# The Normal Distribution

## Using the Normal Distribution

The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x).
The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.
Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.

### Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

### References

Chicago Public Media & Ira Glass. (2013). 403: NUMMI. Retrieved from http://www.thisamericanlife.org/radioarchives/episode/403/nummi
lauramitchell347. (2012, Dec. 28). Smart phone users, by the numbers. Visually. Retrieved from http://visual.ly/smart-phone-users-numbers
Statistics Brain Research Institute. (2013). Facebook company statistics  statistic brain. Retrieved from http://www.statisticbrain.com/facebook-statistics/
Wikipedia (2013). Naegele's rule. Retrieved from http://en.wikipedia.org/wiki/Naegele's_rule
Win at the Lottery. (2013). Scratch-off lottery ticket playing tips. Retrieved  from www.winatthelottery.com/public/department40.cfm

### Chapter Review

The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one.  The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution, is the distribution of z-scores. Its mean is zero, and its standard deviation is one.

### Formula Review

Normal Distribution: X ~ N(µ, σ), where µ is the mean and σ is the standard deviation
Standard Normal Distribution: Z ~ N(0, 1).
Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)
Calculator function for the kth percentile: k =  invNorm (area to the left of k, mean, standard deviation)
Use the following information to answer the next four exercises:
X ~ N(54, 8)

### Homework

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes.


# The Central Limit Theorem

## Introduction

Why are we so concerned with means? Two reasons are they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem.
The central limit theorem (clt) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing a finite samples size n from a population with a known mean, μ, and a known standard deviation, σ. The first alternative says that if we collect samples of size n with a large enough n, calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are large enough, calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell shape. The central limit theorem for sample means is more discussed in the world of statistics, but it is important to note that taking each sample's sum and graphing the sums will also result in a normal histogram. There are instances where one wishes to calculate the sum of a sample, as opposed to its mean.
In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distributions 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 the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement.


# The Central Limit Theorem

## The Central Limit Theorem for Sample Means (Averages)

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose
1.

 = the mean of X
2.

 = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable
, which consists of sample means, tends to be normally distributed and
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done.
To put it more formally, if you draw random samples of size n, the distribution of the random variable
, which consists of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases.
The random variable
 has a different z-score associated with it from that of the random variable X. The mean
 is the value of
 in one sample.
μ is the average of both X and
.

 = standard deviation of
 and is called the standard error of the mean.

### References

Baran, D. (2010). 20 percent of Americans have never used email. WebGuild. Retrieved from http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email
The Flurry Blog. (2013). Retrieved from http://blog.flurry.com
U.S. Department of Agriculture. (n.d.). Retrieved from https://www.usda.gov/

### Chapter Review

In a population whose distribution may be known or unknown, if the size (n) of the sample is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n).

### Formula Review

Central limit theorem for sample means:

 ~ N

Mean
 : μ
Central limit theorem for sample means z-score and standard error of the mean:

Standard error of the mean (standard deviation (
 )):

Use the following information to answer the next six exercises: Yoonie 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 random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let
 be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

### Homework


# The Central Limit Theorem

## The Central Limit Theorem for Sums (Optional)

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
1. μ = the mean of Χ
2. σ = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N[(n)(μ), (
 )(σ)].
The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
The random variable ΣX has the following z-score associated with it:
1. Σx is one sum.
2.

### References

Farago, P.  (2012, Oct. 29). The truth about cats and dogs: Smartphone vs tablet usage differences. Flurry Analytics Blog. Retrieved from http://flurrymobile.tumblr.com/post/113379683050/the-truth-about-cats-and-dogs-smartphone-vs

### Chapter Review

The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution, even if the original population is not normally distributed. Additionally, if the original population has a mean of μ and a standard deviation of σ, the mean of the sums is nμ and the standard deviation is
 (σ), where n is the sample size.

### Formula Review

Central limit theorem for sums: ∑X ~ N[(n)(μ),(
 )(σ)]
Mean for sums (∑X): (n)(μ)
Central limit theorem for sums z-score and standard deviation for sums:

Standard deviation for sums (∑X):
 (σ)
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.
Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.
Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.
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 of interest.
 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.
 An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest.

### Homework


# The Central Limit Theorem

## Using the Central Limit Theorem

It is important for you 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 means. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

### Examples of the Central Limit Theorem

### Law of Large Numbers

The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean
  of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for
  is
 .) This means that the sample mean
  must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.

### Central Limit Theorem for the Mean and Sum Examples

### References

The Wall Street Journal. (n.d.). Retrieved from https://www.wsj.com/
Centers for Disease Control and Prevention. (2017, April 16). National health and nutrition examination survey. National Center for Health Statistics.  Retrieved from http://www.cdc.gov/nchs/nhanes.htm

### Chapter Review

The central limit theorem can be used to illustrate the law of large numbers. The law of large numbers states that the larger the sample size you take from a population, the closer the sample mean,
 , gets to μ.
Use the following information to answer the next 10 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 taken.
 The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.
 A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.

### Homework

 Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long (in hours) past the 10 a.m. start time that individuals wait for their delivery.
Use the following information to answer the next two exercises: 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.
 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.
For each problem, wherever possible, provide graphs and use a calculator.


