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, or expected value, of X is the sum of each outcome multiplied by its corresponding probability: µX = E(X) = x1 · P (x1) + x2 · P (x2) + · · · + xn · P (xn) = n i=1 xi · P (xi) 0−2525−5050−100100+0.000.050.100.150.200.25US household incomes ($1000s)probability 3.4. RANDOM VARIABLES 181 The expected value for a random variable represents the average outcome. For example, E(X) = 117.85 represents the average amount the bookstore expects to make from a single student, which we could also write as µ = 117.85. While the bookstore will make more than this on some students and less than this on other students, the average of many randomly selected students will be near $117.85. It is also possible to compute the expected value of a continuous random variable. However, it requires a little calculus and we save it for a later class.52 In physics, the expectation holds the same meaning as the center of gravity. The distribution can be represented by a series of weights at each outcome, and the mean represents the balancing point. This is represented in Figures 3.18 and 3.24. The idea of a center of gravity also expands to continuous probability distributions. Figure 3.25 shows a continuous probability distribution balanced atop a wedge placed at the mean. Figure 3.24: A weight system representing the probability distribution for X. The string holds the distribution at the mean to keep the system balanced. Figure 3.25: A continuous distribution can also be balanced at its mean. 52µX = xf (x)dx where f (x) represents a function for the density curve. 0137170117.85m 182 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.4.4 Variability in random variables Suppose you ran the university bookstore. Besides how much revenue you expect to generate, you might also want to know the volatility (variability) in your revenue. The variance and standard deviation can be used to describe the variability of a random variable. Section 2.2.2 introduced a method for ﬁnding the variance and standard deviation for a data set. We ﬁrst computed deviations from the mean (xi − µ), squared those deviations, and took
an average to get the variance. In the case of a random variable, we again compute squared deviations. However, we take their sum weighted by their corresponding probabilities, just like we did for the expectation. This weighted sum of squared deviations equals the variance, and we calculate the standard deviation by taking the square root of the variance, just as we did in Section 2.2.2. VARIANCE AND STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE If X takes outcomes x1, x2,..., xn with probabilities P (x1), P (x2),..., P (xn) and expected value µX = E(X), then to ﬁnd the standard deviation of X, we ﬁrst ﬁnd the variance and then take its square root. V ar(X) = σ2 x = (x1 − µX)2 · P (x1) + (x2 − µX)2 · P (x2) + · · · + (xn − µX)2 · P (xn) n = (xi − µX)2 · P (xi) SD(X) = σX = i=1 n i=1 (xi − µX)2 · P (xi) Just as it is possible to compute the mean of a continuous random variable using calculus, we can also use calculus to compute the variance.53 However, this topic is beyond the scope of the AP exam. EXAMPLE 3.67 Compute the expected value, variance, and standard deviation of X, the revenue of a single statistics student for the bookstore. It is useful to construct a table that holds computations for each outcome separately, then add up the results. i xi P (xi) xi · P (xi) 1 $0 0.20 0 2 $137 0.55 75.35 3 $170 0.25 42.50 Total 117.85 Thus, the expected value is µX = 117.85, which we computed earlier. The variance can be constructed using a similar table: i xi P (xi) xi − µX (xi − µX)2 (xi − µX)2 · P (xi) 1 $0 0.20 -117.85 13888.62 2777.7 2 $137 0.55 19.15 366.72 201.7 3 $170 0.25 52.
15 2719.62 679.9 Total 3659.3 The variance of X is σ2 X = 3659.3, which means the standard deviation is σX = √ 3659.3 = $60.49. 53σ2 x = (x − µX )2f (x)dx where f (x) represents a function for the density curve. 3.4. RANDOM VARIABLES 183 GUIDED PRACTICE 3.68 The bookstore also oﬀers a chemistry textbook for $159 and a book supplement for $41. From past experience, they know about 25% of chemistry students just buy the textbook while 60% buy both the textbook and supplement.54 (a) What proportion of students don’t buy either book? Assume no students buy the supplement without the textbook. (b) Let Y represent the revenue from a single student. Write out the probability distribution of Y, i.e. a table for each outcome and its associated probability. (c) Compute the expected revenue from a single chemistry student. (d) Find the standard deviation to describe the variability associated with the revenue from a single student. 3.4.5 Linear transformations of a random variable An online store is selling a limited edition t-shirt. The maximum a person is allowed to buy is 3. Let X be a random variable that represents how many of the t-shirts a t-shirt buyer orders. The probability distribution of X is given in the following table. xi P (xi) 1 0.6 2 0.3 3 0.1 Using the methods of the previous section we can ﬁnd that the mean µX = 1.5 and the standard deviation σX = 0.67. Suppose that the cost of each t-shirt is $30 and that there is ﬂat rate $5 shipping fee. The amount of money a t-shirt buyer pays, then, is 30X + 5, where X is the number of t-shirts ordered. To calculate the mean and standard deviation for the amount of money a t-shirt buyers pays, we could deﬁne a new variable Y as follows: Y = 30X + 5 GUIDED PRACTICE 3.69 Verify that the distribution of Y is given by the table below.55 yi P (yi) $35 0.6 $65 0.3 $95 0.1 54(a) 100%
- 25% - 60% = 15% of students do not buy any books for the class. Part (b) is represented by the ﬁrst two lines in the table below. The expectation for part (c) is given as the total on the line yi · P (yi). The result of part (d) is the square-root of the variance listed on in the total on the last line: σY = V ar(Y ) = 4800 = 69.28. √ i (scenario) yi P (yi) yi · P (yi) yi − µY (yi − µY)2 (yi − µY)2 · P (yi) 5530 × 1 + 5 = 35; 30 × 2 + 5 = 65; 30 × 3 + 5 = 95 1 (noBook) 0.00 0.15 0.00 -159.75 25520.06 3828.0 2 (textbook) 159.00 0.25 39.75 -0.75 0.56 0.1 3 (both) 200.00 0.60 120.00 40.25 1620.06 972.0 Total E(Y ) = 159.75 V ar(Y ) ≈ 4800 184 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Using this new table, we can compute the mean and standard deviation of the cost for tshirt orders. However, because Y is a linear transformation of X, we can use the properties from Section 2.2.7. Recall that multiplying every X by 30 multiplies both the mean and standard deviation by 30. Adding 5 only adds 5 to the mean, not the standard deviation. Therefore, µ30X+5 = E(30X + 5) = 30 × E(X) + 5 = 30 × 1.5 + 5 = 50.00 σ30X+5 = SD(30X + 5) = 30 × SD(X) = 30 × 0.67 = 20.10 Among t-shirt buyers, they spend an average of $50.00, with a standard deviation of $20.10. LINEAR TRANSFORMATIONS OF A RANDOM VARIABLE If X is a random variable, then a linear transformation is given by aX + b, where a and b are some ﬁxed numbers. E(aX + b) = a × E(X) + b
SD(aX + b) = \|a\| × SD(X) 3.4.6 Linear combinations of random variables So far, we have thought of each variable as being a complete story in and of itself. Sometimes it is more appropriate to use a combination of variables. For instance, the amount of time a person spends commuting to work each week can be broken down into several daily commutes. Similarly, the total gain or loss in a stock portfolio is the sum of the gains and losses in its components. EXAMPLE 3.70 John travels to work ﬁve days a week. We will use X1 to represent his travel time on Monday, X2 to represent his travel time on Tuesday, and so on. Write an equation using X1,..., X5 that represents his travel time for the week, denoted by W. His total weekly travel time is the sum of the ﬁve daily values: W = X1 + X2 + X3 + X4 + X5 Breaking the weekly travel time W into pieces provides a framework for understanding each source of randomness and is useful for modeling W. EXAMPLE 3.71 It takes John an average of 18 minutes each day to commute to work. What would you expect his average commute time to be for the week? We were told that the average (i.e. expected value) of the commute time is 18 minutes per day: E(Xi) = 18. To get the expected time for the sum of the ﬁve days, we can add up the expected time for each individual day: E(W ) = E(X1 + X2 + X3 + X4 + X5) = E(X1) + E(X2) + E(X3) + E(X4) + E(X5) = 18 + 18 + 18 + 18 + 18 = 90 minutes The expectation of the total time is equal to the sum of the expected individual times. More generally, the expectation of a sum of random variables is always the sum of the expectation for each random variable. 3.4. RANDOM VARIABLES 185 GUIDED PRACTICE 3.72 Elena is selling a TV at a cash auction and also intends to buy a toaster oven in the auction. If X represents the proﬁt for selling the TV and Y represents the cost of the toaster oven, write an equation that represents the net change in Elena’
s cash.56 GUIDED PRACTICE 3.73 Based on past auctions, Elena ﬁgures she should expect to make about $175 on the TV and pay about $23 for the toaster oven. In total, how much should she expect to make or spend?57 GUIDED PRACTICE 3.74 Would you be surprised if John’s weekly commute wasn’t exactly 90 minutes or if Elena didn’t make exactly $152? Explain.58 Two important concepts concerning combinations of random variables have so far been introduced. First, a ﬁnal value can sometimes be described as the sum of its parts in an equation. Second, intuition suggests that putting the individual average values into this equation gives the average value we would expect in total. This second point needs clariﬁcation – it is guaranteed to be true in what are called linear combinations of random variables. A linear combination of two random variables X and Y is a fancy phrase to describe a combination aX + bY where a and b are some ﬁxed and known numbers. For John’s commute time, there were ﬁve random variables – one for each work day – and each random variable could be written as having a ﬁxed coeﬃcient of 1: 1X1 + 1X2 + 1X3 + 1X4 + 1X5 For Elena’s net gain or loss, the X random variable had a coeﬃcient of +1 and the Y random variable had a coeﬃcient of -1. When considering the average of a linear combination of random variables, it is safe to plug in the mean of each random variable and then compute the ﬁnal result. For a few examples of nonlinear combinations of random variables – cases where we cannot simply plug in the means – see the footnote.59 LINEAR COMBINATIONS OF RANDOM VARIABLES AND THE AVERAGE RESULT If X and Y are random variables, then a linear combination of the random variables is given by aX + bY, where a and b are some ﬁxed numbers. To compute the average value of a linear combination of random variables, plug in the average of each individual random variable and compute the result: E(aX + bY ) = a × E(X) + b × E(Y ) Recall that the expected value is
the same as the mean, i.e. E(X) = µX. 56She will make X dollars on the TV but spend Y dollars on the toaster oven: X − Y. 57E(X − Y ) = E(X) − E(Y ) = 175 − 23 = $152. She should expect to make about $152. 58No, since there is probably some variability. For example, the traﬃc will vary from one day to next, and auction prices will vary depending on the quality of the merchandise and the interest of the attendees. 59If X and Y are random variables, consider the following combinations: X 1+Y, X × Y, X/Y. In such cases, plugging in the average value for each random variable and computing the result will not generally lead to an accurate average value for the end result. 186 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.75 Leonard has invested $6000 in Google Inc. (stock ticker: GOOG) and $2000 in Exxon Mobil Corp. (XOM). If X represents the change in Google’s stock next month and Y represents the change in Exxon Mobil stock next month, write an equation that describes how much money will be made or lost in Leonard’s stocks for the month. For simplicity, we will suppose X and Y are not in percents but are in decimal form (e.g. if Google’s stock increases 1%, then X = 0.01; or if it loses 1%, then X = −0.01). Then we can write an equation for Leonard’s gain as $6000 × X + $2000 × Y If we plug in the change in the stock value for X and Y, this equation gives the change in value of Leonard’s stock portfolio for the month. A positive value represents a gain, and a negative value represents a loss. GUIDED PRACTICE 3.76 Suppose Google and Exxon Mobil stocks have recently been rising 2.1% and 0.4% per month, respectively. Compute the expected change in Leonard’s stock portfolio for next month.60 GUIDED PRACTICE 3.77 You should have found that Leonard expects a positive gain in Guided Practice 3.76. However, would you be surprised if he actually had a loss this month?61 3.4.7 Variability in linear combinations of
