variances of exam grades to be nearly the same among instructors.) The level ofsignificance is 5%.
Solution 14.2 using p-value method
Claim: σ 12 < σ 22 (first instructor variance is smaller than 2nd instructor’s variance).
H0: σ 12 ≥ σ 22 and Ha: σ 12 < σ 22
α = 0.05; left-tailed test
F = 2sampFtest = .5818

NOTE: you can enter the square root
within the 2-SampFTest function

699

Draw the graph labeling and shading appropriately.

p-value = P(F < 0.5818) = 0.0753
Since p-value ≥ α, Do Not Reject H0.
With a 5% level of significance, from the data, there is not sufficient evidence to conclude thatthe variance in grades for the first instructor is smaller.
Solution 14.2 using critical value method
Claim: σ 12 < σ 22 (first instructor variance is smaller than 2nd instructor’s variance).
H0: σ 12 ≥ σ 22 and Ha: σ 12 < σ 22
α = 0.05; left-tailed test
F = 2sampFtest = 0.5818
Distribution for the test is F29,29 where d.f.N. = n1 – 1 = 29 and d.f.D. = n2 – 1 = 29.
Since it is left-tailed, we are looking for the left critical value, FL. Here, when we switch up
the degrees of freedom if we are using the table, it looks the same since in this example d.f.N.
= d.f.D. On the chart, there is no 29 at the top of the chart; therefore, we go to the closest
degrees of freedom which is 30. However, on the degrees of freedom on the side of the chart,
we use 29 since there is one.
FL =

1
1
=
= .5405
chart value 1.85

Or use Excel
FL = F.inv(.05, 29, 29) = .5374
NOTE: due to rounding the
Excel and the table value areslightly not the same.

700

Since the test statistic, 0.5818 is not in the critical region (0.5818 > 0.6173), Do Not Reject
H0.
With a 5% level of significance, from the data, there is not sufficient evidence to conclude thatthe variance in grades for the first instructor is smaller.

Example 14.3
The New York Choral Society divides male singers up into four categories from highest voices
to lowest: Tenor1, Tenor2, Bass1, and Bass2. In the table are heights of the men in the Tenor1
and Bass2 groups. One suspects that taller men will have lower voices, and that the varianceof height may go up with the lower voices as well. Do we have good evidence that the standard
deviations of the heights of singers in each of these two groups (Tenor1 and Bass2) are
different using 10% level of significance? Use the table below.

Tenor 1:
Bass 2:69, 72, 71, 66, 76, 74, 71, 66, 68, 67, 70, 65, 72, 75, 67, 75, 74, 72, 72, 74, 72, 72, 74, 70,
72, 70, 68, 64, 73, 66, 68, 67, 64
66, 68, 75, 68, 70, 72, 67, 70, 70, 69, 72, 71,
74, 75
Solution 14.3 using p-value
H0: σ 1 = σ 2
Ha: σ 1 ≠ σ 2 (claim)
α = 0.10; two-tailed test
F = 2-SampFTest = 1.489

p-value = .343; therefore, Do Not Reject H0.
There is not enough evidence at 10% level of significance to support the claim standard
deviations of the heights of singers in each of these two groups (Tenor1 and Bass2) are
different.

701

Solution 14.3 using critical value
H0: σ 1 = σ 2
Ha: σ 1 ≠ σ 2 (claim)
α = 0.10; two-tailed testF = 2-SampFTest = 1.489 (test value)

Since it is two-tailed, we split α. We look up the critical values using the 0.05 chart of the Fdistribution chart (http://www.socr.ucla.edu/Applets.dir/F_Table.html).
Distribution for the test is F20, 25 where d.f.N. = n1 – 1 = 20 and d.f.D. = n2 – 1 = 25.
FR = 2.01 and FL =1
1=
= .4808
chart value 2.08

NOTE: d.f.N. and d.f.D. switch for left critical value using the table.
Or FL = F.inv(.05, 20, 25) = 0.4822

Do Not Reject H0 because the test statistic, 1.489 is not in neither tail.
There is not enough evidence at 10% level of significance to support the claim standard
deviations of the heights of singers in each of these two groups (Tenor1 and Bass2) are
different.

14.1 The New York Choral Society divides male singers up into four categories from highest
voices to lowest: Tenor1, Tenor2, Bass1, and Bass2. In the table are heights of the men in the
Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that thevariance of height may go up with the lower voices as well. Do we have enough evidence thatthe standard deviation of Tenor 1 heights is larger than the standard deviation of the heights
of Bass 2 singers using 5% level of significance? Use the table below.
Tenor 1:
Bass 2:70, 72, 71, 66, 76, 74, 71, 66, 68, 67, 72, 75, 67, 75, 74, 72, 72, 74, 72, 72, 74, 70, 66, 68,
70, 65, 72, 70, 68, 64, 73, 66, 68, 67, 64 75, 68, 70, 72, 67, 70, 70, 69, 72, 71, 74, 75702

14.2 | One-Way ANOVA
The purpose of a one-way ANOVA test is to determine the existence of a statisticallysignificant difference among several group means. There must be at least three groups. The
test actually uses variances to help determine if the means are equal or not. In order toperform a one- way ANOVA test, there are five basic assumptions to be fulfilled:
1.
2.
3.
4.
5.

Each population from which a sample is taken is assumed to be normal.
All samples are randomly selected and independent.
The populations are assumed to have equal standard deviations (or variances).
The factor is a categorical variable.
The response is a numerical variable.

The Null and Alternative Hypotheses
The null hypothesis is simply that all the group population means are the same. The
alternative hypothesis is that at least one pair of means is different. For example, if there
are k independent samples:
H0: μ1 = μ2 = μ3 = ... = μk
Ha: At least one of the means is different
The graphs, a set of box plots representing the distribution of values with the group means
indicated by a horizontal line through the box, help in the understanding of the hypothesis
test. In the first graph (red box plots), H0: μ1 = μ2 = μ3 and the three populations have thesame distribution if the null hypothesis is true. The variance of the combined data isapproximately the same as the variance of each of the populations.
If the null hypothesis is false, then the variance of the combined data is larger which is causedby the different means as shown in the second graph (green box plots).

NOTE:(a) H0 is true. All means are the same; thedifferences are due to random variation.
(b) H0 is not true. All means are not the same;the differences are too large to be due to randomvariation.

703

To calculate the F ratio for the ANOVA test value, two estimates of the varianceare made
1. Variance between samples: An estimate of σ2 that is the variance of the sample
means multiplied by n (when the sample sizes are the same.). If the samples aredifferent sizes, the variance between samples is weighted to account for the different
sample sizes. The variance is also called variation due to treatment (factor) or
explained variation. This variance is also known as Mean Square Between (MSB).
2. Variance within samples: An estimate of σ2 that is the average of the sample
variances (also known as a pooled variance). When the sample sizes are different, thevariance within samples is weighted. The variance is also called the variation due toerror or unexplained variation. This variance is known as Mean Square Within(MSW).