# Confidence Intervals

## Introduction

Suppose you were trying to determine the mean rent of a two-bedroom apartment in your town. 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 true mean. If you are trying to determine the percentage of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempt. In this case, you would have obtained a point estimate for the true proportion.
We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics.  The sample data help us to make an estimate of a population . We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct interval estimates, called confidence intervals.
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Student's-t, and how it is used with those intervals. Throughout the chapter, it is important to keep in mind that the confidence interval is a random variable. It is the population parameter that is fixed.
If you worked in the marketing department of an entertainment company, you might be interested in the mean number of songs a consumer downloads a month from an internet music store. If so, you could conduct a survey and calculate the sample mean,
  and the sample standard deviation, s. You would use
  to estimate the population mean and s to estimate the population standard deviation. The sample mean,
 , is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ.
Each instance of
  and s is called a statistic.
A confidence interval is another type of estimate but, instead of being just one number, it is an 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.
Suppose, for the internet music example, we do not know the population mean, μ, but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then, by the central limit theorem, the standard deviation for the sample mean is
The Empirical Rule, which applies to bell-shaped distributions, says that in approximately 95 percent of the samples, the sample mean,
 , will be within two standard deviations of the population mean, μ. For our internet music example, two standard deviations would be calculated as (2)(0.1) = 0.2. The sample mean,
  is likely to be within 0.2 units of μ.
In this example, we do not know the true population mean μ (because we do not have information from all the internet music users!), but we can compute the sample mean

   based on our sample of 100 individuals. Because the sample mean is likely to be within 0.2 units of the true population mean 95 percent of the times that we take a sample of 100 users, we can say with 95 percent confidence that μ is within 0.2 units of

  . In other words, μ is somewhere between

   and

  .
Suppose that from the sample of 100 internet music customers, we compute a sample mean download of

   songs per month. Since we know that the population standard deviation is

  , according to the central limit theorem, the standard deviation for the sample means is

  .
We know that there is a 95 percent chance that the true population mean value μ is between two standard deviations from the sample mean. That is, with 95 percent confidence we can say that μ is between

     and

  .
Replacing the symbols for their values in this example, we say that we are 95 percent confident that the true average number of songs downloaded from an internet music store per month is between
The 95 percent confidence interval for μ is (1.8, 2.2).
The 95 percent confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean, μ, or our sample produced an
  that is not within 0.2 units of the true mean μ. The second possibility happens for only 5 percent of all the samples (95–100 percent).
Remember that a confidence interval is created for an unknown population parameter like the population mean, μ. Confidence intervals for some parameters have the form
(point estimate – margin of error, point estimate +  margin of error).
The margin of error depends on the confidence level or percentage of confidence and the standard error of the mean.
When you read newspapers and journals, you might notice that some reports use the phrase margin of error. Other reports will not use that phrase, but include a confidence interval as the point estimate plus or minus the margin of error. Those are two ways of expressing the same concept.


# Confidence Intervals

## A Single Population Mean Using the Normal Distribution

A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of
      and we have constructed the 90 percent confidence interval (5, 15), where the margin of error = 5.

### Calculating the Confidence Interval

To construct a confidence interval for a single unknown population mean, μ, where the population standard deviation is known, we need
  as an estimate for μ, and we need the margin of error. The margin of error for the population mean is called the error bound for a population mean (EBM). The sample mean,
 is the point estimate of the unknown population mean, μ.
The confidence interval (CI) estimate will have the form:
(point estimate – error bound, point estimate + error bound) or, in symbols, (
 ).
The margin of error (EBM) depends on the confidence level (). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percentage of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, the person constructing the confidence interval will choose a confidence level of 90 percent or higher, because that person wants to be reasonably certain of his or her conclusions.
Another probability, which is called alpha

   is related to the confidence level, CL. Alpha is the probability that the confidence interval does not contain the unknown population parameter. Mathematically, alpha can be computed as

  .
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of
  = 10, and we have constructed the 90 percent confidence interval (5, 15) where EBM = 5.
To get a 90 percent confidence interval, we must include the central 90 percent of the probability of the normal distribution. If we include the central 90 percent, we leave out a total of α = 10 percent in both tails, or 5 percent in each tail, of the normal distribution.
The critical value 1.645 is the  in a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail. To capture the central 90 percent, we must go out 1.645 standard deviations on either side of the calculated sample mean. The critical value will change depending on the confidence level of the interval.
It is important that the standard deviation used be appropriate for the parameter we are estimating, so in this section, we need to use the standard deviation that applies to sample means, which is
     . The fraction
      is commonly called the standard error of the mean in order to distinguish clearly the standard deviation for a mean from the population standard deviation, σ.
2.
      is normally distributed, that is,
      ~ N
     .
3. When the population standard deviation

### Calculating the Confidence Interval

To construct a confidence interval estimate for an unknown population mean, we need data from a random sample.  The steps to construct and interpret the confidence interval are as follows:
1.  Calculate the sample mean,
      from the sample data. Remember, in this section, we already know the population standard deviation, σ.
2. Find the z-score that corresponds to the confidence level.
3. Calculate the error bound EBM.
4. Construct the confidence interval.
5. If we denote the critical z-score by
, and the sample size by n, then the formula for the confidence interval with confidence level

  , is given by

6. 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.)
We will first examine each step in more detail and then illustrate the process with some examples.