random variables Quantifying the average outcome from a linear combination of random variables is helpful, but it is also important to have some sense of the uncertainty associated with the total outcome of that combination of random variables. The expected net gain or loss of Leonard’s stock portfolio was considered in Guided Practice 3.76. However, there was no quantitative discussion of the volatility of this portfolio. For instance, while the average monthly gain might be about $134 according to the data, that gain is not guaranteed. Figure 3.26 shows the monthly changes in a portfolio like Leonard’s during the 36 months from 2009 to 2011. The gains and losses vary widely, and quantifying these ﬂuctuations is important when investing in stocks. Figure 3.26: The change in a portfolio like Leonard’s for the 36 months from 2009 to 2011, where $6000 is in Google’s stock and $2000 is in Exxon Mobil’s. Just as we have done in many previous cases, we use the variance and standard deviation to describe the uncertainty associated with Leonard’s monthly returns. To do so, the standard deviations and variances of each stock’s monthly return will be useful, and these are shown in Figure 3.27. The stocks’ returns are nearly independent. 60E($6000 × X + $2000 × Y ) = $6000 × 0.021 + $2000 × 0.004 = $134. 61No. While stocks tend to rise over time, they are often volatile in the short term. Monthly returns (2009−2011)−1000−50005001000 3.4. RANDOM VARIABLES 187 GOOG XOM Mean (¯x) 0.0210 0.0038 Standard deviation (s) Variance (s2) 0.0072 0.0849 0.0027 0.0520 Figure 3.27: The mean, standard deviation, and variance of the GOOG and XOM stocks. These statistics were estimated from historical stock data, so notation used for sample statistics has been used. We want to describe the uncertainty of Leonard’s monthly returns by ﬁnding the standard deviation of the return on his combined portfolio. First, we note that the variance of a sum has a nice property: the variance of a sum is the sum of the variances. That is, if X and Y are independent random variables: V ar(X + Y ) = V ar(X)
+ V ar(Y ) Because the standard deviation is the square root of the variance, we can rewrite this equation using standard deviations: (SDX+Y )2 = (SDX )2 + (SDY )2 This equation might remind you of a theorem from geometry: c2 = a2 + b2. The equation for the standard deviation of the sum of two independent random variables looks analogous to the Pythagorean Theorem. Just as the Pythagorean Theorem only holds for right triangles, this equation only holds when X and Y are independent.62 STANDARD DEVIATION OF THE SUM AND DIFFERENCE OF RANDOM VARIABLES If X and Y are independent random variables: SDX+Y = SDX−Y = (SDX )2 + (SDY )2 Because SDY = SD−Y, the standard deviation of the diﬀerence of two variables equals the standard deviation of the sum of two variables. This property holds for more than two variables as well. For example, if X, Y, and Z are independent random variables: SDX+Y +Z = SDX−Y −Z = (SDX )2 + (SDY )2 + (SDZ)2 If we need the standard deviation of a linear combination of independent variables, such as aX + bY, we can consider aX and bY as two new variables. Recall that multiplying all of the values of variable by a positive constant multiplies the standard deviation by that constant. Thus, SDaX = a × SDX and SDbY = b × SDY. It follows that: SDaX+bY = (a × SDX )2 + (b × SDY )2 This equation can be used to compute the standard deviation of Leonard’s monthly return. Recall that Leonard has $6,000 in Google stock and $2,000 in Exxon Mobil’s stock. From Figure 3.27, the standard deviation of Google stock is 0.0849 and the standard deviation of Exxon Mobile stock is 0.0520. SD6000X+2000Y = (6000 × SDX )2 + (2000 × SDY )2 = (6000 × 0.0849)2 + (2000 ×.0520)2 = 270,304 = 520 The standard deviation of the total is $520. While an average monthly return of $134 on an $8000 investment is nothing to scoﬀ at
, the monthly returns are so volatile that Leonard should not expect this income to be very stable. 62Another word for independent is orthogonal, meaning right angle! When X and Y are dependent, the equation for SDX+Y becomes analogous to the law of cosines. 188 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS STANDARD DEVIATION OF LINEAR COMBINATIONS OF RANDOM VARIABLES To ﬁnd the standard deviation of a linear combination of random variables, we ﬁrst consider aX and bY separately. We ﬁnd the standard deviation of each, and then we apply the equation for the standard deviation of the sum of two variables: SDaX+bY = (a × SDX )2 + (b × SDY )2 This equation is valid as long as the random variables X and Y are independent of each other. EXAMPLE 3.78 Suppose John’s daily commute has a standard deviation of 4 minutes. What is the uncertainty in his total commute time for the week? The expression for John’s commute time is X1 + X2 + X3 + X4 + X5 Each coeﬃcient is 1, so the standard deviation of the total weekly commute time is SD = (1 × 4)2 + (1 × 4)2 + (1 × 4)2 + (1 × 4)2 + (1 × 4)2 = 5 × (4)2 = 8.94 The standard deviation for John’s weekly work commute time is about 9 minutes. GUIDED PRACTICE 3.79 The computation in Example 3.78 relied on an important assumption: the commute time for each day is independent of the time on other days of that week. Do you think this is valid? Explain.63 GUIDED PRACTICE 3.80 Consider Elena’s two auctions from Guided Practice 3.72 on page 185. Suppose these auctions are approximately independent and the variability in auction prices associated with the TV and toaster oven can be described using standard deviations of $25 and $8. Compute the standard deviation of Elena’s net gain.64 Consider again Guided Practice 3.80. The negative coeﬃcient for Y in the linear combination was eliminated when we squared the coeﬃcients. This generally holds true: negatives in a linear combination will have no impact on the variability
computed for a linear combination, but they do impact the expected value computations. 63One concern is whether traﬃc patterns tend to have a weekly cycle (e.g. Fridays may be worse than other days). If that is the case, and John drives, then the assumption is probably not reasonable. However, if John walks to work, then his commute is probably not aﬀected by any weekly traﬃc cycle. 64The equation for Elena can be written as: (1) × X + (−1) × Y. To ﬁnd the SD of this new variable we do: SD(1)×X+(−1)×Y = (1 × SDX )2 + (−1 × SDY )2 = (1 × 25)2 + (−1 × 8)2 = 26.25 The SD is about $26.25. 3.4. RANDOM VARIABLES 189 3.4.8 Normal approximation for sums of random variables We have seen that many distributions are approximately normal. The sum and the diﬀerence of normally distributed variables is also normal. While we cannot prove this here, the usefulness of it is seen in the following example. EXAMPLE 3.81 Three friends are playing a cooperative video game in which they have to complete a puzzle as fast as possible. Assume that the individual times of the 3 friends are independent of each other. The individual times of the friends in similar puzzles are approximately normally distributed with the following means and standard deviations. Friend 1 Friend 2 Friend 3 Mean 5.6 5.8 6.1 SD 0.11 0.13 0.12 To advance to the next level of the game, the friends’ total time must not exceed 17.1 minutes. What is the probability that they will advance to the next level? Because each friend’s time is approximately normally distributed, the sum of their times is also approximately normally distributed. We will do a normal approximation, but ﬁrst we need to ﬁnd the mean and standard deviation of the sum. We learned how to do this in Section 3.4. Let the three friends be labeled X, Y, Z. We want P (X + Y + Z < 17.1). The mean and standard deviation of the sum of X, Y, and Z is given by: µsum = E(X + Y + Z) = E(X) + E(Y ) +
E(Z) = 5.6 + 5.8 + 6.1 = 17.5 Now we can ﬁnd the Z-score. σsum = (SDX )2 + (SDY )2 + (SDZ)2 = (0.11)2 + (0.13)2 + (0.12)2 = 0.208 Z = = xsum − µsum σsum 17.1 − 17.5 0.208 = −1.92 Finally, we want the probability that the sum is less than 17.5, so we shade the area to the left of Z = −1.92. Using technology, we get There is a 2.7% chance that the friends will advance to the next level. P (Z < −1.92) = 0.027 GUIDED PRACTICE 3.82 What is the probability that Friend 2 will complete the puzzle with a faster time than Friend 1? Hint: ﬁnd P (Y < X), or P (Y − X < 0).65 65First ﬁnd the mean and standard deviation of Y − X. The mean of Y − X is µY −X = 5.8 − 5.6 = 0.2. The 0.170 = −1.18 and P (Z < −1.18) =.119. standard deviation is SDY −X = (0.13)2 + (0.11)2 = 0.170. Then Z = 0−0.2 There is an 11.9% chance that Friend 2 will complete the puzzle with a faster time than Friend 1. 190 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • A discrete probability distribution can be summarized in a table that consists of all possible outcomes of a random variable and the probabilities of those outcomes. The outcomes must be disjoint, and the sum of the probabilities must equal 1. • A probability distribution can be represented with a histogram and, like the distributions of data that we saw in Chapter 2, can be summarized by its center, spread, and shape. • When given a probability distribution table, we can calculate the mean (expected value) and standard deviation of a random variable using the following formulas. E(X) = µX = xi · P (xi) = x1 · P (x1) + x2 · P (x2) + · · · +
xn · P (xn) V ar(X) = σ2 x = (xi − µX)2 · P (xi) SD(X) = σX = (xi − µX)2 · P (xi) = (x1 − µX)2 · P (x1) + (x2 − µX)2 · P (x2) + · · · + (xn − µX)2 · P (xn) We can think of P (xi) as the weight, and each term is weighted its appropriate amount. • The mean of a probability distribution does not need to be a value in the distribution. It represents the average of many, many repetitions of a random process. The standard deviation represents the typical variation of the outcomes from the mean, when the random process is repeated over and over. • Linear transformations. Adding a constant to every value in a probability distribution adds that value to the mean, but it does not aﬀect the standard deviation. When multiplying every value by a constant, this multiplies the mean by the constant and it multiplies the standard deviation by the absolute value of the constant. • Combining random variables. Let X and Y be random variables and let a and b be constants. – The expected value of the sum is the sum of the expected values. E(X + Y ) = E(X) + E(Y ) E(aX + bY ) = a × E(X) + b × E(Y ) – When X and Y are independent: The standard deviation of a sum or a diﬀerence is the square root of the sum of each standard deviation squared. SD(X + Y ) = (SD(X))2 + (SD(Y ))2 SD(X − Y ) = (SD(X))2 + (SD(Y ))2 SD(aX + bY ) = (a × SD(X))2 + (b × SD(Y ))2 The SD properties require that X and Y be independent. The expected value properties hold true whether or not X and Y are independent. • Because the sum or diﬀerence of two normally distributed variables is itself a normally distributed variable, the normal approximation is also used in the following type of problem. Find the probability that a sum X + Y or a diﬀerence X − Y is greater/less than some value. 1. Verify that the