Test Statistic (Value):

F=

MSB
MSW

Always right-tailed test
SSbetween (SSB) = the sum of squares that represents the variation among the different samples
SSwithin (SSW) = the sum of squares that represents the variation within samples that is due
to chance.
To find a "sum of squares" means to add together squared quantities that, in some cases, may
be weighted. We used sum of squares to calculate the sample variance and the sample
standard deviation in Descriptive Statistics.
Calculation of Sum of Squares:••••

k = the number of independent samples (groups)
nj = the size of the jth groupsj = the sum of the values in the jth group
N = total sample size: ∑ n j•

x = one value ∑ x = ∑ s j•

Sum of squares of all values from every group combined: ∑ x 2•

Total Sum of Squares: SS
=
total

∑

(∑ x)
x −
2

n

704

2

•
•

 ( s )2  ( s )2j− ∑ jExplained variation:=
SSbetween ∑  nj 
nUnexplained variation: SS=
SStotal − SSbetween
withinCalculation of Mean Square:••••

Degrees of freedom of numerator: d.f.N. = k – 1
Degrees of freedom of denominator: d.f.D. = N – k

SSB
d.f.N
SSW
Mean Square Within: MSW =d.f.D
Mean Square Between: MSB =The one-way ANOVA test depends on the fact that MSB can be influenced by populationdifferences among means of the several groups. Since MSW compares values of each group to
its own group mean, the fact that group means might be different does not affect MSW.
The null hypothesis says that all groups are samples from populations having the same
normal distribution. The alternate hypothesis says that at least one of the means of thesample groups come from populations with different normal distributions. If the null
hypothesis is true, MSB and MSW should both estimate the same value.
NOTE: The null hypothesis says that all the group population means are equal. The
hypothesis of equal means implies that the populations have the same normal distribution,
because it is assumed that the populations are normal and that they have equal variances.
Test Statistic (Value):
F=

MSB
MSW

If MSB and MSW estimate the same value (following the belief that H0 is true), then the Fratio should be approximately equal to one. Mostly, just sampling errors would contribute tovariations away from one. As it turns out, MSB consists of the population variance plus avariance produced from the differences between the samples. MSW is an estimate of the
population variance. Since variances are always positive, if the null hypothesis is false, MSB
will generally be larger than MSW. Then the F-ratio will be larger than one. However, if the
population effect is small, it is not unlikely that MSW will be larger in a given sample.

705

ANOVA Summary Table
Variation
Between(Factor)Within(Error)Sum ofSquares(SS)

Degrees ofFreedom(df)

Mean
Squares(MS)

Test Value
(F)

SSB

d.f.N = k - 1

MSB =

SSB
d.f.N

F=

SSW

d.f.D = N – k

MSW =

SSW
d.f.D

MSB
MSW

p-value
Fcdf bycalculator

N-1

Total

Using the TI-83, 83+, 84, 84+ Calculator
To calculate p-value of F-distribution_

p-value = Fcdf(Test Value, 10^99, d.f.N., d.f.D.)

Example 14.4
Three different diet plans are to be tested for mean weight loss. The entries in the table are
the weight losses for the different plans. The one-way ANOVA results are shown in thesummary table:
Variation

Sum ofSquares (SS)

Degrees ofFreedom (df)

Between(Factor)

2.24582

Within(Error)20.8542

7

Mean
Squares(MS)

Test Value(F)

p-valueTotal
a. Complete the ANOVA summary tableb. Using 5% level of significance, test the claim that the mean weight loss for the threedifferent plans are the same.

706

Solution 14.4 part a
Variation
Between(Factor)Sum ofSquares(SS)
2.2458Degrees ofMean Squares
Freedom (df) (MS)2.24582
MSB =1.1229
20.8542MSW =7MSW =2.9792
MSB =

2

Within(Error)20.8542

7

Total

23.1

9

Test Value(F)
F=

1.12292.9792p-value.6991

F = .3769

Solution 14.4 part b
H0: μ1 = μ2 = μ3 (claim)
Ha: At least one of the means is different
α = 0.05; right-tailed test
F=

MSB
= .3769
MSW

p-value = .6991; therefore Do Not Reject H0
There is not sufficient evidence at 5% level of significance to reject the claim that the mean
weight loss for the three different plans are the same.

Using the TI-83, 83+, 84, 84+ Calculator
To calculate test statistic given the data_

If we are given the data of 3 or more groups, we can use the List feature of the calculator toenter the data. We can also use the ANOVA(L1, L2, L3, …) function to find the test statistic.

707

Example 14.5
A car manufacturing company needs to purchase steel to make roll bar frames for dune buggy
style autos. The purchasing manager looks at 4 steel makers and took a random sample ofsteel bars and tested their strength (in MPa). The results are shown in the following table:
Maker 1
980
990
960
970
950

Maker 2
960
970
980
920
860

Maker 3
1000
900
980
1020
960

Maker 4
840
880
900
890
920

a. Using a significance level of 1%, is there a difference in the mean steel strength
among the 4 steel makers.
b. Create an ANOVA summary table.
Solution 14.5 part a
H0: μ1 = μ2 = μ3 = μ4
Ha: At least one of the means is different (claim)
α = 0.01; right-tailed test
F = ANOVA(L1, L2, L3, L4) = 5.68
p-value = 0.00759; therefore Reject H0 since p-value ≤ α.

There is sufficient evidence at 1% level of significance to support the claim that the meansteel strength is different among the 4 steel makers.
Solution 14.5 part b
The calculator will help us fill in the ANOVA Summary table

708

Variation

Sum ofSquares (SS)

Degrees ofFreedom (df)MeanSquares(MS)

Between(Factor)24175

38058.33

Within(Error)22680

161417.5

Total

46855

19

Test Value(F)

p-value

F = 5.68

0.00759

To recap, MSB = 8058.33 is an estimate of σ2 that is based on the variability among thesample means. MSW = 1417.5 is an estimate of σ2 that is based on the sample variances.

Example 14.6
A medical researcher wishes to try three different techniques to lower blood pressure ofpatients with high blood pressure. The subjects are randomly selected and assigned to oneof three groups. Group 1 is given medication, Group 2 is given an exercise program, andGroup 3 is assigned a diet program. At the end of six weeks, each subject's blood pressure isrecorded, with data as shown below. Using p-value and the level of significance of 1%, test
the claim that the mean is the same for the different groups.

709

Solution 14.6
H0: μ1 = μ2 = μ3 (claim);
Ha: At least one of the means is different
α = 0.01; right-tailed test
F = ANOVA(L1, L2, L3) = 11.09; p-value = 0.0011
Reject H0 because 0.0011 < α
There is enough evidence at 1% level of significance to reject the claim that the mean is thesame for the different groups.

14.2 Four different types of fertilizers are used on raspberry plants. The number ofraspberries on each randomly selected plant is given below.
A suitable hypothesis test will be conducted to test the claim that the type of fertilizer
makes no difference in the mean number of raspberries per plant. Use α = .05 to test theclaim.

Key Terms
Analysis of Variance also referred to as ANOVA, is a method of testing whether or not
the means of three or more populations are equal. The method is applicable if:•••

all populations of interest are normally distributed.
the populations have equal standard deviations.
samples (not necessarily of the same size) are randomly and independently
selected from each population. The test statistic for analysis of variance is the Fratio.

710

Review of Tests
You have seen the F test statistic used in two different circumstances. The following bulleted
list is a summary that will help you decide which F test is the appropriate one to use.
•

Testing 2 Variances:
Test statistic: F =s12(calculator: 2-SampFTest)
s22

Degrees of freedom of numerator: d.f.N. = n1 – 1
Degrees of freedom of denominator: d.f.D. = n2 – 1
Test may be left-, right-, or two-tailed•

Testing 3 or more Means:
Test statistic (value): F =MSB(calculator: ANOVA(L1, L2, …Lk))
MSW

P-value: Fcdf(Test Value, 10^99, d.f.N., d.f.D.)