### Finding the z-Score for the Stated Confidence Level

When we know the population standard deviation, σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1).
The confidence level, CL, is the area in the middle of the standard normal distribution. CL = 1 – α, so α is the area that is split equally between the two tails. Each of the tails contains an area equal to
     .
The z-score that has an area to the right of
      is denoted by
     .
For example, when CL = 0.95, α = 0.05, and
      = 0.025, we write
      = 0.025.
The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975.

 , using a calculator, computer, or standard normal probability table.
Normal table (see appendices) shows that the probability for 0 to 1.96 is 0.47500, and so the probability to the right tail of the critical value 1.96 is 0.5 – 0.475 = 0.025

### Calculating the Margin of Error EBM

The error bound formula for an unknown population mean, μ, when the population standard deviation, σ, is known is
Margin of error =

### Constructing the Confidence Interval

The confidence interval estimate has the format
    sample mean plus or minus the margin of error.
The graph gives a picture of the entire situation
CL +
      +
      = CL + α = 1.

### Writing the Interpretation

The interpretation should clearly state the confidence level (CL), explain which population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints): "We estimate with ___percent confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)."
Notice the difference in the confidence intervals calculated in  and the following Try It exercise. These intervals are different for several reasons: they are calculated from different samples, the samples are different sizes, and the intervals are calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.

### Changing the Confidence Level or Sample Size

### Working Backward to Find the Error Bound or Sample Mean

When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backward to find both the error bound and the sample mean.
2. From the upper value for the interval, subtract the sample mean,
3. Or, from the upper value for the interval, subtract the lower value.  Then divide the difference by 2.
2. Subtract the error bound from the upper value of the confidence interval,
3. Or, average the upper and lower endpoints of the confidence interval.
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.

### Calculating the Sample Size n

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. In this situation, we are given the desired margin of error, EBM, and we need to compute the sample size n.
The formula for sample size is n =
     , found by solving the error bound formula for n. Always round up the value of n to the closest integer.
In this formula, z is the critical value
     , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.

### References

Centers for Disease Control and Prevention. (n.d.). National health and nutrition examination survey. Retrieved from http://www.cdc.gov/nchs/nhanes.htm
Foothill De Anza Community College District. (n.d.). Headcount enrollment trends by student demographics ten-year fall trends to most recently completed fall. Retrieved from http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm
Kuczmarski, R. J., et al. (2002). 2000 CDC growth charts for the United States: Methods and development. Retrieved from http://www.cdc.gov/growthcharts/2000growthchart-us.pdf
La, L., and; German, K. (n.d.). Cell phones with the highest radiation levels. CNET. Retrieved from http://reviews.cnet.com/cell-phone-radiation-levels/
U.S. Census Bureau. (n.d.). American FactFinder. Retrieved from http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t
U.S. Census Bureau. (2011). Mean income in the past 12 months (in 2011 inflation-adjusted dollars): 2011 American Community Survey 1-year estimates. Retrieved from http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table
U.S. Federal Election Commission. (n.d.). Disclosure data catalog: Candidate summary report 2012. Retrieved from http://www.fec.gov/data/CandidateSummary.do?format=html&election_yr=2012
U.S. Federal Election Commission. (n.d.). Metadata description of candidate summary file. Retrieved from http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml

### Chapter Review

In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean (EBM). A confidence interval has the general form
(lower bound, upper bound) = (point estimate – EBM, point estimate + EBM).
The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percentage of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem,

Given a confidence interval, you can work backward to find the error bound (EBM) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half of the difference of the upper and lower bounds.  To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean.
Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal:

### Formula Review

  The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.
The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
    (lower bound, upper bound) = (point estimate – EBM, point estimate + EBM)

    EBM =
      = the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.
CL = confidence level, or the proportion of confidence intervals created that is expected to contain the true population parameter
α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter

      = the z-score with the property that the area to the right of the z-score is
     ;	 this is the z-score, used in the calculation of EBM, where α = 1 – CL.
n =
      = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence
General form of a confidence interval
(lower value, upper value) = (point estimate error bound, point estimate + error bound)
To find the error bound when you know the confidence interval,
error bound = upper value point estimate or error bound =

Single population mean, known standard deviation, normal distribution
Use the normal distribution for means; population standard deviation is known: EBM = z

The confidence interval has the format (
      − EBM,
      + EBM).
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb.
Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
Use the following information to answer the next 14 exercises: 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 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student.
Construct a 95 percent confidence interval for the true mean age of winter Foothill College students  by working out and then answering the next eight exercises.

### Homework


# Confidence Intervals

## A Single Population Mean Using the Student's t-Distribution

In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this unknown number 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 this  problem. His experiments with hops and barley produced very few samples. Just replacing σ with 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 the actual distribution depends on the sample size. This problem led him to discover what is called the Student's . 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 for large 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 t-distribution whenever s is used as an estimate for σ.
If you draw a simple random sample of size n from a population that has an approximately normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t =
 , then the t-scores follow a Student's . The t-score has the same interpretation as the : It measures how far
  is from its mean μ. For each sample size n, there is a different Student's t-distribution.
The degrees of freedom (, n - 1, are the sample size minus 1.
2. The graph for the Student's t-distribution is similar to the standard normal curve.
3. The mean for the Student's t-distribution is zero, and the distribution is symmetric about zero.
4. The Student's t-distribution has more probability in its tails than the standard normal distribution.  shows the graphs of the student t-distribution for 1, 2 and 5 degrees of freedom: (v), compare to the standard normal distribution (in black).