distribution of X and the distribution of Y are approximately normal. 2. Find the mean of the sum or diﬀerence. Recall: the mean of a sum is the sum of the means. The mean of a diﬀerence is the diﬀerence of the means. Find the SD of the sum or diﬀerence using: SD(X + Y ) = SD(X − Y ) = (SD(X))2 + (SD(Y ))2. 3. Calculate the Z-score. Use the calculated mean and SD to standardize the given sum or diﬀerence. 4. Find the appropriate area under the normal curve. 3.4. RANDOM VARIABLES 191 Exercises 3.31 Patreon contributions. The distribution of monthly contributions made to a particular Patreon creator is given as follows: Monthly Contribution Proportion $3 0.50 $5 0.30 $10 0.15 $25 0.05 (a) Compute the average monthly contribution made to this creator. (b) Compute the standard deviation of the monthly contributions made to this creator. 3.32 Experiment launch impact. An online shopping website develops new features and iterates on its product recommendations on a regular basis. The team also uses an experiment when launching each new feature to assess the impact of the change on the total sales. Below is a summary table of the impacts that the team observes: Number of Sales From Launch Proportion of Launches -200 0.1 0 0.5 +200 +500 0.3 0.1 (a) Compute the average impact for the experiments. (b) Compute the standard deviation of the impact from these experiments. 3.33 Hearts win. In a new card game, you start with a well-shuﬄed full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing. (a) Create a probability model for the amount you win at this game, and ﬁnd the expected winnings. Also compute the standard deviation of this distribution. (b) If the game costs $5 to play, what would be the expected value and standard deviation of the net proﬁt (or loss)? (Hint: proﬁt = winnings − cost; X − 5) (c) If the game
costs $5 to play, should you play this game? Explain. 3.34 Ace of clubs wins. Consider the following card game with a well-shuﬄed deck of cards. If you draw a red card, you win nothing. If you get a spade, you win $5. For any club, you win $10 plus an extra $20 for the ace of clubs. (a) Create a probability model for the amount you win at this game. Also, ﬁnd the expected winnings for a single game and the standard deviation of the winnings. (b) What is the maximum amount you would be willing to pay to play this game? Explain your reasoning. 3.35 Portfolio return. A portfolio’s value increases by 18% during a ﬁnancial boom and by 9% during normal times. It decreases by 12% during a recession. What is the expected return on this portfolio if each scenario is equally likely? 3.36 Baggage fees. An airline charges the following baggage fees: $25 for the ﬁrst bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags. (a) Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation. (b) About how much revenue should the airline expect for a ﬂight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justiﬁed. 3.37 American roulette. The game of American roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. Suppose you bet $1 on red. What’s the expected value and standard deviation of your winnings? 192 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.38 European roulette. The game of European roulette involves spinning a wheel with 37 slots: 18 red, 18 black, and 1 green. A ball is spun onto the wheel and will eventually
land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. (a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings? (b) Suppose you bet $1 in three diﬀerent rounds. What is the expected value and standard deviation of your total winnings? (c) How do your answers to parts (a) and (b) compare? What does this say about the riskiness of the two games? 3.39 Lemonade at The Cafe. Drink pitchers at The Cafe are intended to hold about 64 ounces of lemonade and glasses hold about 12 ounces. However, when the pitchers are ﬁlled by a server, they do not always ﬁll it with exactly 64 ounces. There is some variability. Similarly, when they pour out some of the lemonade, they do not pour exactly 12 ounces. The amount of lemonade in a pitcher is normally distributed with mean 64 ounces and standard deviation 1.732 ounces. The amount of lemonade in a glass is normally distributed with mean 12 ounces and standard deviation 1 ounce. (a) How much lemonade would you expect to be left in a pitcher after pouring one glass of lemonade? (b) What is the standard deviation of the amount left in a pitcher after pouring one glass of lemonade? (c) What is the probability that more than 50 ounces of lemonade is left in a pitcher after pouring one glass of lemonade? 3.40 Spray paint, Part I. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. Suppose also that you buy three cans of spray paint. (a) How much area would you expect to cover with these three cans of spray paint? (b) What is the standard deviation of the area you expect to cover with these three cans of spray paint? (c) The area you wanted to cover is 80 square feet. What is the probability that you will be able to cover this entire area with these three cans of spray paint? In Exercises 2.27 and 2.29 we saw two distributions for GRE scores: 3.41 GRE scores,
Part III. N (µ = 151, σ = 7) for the verbal part of the exam and N (µ = 153, σ = 7.67) for the quantitative part. Suppose performance on these two sections is independent. Use this information to compute each of the following: (a) The probability of a combined (verbal + quantitative) score above 320. (b) The score of a student who scored better than 90% of the test takers overall. 3.42 Betting on dinner, Part I. Suppose a restaurant is running a promotion where prices of menu items are random following some underlying distribution. If you’re lucky, you can get a basket of fries for $3, or if you’re not so lucky you might end up having to pay $10 for the same menu item. The price of basket of fries is drawn from a normal distribution with mean $6 and standard deviation of $2. The price of a fountain drink is drawn from a normal distribution with mean $3 and standard deviation of $1. What is the probability that you pay more than $10 for a dinner consisting of a basket of fries and a fountain drink? 3.5. GEOMETRIC DISTRIBUTION 193 3.5 Geometric distribution How many times should we expect to roll a die until we get a 1? How many people should we expect to see at a hospital until we get someone with blood type O+? These questions can be answered using the geometric distribution. Learning objectives 1. Determine if a scenario is geometric. 2. Calculate the probabilities of the possible values of a geometric random variable. 3. Find and interpret the mean (expected value) and standard deviation of a geometric distribu- tion. 4. Understand the shape of the geometric distribution. 3.5.1 Bernoulli distribution Many health insurance plans in the United States have a deductible, where the insured individual is responsible for costs up to the deductible, and then the costs above the deductible are shared between the individual and insurance company for the remainder of the year. Suppose a health insurance company found that 70% of the people they insure stay below their deductible in any given year. Each of these people can be thought of as a trial. We label a person a success if her healthcare costs do not exceed the deductible. We label a person a failure if she does exceed her deductible in the year. Because 70% of the individuals will not exceed their deductible, we denote the probability of a success
as p = 0.7. The probability of a failure is sometimes denoted with q = 1 − p, which would be 0.3 in for the insurance example. When an individual trial only has two possible outcomes, often labeled as success or failure, it is called a Bernoulli random variable. We chose to label a person who does not exceed her deductible as a “success” and all others as “failures”. However, we could just as easily have reversed these labels. The mathematical framework we will build does not depend on which outcome is labeled a success and which a failure, as long as we are consistent. 194 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Bernoulli random variables are often denoted as 1 for a success and 0 for a failure. In addition to being convenient in entering data, it is also mathematically handy. Suppose we observe ten trials Then the sample proportion, ˆp, is the sample mean of these observations: ˆp = # of successes # of trials = 10 = 0.6 This mathematical inquiry of Bernoulli random variables can be extended even further. Because 0 and 1 are numerical outcomes, we can deﬁne the mean and standard deviation of a Bernoulli random variable.66 BERNOULLI RANDOM VARIABLE If X is a random variable that takes value 1 with probability of success p and 0 with probability 1 − p, then X is a Bernoulli random variable with mean and standard deviation µ = p σ = p(1 − p) In general, it is useful to think about a Bernoulli random variable as a random process with only two outcomes: a success or failure. Then we build our mathematical framework using the numerical labels 1 and 0 for successes and failures, respectively. 66See Exercises 3.47 and 3.48 3.5. GEOMETRIC DISTRIBUTION 195 3.5.2 Geometric distribution The geometric distribution is used to describe how many trials it takes to observe a success. Let’s ﬁrst look at an example. EXAMPLE 3.83 Suppose we are working at the insurance company and need to ﬁnd a case where the person did not exceed her deductible as a case study. If the probability a person will not exceed her deductible is 0.7 and we are drawing people at random, what are the chances that the ﬁrst person will
not have exceeded her deductible, i.e. be a success? The second person? The third? What about the probability that we pull x − 1 cases before we ﬁnd the ﬁrst success, i.e. the ﬁrst success is the xth person? (If the ﬁrst success is the ﬁfth person, then we say x = 5.) The probability of stopping after the ﬁrst person is just the chance the ﬁrst person will not exceed her deductible: 0.7. The probability the second person is the ﬁrst to exceed her deductible is P (second person is the ﬁrst to exceed deductible) = P (the ﬁrst won’t, the second will) = (0.3)(0.7) = 0.21 Likewise, the probability it will be the third case is (0.3)(0.3)(0.7) = 0.063. If the ﬁrst success is on the xth person, then there are x − 1 failures and ﬁnally 1 success, which corresponds to the probability (0.3)x−1(0.7). This is the same as (1 − 0.7)x−1(0.7). Example 3.83 illustrates what the geometric distribution, which describes the waiting time until a success for independent and identically distributed (iid) Bernoulli random variables. In this case, the independence aspect just means the individuals in the example don’t aﬀect each other, and identical means they each have the same probability of success. The geometric distribution from Example 3.83 is shown in Figure 3.28. In general, the proba- bilities for a geometric distribution decrease exponentially fast. Figure 3.28: The geometric distribution when the probability of success is p = 0.7. While this text will not derive the formulas for the mean (expected) number of trials needed to ﬁnd the ﬁrst success or the standard deviation of this distribution, we present general formulas for each. ProbabilityNumber of Trials Until a Success for p = 0.7123456780.00.20.40.6 196 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS GEOMETRIC DISTRIBUTION Let X have a geometric distribution with one