SSB
d.f.N
SSW
Mean Square Within: MSW =d.f.D

Mean Square Between: MSB =Total Sum of Squares: SS
=
total

∑

(∑ x )x −22n

 ( s )  ( s )2j− ∑ jExplained variation:=
SS B ∑  nj 
nUnexplained variation: =SSW SStotal − SS B
2

Degrees of freedom of numerator: d.f.N. = k – 1
Degrees of freedom of denominator: d.f.D. = N – k711

Exercises for Chapter 14
1. Find the right critical value for F-distribution with α = 0.10 and d.f.n = 8 and d.f.d = 14
2. Find the right critical value for F-distribution with α = 0.05 and d.f.n. = 6 and d.f.d = 8
3. Find the left critical value for F-distribution with α = 0.025 and d.f.n. = 12 and d.f.d = 15
4. Find the left critical value for F-distribution with α = 0.10 and d.f.n. = 10 and d.f.d = 20
5. State the two assumptions that must be true in order to perform an F- test of two
variances.
6. Suppose that we test the hypotheses H0: σ12 = σ22 vs. Ha: σ12 ≠ σ22 using the followingdata:s12 = 9.4
n1 = 12s22 = 10.6
n2 = 11

Find the test statistic and p-value for the test.
7. Suppose that we test the hypotheses H0: σ12 = σ22 vs. Ha: σ12 ≠ σ22 using the followingdata:s12 = 22.5
n1 = 19s22 = 31.3
n2 = 23

Find the test statistic and p-value for the test.
8. Two coworkers commute from the same building. They are interested in comparing thevariability of their driving times to work. They each record their times for 20 commutes. The
first worker’s times have a variance of 12.1. The second worker’s times have a variance of
16.9. The first worker claims that his commuting time is less variable than that of hiscolleague. Test this claim at the 10% level.
a. State the null and alternative hypotheses.
b. What test should be used in this problem?
c. What is the F statistic?
d. What is the p-value?
e. Using α = .10, what is the decision about H0?
f. Does this data support the first worker’s claim?

712

9. Two students are interested in comparing the amount of variation in their test scores formath class. There are 15 total math tests they have taken so far. The first student’s grades
have a standard deviation of 38.1. The second student’s grades have a standard deviation of
22.5. The second student thinks his scores are less variable. Assuming that the populationsof test scores are normally distributed, we will test her claim at the α = .05 significance level.
a. State the null and alternative hypotheses.
b. What test should be used in this problem?
c. What are the F statistic and p-value?
d. At the 5% significance level, would we reject the null hypothesis?
e. What does this say about the student’s claim?
10. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist
records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8 and thesecond cyclist has a variance of 32.1. Use this data along with a .05 significance level to test
the claim that there is a difference in the variances.
11. List the five basic assumptions that must be fulfilled in order to perform a one-way
ANOVA
test.
12. State the hypotheses for a one-way ANOVA test if there are three groups.
13. State the hypotheses for a one-way ANOVA test if there are four groups.
14. Groups of men from three different areas of the country are to be tested for mean
weight. The entries in the table are the weights for the different groups. The one-way
ANOVA results are shown in the table:
Group 1 Group 2 Group 3
216
202
170
198
213
165
240
284
182
187
228
197
176
210
201
a. What is the Sum of Squares Factor?
b. What is the Sum of Squares Error?
c. What is the degrees of freedom for the numerator?
d. What is the degrees of freedom for the denominator?
e. What is the Mean Square Factor?
f. What is the Mean Square Error?
g. What is the F statistic?

713

15. Girls from four different soccer teams are to be tested for mean goals scored per game.
The entries in the table are the goals per game for the different teams. The one-way
ANOVA results are shown in the table below:
Team 1 Team 2 Team 3 Team 4
1
2
0
3
2
3
1
4
0
2
1
4
3
4
0
3
2
4
0
2
a. What is SSbetween?
b. What is the degrees of freedom for the numerator?
c. What is MSbetween?
d. What is SSwithin?
e. What is the degrees of freedom for the denominator?
f. What is MSwithin?
g. What is the F statistic?
h. Judging by the F statistic, do you think it is likely that you will reject the null
hypothesis?

16. Five basketball teams each took a random sample of players regarding how high eachplayer can jump (in inches). The results are shown in the table below:.

Team 1 Team 2 Team 3 Team 4 Team 5
36
32
48
38
41
42
35
50
44
39
51
38
39
46
40

a. What are the Sum of Squares and Mean Squares Factors?
b. What are the Sum of Squares and Mean Squares Errors?
c. What is the F statistic?
d. What is the p-value?
e. At the 5% significance level, is there a difference in the mean jump heights among theteams?

714

17. A video game developer is testing a new game on three different groups. Each grouprepresents a different target market for the game. The developer collects scores from arandom sample from each group. The results are shown in the table:
Group A
101
108
98
107
111

Group B
151
149
160
112
126

Group C
101
109
198
186
160

This data is used to test whether there is a significant difference among the means for eachgroup.
a. What are the hypotheses for the test?
b. What are the SSbetween and MSbetween?|c. What are the SSwithin and MSwithin?
d. What is the F Statistic?
e. What is the p-value?
f. At the 10% significance level, is there evidence that the mean scores among thedifferent groups are different?
18. Three different traffic routes are tested for mean driving time. The entries in the table
are the driving times in minutes on the three different routes. The data from three random
samples are shown in the table below:
Route 1
30
32
27
35

Route 2
27
29
28
36

Route 3
16
41
22
31

Use this data to test the claim that the routes have the same mean driving time.

715

19. Three students, Linda, Tuan, and Javier, are each given five laboratory rats for anutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula
A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a
specified time period, each rat is weighed again, and the net gain in grams is recorded. Usinga significance level of 10%, test the hypothesis that the three formulas produce the samemean weight gain.
Linda's rats
43.5
39.4
41.3
46.0
38.2

Tuan's rats
47.0
40.5
38.9
46.3
44.2

Javier's rats
51.2
40.9
37.9
45.0
48.6

20. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase
would hurt working-class people the most, since they commute the farthest to work. Supposethat the group randomly surveyed 24 individuals and asked them their daily one-way
commuting mileage. The results are in the table below. Using a 5% significance level, test
the hypothesis that the three mean commuting mileages are the same.
Working Class
17.8
26.7
49.4
9.4
65.4
47.1
19.5
51.2

Professional (middle incomes)
16.5
17.4
22.0
7.4
9.4
2.1
6.4
13.9

Professional (wealthy)
8.5
6.3
4.6
12.6
11.0
28.6
15.4
9.3

21. The table below lists the number of pages in four different types of magazines. Use thisdata and a 5% level of significance to test the hypothesis that the four magazine types havethe same mean number of pages.
Home Decorating
172
286
163
205
197

News
87
94
123
106
101

Health
82
153
87
103
96

716

Computer
104
136
98
207
146

22. A researcher wants to know if the mean times (in minutes) that people watch theirfavorite news station are the same. The table below shows the results of a study.
CNN
45
12
18
38
23
35

FOX
15
43
68
50
31
22

LOCAL
72
37
56
60
51

Assume that all distributions are normal, the four population standard deviations areapproximately the same, and the data were collected independently and randomly. Use a .05
level of significance.
23. Are the means for the final exams the same for all statistics class delivery types? The
table below shows the scores on final exams from randomly selected classes using thedifferent delivery types.
Online
72
84
77
80
81