5. The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increase, the graph of the Student's t-distribution becomes more like the graph of the standard normal distribution.
6. The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ. The size of the underlying population is generally not relevant unless it is very small. If it is bell-shaped (normal), then the assumption is met and does not need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.
Calculators and computers can easily calculate any Student's t-probabilities. The TI-83, 83+, and 84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However, for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
For the TI-84+, you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom). The output is the t-score that corresponds to the area we specified.
 The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.)
A probability table for the Student's t-distribution can also be used.  The table gives critical t-values that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-distribution.) When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.

 A Student's t-table (see ) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's  t-probabilities.
If the population standard deviation is  not known,  the error bound for a population mean is
1.
 ,
2.
  is the t-score with area to the right equal to
 ,
3. use df = n – 1 degrees of freedom, and
4. s = sample standard deviation.
The format for the confidence interval is

### References

Bloomberg Businessweek. (n.d.). Retrieved from http://www.businessweek.com/
Environmental Working Group. (n.d.). Human toxome project: Mapping the pollution in people. Retrieved from http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn
Federal Election Commission. (n.d.). Disclosure data catalog: 2012 leadership PACs and sponsors. Retrieved from http://www.fec.gov/data/index.jsp
Federal Election Commission. (n.d.). Metadata description of leadership PAC list. Retrieved from http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml
Forbes. (2013). Americas best small companies. Retrieved from http://www.forbes.com/best-small-companies/list/
Forbes. (n.d.). Retrieved from http://www.forbes.com/
Microsoft Bookshelf. (n.d.).

### Chapter Review

In many cases, the researcher does not know the population standard deviation, σ, of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of σ. The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
The t-score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM =
  where
  is the t-score with area to the right equal to
 , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find
  for a given α.

### Formula Review

s = the standard deviation of sample values

  is the formula for the t-score, which measures how far away a measure is from the population mean in the Student’s t-distribution.
df = n – 1; the degrees of freedom for a Student’s t-distribution, where n represents the size of the sample
T~t the random variable, T, has a Student’s t-distribution with df degrees of freedom

  = the error bound for the population mean when the population standard deviation is unknown

  is the t-score in the Student’s t-distribution with area to the right equal to

The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by  (lower bound, upper bound)
= (point estimate – EBM, point estimate + EBM)
Use the following information to answer the next five exercises: A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hr, with a sample standard deviation of 0.5 hr.
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Use the following information to answer the next 13 exercises: The data in  are the result of a random survey of 39 national flags (with replacement 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.
Construct a 95 percent confidence interval for the true mean number of colors on national flags.

### Homework

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16-oz serving size. The sample mean is 13.30, with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.


# Confidence Intervals

## A Population Proportion

During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40 percent of the vote within 3 percentage points  (if the sample is large enough). Often, election polls are calculated with 95 percent confidence, so the pollsters would be 95 percent 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).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound for a population (, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.
How do you know you are dealing with a proportion problem? First, the data that you are collecting is categorical, consisting of two categories: Success or Failure, Yes or No. Examples of situations where you are the following trying to estimate the true population proportion are the following: What proportion of the population smoke? What proportion of the population will vote for candidate A? What proportion of the population has a college-level education?
The distribution of the sample proportions (based on samples of size n) is denoted by P′ (read “P prime”).
The central limit theorem for proportions asserts that the sample proportion distribution P′ follows a normal distribution with mean value p, and standard deviation

  , where p is the population proportion and q = 1 - p.
The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion.
p′ = the estimated proportion of successes (p′ is a point estimate for p, the true proportion.)
x = the number of successes
n = the size of the sample
The error bound for a proportion is

  where q′ = 1 – p′.
This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is
. For a proportion, the appropriate standard deviation is
.
However, in the error bound formula, we use

  as the standard deviation, instead of
.
In the error bound formula, the sample proportions . The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures.
The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five.
That is, in order to use the formula for confidence intervals for proportions, you need to verify that both

   and

  .

### 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 the sampling 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 these observations 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 the plus-four confidence interval for . It should be used when the confidence level desired is at least 90 percent and the sample size is at least ten.

### Calculating the Sample Size n

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.

The margin of error formula for a population proportion is

1.

  , where p′ is the sample proportion, q′ = 1 – p′, and n is the sample size.
2. Solving for n gives you an equation for the sample size.
3.

 . This formula tells us that we can compute the sample size n required for a confidence level of

   by taking the square of the critical value
, multiplying by the point estimate p′, and by q′ = 1 – p′ and finally dividing the result by the square of the margin of error. Always remember to round up the value of n.