parameter p, where p is the probability of a success in one trial. Then the probability of ﬁnding the ﬁrst success in the xth trial is given by P (X = x) = (1 − p)x−1p where x = 1, 2, 3,... The mean (i.e. expected value) and standard deviation of this wait time are given by µX = 1 p σX = √ 1 − p p It is no accident that we use the symbol µ for both the mean and expected value. The mean and the expected value are one and the same. It takes, on average, 1/p trials to get a success under the geometric distribution. This mathematical result is consistent with what we would expect intuitively. If the probability of a success is high (e.g. 0.8), then we don’t usually wait very long for a success: 1/0.8 = 1.25 trials on average. If the probability of a success is low (e.g. 0.1), then we would expect to view many trials before we see a success: 1/0.1 = 10 trials. GUIDED PRACTICE 3.84 The probability that a particular case would not exceed their deductible is said to be 0.7. If we were to examine cases until we found one that where the person did not exceed her deductible, how many cases should we expect to check?67 EXAMPLE 3.85 What is the chance that we would ﬁnd the ﬁrst success within the ﬁrst 3 cases? This is the chance the ﬁrst (X = 1), second (X = 2), or third (X = 3) case is the ﬁrst success, which are three disjoint outcomes. Because the individuals in the sample are randomly sampled from a large population, they are independent. We compute the probability of each case and add the separate results: P (X = 1, 2, or 3) = P (X = 1) + P (X = 2) + P (X = 3) = (0.3)1−1(0.7) + (0.3)2−1(0.7) + (0.3)3−1(0.7) = 0.973 There is a probability of 0.973 that we would ﬁ
nd a successful case within 3 cases. GUIDED PRACTICE 3.86 Determine a more clever way to solve Example 3.85. Show that you get the same result.68 67We would expect to see about 1/0.7 ≈ 1.43 individuals to ﬁnd the ﬁrst success. 68First ﬁnd the probability of the complement: P (no success in ﬁrst 3 trials) = 0.33 = 0.027. Next, compute one minus this probability: 1 − P (no success in 3 trials) = 1 − 0.027 = 0.973. 3.5. GEOMETRIC DISTRIBUTION 197 EXAMPLE 3.87 Suppose a car insurer has determined that 88% of its drivers will not exceed their deductible in a given year. If someone at the company were to randomly draw driver ﬁles until they found one that had not exceeded their deductible, what is the expected number of drivers the insurance employee must check? What is the standard deviation of the number of driver ﬁles that must be drawn? In this example, a success is again when someone will not exceed the insurance deductible, which has probability p = 0.88. The expected number of people to be checked is 1/p = 1/0.88 = 1.14 and the standard deviation is √ √ 1−p p = 1−0.88 0.88 = 0.39. GUIDED PRACTICE 3.88 Using the results from Example 3.87, where µX = 1.14 and σX = 0.39, would it be appropriate to use the empirical rule to ﬁnd what proportion of experiments would end in 1.14 ± 0.39 trials?69 The independence assumption is crucial to the geometric distribution’s accurate description of a scenario. Mathematically, we can see that to construct the probability of the success on the xth trial, we had to use the General Multiplication Rule for independent processes. It is no simple task to generalize the geometric model for dependent trials. 3.5.3 Technology: geometric probablities Get started quickly with this Desmos Geometric Calculator (available at openintro.org/ahss/desmos). 69No. The geometric distribution is always right skewed and can never be well-approximated by a normal model. 198 CHAPTER 3. PROBABILITY AND PROBABILITY DISTR
IBUTIONS Section summary • It is useful to model yes/no, success/failure with the values 1 and 0, respectively. We call the prob- ability of success p and the probability of failure 1 − p. • When the trials are independent and the value of p is constant, the probability of ﬁnding the ﬁrst success on the xth trial is given by (1 − p)x−1p. We can see the reasoning behind this formula as follows: for the ﬁrst success to happen on the xth trial, it has to not happen the ﬁrst x − 1 trials (with probability 1 − p), and then happen on the xth trial (with probability p). • When we consider the entire distribution of possible values for the how long until the ﬁrst success, we get a discrete probability distribution known as the geometric distribution. The geometric distribution describes the waiting time until the ﬁrst success, when the trials are independent and the probability of success, p, is constant. If X has a geometric distribution with parameter p, then P (X = x) = (1 − p)x−1p, where x = 1, 2, 3.... • The geometric distribution is always right skewed and, in fact, has no maximum value. The probabil- ities, though, decrease exponentially fast. • Even though the geometric distribution has an inﬁnite number of values, it has a well-deﬁned mean: 10, then on average. If the probability of success is 1 1−p µX = 1 p and standard deviation: σX = p it takes 10 trials until we see the ﬁrst success. √ • Note that when the trials are not independent, we can modify the geometric formula to ﬁnd the probability that the ﬁrst success happens on the xth trial. Instead of simply raising (1 − p) to the x − 1, multiply the appropriate conditional probabilities. 3.5. GEOMETRIC DISTRIBUTION 199 Exercises 3.43 Is it Bernoulli? Determine if each trial can be considered an independent Bernoulli trial for the following situations. (a) Cards dealt in a hand of poker. (b) Outcome of each roll of a die. 3.44 With and without replacement. In the following situations assume that
half of the speciﬁed population is male and the other half is female. (a) Suppose you’re sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you’re sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your ﬁndings from parts (a) and (b), explain whether or not this assumption is reasonable. 3.45 Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75), blue eyes (0.125), or green eyes (0.125). (a) What is the probability the ﬁrst blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. (b) On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the ﬁrst blue-eyed child? 3.46 Defective rate. A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the 10th transistor produced is the ﬁrst with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of 100? (c) On average, how many transistors would you expect to be produced until the ﬁrst with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the ﬁrst with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event aﬀect the mean and standard deviation of the wait time until success
? 3.47 Bernoulli, the mean. Use the probability rules from Section 3.4 to derive the mean of a Bernoulli random variable, i.e. a random variable X that takes value 1 with probability p and value 0 with probability 1 − p. That is, compute the expected value of a generic Bernoulli random variable. 3.48 Bernoulli, the standard deviation. Use the probability rules from Section 3.4 to derive the standard deviation of a Bernoulli random variable, i.e. a random variable X that takes value 1 with probability p and value 0 with probability 1 − p. That is, compute the square root of the variance of a generic Bernoulli random variable. 200 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.6 Binomial distribution What is the probability of exactly 50 heads in 100 coin tosses? Or the probability of randomly sampling 12 people and having more than 9 of them identify as male? If the probability of a defective part is 1%, how many defective items would we expect in a random shipment of 200 of those parts? We can model these scenarios and answer these questions using the binomial distribution. Learning objectives 1. Calculate the number of possible scenarios for obtaining x successes in n trials. 2. Determine whether a scenario is binomial or not. 3. Calculate the probabilities of the possible values of a binomial random variable using the binomial formula. 4. Recognize that the binomial formula uses the special Addition Rule for mutually exclusive events. 5. Find probabilities of the form “at least” or “at most” by applying the binomial formula multiple times. 6. Calculate and interpret the mean (expected value) and standard deviation of the number of successes in n binomial trials. 7. Determine whether a binomial distribution can be modeled as approximately normal. If so, use normal approximation to estimate cumulative binomial probabilities. 3.6.1 Introducing the binomial formula Let’s again imagine ourselves back at the insurance agency where 70% of individuals do not exceed their deductible. Each people is thought of as a trial. We label a trial a success if the individual healthcare costs do not exceed the deductible. We label a trial a failure if the individual healthcare costs do exceed the deductible. Because 70% of the individuals will not exceed their deductible, we denote the probability of a success as p = 0.7. EXAMPLE 3.89
Suppose the insurance agency is considering a random sample of four individuals they insure. What is the chance exactly one of them will exceed the deductible and the other three will not? Let’s call the four people Ariana (A), Brittany (B), Carlton (C), and Damian (D) for convenience. Let’s consider a scenario where one person exceeds the deductible: P (A = exceed, B = not, C = not, D = not) = P (A = exceed) P (B = not) P (C = not) P (D = not) = (0.3)(0.7)(0.7)(0.7) = (0.7)3(0.3)1 = 0.103 But there are three other scenarios: Brittany, Carlton, or Damian could have been the one to exceed the deductible. In each of these cases, the probability is again (0.7)3(0.3)1. These four scenarios exhaust all the possible ways that exactly one of these four people could have exceeded the deductible, so the total probability is 4 × (0.7)3(0.3)1 = 0.412. GUIDED PRACTICE 3.90 Verify that the scenario where Brittany is the only one to exceed the deductible has probability (0.7)3(0.3)1. 70 70 P (A = not, B = exceed, C = not, D = not) = (0.7)(0.3)(0.7)(0.7) = (0.7)3(0.3)1. 3.6. BINOMIAL DISTRIBUTION 201 The binomial distribution describes the probability of having exactly x successes in n independent trials with probability of a success p (in Example 3.89, n = 4, x = 3, p = 0.7). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use n, x, and p to obtain the probability. To do this, we reexamine each part of Example 3.89. There were four individuals who could have been the one to exceed the deductible, and each of these four scenarios had the same probability. Thus, we could identify the ﬁnal probability as [# of scenarios] × P (single scenario) The ﬁrst component of this equation is the number of ways
to arrange the x = 3 successes among the n = 4 trials. The second component is the probability of any of the four (equally probable) scenarios. Consider P (single scenario) under the general case of x successes and n − x failures in the n trials. In any such scenario, we apply the Multiplication Rule for independent events: px(1 − p)n−x This is our general formula for P (single scenario). Secondly, we introduce the binomial coeﬃcient, which gives the number of ways to choose x successes in n trials, i.e. arrange x successes and n − x failures: n x = n! x!(n − x)! The quantity n expression. x is read n choose x.71 The exclamation point notation (e.g. n!) denotes a factorial 0! = 1 1! = 1 2! = 2 × 1 = 2 3! = 4 × 3 × 2 × 1 = 24... n! = n × (n − 1) ×... × 3 × 2 × 1 Using the formula, we can compute the number of ways to choose x = 3 successes in n = 4 trials: 4 3 = 4! 3!(4 − 3)! = 4! 3!13 × 2 × 1)(1) = 4 This result is exactly what we found by carefully thinking of each possible scenario in Example 3.89. Substituting n choose x for the number of scenarios and px(1 − p)n−x for the single scenario probability yields the binomial formula. BINOMIAL FORMULA Suppose the probability of a single trial being a success is p. Then the probability of observing exactly x successes in n independent trials is given by P (X = x) = n x px(1 − p)n−x = n! x!(n − x)! px(1 − p)n−x 71Other notations for n choose x includes nCx, Cx n, and C(n, x). 202 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.6.2 When and how to apply the formula IS IT BINOMIAL? FOUR CONDITIONS TO CHECK. (1) The trials are independent. (2) The number of trials, n, is ﬁxed. (3) Each trial outcome can be classiﬁed as a success or failure.