Hybrid8373
84
81

Face-to-face
80
78
84
81
86
79
82

Assume that all distributions are normal, the population standard deviations areapproximately the same, and the data were collected independently and randomly. Use a .05
level of significance.
24. Are the mean number of times a month a person eats out the same for whites, blacks,
Hispanics and Asians? Suppose that the table below shows the results of a study:
White
6
8
2
4
6

Black
4
1
5
2

Hispanic
7
3
5
4
6

Asian
8
3
5
1
7

Use this data, along with a .05 significance level, to test the claim that the mean is thesame for all four populations. Assume that all distributions are normal, the four population

717

standard deviations are approximately the same, and the data were collected independently
and randomly.
25. Are the mean numbers of daily visitors to a ski resort the same for the three types ofsnow conditions? Suppose that the data below shows the results of a study:
Powder
1210
1080
1537
941

Machine Made
2107
1149
862
1870
1528
1382

Hard Packed
2846
1638
2019
1178
2233

Use this data and a .05 significance level to test the claim that the mean number of visitors
is the same for all three types of snow conditions. Assume that all distributions are normal,
the three population standard deviations are approximately the same, and the data werecollected independently and randomly.
26. DDT is a pesticide that has been banned from use in the United States and most otherareas of the world. It is quite effective, but persisted in the environment and over time became
seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors
were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in
the food chain of the birds.
An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies.
There are three groups of flies. One group was bred to be resistant to DDT (the RS group).
Another was bred to be especially susceptible to DDT (SS). Finally there was a control line ofnon-selected or typical fruitflies (NS). Here are the data:
RS
12.8
21.6
14.8
23.1
34.6
19.7
22.6
29.6
16.4
20.3
29.3
14.9
27.3

SS
38.4
32.9
48.5
20.9
11.6
22.3
30.2
33.4
26.7
39
12.8
14.6
12.2

NS
RS
SS
NS
35.4
22.4
23.1 22.6
27.4
27.5
29.4 40.4
19.3
20.3
16
34.4
41.8
38.7
20.1 30.4
20.3
26.4
23.3 14.9
37.6
23.7
22.9 51.8
36.9
26.1
22.5 33.8
37.3
29.5
15.1 37.9
28.2
38.6
31
29.5
23.4
44.4
16.9 42.4
33.7
23.2
16.1 36.6
29.2
23.6
10.8 47.4
41.7

718

The values are the average number of eggs laid daily for each of 75 flies (25 in each group)
over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg
selection for the three strains of fruitfly different? If so, in what way? Specifically, theresearchers were interested in whether or not the selectively bred strains were different fromthe non-selected line, and whether the two selected lines were different from each other.
Here is a chart of the three groups:

27. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for anutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula
A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a
specified time period, each rat is weighed again and the net gain in grams is recorded.
Linda's rats
43.5
39.4
41.3
46.0
38.2

Tuan's rats
47.0
40.5
38.9
46.3
44.2

Javier's rats
51.2
40.9
37.9
45.0
48.6

Determine whether or not there is a significant difference in the variance among Javier’s and
Linda’s rats. Test at a significance level of 10%.
28. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase
would hurt working-class people the most, since they commute the farthest to work. Supposethat the group randomly surveyed 24 individuals and asked them their daily one-way
commuting mileage. The results are as follows.
Working Class
Professional(middle income)
Professional(Wealthy)

17.8
16.5

26.7 49.4 9.4
17.4 22
7.4

8.5

6.3

4.6

65.4
9.4

12.6 11

719

47.1
2.1

19.5
6.4

51.2
13.9

28.6

15.4

9.3

Determine whether or not the variance in mileage driven is statistically the same among theworking class and professional (middle income) groups. Use a 5% significance level.
29. The following table lists the number of pages in four different types of magazines:
Home Decorating
172
286
163
205
197

News
87
94
123
106
101

Health
82
153
87
103
96

Computer
104
136
98
207
146

Use this data to test whether there is a significant difference in the variance for homedecorating magazines and news magazines.
30. Is the variance for the amount of money (in dollars) that shoppers spend on Saturdays
at the mall the same as the variance for the amount of money that shoppers spend on
Sundays at the mall? Suppose that the table shows the results of a study.
Saturday Sunday Saturday Sunday
75

44

62

137

18

58

0

82

150

61

124

39

94

19

50

127

62

99

31

141

73

60

118

73

89
Use this data, along with a 5% significance level to test the claim that the variances areequal.
31. Are the variances for incomes on the East Coast and the West Coast the same? Supposethat the table below shows the results of a study; income is shown in thousands of dollars.
Use this data to test the hypothesis that the variances are the same. Assume that both
distributions are normal, and the samples are randomly and independently selected.
East
West

38
71

47
126

30
42

82
51

720

75
44

52
90

115
88

67

32. Thirty men in college were taught a method of finger tapping. They were randomly
assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100
mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours
after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The
experiment was double blind, so neither the recorders nor the students knew which groupthey were in. Does caffeine affect the rate of tapping, and if so how?
Here are the data:
0 mg
100
mg
200
mg

242
248

244
245

247
248

242
247

246
243

245
246

248
247

248
250

244
246

242
244

246

250

248

246

245

248

252

250

248

250

33. King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul)
during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but
declined significantly afterwards. Coins minted during his era were found in Cyprus, anisland in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven fromthe second, four from the third, and seven from a fourth. These spanned most of his reign.
We have data on the silver content of the coins:First Coinage Second
C6.9
i
5.9

Third
C4.9
i

Fourth
C5.3
i

6.8

9.0

5.5

5.6

6.4

6.6

4.6

5.5

7.0

8.1

4.5

5.1

6.6

9.3

6.2

7.7

9.2

5.8

7.2

8.6

5.8

6.9
6.2
Did the silver content of the coins change over the course of Manuel’s reign? Here are themeans and variances of each coinage. The data are unbalanced.
Mean
Variance

First6.74440.2953

Second8.24291.2095

721

Third
4.875
0.2025

Fourth
5.61430.1314

34. Four different diet plans are to be tested for mean weight loss. The entries in the table
are the weight losses for the different plans. The one-way ANOVA results are shown in thesummary table:
Variation

Sum ofSquares (SS)

Degrees ofFreedom (df)

Between(Factor)60.37

3

Within(Error)101.8153

Mean
Squares(MS)

Test Value(F)

p-valueTotal
a. Complete the ANOVA summary tableb. Using 5% level of significance, test the claim that the mean weight loss for the four
different plans are the same.

35. We are interested in testing the mean compression strength of four different box types.
The one-way ANOVA results are shown in the summary table:
Variation

Sum ofSquares (SS)

Degrees ofFreedom (df)

Between(Factor)32.1383

Within(Error)32.901

Mean
Squares(MS)

Test Value(F)

p-value

Total
23
a. Complete the ANOVA summary tableb. Using 1% level of significance, test the claim that the mean compression strength forthe four different box types is different.

722

REFERENCES
14.1 Test of Two Variances“MLB Vs. Division Standings – 2012.” Available online at
http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/order/true.
Tomato Data, Marist College School of Science (unpublished student research)Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets:
Data for Fruitfly
14.2 ANOVA
Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA.
Fecundity. London: Chapman & Hall, 1994.
Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets.
London: Chapman & Hall, 1994, pg. 50.
Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets.
London: Chapman & Hall, 1994, pg. 118.
“MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012.
Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees
F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August
Wunderlich," Journal of the American Medical Association, 268, 1578-1580.