### References

Jensen, T. (2013). Democrats, Republicans divided on opinion of music icons. Retrieved from http://www.publicpolicypolling.com/Day2MusicPoll.pdf
Madden, M., et al. (2013). Teens, social media, and privacy. Retrieved from http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx
Princeton Survey Research Associates International. (2012). 2012 teens and privacy management survey. Retrieved from http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf
Rasmussen Reports. (2013). 52% say big-time college athletics corrupt education process. Retrieved from http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process
Saad, L. (2013). Three in four U.S. workers plan to work past retirement age. Retrieved from http://www.gallup.com/poll/162758/three-fourworkers-plan-work-past-retirement-age.aspx
The Field Poll. (n.d.). Retrieved from http://field.com/fieldpollonline/subscribers/
Zogby Analytics. (2013). New SUNYIT/Zogby analytics poll: Few Americans worry about emergency situations occurring in their community; Only one in three have an emergency plan; 70% support infrastructure investment for national security. Retrieved from http://www.zogbyanalytics.com/news/299-americans-neither-worried-norprepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll

### Chapter Review

Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion by following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning.
Let p′ represent the sample proportion, x/n, where x represents the number of successes, and n represents the sample size. Let q′ = 1 – p′. Then the confidence interval for a population proportion is given by the following formula:
(lower bound, upper bound)

The plus–four method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate
 , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples.

### Formula Review

p′ = x/n, where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion.

  The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here,
Confidence interval for a proportion:

  provides the number of participants needed to estimate the population proportion with confidence 1 – α and margin of error EBP.
Use the normal distribution for a single population proportion

The confidence interval has the format (p′ – EBP, p′ + EBP).

  is a point estimate for μ.
p′ is a point estimate for ρ.
s is a point estimate for σ.
Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percentage of women who make the majority of household purchasing decisions.
Use the following information to answer the next five exercises: Suppose a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.
Use the following information to answer the next five exercises: 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 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks.

Use the following information to answer the next five exercises: 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 believe the economy is the most important.
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

### Homework

Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California.
Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not.


# Hypothesis Testing with One Sample

## Introduction

One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about a parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90 percent of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.
A statistician will make a decision about these claims. This process is called hypothesis testing. A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis.
In this chapter, you will conduct hypothesis tests on single means and single proportions.
    You will also learn about the errors associated with these tests.
Hypothesis testing consists of two contradictory hypotheses or statements, a decision based on the data, and a conclusion. To perform a hypothesis test, a statistician will do the following:


# Hypothesis Testing with One Sample

## Null and Alternative Hypotheses

The actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints.
H, the null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.
H, the alternative hypothesis: a claim about the population that is contradictory to H and what we conclude when we reject H.
Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.
After you have determined which hypothesis the sample supports, you make a decision.
There are two options for a decision. They are reject H if the sample information favors the alternative hypothesis or do not reject H or decline to reject H if the sample information is insufficient to reject the null hypothesis.
Mathematical Symbols Used in H and H:

### Chapter Review

In a hypothesis test, sample data are evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we do the following:

### Formula Review

H and H are contradictory.
If α ≤ p-value, then do not reject H.
If α > p-value, then reject H.
α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data.

### Homework

### References

Centers for Disease Control and Prevention. (n.d.). Physical activity facts. Retrieved from http://www.cdc.gov/healthyschools/physicalactivity/facts.htm
National Institute of Mental Health. (n.d.). Publications about depression. Retrieved from http://www.nimh.nih.gov/publicat/depression.cfm


# Hypothesis Testing with One Sample

## Outcomes and the Type I and Type II Errors

When you perform a hypothesis test, there are four possible outcomes depending on the actual truth, or falseness, of the null hypothesis H and the decision to reject or not. The outcomes are summarized in the following table:
The four possible outcomes in the table are as follows:

Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.
α = probability of a Type I error =  = probability of rejecting the null hypothesis when the null hypothesis is true.
β = probability of a Type II error =  = probability of not rejecting the null hypothesis when the null hypothesis is false.
α and β should be as small as possible because they are probabilities of errors. They are rarely zero.
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.

### Chapter Review

In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected.
The probabilities of these errors are denoted by the Greek letters α and β, for a Type I and a Type II error respectively. The power of the test, 1 – β, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable.

### Formula Review

α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true.
β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false.

### Homework


# Hypothesis Testing with One Sample

## Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's . (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large).

### Assumptions

When you perform a hypothesis test of a single population mean  using a
Student's  (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed.
When you perform a hypothesis test of a single population mean  using a
normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.
When you perform a hypothesis test of a single population proportion , you
  take a simple random sample from the population. You must meet the conditions for a binomial distribution, which are the following: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and
    .

Remember that q = 1 – p.

### Chapter Review

In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.
When testing for a single population mean:
1. A Student's t-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
2. The normal test will work if the data come from a simple, random sample and the population
      is approximately normally distributed, or the sample size is large, with a known standard
      deviation.
When testing a single population proportion use a normal test for a single population proportion if the data come from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure.

### Formula Review

If there is no given preconceived α, then use α = 0.05.
2. Single population mean, known population variance (or standard deviation): Normal
    test.
3. Single population mean, unknown population variance (or standard deviation): Student's .
4. Single population proportion: Normal test.
5. For a single population mean, we may use a normal distribution with the following mean and standard deviation. Means:
         and

6. For a single population proportion, we may use a normal distribution with the following mean and standard deviation. Proportions:  µ =  and
        .

### Homework


# Hypothesis Testing with One Sample

## Rare Events, the Sample, and the Decision and Conclusion

Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.

### Rare Events

The thinking process in hypothesis testing can be summarized as follows: You want to test whether or not a particular property of the population is true. You make an assumption about the true population mean for numerical data or the true population proportion for categorical data. This assumption is the null hypothesis. Then you gather sample data that is representative of the population. From this sample data you compute the sample mean (or the sample proportion). If the value that you observe is very unlikely to occur (a rare event) if the null hypothesis is true, then you wonder why this is happening. A plausible explanation is that the null hypothesis is false.
For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket, and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is  = 0.005. Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A rare event has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket.