(4) The probability of a success, p, is the same for each trial. EXAMPLE 3.91 What is the probability that 3 of 8 randomly selected individuals will have exceeded the insurance deductible, i.e. that 5 of 8 will not exceed the deductible? Recall that 70% of individuals will not exceed the deductible. We would like to apply the binomial model, so we check the conditions. The number of trials is ﬁxed (n = 8) (condition 2) and each trial outcome can be classiﬁed as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4). In the outcome of interest, there are x = 5 successes in n = 8 trials (recall that a success is an individual who does not exceed the deductible, and the probability of a success is p = 0.7. So the probability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by (0.7)5(1 − 0.7)8−5 = 8 5 8! 5!(5 − 3)! (0.7)5(1 − 0.7)8−5 = 8! 5!3! (0.7)5(0.3)3 Dealing with the factorial part: 8! 5!35 × 4 × 3 × 2 × 1)(3 × 2 × 1 = 56 Using (0.7)5(0.3)3 ≈ 0.00454, the ﬁnal probability is about 56 × 0.00454 ≈ 0.254. If you must calculate the binomial coeﬃcient by hand, it’s often useful to cancel out as many terms as possible in the top and bottom. See Section 3.6.3 for how to evaluate the binomial coeﬃcient and the binomial formula using a calculator. COMPUTING BINOMIAL PROBABILITIES The ﬁrst step in using the binomial model is to check that the model is appropriate. The second step is to identify n, p, and x. Finally, apply the binomial formula to determine the probability and interpret the results. 3.6. BINOMIAL DISTRIBUTION 203 EXAMPLE 3.92 Approximately 35% of a population has blood type O+. Suppose
four people show up at a hospital and we want to ﬁnd the probability that exactly one of them has blood type O+. Can we use the binomial formula? To check if the binomial model is appropriate, we must verify the conditions. 1. We will suppose that these 4 people comprise a random sample. This seems reasonable, since one person with a particular blood type showing up at a hospital seems unlikely to aﬀect the chance that other people with that blood type would show up at the hospital. This sample is without replacement, though, not with replacement, so the observations are not entirely independent. However, when the sample size is very small compared to the population size, we can treat the observations as if they were with replacement, since the composition of the population changes very little with each additional person sampled. Since we have a random sample of a very small percent of the population, we will consider the independence condition met. 2. We have a ﬁxed number of trials (n = 4). 3. Each outcome is a success or failure (blood type O+ or not blood type O+). 4. The probability of a success is the same for each trial since the individuals are like a random sample (p = 0.35 if we say a “success” is someone having blood type O+). SAMPLING WITHOUT REPLACEMENT When randomly sampling without replacement, if the sample size is small relative to the population size (rule of thumb: sample size less than 1/10 of the population size), we will consider the observations to be independent. EXAMPLE 3.93 Given that 35% of a population has blood type O+, what is the probabilty that in a random sample of 4 people: (a) none of them have blood type O+? (b) one will have blood type O+? (c) no more than one will have blood type O+? (a) P (X = 0) = 4 (0.35)0(0.65)4 = 1 × 1 × 0.654 = 0.654 = 0.179 0 Note that we could have answered this question without the binomial formula, using methods from the previous section. (b) P (X = 1) = 4 (c) This can be computed as the sum of parts (a) and (b): P (X = 0) + P (X = 1) = 0.179 + 0.384 = 0.563.
(0.35)1(0.65)3 = 0.384. 1 That is, there is about a 56.3% chance that no more than one of them will have blood type O+. GUIDED PRACTICE 3.94 What is the probability that at least 3 of 4 people in a random sample will have blood type O+ if 35% of the population has blood type O+?72 GUIDED PRACTICE 3.95 The probability that a random smoker will develop a severe lung condition in her lifetime is about 0.3. If you have 4 friends who smoke and you want to ﬁnd the probability that 1 of them will develop a severe lung condition in her lifetime, can you apply the binomial formula?73 72P (at least 3 of 4 have blood type O+) = P (X = 3) + P (X = 4) = 4 3 0.126 (0.35)3(0.65)1 + (0.35)4 = 0.111 + 0.015 = 73While conditions (2) and (3) are met, most likely the friends know each other, so the independence assumption (1) is probably not satisﬁed. For example, acquaintances may have similar smoking habits, or those friends might make a pact to quit together. Condition (4) is also not satisﬁed since this is not a random sample of people. 204 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.96 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles without replacement. Find the probability you get exactly 3 blue marbles. Because we are drawing without replacement the probability of success p is not the same for each trial. Also, the sample size is large compared to the population size (much greater than 1/10 of the population size), so we cannot treat these observations as independent and we cannot use the binomial formula. However, we can use the same logic to arrive at the following answer. P (X = 3) = (# of combinations with 3 blue) × P (3 blue and 2 red in a speciﬁc order) 5 3 × P (BBBRR) 5 3 4 13 × 3 12 × 2 11 × 9 10 × 8 9 = = = 0.1119 GUIDED PRACTICE 3.97 Draw 4 cards without replacement
from a deck of 52 cards. What is the probability that you get at least two hearts?74 Lastly, we consider the binomial coeﬃcient,, n choose x, under some special scenarios. GUIDED PRACTICE 3.98 Why is it true that n 0 = 1 and n = 1 for any number n?75 n GUIDED PRACTICE 3.99 How many ways can you arrange one success and n − 1 failures in n trials? How many ways can you arrange n − 1 successes and one failure in n trials?76 74P (at least 2 hearts in 4 draws from a deck) = 1 − [P (X = 0) + P (X = 1)] = 1 − [( 39 ( 13 52 )( 39 75Frame these expressions into words. How many diﬀerent ways are there to arrange 0 successes and n failures in 49 )] = 1 − [.0.3038 + 0.4388] = 0.2574. 50 )( 36 52 )( 38 51 )( 37 50 )( 37 51 )( 38 49 ) + 4 1 n trials? (1 way.) How many diﬀerent ways are there to arrange n successes and 0 failures in n trials? (1 way.) 76One success and n − 1 failures: there are exactly n unique places we can put the success, so there are n ways to arrange one success and n − 1 failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations: n 1 = n, n n − 1 = n 3.6. BINOMIAL DISTRIBUTION 205 3.6.3 Technology: binomial probabilities Get started quickly with this Desmos Binomial Calculator (available at openintro.org/ahss/desmos). Calculator instructions n n TI-83/84: COMPUTING THE BINOMIAL COEFFICIENT n x x x Use MATH, PRB, nCr to evaluate n choose r. Here r and x are diﬀerent letters for the same quantity. 1. Type the value of n. 2. Select MATH. 3. Right arrow to PRB. 4. Choose 3:nCr. 5. Type the value of x. 6. Hit ENTER. Example: 5 nCr 3 means 5 choose 3. n n CASIO FX-9750GII: COMPUTING THE BINOMIAL
COEFFICIENT n x x x 1. Navigate to the RUN-MAT section (hit MENU, then hit 1). 2. Enter a value for n. 3. Go to CATALOG (hit buttons SHIFT and then 7). 4. Type C (hit the ln button), then navigate down to the bolded C and hit EXE. 5. Enter the value of x. Example of what it should look like: 7C3. 6. Hit EXE. 206 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS P (X = x) = n P (X = x) = n TI-84: COMPUTING THE BINOMIAL FORMULA, P (X = x) = n px(1 − p)n−x px(1 − p)n−x px(1 − p)n−x x x x Use 2ND VARS, binompdf to evaluate the probability of exactly x occurrences out of n independent trials of an event with probability p. 1. Select 2ND VARS (i.e. DISTR) 2. Choose A:binompdf (use the down arrow to scroll down). 3. Let trials be n. 4. Let p be p 5. Let x value be x. 6. Select Paste and hit ENTER. TI-83: Do step 1, choose 0:binompdf, then enter n, p, and x separated by commas: binompdf(n, p, x). Then hit ENTER. P (X ≤ x) = n TI-84: COMPUTING P (X ≤ x) = n P (X ≤ x) = n p0(1 − p)n−0 +... + n p0(1 − p)n−0 +... + n p0(1 − p)n−0 +... + n Use 2ND VARS, binomcdf to evaluate the cumulative probability of at most x occurrences out of n independent trials of an event with probability p. px(1 − p)n−x px(1 − p)n−x px(1 − p)n−x x x x 0 0 0 1. Select 2ND VARS (i.e. DISTR) 2. Choose B:binomcdf (use the down arrow). 3. Let trials be n. 4. Let p be p 5. Let x value be x