723

CHAPTER 14 SOLUTIONS:
1)

using Excel FR = F.inv.rt(0.10, 8, 14) = 2.1539

3)

using Excel FL = F.inv(0.025, 12, 15) = 0.3147 Using chart, switch DF dfn = 15, dfd= 12 FL =1/3.1772 = 0.3147

5)i) The populations from which we are sampling are normally distributed.
ii) The samples are randomly selected and independent of one another.

7)

From the calculator with 2SampFTest, F = .72 and p = .4810 Note: If the subscripts
are switched, then we get F = 1.39, with the same p-value.

9)a. Ho: σ1 < σ2; Ha: σ1 > σ2. b. Use a two-sample F-test (test for two variances)c. From the calculator, F = 2.87 and p = 0.0291. d. Since p < .05, reject Ho. e.
There is sufficient evidence to conclude that the second student’s scores are lessvariable.

11)i) Each population from which we are sampling is normally distributed.
ii) All samples are random and independent of one another.
iii) The populations are assumed to have equal variances.
iv) The factor is a categorical variablev) The response is a numerical variable.
Ho: µ1 = µ2 = µ3 = µ4; Ha: At least one of the means is different from the others.
a. SSbetween = 25.75 b. dfnum = 3
c. MSbetween = 8.58
d. SSwithin = 13.2
e. dfdenom = 16
f. MSwithin = .825
g. F = 10.40
h. The test is right-tailed, so a large F-statistic suggests that Ho would be rejected.
a. Ho: µ1 = µ2 = µ3; Ha: At least one of the means is different from the others.
b. SSbetween = 5700.4 and MSbetween = 2850.2
d. F = 3.61 e. p = 0.0592
c. SSwithin = 9474 and MSwithin = 789.5
f. Since p < .10, we reject Ho. There is sufficient evidence to conclude that there
is a difference in the mean scores for the different groups.
Ho: µ1 = µ2 = µ3; Ha: At least one of the means is different from the others.
Use ANOVA F-test. From the calculator, F = 0.67 and p = 0.5305.

13)
15)

17)

19)

Since p > .05, we do not reject Ho. There is not enough evidence to conclude that
there is a difference in the mean weight gains among the three groups of rats.
21)

Ho: µ1 = µ2 = µ3= µ4; Ha: At least one of the means is different from the others.Use
ANOVA F-test; from the calculator, F = 8.69 and p = .0012. Since p < .05, we reject
Ho. There is sufficient evidence to conclude that there is a difference in the mean
number of pages for different magazine types.

724

23)

Ho: µ1 = µ2 = µ3; Ha: At least one of the means is different from the others. Use
ANOVA F-test; from the calculator, F = .64 and p = .5437. Since p > .05, we do not
reject Ho. There is not enough evidence to reject the claim that the mean final exam
scores are the same for all three delivery types.

25)

Ho: µ1 = µ2 = µ3; Ha: At least one of the means is different from the others. Use
ANOVA F-test. From the calculator, F = 3.10 and p = 0.082. Since p > .05, we do
not reject Ho. There is not enough evidence to reject the claim that the mean
number of visitors is the same for three types of snow conditions.

27)

Ho: σ12 = σ22; Ha: σ12 ≠ σ22. Use 2SampFTest; from the calculator, we get F = .33
and p = 0.3127. Since p > .05, do not reject Ho. There is not enough evidence toconclude that there is a difference in the variances among Javier’s and Linda’s rats.
NOTE: Here we used σ12 to represent the variance for Linda’s rats, and σ22 torepresent the variance for Javier’s rats. Had these been entered in the opposite
order, we would have gotten F = 3.0; however, because it is a two-tailed test, the pvalue would be the same.

29)

Ho: σ12 = σ22; Ha: σ12 ≠ σ22. Use 2SampFTest; from the calculator, we get F = 12.69
and p = 0.0305. Since p < .05, do reject Ho. There is sufficient evidence to conclude
that the variance is different for the two types of magazines. Again, had we
switched the subscripts, we would get a different F-statistic, F = 0.079. However,
because it is a two-tailed test, the p – value would remain the same.

31)

Ho: σ12 = σ22; Ha: σ12 ≠ σ22. Use 2SampFTest; from the calculator, we get F = 0.812
and p = 0.7825. Since p > .05, do reject Ho. There is not enough evidence to reject
the claim that the variances for incomes are the same on the two coasts.

33)

Ho: µ1 = µ2 = µ3 = µ4; Ha: At least one of the means is different from the others. Use
ANOVA F-test. From the calculator, F = 26.27 and p = 0.0000001. Since p < .05, we
reject Ho. There is very strong evidence that the mean silver content of the different
coinages are not all the same.

35)

a.

Variation
Between(Factor)Within(Error)
Total

(SS)

(df)

(MS)

32.1383

10.71332.90120

1.645

65.03923

Test Value (F)6.512

p-value.00298c. Ho: µ1 = µ2 = µ3 = µ4; Ha: at least 1 mean is different; F = 6.512; p-value = .00298;
Reject Ho. There is enough evidence to support claim that the means are different.
725This page is purposely left blank.

726

Z

0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0

0
0.5
0.53980.57930.61790.65540.69150.72570.758
0.78810.81590.84130.86430.88490.90320.91920.93320.94520.95540.9641
0.97130.97720.98210.98610.98930.99180.99380.99530.99650.9974
0.9981
0.99870.01
0.504
0.54380.58320.62170.6591
0.695
0.7291
0.7611
0.791
0.8186
0.84380.86650.88690.90490.92070.93450.94630.95640.96490.97190.97780.9826
0.98640.98960.992
0.994
0.99550.9966
0.99750.99820.99870.02
0.508
0.54780.58710.62550.66280.69850.73240.7642
0.79390.82120.84610.86860.88880.90660.92220.93570.94740.95730.9656
0.97260.97830.983
0.98680.98980.9922
0.99410.9956
0.99670.9976
0.99820.9987

Standard Normal Distribution
0.05
0.03
0.04
0.512
0.516
0.51990.55170.55570.55960.591
0.59480.59870.62930.63310.63680.66640.67
0.67360.70540.70880.70190.73570.73890.74220.76730.77040.77340.79950.80230.79670.82380.82640.82890.84850.85080.85310.87080.87290.87490.89440.89070.89250.90820.90990.91150.92360.92510.92650.93940.937
0.9382
0.94840.94950.95050.95820.95910.95990.96640.9671
0.96780.9732
0.97380.97440.97880.97930.97980.98340.98380.98420.98780.98710.98750.99010.99040.9906
0.99250.9927
0.99290.9946
0.99430.99450.99570.99590.996
0.99680.99690.997
0.99770.99770.99780.99830.9984
0.9984
0.99880.99880.9989A1

0.06
0.52390.56360.60260.6406
0.67720.71230.74540.77640.80510.83150.85540.877
0.89620.91310.92790.9406
0.95150.96080.96860.975
0.98030.9846
0.98810.99090.9931
0.99480.9961
0.9971
0.99790.99850.99890.07
0.52790.56750.60640.64430.68080.71570.74860.77940.80780.834
0.85770.879
0.898
0.91470.9292
0.94180.95250.9616
0.96930.97560.98080.985
0.9884
0.9911
0.9932
0.99490.99620.99720.99790.99850.99890.08
0.53190.57140.61030.648
0.68440.719
0.75170.78230.81060.83650.85990.881
0.89970.91620.9306
0.94290.95350.96250.96990.9761
0.98120.98540.9887
0.99130.9934
0.9951
0.99630.99730.998
0.99860.999