### Using the Sample to Test the Null Hypothesis

After you collect data and obtain the test statistic (the sample mean, sample proportion, or other test statistic), you can determine the probability of obtaining that test statistic when the null hypothesis is true. This probability is called the .
When the p-value is very small, it means that the observed test statistic is very unlikely to happen if the null hypothesis is true. This gives significant evidence to suggest that the null hypothesis is false, and to reject it in favor of the alternative hypothesis. In practice, to reject the null hypothesis we want the p-value to be smaller than 0.05 (5 percent) or sometimes even smaller than 0.01 (1 percent).

### Decision and Conclusion

A systematic way to make a decision of whether to reject or not reject the null hypothesis
is to compare the p-value and a preset or preconceived α, also called the level of significance of the test. A preset α is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.
When you make a decision to reject or not reject H, do as follows:

Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.

### Chapter Review

When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:

Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is  2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was  three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing  homework each week has increased. Assume the distribution of homework times is approximately normal.

### Homework


# Hypothesis Testing with One Sample

## Additional Information and Full Hypothesis Test Examples

1. In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is the preconceived or preset α.
2. The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
3. If no level of significance is given, a common standard to use is
4. When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
5. The alternative hypothesis, , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
6. H  never has a symbol that contains an equal sign.
7. Thinking about the meaning of the : A data analyst should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
The following examples illustrate a left-, right-, and two-tailed test.

### Full Hypothesis Test Examples

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95 percent confidence interval computation. The poem is clever and humorous, so please enjoy it!

### Chapter Review

The hypothesis test itself has an established process. This can be summarized as follows:

Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that are used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when it should be.

### Homework

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in
Hypothesis testing: For the following 10 exercises, answer each question.
1. State the null and alternate hypotheses.
2. State the p-value.
3. State alpha.
4. What is your decision?
5. Write a conclusion.
6. Answer any other questions asked in the problem.

### References

American Automobile Association. (n.d.). Retrieved from www.aaa.com
American Library Association. (n.d.). Retrieved from www.ala.org
Amit Schitai. (n.d.). Data.
Bureau of Labor Statistics. (n.d.). Occupational employment statistics. Retrieved from http://www.bls.gov/oes/current/oes291111.htm
Centers for Disease Control and Prevention. (n.d.). Retrieved from www.cdc.gov
De Anza College. (2006). Foothill-De Anza Community College District. Retrieved from http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf
Federal Bureau of Investigation. (n.d.). Uniform crime reports and index of crime in Daviess in the state of Kentucky enforced by Daviess County from 1985 to 2005. Retrieved from http://www.disastercenter.com/kentucky/crime/3868.htm
Gallup. (n.d.). Retrieved from www.gallup.com
Johansen, C., Boice, Jr. J., McLaughlin, J., & Olsen, J. (2001, Feb. 7). Cellular telephones and cancera nationwide cohort study in Denmark. Journal of National Cancer Institute, 93(3), 2037. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/11158188
La Leche League International. (n.d.). Retrieved from http://www.lalecheleague.org/Law/BAFeb01.html
Online Learning Consortium. (2005 Nov.). Growing by degrees: Online education in the United States, 2005. Newburyport, MA: Allen, I. E., & Seaman, J. Available at http://files.eric.ed.gov/fulltext/ED530062.pdf
Toastmasters International. (n.d.). Retrieved from http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1
U.S. Census Bureau. (n.d.). Language use. Retrieved from https://www.census.gov/topics/population/language-use.html
U.S. Census Bureau. (n.d.). QuickFacts. Retrieved from https://www.census.gov/quickfacts/table/PST045216/00
U.S. Department of Energy. (n.d.). Retrieved from http://energy.gov
Weather Underground. (n.d.). Retrieved from www.wunderground.com


# Hypothesis Testing with Two Samples

## Introduction

Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied 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 of individuals from different income brackets who might vote for them. Students are interested in whether the SAT or GRE preparatory courses really help raise their scores.
You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is the same, just expanded.
To compare two means or two proportions, you work with two groups. The groups are classified as independent groups or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions.
This chapter deals with the following hypothesis tests:
1. Independent groups (samples are independent)

2. Matched or paired samples (samples are dependent)


# Hypothesis Testing with Two Samples

## Two Population Means with Unknown Standard Deviations

1. The two independent samples are simple random samples from two distinct populations.
2. For the two distinct populations

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. To account for the variation, we take the difference of the sample means,

  , and divide by the standard error to standardize the difference. The result is a t-score test statistic.
Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means,

The standard error is calculated as follows:
The test statistic (t-score) is calculated as follows:

The number of degrees of freedom ( requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t-distribution with df as follows:
Degrees of freedom
When both sample sizes n1 and n2 are five or larger, the Student’s t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.)

Cohen’s Standards for Small, Medium, and Large Effect SizesCohen’s  is a measure of effect size based on the differences between two means. Cohen’s d, named for U.S. statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.
Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation:
 where
.