. 6. Select Paste and hit ENTER. TI-83: Do steps 1-2, then enter the values for n, p, and x separated by commas as follows: binomcdf(n, p, x). Then hit ENTER. CASIO FX-9750GII: BINOMIAL CALCULATIONS 1. Navigate to STAT (MENU, then hit 2). 2. Select DIST (F5), and then BINM (F5). 3. Choose whether to calculate the binomial distribution for a speciﬁc number of successes, P (X = k), or for a range P (X ≤ k) of values (0 successes, 1 success,..., x successes). • For a speciﬁc number of successes, choose Bpd (F1). • To consider the range 0, 1,..., x successes, choose Bcd(F1). 4. If needed, set Data to Variable (Var option, which is F2). 5. Enter the value for x (x), Numtrial (n), and p (probability of a success). 6. Hit EXE. GUIDED PRACTICE 3.100 Find the number of ways of arranging 3 blue marbles and 2 red marbles.77 GUIDED PRACTICE 3.101 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get exactly 3 blue marbles.78 GUIDED PRACTICE 3.102 There are 13 marbles in a bag. 4 are blue and 9 are red. Randomly draw 5 marbles with replacement. Find the probability you get at most 3 blue marbles (i.e. less than or equal to 3 blue marbles).79 77Here n = 5 and x = 3. Doing 5 nCr 3 gives the number of combinations as 10. 78Here, n = 5, p = 4/13, and x = 3, so set trials = 5, p = 4/13 and x value = 3. The probability is 0.1396. 79Similarly, set trials = 5, p = 4/13 and x value = 3. The cumulative probability is 0.9662. 3.6. BINOMIAL DISTRIBUTION 207 3.6.4 An example of a binomial distribution In Guided Practice 3.93, we asked various probability questions regarding
the number of people out of 4 with blood type O+. We veriﬁed that the scenario was binomial and that each problem could be solved using the binomial formula. Instead of looking at it piecewise, we could describe the entire distribution of possible values and their corresponding probabilities. Since there are 4 people, there are several possible outcomes for the number who might have blood type O+: 0, 1, 2, 3, 4. We can make a distribution table with these outcomes. Recall that the probability of a randomly sampled person being blood type O+ is about 0.35. The binomial distribution is used to describe the number of successes in a ﬁxed number of trials. This is diﬀerent from the geometric distribution, which described the number of trials we must wait before we observe a success. 1 xi P (xi0.35)0(0.65)4 = 0.179 (0.35)1(0.65)3 = 0.384 (0.35)2(0.65)2 = 0.311 (0.35)3(0.65)1 = 0.111 (0.35)4(0.65)0 = 0.015 2 3 4 Figure 3.29: Probability distribution for the number with blood type O+ in a random sample of 4 people. This is a binomial distribution. Correcting for rounding error, the probabilities add up to 1, as they must for any probability distribution. 3.6.5 The mean and standard deviation of a binomial distribution Since this is a probability distribution we could ﬁnd its mean and standard deviation using the formulas from Chapter 3. Those formulas require a lot of calculations, so it is fortunate there’s an easier way to compute the mean and standard deviation for a binomial random variable. MEAN AND STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION For a binomial distribution with parameters n and p, where n is the number of trials and p is the probability of a success, the mean and standard deviation of the number of observed successes are µX = np σX = np(1 − p) EXAMPLE 3.103 If the probability that a person has blood type O+ is 0.35 and you have 40 randomly selected people, about how many would you expect to have blood type O+? What is the standard deviation of the number of people who would have blood
type O+ among the 40 people? We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas above. µX = np = 40(0.35) = 14 σX = np(1 − p) = 40(0.35)(0.65) = 3.0 The exact distribution is shown in Figure 3.30. 012340.00.10.20.30.4ProbabilityNumber With Blood Type O+ in a Random Sample of Size 4 208 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Figure 3.30: Distribution for the number of people with blood type O+ in a random sample of size 40, where p = 0.35. The distribution is binomial and is centered on 14 with a standard deviation of 3. 3.6.6 Normal approximation to the binomial distribution The binomial formula is cumbersome when the sample size (n) is large, particularly when we consider a range of observations. EXAMPLE 3.104 Find the probability that fewer than 12 out of 40 randomly selected people would have blood type O+, where probability of blood type O+ is 0.35. This is equivalent to asking, what is the probability of observing X = 0, 1, 2,..., or 11 with blood type O+ in a sample of size 40 when p = 0.35? We previously veriﬁed that this scenario is binomial. We can compute each of the 12 probabilities using the binomial formula and add them together to ﬁnd the answer: P (X = 0 or X = 1 or · · · or X = 11) = P (X = 0) + P (X = 1) + · · · + P (X = 11) (0.35)0(0.65)40 + 40 1 (0.35)1(0.65)39 + · · · + 40 11 (0.35)11(0.65)29 40 0 = = 0.21 If the true proportion with blood type O+ in the population is p = 0.35, then the probability of observing fewer than 12 in a sample of n = 40 is 0.21. The computations in Example 3.6.6 are tedious and long. In general, we should avoid such work if an alternative method exists that is faster, easier, and still accurate. Recall that calculating
probabilities of a range of values is much easier in the normal model. In some cases we may use the normal distribution to estimate binomial probabilities. While a normal approximation for the distribution in Figure 3.29 when the sample size was n = 4 would not be appropriate, it might not be too bad for the distribution in Figure 3.30 where n = 40. We might wonder, when is it reasonable to use the normal model to approximate a binomial distribution? GUIDED PRACTICE 3.105 Here we consider the binomial model when the probability of a success is p = 0.10. Figure 3.31 shows four hollow histograms for simulated samples from the binomial distribution using four diﬀerent sample sizes: n = 10, 30, 100, 300. What happens to the shape of the distributions as the sample size increases? How does the binomial distribution change as n gets larger?80 The shape of the binomial distribution depends upon both n and p. Here we introduce a rule of thumb for when normal approximation of a binomial distribution is reasonable. We will use this rule of thumb in many applications going forward. 80The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normal distribution in the last hollow histogram. Number With Blood Type O+ in a Random Sample of Size 40Probability8111417200.000.050.10 3.6. BINOMIAL DISTRIBUTION 209 Figure 3.31: Hollow histograms of samples from the binomial model when p = 0.10. The sample sizes for the four plots are n = 10, 30, 100, and 300, respectively. NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION The binomial distribution with probability of success p is nearly normal when the sample size n is suﬃciently large that np ≥ 10 and n(1 − p) ≥ 10. The approximate normal distribution has parameters corresponding to the mean and standard deviation of the binomial distribution: µ = np σ = np(1 − p) The normal approximation may be used when computing the range of many possible successes. For instance, we may apply the normal distribution to the setting described in Figure 3.30. EXAMPLE 3.106 Use the normal approximation to estimate the probability of observing fewer than 12 with blood type O+ in a random sample of 40, if the true proportion with blood type O+ in the population is p = 0.35.