0.09
0.53590.57530.61410.65170.68790.72240.75490.78520.81330.83890.86210.883
0.90150.91770.93190.9441
0.95450.96330.9706
0.97670.98170.98570.989
0.9916
0.9936
0.99520.9964
0.9974
0.9981
0.99860.999

The critical values of t distribution are calculated according to the probabilities of
two alpha values and the degrees of freedom. The Alpha(α) values 0.05 one tailed
and 0.1 two tailed are the two columns to be compared with the degrees of
freedom in the row of the table.
0.005 0.0025
0.001
0.00050.01
α (1 tail)0.05 0.025
α (2 tail)
0.1
0.05
0.02
0.01
0.005
0.002
0.001
1
6.3138 12.7062 31.8205 63.6567 127.3213 318.3088 636.6192
2
2.9200 4.3027 6.9646 9.9248 14.0890 22.3271 31.5991
3
2.3534 3.1824 4.5407 5.8409
7.4533 10.2145 12.9240
42.1318 2.7764 3.7469 4.6041
5.59767.17328.61034.7733
5.89346.868852.0150 2.5706 3.3649 4.0321
6
1.9432 2.4469 3.1427 3.7074
4.31685.2076
5.95887
1.8946 2.3646 2.9980 3.4995
4.02934.78535.40793.83254.50085.04138
1.8595 2.3060 2.8965 3.3554
9
1.8331 2.2622 2.8214 3.2498
3.68974.29684.780910
1.8125 2.2281 2.7638 3.1693
3.58144.14374.58693.49664.02474.437011
1.7959 2.2010 2.7181 3.1058
12
1.7823 2.1788 2.6810 3.0545
3.42843.9296
4.3178131.7709 2.1604 2.6503 3.0123
3.37253.85204.22083.3257
3.78744.140514
1.7613 2.1448 2.6245 2.9768
15
1.7531 2.1314 2.6025 2.9467
3.28603.73284.0728161.7459 2.1199 2.5835 2.9208
3.25203.68624.01503.22243.64583.965117
1.7396 2.1098 2.5669 2.8982
18
1.7341 2.1009 2.5524 2.8784
3.19663.61053.921619
1.7291 2.0930 2.5395 2.8609
3.1737
3.57943.8834
3.15343.55183.849520
1.7247 2.0860 2.5280 2.8453
21
1.7207 2.0796 2.5176 2.8314
3.1352
3.52723.819322
1.7171 2.0739 2.5083 2.8188
3.11883.50503.792123
1.7139 2.0687 2.4999 2.8073
3.10403.48503.767624
1.7109 2.0639 2.4922 2.7969
3.09053.46683.745425
1.7081 2.0595 2.4851 2.7874
3.07823.45023.725126
1.7056 2.0555 2.4786 2.7787
3.06693.43503.706627
1.7033 2.0518 2.4727 2.7707
3.05653.42103.689628
1.7011 2.0484 2.4671 2.7633
3.04693.4082
3.67393.03803.39623.659429
1.6991 2.0452 2.4620 2.7564
30
1.6973 2.0423 2.4573 2.7500
3.02983.38523.646031
1.6955 2.0395 2.4528 2.7440
3.02213.37493.6335
1.6939 2.0369 2.4487 2.7385
3.01493.36533.621832
33
1.6924 2.0345 2.4448 2.7333
3.00823.35633.610934
1.6909 2.0322 2.4411 2.7284
3.00203.34793.60071.6896 2.0301 2.4377 2.7238
2.99603.34003.591135

B1

The critical values of t distribution are calculated according to the probabilities of
two alpha values and the degrees of freedom. The Alpha(α) values 0.05 one tailed
and 0.1 two tailed are the two columns to be compared with the degrees of
freedom in the row of the table.
0.01
0.005 0.0025
0.001
0.0005α (1 tail)0.05 0.025
α (2 tail)
0.1
0.05
0.02
0.01
0.005
0.002
0.001
36
1.6883 2.0281 2.4345 2.7195
2.99053.3326
3.58212.98523.3256
3.573737
1.6871 2.0262 2.4314 2.7154
38
1.6860 2.0244 2.4286 2.7116
2.98033.31903.565739
1.6849 2.0227 2.4258 2.7079
2.9756
3.31283.55812.97123.30693.551040
1.6839 2.0211 2.4233 2.7045
41
1.6829 2.0195 2.4208 2.7012
2.96703.30133.544242
1.6820 2.0181 2.4185 2.6981
2.96303.29603.53772.95923.29093.531643
1.6811 2.0167 2.4163 2.6951
44
1.6802 2.0154 2.4141 2.6923
2.95553.2861
3.525845
1.6794 2.0141 2.4121 2.6896
2.95213.28153.52033.27713.515046
1.6787 2.0129 2.4102 2.6870
2.948847
1.6779 2.0117 2.4083 2.6846
2.94563.27293.509948
1.6772 2.0106 2.4066 2.6822
2.9426
3.26893.50512.93973.26513.500449
1.6766 2.0096 2.4049 2.6800
50
1.6759 2.0086 2.4033 2.6778
2.93703.26143.496051
1.6753 2.0076 2.4017 2.6757
2.93433.25793.49182.93183.25453.487752
1.6747 2.0066 2.4002 2.6737
53
1.6741 2.0057 2.3988 2.6718
2.92933.25133.483854
1.6736 2.0049 2.3974 2.6700
2.92703.24813.48002.92473.24513.476455
1.6730 2.0040 2.3961 2.6682
56
1.6725 2.0032 2.3948 2.6665
2.92253.24233.472957
1.6720 2.0025 2.3936 2.6649
2.9204
3.23953.469658
1.6716 2.0017 2.3924 2.6633
2.9184
3.23683.466359
1.6711 2.0010 2.3912 2.6618
2.9164
3.2342
3.463260
1.6706 2.0003 2.3901 2.6603
2.91463.23173.460261
1.6702 1.9996 2.3890 2.6589
2.91273.22933.457362
1.6698 1.9990 2.3880 2.6575
2.91103.22703.454563
1.6694 1.9983 2.3870 2.6561
2.90933.22473.451864
1.6690 1.9977 2.3860 2.6549
2.9076
3.22253.449165
1.6686 1.9971 2.3851 2.6536
2.90603.22043.446666
1.6683 1.9966 2.3842 2.6524
2.9045
3.21843.444167
1.6679 1.9960 2.3833 2.6512
2.90303.21643.441768
1.6676 1.9955 2.3824 2.6501
2.9015
3.21453.439469
1.6672 1.9949 2.3816 2.6490
2.90013.21263.437270
1.6669 1.9944 2.3808 2.6479
2.89873.21083.4350