### References

Baseball-Almanac. (2013). World series history. Retrieved from http://www.baseball-almanac.com/ws/wsmenu.shtml
Graduating Engineer + Computer Careers. (n.d.). Retrieved from http://www.graduatingengineer.com
Microsoft Bookshelf. (n.d.).
Nasdaq. (n.d.). Sectoring by industry groups. Retrieved from http://www.nasdaq.com/markets/barchart-sectors.aspx
U.S. Senate. (n.d.). Retrieved from www.senate.gov
Wikipedia. (n.d.). List of current United States Senators by age. Retrieved from http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age

### Chapter Review

Two population means from independent samples where the population standard deviations are not known
1. Random variable:
  = the difference of the sampling means
2. Distribution: Student’s t-distribution with degrees of freedom (variances not pooled)

### Formula Review

Standard error: SE =

Test statistic (t-score): t =

Degrees of freedom:

where:
s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

 and
 are the sample means.
Cohen’s d is the measure of effect size:

where

Use the following information to answer the next 15 exercises. Indicate if the hypothesis test is for
1. independent group means, population standard deviations, and/or variances known,
2. independent group means, population standard deviations, and/or variances unknown,
3. matched or paired samples,
4. single mean,
5. two proportions, or
6. single proportion.
Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions.
Use the following information to answer the next 12 exercises. The U.S. Centers 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.

### Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in
Use the information from
Use the following information to answer the next two exercises. The Eastern and Western 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.
Conduct a hypothesis test to answer the next two exercises.


# Hypothesis Testing with Two Samples

## Two Population Means with Known Standard Deviations

Even though this situation is not likely (knowing the population standard deviations), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal, and both populations must be normal. The random variable is

 . The normal distribution has the following format:
Normal distribution is

### References

Centers for Disease Control and Prevention. (2008, July 18). State-specific prevalence of obesity among adultsUnited States, 2007. MMWR, 57(28), 765768. Retrieved from http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm
Federal Bureau of Investigation. (n.d.). Texas Crime Rates 19601012. Available at http://www.disastercenter.com/crime/txcrime.htm
Hinduja, S. (2013). Sexting Research and Gender Differences. Cyberbulling Research Center. Retrieved from http://cyberbullying.us/blog/sexting-research-and-gender-differences/
Humes, K. R., Jones, N. A., & Ramirez, R. R. (2011 March). Overview of race and Hispanic origin: 2010 (2010 Census Briefs). Washington, DC: U.S. Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf
Smith, A. (2011, July 11). 35% of American adults own a smartphone. Pew Internet. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf
Visually. (2013). Smart phone users, by the numbers. Retrieved from http://visual.ly/smart-phone-users-numbers

### Chapter Review

A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations) will have these characteristics:

### Formula Review

Normal distribution:
 .
Generally,
Test statistic (z-score):

Generally,
where

σ1 and σ2 are the known population standard deviations, n1 and n2 are the sample sizes,
  and
  are the sample means, and μ1 and μ2 are the population means.
Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions.  shows the result. Scouters believe that Rodriguez pitches a speedier fastball.
Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.
Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.

### Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in


# Hypothesis Testing with Two Samples

## Comparing Two Independent Population Proportions

When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:
1. The two independent samples are simple random samples that are independent.
2.  The number of successes is at least five, and the number of failures is at least five, for each of the samples.
3. Growing literature states that the population must be at least 10 or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results.
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H: p = p. To conduct the test, we use a pooled proportion, p.
The pooled proportion is calculated as follows:
The distribution for the differences is
The test statistic (z-score) is

### References

American Cancer Society. (n.d.). Retrieved from http://www.cancer.org/index
Centers for Disease Control and Prevention. (n.d.). West Nile virus. Retrieved from http://www.cdc.gov/ncidod/dvbid/westnile/index.htm
Chancellors Office, California Community Colleges. (1994, Nov.).
Educational Resources. (n.d.).
Gallup. (2013). State of the states. Retrieved from http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive
Hilton Hotels. (n.d.). Retrieved from http://www.hilton.com
Hyatt Hotels. Retrieved from http://hyatt.com
San Jose Museum of Art. (n.d.). Whitney exhibit (on loan).
U.S. Department of Health and Human Services. (n.d). Statistics. Retrieved from https://www.hhs.gov/

### Chapter Review

Test of two population proportions from independent samples

### Formula Review

Pooled proportion: p =

Distribution for the differences:

where the null hypothesis is H: p = p or H: p – p = 0
Test statistic (z-score):

where the null hypothesis is H: p = p or H: p − p = 0
and where
p′
 and p′
 are the sample proportions, p and p are the population proportions,
P is the pooled proportion, and n and n are the sample sizes.
Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in 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 another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.
Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only 9 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 percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.

### Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test.  The solution sheet is found in
Use the following information to answer the next three exercises. Neuroinvasive 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 of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1 percent level of significance, conduct an appropriate hypothesis test.


# Hypothesis Testing with Two Samples

## Matched or Paired Samples (Optional)

When using a hypothesis test for matched or paired samples, the following characteristics should be present:
1. Simple random sampling is used.
2. Sample sizes are often small.
3. Two measurements (samples) are drawn from the same pair of individuals or objects.
4. Differences are calculated from the matched or paired samples.
5. The differences form the sample that is used for the hypothesis test.
6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.
In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ, is then tested using a Student’s-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences.
The test statistic (t-score) is

### Chapter Review

A hypothesis test for matched or paired samples (t-test) has these characteristics:
1. Test the differences by subtracting one measurement from the other measurement
2. Random variable:
 = mean of the differences.
3. Distribution:  Student’s t distribution with n – 1 degrees of freedom.
4. If the number of differences is small (less than 30), the differences must follow a normal distribution.
5. Two samples are drawn from the same set of objects.
6. Samples are dependent.