First we verify that np and n(1 − p) are at least 10 so that we can apply the normal approximation to the binomial model: np = 40(0.35) = 14 ≥ 10 n(1 − p) = 40(0.65) = 26 ≥ 10 With these conditions checked, we may use the normal distribution to approximate the binomial distribution with the following mean and standard deviation: µ = np = 40(0.35) = 14 σ = np(1 − p) = 40(0.35)(0.65) = 3.0 We want to ﬁnd the probability of observing fewer than 12 with blood type O+ using this model. We note that 12 is less than 1 standard deviation below the mean: Next, we compute the Z-score as Z = 12−14 3 = −0.67 to ﬁnd the shaded area in the picture: P (Z < −0.67) = 0.25. This probability of 0.25 using the normal approximation is reasonably close to the true probability of 0.21 computed using the binomial distribution. n = 100246n = 300246810n = 10005101520n = 3001020304050581114172023 210 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS EXAMPLE 3.107 Use the normal approximation to estimate the probability of observing fewer than 120 people with blood type O+ in a random sample of 400, if the true proportion with blood type O+ in the population is p = 0.35. We have previously veriﬁed that the binomial model is reasonable for this context. Now we will verify that both np and n(1 − p) are at least 10 so we can apply the normal approximation to the binomial model: np = 400(0.35) = 140 ≥ 10 n(1 − p) = 400(0.35) = 260 ≥ 10 With these conditions checked, we may use the normal approximation in place of the binomial distribution with the following mean and standard deviation: µ = np = 400(0.35) = 140 σ = np(1 − p) = 400(0.35)(0.65) = 9.5 We want to ﬁnd the probability of observing fewer than 120 with blood type O+ using this model. We note that 120 is just over 2 standard deviations below the mean: Next, we
compute the Z-score as Z = 120−140 9.5 = −2.1 to ﬁnd the shaded area in the picture: P (Z < −2.1) = 0.0179. This probability of 0.0179 using the normal approximation is very close to the true probability of 0.0196 from the binomial distribution. GUIDED PRACTICE 3.108 Use normal approximation, if applicable, to estimate the probability of getting greater than 15 sixes in 100 rolls of a fair die.81 3.6.7 Normal approximation breaks down on small intervals (special topic) THE NORMAL APPROXIMATION MAY FAIL ON SMALL INTERVALS The normal approximation to the binomial distribution tends to perform poorly when estimating the probability of a small range of counts, even when the conditions are met. We consider again our example where 35% of people are blood type O+. Suppose we want to ﬁnd the probability that between 129 and 131 people, inclusive, have blood type O+ in a random sample of 400 people. We want to compute the probability of observing 129, 130, or 131 people with blood type O+ when p = 0.20 and n = 400. With such a large sample, we might be tempted to apply the normal approximation and use the range 129 to 131. However, we would ﬁnd that the binomial solution and the normal approximation notably diﬀer: Binomial: 0.0732 Normal: 0.0483 81np = 100(1/6) = 16.7 ≥ 10 and n(1 − p) = 100(5/6) = 83.3 ≥ 10 µ = np = 100(1/6) = 16.7; σ = np(1 − p) = 100(1/6)(5/6) = 3.7 Z = 15−16.7 3.7 = −0.46. P (Z > −0.46) = 0.677 111.5121130.5140149.5159168.5 3.6. BINOMIAL DISTRIBUTION 211 We can identify the cause of this discrepancy using Figure 3.32, which shows the areas representing the binomial probability (outlined) and normal approximation (shaded). Notice that the width of the area under the normal distribution is 0.5 units too slim on both sides of the interval. The binomial distribution is a discrete distribution,
and the each bar is centered over an integer value. Looking closely at Figure 3.32, we can see that the bar corresponding to 129 begins at 128.5 and ends at 129.5, the bar corresponding to 131 begins at 130.5 and ends at 131.5, etc. Figure 3.32: A normal curve with the area between 129 and 131 shaded. The outlined area from 128.5 to 131.5 represents the exact binomial probability. IMPROVING ACCURACY OF THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION The normal approximation to the binomial distribution for intervals of values is usually improved if cutoﬀ values for the lower end of a shaded region are reduced by 0.5 and the cutoﬀ value for the upper end are increased by 0.5. This correction is called the continuity correction and accounts for the fact that the binomial distribution is discrete. EXAMPLE 3.109 Use the method described to ﬁnd a more accurate estimate for the probability of observing 129, 130, or 131 people with blood type O+ in 400 randomly selected people when p = 0.35. Instead of standardizing 129 and 131, we will standardize 128.5 and 131.5: Zlef t = Zright = 128.5 − 140 9.5 131.5 − 140 9.5 = −1.263 = −0.895 P (−1.263 < Z < −0.895) = 0.0772 The probability 0.0772 is much closer to the true value of 0.0732 than the previous estimate of 0.0483 we calculated using normal approximation without the continuity correction. It is always possible to apply the continuity correction when ﬁnding a normal approximation to the binomial distribution. However, when n is very large or when the interval is wide, the beneﬁt of the modiﬁcation is limited since the added area becomes negligible compared to the overall area being calculated. 120130140150160 212 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Section summary • n x, the binomial coeﬃcient, describes the number of combinations for arranging x successes among n trials. n = x n! x!(n−x)!, where n! = 1 × 2 × 3 ×...n, and 0!=0. • The binomial formula can
be used to ﬁnd the probability that something happens exactly x times in n trials. Suppose the probability of a single trial being a success is p. Then the probability of observing exactly x successes in n independent trials is given by n x px(1 − p)n−x = n! x!(n − x)! px(1 − p)n−x • To apply the binomial formula, the events must be independent from trial to trial. Additionally, n, the number of trials must be ﬁxed in advance, and p, the probability of the event occurring in a given trial, must be the same for each trial. • To use the binomial formula, ﬁrst conﬁrm that the binomial conditions are met. Next, identify the number of trials n, the number of times the event is to be a “success” x, and the probability that a single trial is a success p. Finally, plug these three numbers into the formula to get the probability of exactly x successes in n trials. • To ﬁnd a probability involving at least or at most, ﬁrst determine if the scenario is binomial. If so, apply the binomial formula as many times as needed and add up the results, e.g. P (at least 3 Heads in 5 tosses of a fair coin) = P (exactly 3 Heads) + P (exactly 4 Heads) + P (exactly 5 Heads), where each probability can be found using the binomial formula. • The distribution of the number of successes in n independent trials gives rise to a binomial distri- bution. If X has a binomial distribution with parameters n and p, then px(1 − p)n−x, where x = 0, 1, 2, 3..., n. P (X = x) = n x • To write out a binomial probability distribution table, list all possible values for x, the number of successes, then use the binomial formula to ﬁnd the probability of each of those values. • If X follows a binomial distribution with parameters n and p, then: – The mean is given by µX = np. – The standard deviation is given by σX = np(1 − p). – When np ≥ 10 and n(1 − p) ≥ 10, the binomial distribution is approximately normal. (spread ) (
center ) (shape) 3.6. BINOMIAL DISTRIBUTION 213 Exercises 3.49 Exploring combinations. A coin is tossed 5 times. How many sequences / combinations of Heads/Tails are there that have: (a) Exactly 1 Tail? (b) Exactly 4 Tails? (c) Exactly 3 Tails? (d) At least 3 Tails? 3.50 Political afﬁliation. Suppose that in a large population, 51% identify as Democrat. A researcher takes a random sample of 3 people. (a) Use the binomial model to calculate the probability that two of them identify as Democrat. (b) Write out all possible orderings of 3 people, 2 of whom identify as Democrat. Use these scenarios to calculate the same probability from part (a) but using the Addition Rule for disjoint events. Conﬁrm that your answers from parts (a) and (b) match. (c) If we wanted to calculate the probability that a random sample of 8 people will have 3 that identify as Democrat, brieﬂy describe why the approach from part (b) would be more tedious than the approach from part (a). 3.51 Underage drinking, Part I. Data collected by the Substance Abuse and Mental Health Services Administration (SAMSHA) suggests that 69.7% of 18-20 year olds consumed alcoholic beverages in any given year.82 (a) Suppose a random sample of ten 18-20 year olds is taken. Is the use of the binomial distribution appropriate for calculating the probability that exactly six consumed alcoholic beverages? Explain. (b) Calculate the probability that exactly 6 out of 10 randomly sampled 18- 20 year olds consumed an alcoholic drink. (c) What is the probability that exactly four out of ten 18-20 year olds have not consumed an alcoholic beverage? (d) What is the probability that at most 2 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? (e) What is the probability that at least 1 out of 5 randomly sampled 18-20 year olds have consumed alcoholic beverages? 3.52 Chickenpox, Part I. Boston Children’s Hospital estimates that 90% of Americans have had chickenpox by the time they reach adulthood.83 (a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox
during childhood? Explain. (b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood. (c) What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood? (d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox? (e) What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox? 82SAMHSA, Oﬃce of Applied Studies, National Survey on Drug Use and Health, 2007 and 2008. 83Boston Children’s Hospital, Chickenpox summary page, referenced April 29, 2021. 214 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.53 Game of dreidel. A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting (a) at least one nun? (b) exactly 2 nuns? (c) exactly 1 hei? (d) at most 2 gimel s? Photo by Staccabees, cropped (http://ﬂic.kr/p/7gLZTf) CC BY 2.0 license 3.54 Sickle cell anemia. Sickle cell anemia is a genetic blood disorder where red blood cells lose their ﬂexibility and assume an abnormal, rigid, “sickle” shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that (a) two will have the disease? (b) none will have the disease? (c) at least one will neither have the disease nor be a carrier? (d) the ﬁrst child with the disease will the be 3rd child? We learned in Exercise 3.51 that about 70% of 18-20 year olds 3.55 Underage drinking, Part II. consumed alcoholic beverages in any given year.