B2

The critical values of t distribution are calculated according to the probabilities of
two alpha values and the degrees of freedom. The Alpha(α) values 0.05 one tailed
and 0.1 two tailed are the two columns to be compared with the degrees of
freedom in the row of the table.
0.01
0.005 0.0025
0.001
0.0005α (1 tail)0.05 0.025
α (2 tail)
0.1
0.05
0.02
0.01
0.005
0.002
0.001
71
1.6666 1.9939 2.3800 2.6469
2.89743.20903.43292.89613.20733.430872
1.6663 1.9935 2.3793 2.6459
73
1.6660 1.9930 2.3785 2.6449
2.89493.20573.428974
1.6657 1.9925 2.3778 2.6439
2.8936
3.20413.42692.89243.20253.425075
1.6654 1.9921 2.3771 2.6430
76
1.6652 1.9917 2.3764 2.6421
2.89133.20103.423277
1.6649 1.9913 2.3758 2.6412
2.8902
3.19953.42142.8891
3.19803.419778
1.6646 1.9908 2.3751 2.6403
79
1.6644 1.9905 2.3745 2.6395
2.88803.1966
3.418080
1.6641 1.9901 2.3739 2.6387
2.88703.19533.41633.19393.414781
1.6639 1.9897 2.3733 2.6379
2.886082
1.6636 1.9893 2.3727 2.6371
2.88503.1926
3.413283
1.6634 1.9890 2.3721 2.6364
2.88403.19133.41162.88313.19013.410284
1.6632 1.9886 2.3716 2.6356
85
1.6630 1.9883 2.3710 2.6349
2.8822
3.18893.408786
1.6628 1.9879 2.3705 2.6342
2.88133.18773.40732.88043.18663.405987
1.6626 1.9876 2.3700 2.6335
88
1.6624 1.9873 2.3695 2.6329
2.87953.18543.404589
1.6622 1.9870 2.3690 2.6322
2.8787
3.18433.40322.87793.18333.401990
1.6620 1.9867 2.3685 2.6316
91
1.6618 1.9864 2.3680 2.6309
2.8771
3.18223.400792
1.6616 1.9861 2.3676 2.6303
2.87633.1812
3.399493
1.6614 1.9858 2.3671 2.6297
2.8755
3.1802
3.398294
1.6612 1.9855 2.3667 2.6291
2.87483.1792
3.397195
1.6611 1.9853 2.3662 2.6286
2.8741
3.1782
3.395996
1.6609 1.9850 2.3658 2.6280
2.8734
3.17733.394897
1.6607 1.9847 2.3654 2.6275
2.87273.1764
3.393798
1.6606 1.9845 2.3650 2.6269
2.87203.17553.392699
1.6604 1.9842 2.3646 2.6264
2.87133.1746
3.3915100
1.6602 1.9840 2.3642 2.6259
2.87073.17373.3905

B3

The LEFT critical values of chi-distribution are calculated according to theprobabilities of one-tail alpha value and the degrees of freedom.
α (1 tail)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100

0.05
0.00390.1026
0.35180.71071.14551.63542.16732.7326
3.32513.9403
4.57485.22605.89196.57067.26097.9616
8.67189.390510.1170
10.8508
11.591312.3380
13.0905
13.8484
14.611415.3792
16.151416.9279
17.7084
18.492722.4650
26.509330.612334.764338.9580
43.1880
47.4496
51.739356.0541
60.391564.749469.1260
73.5198
77.9295

0.025
0.00100.05060.21580.48440.8312
1.2373
1.68992.17972.7004
3.24703.81574.40385.00885.62876.2621
6.90777.56428.23078.90659.590810.2829
10.9823
11.6886
12.4012
13.1197
13.8439
14.5734
15.3079
16.0471
16.7908
20.569424.4330
28.366232.3574
36.398140.481744.603048.7576
52.941957.1532
61.3888
65.6466
69.9249
74.22190.01
0.00020.02010.11480.29710.55430.8721
1.23901.64652.08792.5582
3.05353.57064.10694.66045.22935.8122
6.40787.01497.63278.26048.89729.542510.1957
10.8564
11.5240
12.1981
12.8785
13.5647
14.2565
14.9535
18.508922.164325.901329.7067
33.5705
37.4849
41.443645.441749.4750
53.5401
57.633961.754165.8984
70.0649
C1

0.005 0.0025
0.0000 0.0000
0.0100 0.0050
0.0717 0.0449
0.2070 0.1449
0.4117 0.3075
0.6757 0.5266
0.9893 0.7945
1.3444 1.1043
1.7349 1.4501
2.1559 1.8274
2.6032 2.2321
3.0738 2.6612
3.5650 3.1119
4.0747 3.5820
4.6009 4.0697
5.1422 4.5734
5.6972 5.0917
6.2648 5.6233
6.8440 6.1674
7.4338 6.7228
8.0337 7.2889
8.6427 7.8649
9.2604 8.4502
9.8862 9.0442
10.5197 9.6463
11.1602 10.2562
11.8076 10.8733
12.4613 11.4973
13.1211 12.1279
13.7867 12.7646
17.1918 16.0317
20.7065 19.4171
24.3110 22.8996
27.9907 26.4636
31.7348 30.0971
35.5345 33.7911
39.3831 37.5382
43.2752 41.3323
47.2060 45.1686
51.1719 49.0429
55.1696 52.9517
59.1963 56.8921
63.2496 60.8614
67.3276 64.8574

0.001 0.0005
0.0000 0.0000
0.0020 0.0010
0.0243 0.0153
0.0908 0.0639
0.2102 0.1581
0.3811 0.2994
0.5985 0.4849
0.8571 0.7104
1.1519 0.9717
1.4787 1.2650
1.8339 1.5868
2.2142 1.9344
2.6172 2.3051
3.0407 2.6967
3.4827 3.1075
3.9416 3.5358
4.4161 3.9802
4.9048 4.4394
5.4068 4.9123
5.9210 5.3981
6.4467 5.8957
6.9830 6.4045
7.5292 6.9237
8.0849 7.4527
8.6493 7.9910
9.2221 8.5379
9.8028 9.0932
10.3909 9.6563
10.9861 10.2268
11.5880 10.8044
14.6878 13.7875
17.9164 16.9062
21.2507 20.1366
24.6739 23.4610
28.1731 26.8658
31.7383 30.3405
35.3616 33.8767
39.0364 37.4674
42.7573 41.1072
46.5199 44.7910
50.3203 48.5151
54.1552 52.2758
58.0220 56.0702
61.9179 59.8957

The RIGHT critical values of chi-distribution are calculated according to the probabilities
of one-tail alpha value and the degrees of freedom.
α (1 tail)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100