### Formula Review

Test statistic (t-score): t =

where:

 is the mean of the sample differences, μd is the mean of the population differences, s is the sample standard deviation of the differences, and n is the sample size.
Use the following information to answer the next five exercises. 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 . The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level.
Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, 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 are calculated. The differences have a normal distribution. Test at the 1 percent significance level.
Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. 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 of concern. Test at the 1 percent significance level.

### Homework

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in
Use the following information to answer the next two exercises. A new preventative medication was tried on a group of 224 patients who had the same risk factors for a disease. 45 patients developed the disease after four years. In a control group of 224 patients, 68 developed the disease after four years. We want to test whether the method of treatment reduces the proportion of patients who develop the disease after four years.
Let the subscript t = treated patient and ut = untreated patient.
Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a biofeedback exercise program. Six subjects were randomly selected, and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after – before), producing the following results:
 = −10.2 s = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training.

### Bringing It Together

Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test.
1. Independent group means, population standard deviations and/or variances known
2. Independent group means, population standard deviations and/or variances unknown
3. Matched or paired samples
4. Single mean
5. Two proportions
6. Single proportion


# The Chi-Square Distribution

## Introduction

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 were different 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 by conducting a hypothesis test.
You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution.
In this chapter, you will learn the three major applications of the chi-square distribution:


# The Chi-Square Distribution

## Facts About the Chi-Square Distribution

The notation for the chi-square distribution is
where df = degrees of freedom, which depends on how chi-square is being used. If you want to practice calculating chi-square probabilities then use df = n – 1. The degrees of freedom for the three major uses are calculated differently.
For the χ distribution, the population mean is μ = df, and the population standard deviation is
 .
The random variable is shown as χ, but it may be any uppercase letter.
The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables is
χ2 = (Z1)2 + (Z2)2 + ... + (Z)2, where the following are true:
1. The curve is nonsymmetrical and skewed to the right.
2. There is a different chi-square curve for each df.

3. The test statistic for any test is always greater than or equal to zero.
4. When df > 90, the chi-square curve approximates the normal distribution. For X ~
 , the mean, μ = df = 1,000 and the standard deviation, σ =
  = 44.7. Therefore, X ~ N(1,000, 44.7), approximately.
5. The mean, μ, is located just to the right of the peak.

### References

Parade Magazine. (n.d.). Retrieved from https://parade.com/
Santa Clara County Public Health Department. (2011, May). HIV/AIDS epidemiology Santa Clara County. Retrieved from http://sccgov.iqm2.com/Citizens/FileOpen.aspx?Type=4&ID=32762

### Chapter Review

The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population.
An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom.
The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests.

### Formula Review

χ2 = (Z1)2 + (Z2)2 + . . . (Z)2
chi-square distribution random variable
μ = df chi-square distribution population mean

  chi-square distribution population standard deviation

### Homework

Decide whether the following statements are true or false.


# The Chi-Square Distribution

## Goodness-of-Fit Test

In this type of hypothesis test, you determine whether the data fit a particular distribution. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test, meaning the distribution for the hypothesis test is chi-square, to determine if there is a fit. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.
The test statistic for a goodness-of-fit test is:
where
1. O = observed values (data),
2. E = expected values (from theory), and
3. 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. There are n terms of the form
 .
The number of degrees of freedom is df = (number of categories – 1).
The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.

### References

College Board. (n.d.). Retrieved from http://www.collegeboard.com
Ma, Y., et al. (2003). Association between eating patterns and obesity in a free-living US adult population. American Journal of Epidemiology 158(1), 8592.
Ogden, C. L., et al. (2012, January). Prevalence of obesity in the United States, 20092010 (NCHS Data Brief No. 82). Hyattsville, MD: National Center for Health Statistics. Retrieved from http://www.cdc.gov/nchs/data/databriefs/db82.pdf
Stevens, B. J. (n.d.). Multi-family and commercial solid waste and recycling survey. Arlington County, VA. Retrieved from http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf
U.S. Census Bureau. (n.d.). Current population reports. Retrieved from https://www.census.gov/main/www/cprs.html
U.S. Census Bureau. (n.d). Retrieved from https://www.census.gov/

### Chapter Review

To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five.

### Formula Review

 goodness-of-fit test statistic where

    O: observed values
E: expected values

k: number of different data cells or categories

df = k − 1 degrees of freedom
Determine the appropriate test to be used  in the next three exercises.
Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in .
The actual distribution for a class of 20 is in .
Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County is broken down by ethnicity as in .
The percentage of each ethnic group in Santa Clara County is as in .

### Homework

For each problem, use a solution sheet to solve the hypothesis test problem. Go to
Use the following information to answer the next two exercises. The columns in  contain the Race/Ethnicity of U.S. Public Schools for a recent year,  the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the results of a survey of 1,000 local students from that year who took an AP exam.
Use the following information to answer the next two exercises. UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students’ expected majors by gender were reported in . Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for  and . The second column in each table does not add to 100 percent because of rounding.
Read the statement and decide whether it is true or false.