0.05
0.025
0.01
0.005
0.00250.001
0.00053.84155.02396.63497.87949.1406 10.8276 12.1157
5.99157.37789.2103 10.5966 11.9829 13.8155 15.2018
7.81479.3484 11.3449 12.8382 14.3203 16.2662 17.7300
9.4877 11.1433 13.2767 14.8603 16.4239 18.4668 19.9974
11.0705 12.8325 15.0863 16.7496 18.3856 20.5150 22.1053
12.5916 14.4494 16.8119 18.5476 20.2494 22.4577 24.1028
14.0671 16.0128 18.4753 20.2777 22.0404 24.3219 26.0178
15.5073 17.5345 20.0902 21.9550 23.7745 26.1245 27.8680
16.9190 19.0228 21.6660 23.5894 25.4625 27.8772 29.6658
18.3070 20.4832 23.2093 25.1882 27.1122 29.5883 31.4198
19.6751 21.9200 24.7250 26.7568 28.7293 31.2641 33.1366
21.0261 23.3367 26.2170 28.2995 30.3185 32.9095 34.8213
22.3620 24.7356 27.6882 29.8195 31.8831 34.5282 36.4778
23.6848 26.1189 29.1412 31.3193 33.4260 36.1233 38.1094
24.9958 27.4884 30.5779 32.8013 34.9496 37.6973 39.7188
26.2962 28.8454 31.9999 34.2672 36.4557 39.2524 41.3081
27.5871 30.1910 33.4087 35.7185 37.9461 40.7902 42.8792
28.8693 31.5264 34.8053 37.1565 39.4221 42.3124 44.4338
30.1435 32.8523 36.1909 38.5823 40.8850 43.8202 45.9731
31.4104 34.1696 37.5662 39.9968 42.3357 45.3147 47.4985
32.6706 35.4789 38.9322 41.4011 43.7751 46.7970 49.0108
33.9244 36.7807 40.2894 42.7957 45.2041 48.2679 50.5111
35.1725 38.0756 41.6384 44.1813 46.6235 49.7282 52.0002
36.4150 39.3641 42.9798 45.5585 48.0337 51.1786 53.4788
37.6525 40.6465 44.3141 46.9279 49.4354 52.6197 54.9475
38.8851 41.9232 45.6417 48.2899 50.8291 54.0520 56.4069
40.1133 43.1945 46.9629 49.6449 52.2153 55.4760 57.8576
41.3371 44.4608 48.2782 50.9934 53.5943 56.8923 59.3000
42.5570 45.7223 49.5879 52.3356 54.9666 58.3012 60.7346
43.7730 46.9792 50.8922 53.6720 56.3325 59.7031 62.1619
49.8018 53.2033 57.3421 60.2748 63.0764 66.6188 69.1986
55.7585 59.3417 63.6907 66.7660 69.6991 73.4020 76.0946
61.6562 65.4102 69.9568 73.1661 76.2229 80.0767 82.8757
67.5048 71.4202 76.1539 79.4900 82.6640 86.6608 89.5605
73.3115 77.3805 82.2921 85.7490 89.0346 93.1675 96.1632
79.0819 83.2977 88.3794 91.9517 95.3440 99.6072 102.6948
84.8206 89.1771 94.4221 98.1051 101.5999 105.9881 109.1639
90.5312 95.0232 100.4252 104.2149 107.8082 112.3169 115.5776
96.2167 100.8393 106.3929 110.2856 113.9741 118.5991 121.9418
101.8795 106.6286 112.3288 116.3211 120.1017 124.8392 128.2613
107.5217 112.3934 118.2357 122.3246 126.1945 131.0412 134.5403
113.1453 118.1359 124.1163 128.2989 132.2556 137.2084 140.7823
118.7516 123.8580 129.9727 134.2465 138.2876 143.3435 146.9903
124.3421 129.5612 135.8067 140.1695 144.2928 149.4493 153.1670
C2

The critical values of r distribution are calculated according to the probabilities of
two alpha values and the degrees of freedom. The Alpha(α) values 0.05 one tailed
and 0.1 two tailed are the two columns to be compared with the degrees of
freedom in the row of the table.
0.005 0.0025
0.01
α (1 tail)0.05 0.025
α (2 tail)
0.1
0.05
0.02
0.01
0.005
1
0.9877 0.9969 0.9995 0.9999
1.00000.995020.9000 0.9500 0.9800 0.9900
30.8054 0.8783 0.9343 0.9587
0.97404
0.7293 0.8114 0.8822 0.9172
0.94170.9056
5
0.6694 0.7545 0.8329 0.8745
6
0.6215 0.7067 0.7887 0.8343
0.8697
70.5822 0.6664 0.7498 0.7977
0.83590.80468
0.5494 0.6319 0.7155 0.7646
90.5214 0.6021 0.6851 0.7348
0.7759100.4973 0.5760 0.6581 0.7079
0.74960.725511
0.4762 0.5529 0.6339 0.6835
12
0.4575 0.5324 0.6120 0.6614
0.7034130.4409 0.5140 0.5923 0.6411
0.68310.6643140.4259 0.4973 0.5742 0.6226
15
0.4124 0.4821 0.5577 0.6055
0.6470160.4000 0.4683 0.5425 0.5897
0.63080.6158170.3887 0.4555 0.5285 0.5751
18
0.3783 0.4438 0.5155 0.5614
0.6018190.3687 0.4329 0.5034 0.5487
0.5886
0.576320
0.3598 0.4227 0.4921 0.5368
21
0.3515 0.4132 0.4815 0.5256
0.5647220.3438 0.4044 0.4716 0.5151
0.553723
0.3365 0.3961 0.4622 0.5052
0.543424
0.3297 0.3882 0.4534 0.4958
0.533625
0.3233 0.3809 0.4451 0.4869
0.524326
0.3172 0.3739 0.4372 0.4785
0.515427
0.3115 0.3673 0.4297 0.4705
0.507028
0.3061 0.3610 0.4226 0.4629
0.49900.491429
0.3009 0.3550 0.4158 0.4556
30
0.2960 0.3494 0.4093 0.4487
0.484031
0.2913 0.3440 0.4032 0.4421
0.47700.2869 0.3388 0.3972 0.4357
0.470332
33
0.2826 0.3338 0.3916 0.4296
0.463934
0.2785 0.3291 0.3862 0.4238
0.45770.2746 0.3246 0.3810 0.4182
0.451835

D1

The critical values of r distribution are calculated according to the probabilities of
two alpha values and the degrees of freedom. The Alpha(α) values 0.05 one tailed
and 0.1 two tailed are the two columns to be compared with the degrees of
freedom in the row of the table.
0.01
0.005 0.0025
α (1 tail)0.05 0.025
α (2 tail)
0.1
0.05
0.02
0.01
0.005
36
0.2709 0.3202 0.3760 0.4128
0.44610.440637
0.2673 0.3160 0.3712 0.4076
38
0.2638 0.3120 0.3665 0.4026
0.4353390.2605 0.3081 0.3621 0.3978
0.43010.425240
0.2573 0.3044 0.3578 0.3932
41
0.2542 0.3008 0.3536 0.3887
0.4204420.2512 0.2973 0.3496 0.3843
0.41580.411343
0.2483 0.2940 0.3457 0.3801
44
0.2455 0.2907 0.3420 0.3761
0.407045
0.2429 0.2876 0.3384 0.3721
0.40280.3987460.2403 0.2845 0.3348 0.3683
47
0.2377 0.2816 0.3314 0.3646
0.394848
0.2353 0.2787 0.3281 0.3610
0.39090.3872490.2329 0.2759 0.3249 0.3575
50
0.2306 0.2732 0.3218 0.3542
0.383651
0.2284 0.2706 0.3188 0.3509
0.38010.376652
0.2262 0.2681 0.3158 0.3477
53
0.2241 0.2656 0.3129 0.3445
0.373354
0.2221 0.2632 0.3102 0.3415
0.37000.366955
0.2201 0.2609 0.3074 0.3385
56
0.2181 0.2586 0.3048 0.3357
0.363857
0.2162 0.2564 0.3022 0.3328
0.36080.357858
0.2144 0.2542 0.2997 0.3301
59
0.2126 0.2521 0.2972 0.3274
0.3550600.2108 0.2500 0.2948 0.3248
0.35220.349461
0.2091 0.2480 0.2925 0.3223
62
0.2075 0.2461 0.2902 0.3198
0.346863
0.2058 0.2441 0.2880 0.3173
0.3441640.2042 0.2423 0.2858 0.3150
0.341665
0.2027 0.2404 0.2837 0.3126
0.339166
0.2012 0.2387 0.2816 0.3104
0.3366
0.334367
0.1997 0.2369 0.2796 0.3081
68
0.1982 0.2352 0.2776 0.3060
0.331969
0.1968 0.2335 0.2756 0.3038
0.329670
0.1954 0.2319 0.2737 0.3017
0.3274D2

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