We now consider a random sample of ﬁfty 18-20 year olds. (a) How many people would you expect to have consumed alcoholic beverages? And with what standard deviation? (b) Would you be surprised if there were 45 or more people who have consumed alcoholic beverages? (c) What is the probability that 45 or more people in this sample have consumed alcoholic beverages? How does this probability relate to your answer to part (b)? 3.56 Chickenpox, Part II. We learned in Exercise 3.52 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults. (a) How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation? (b) Would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? How does this probability relate to your answer to part (b)? 3.6. BINOMIAL DISTRIBUTION 215 Chapter highlights This chapter focused on understanding likelihood and chance variation, ﬁrst by solving individual probability questions and then by investigating probability distributions. The main probability techniques covered in this chapter are as follows: • The General Multiplication Rule for and probabilities (intersection), along with the special case when events are independent. • The General Addition Rule for or probabilities (union), along with the special case when events are mutually exclusive. • The Conditional Probability Rule. • Tree diagrams and Bayes’ Theorem to solve more complex conditional problems. • Simulations and the use of random digits to estimate probabilities. Fundamental to all of these problems is understanding when events are independent and when they are mutually exclusive. Two events are independent when the outcome of one does not aﬀect the outcome of the other, i.e. P (A\|B) = P (A). Two events are mutually exclusive when they cannot both happen together, i.e. P (A and B) = 0. Moving from solving individual probability questions to studying probability distributions helps us better understand chance processes and quantify expected chance variation. • For a discrete probability distribution, the sum of the probabilities must equal 1. • As with any distribution, one can calculate the mean and standard deviation of a probability distribution. In the context of a probability distribution, the mean and standard deviation describe the average and the typical deviation
from the average, respectively, after many, many repetitions of the chance process. • A probability distribution can be summarized by its center (mean, median), spread (SD, IQR), and shape (right skewed, left skewed, approximately symmetric). • Adding a constant to every value in a probability distribution adds that value to the mean, but it does not aﬀect the standard deviation. When multiplying every value by a constant, this multiplies the mean by the constant and it multiplies the standard deviation by the absolute value of the constant. • The mean of the sum of two random variables equals the sum of the means. However, this is not true for standard deviations. Instead, when ﬁnding the standard deviation of a sum or diﬀerence of random variables, take the square root of the sum of each of the standard deviations squared. • The geometric distribution provides a model for the number of trials until the ﬁrst success, when the trials are independent. • The binomial distribution provides a model for the number of successes in n independent trials. • The geometric distribution is always right skewed. However, when the success-failure rule is met (at least 10 success and 10 failures), the binomial distribution can be modeled using a normal distribution with mean = np and standard deviation np(1 − p). The study of probability is useful for measuring uncertainty and assessing risk. In addition, probability serves as the foundation for inference, providing a framework for evaluating when an outcome falls outside of the range of what would be expected by chance alone. 216 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS Chapter exercises 3.57 Grade distributions. Each row in the table below is a proposed grade distribution for a class. Identify each as a valid or invalid probability distribution, and explain your reasoning. A 0.3 0 0.3 0.3 0.2 0 (a) (b) (c) (d) (e) (f) Grades C 0.3 1 0.3 0.2 0.2 1.1 B 0.3 0 0.3 0.5 0.4 -0.1 D 0.2 0 0 0.1 0.1 0 F 0.1 0 0 -0.1 0.1 0 3.58 Health coverage, frequencies. The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and
report emerging health trends. The following table summarizes two variables for the respondents: health status and health coverage, which describes whether each respondent had health insurance.84 Health Coverage No Yes Total Health Status Excellent Very good Good 854 4,821 5,675 459 4,198 4,657 727 6,245 6,972 Fair Poor 99 385 578 1,634 677 2,019 Total 2,524 17,476 20,000 (a) If we draw one individual at random, what is the probability that the respondent has excellent health and doesn’t have health coverage? (b) If we draw one individual at random, what is the probability that the respondent has excellent health or doesn’t have health coverage? 3.59 HIV in Eswatini. Eswatini (formerly named in English as Swaziland) has the highest HIV prevalence in the world: 25.9% of this country’s population is infected with HIV.85 The ELISA test is one of the ﬁrst and most accurate tests for HIV. For those who carry HIV, the ELISA test is 99.7% accurate. For those who do not carry HIV, the test is 92.6% accurate. If an individual from Eswatini has tested positive, what is the probability that he carries HIV? 3.60 Twins. About 30% of human twins are identical, and the rest are fraternal. Identical twins are necessarily the same sex – half are males and the other half are females. One-quarter of fraternal twins are both male, one-quarter both female, and one-half are mixes: one male, one female. You have just become a parent of twins and are told they are both girls. Given this information, what is the probability that they are identical? 3.61 Cost of breakfast. Sally gets a cup of coﬀee and a muﬃn every day for breakfast from one of the many coﬀee shops in her neighborhood. She picks a coﬀee shop each morning at random and independently of previous days. The average price of a cup of coﬀee is $1.40 with a standard deviation of 30¢ ($0.30), the average price of a muﬃn is $2.50 with a standard deviation of 15¢, and the two prices are independent of each other. (a) What is the mean and standard deviation
of the amount she spends on breakfast daily? (b) What is the mean and standard deviation of the amount she spends on breakfast weekly (7 days)? 84Oﬃce of Surveillance, Epidemiology, and Laboratory Services Behavioral Risk Factor Surveillance System, BRFSS 2010 Survey Data. 85Source: CIA Factbook, Country Comparison: HIV/AIDS - Adult Prevalence Rate. 3.6. BINOMIAL DISTRIBUTION 217 3.62 Scooping ice cream. Ice cream usually comes in 1.5 quart boxes (48 ﬂuid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y. Suppose these random variables have the following means, standard deviations, and variances: mean 48 2 SD variance 1 0.25 1 0.0625 X Y (a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served? (b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, ﬁnd the expected value of X − Y. What is the standard deviation of the amount left in the box? (c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another. 3.63 College smokers. At a university, 13% of students smoke. (a) Calculate the expected number of smokers in a random sample of 100 students from this university. (b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are 27 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 27 students? 3.64 Speeding on the I-5, Part II. Exercise 2.53 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70
miles/hour. (a) A highway patrol oﬃcer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (b) On average, how many cars would the highway patrol oﬃcer expect to watch until the ﬁrst car that is speeding? What is the standard deviation of the number of cars he would expect to watch? 3.65 Roulette winnings. In the game of roulette, a wheel is spun and you place bets on where it will stop. One popular bet is that it will stop on a red slot; such a bet has an 18/38 chance of winning. If it stops on red, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, each time with a $1 bet. Let Y represent the total amount won or lost. Write a probability model for Y. 3.66 Multiple choice quiz. In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that (a) the ﬁrst question she gets right is the 3rd question? (b) she gets exactly 3 or exactly 4 questions right? (c) she gets the majority of the questions right? 218 Chapter 4 Sampling distributions 4.1 Sampling distribution for a sample proportion 4.2 Sampling distribution for a sample mean 4.3 Sampling distribution for a difference of proportions or means 219 At the end of the previous chapter, we explored the Geometric Distribution and the Binomial Distribution. In this chapter, we consider what is called the sampling distribution, which is a term for describing the distribution of a statistic. For example, if we take a survey of a random sample of a population and compute the proportion of “yes” responses to a particular question, then it’s helpful to consider the distribution of all possible sample proportions. We explore sampling distributions for two common statistics: the sample proportion and the sample mean. We then consider the distributions of a diﬀerence of sample proportions and a diﬀerence of sample means. In each case, the goal is the same – to describe the center, spread, and shape of the sampling distribution and determine when the normal approximation to
the sampling distribution is reasonable. For videos, slides, and other resources, please visit www.openintro.org/ahss 220 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1 Sampling distribution for a sample proportion Often, instead of the number of successes in n trials, we are interested in the proportion of successes in n trials. We can use the sampling distribution for a sample proportion to answer questions such as the following: • Given a fair coin, what is the probability that in 200 tosses you would get greater than 52% Tails just by random variation? • In a particular state, 48% support a controversial measure. When estimating the percent through polling, what is the probability that a random sample of size 200 will mistakenly estimate the percent support to be greater than 50%? Learning objectives 1. Understand the concept of a sampling distribution. 2. Describe the center, spread, and shape of the sampling distribution for a sample proportion. 3. Recognize the relationship between the distribution of a sample proportion and the corre- sponding binomial distribution. 4. Explain the Central Limit Theorem and what it says about the shape of the sampling distri- bution for a sample proportion. 5. Verify appropriate conditions and, if met, carry out normal approximation for a sample pro- portion or sample count. 4.1.1 The mean and standard deviation of ˆpˆpˆp To answer the two questions posed at the beginning of this section, we investigate the distribution of the sample proportion ˆp. In the previous section, we saw that the number of people with blood type O+ in a random sample of size 40 follows a binomial distribution with n = 40 and p = 0.35 that is centered on 14 and has standard deviation 3.0. What does the distribution of the proportion of people with blood type O+ in a sample of size 40 look like? To convert from a count to a proportion, we divide the count (i.e. number of yeses) by the sample size, n = 40. For example, 8 becomes 8/40 = 0.20 as a proportion and 11 becomes 11/40 = 0.275. We can ﬁnd the general formula for the mean (expected value) and standard deviation of a sample proportion ˆp using our tools that we’ve learned so far. To get the sample mean for ˆp, we divide the binomial mean µbinomial
= np by n: µ ˆp = µbinomial n = np n = p As one might expect, the sample proportion ˆp is centered on the true proportion p. Likewise, the standard deviation of ˆp is equal to the standard deviation of the binomial distribution divided by n: σ ˆp = σbinomial n = np(1 − p) n = p(1 − p) n 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 221 MEAN AND STANDARD DEVIATION OF A SAMPLE PROPORTION The mean and standard deviation of the sample proportion describe the center and spread of the distribution of all possible sample proportions ˆp from a random sample of size n with true population proportion p. µ ˆp = p σ ˆp = p(1 − p) n The standard deviation of a sample proportion, σ ˆp, tells us about the “typical” deviation in the sample proportions from the true population proportion. In analyses, we think of the standard deviation of a sample proportion as describing the uncertainty associated with the estimate ˆp. That is, σ ˆp can be thought of as a way to quantify the typical error in our sample estimate ˆp of the true proportion p. Understanding the variability of statistics such as ˆp is a central component in the study of statistics. In our blood type O+ example, we have n = 40 and p = 0.35. If we look at the distribution of all possible values of a sample proportion for random samples of size 40 from this population, it is = 0.075. We see in Figure 4.1 centered on µ ˆp = 0.35 and has standard deviation σ ˆp = that the distribution of proportion of people in a sample of size 40 with blood type O+ is equivalent to the distribution of number of people in a sample with blood type O+ out of 40, but with a change of scale. Instead of counts along the horizontal axis, we have proportions. 0.35(1−0.35) 40 (a) (b) Figure 4.1: Two distributions where p = 0.35 and n = 40: the binomial distribution for the number with blood type O+ and the sampling distribution for the proportion with blood type O+. EXAMPLE 4.1 If the
proportion of people in the county with blood type O+ is really 35%, ﬁnd and interpret the mean and standard deviation of the sample proportion for a random sample of size 400. The mean of the distribution of the sample proportion is the population proportion: 0.35. That is, the distribution of all possible values for the sample proportion is centered on 0.35. In other words, if we took many, many samples and calculated ˆp, these values would average out to 0.35. The standard deviation of ˆp is described by the standard deviation for the proportion: σ ˆp = p(1 − p) n = 0.35(0.65) 400 = 0.024 The sample proportion will typically be about 0.024 or 2.4% away from the true proportion of p = 0.35. We’ll become more rigorous about quantifying how close ˆp will tend to be to p in Chapter 5. Number With Blood Type O+ in a Random Sample of Size 40Probability8111417200.000.050.10Proportion With Blood Type O+ in a Random Sample of Size 40Probability0.200.2750.350.4250.500.000.050.10 222 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1.2 Central Limit Theorem The distribution in Figure 4.1 looks an awful lot like a normal distribution. That is no anomaly; it is the result of a general principle called the Central Limit Theorem. CENTRAL LIMIT THEOREM AND THE SUCCESS-FAILURE CONDITION When observations are independent and the sample size is suﬃciently large, the sample proportion ˆp will tend to follow a normal distribution with the following mean and standard deviation: µ ˆp = p σ ˆp = p(1 − p) n In order for the Central Limit Theorem to hold, the sample size is typically considered suﬃciently large when np ≥ 10 and n(1 − p) ≥ 10, which is called the success-failure condition. The Central Limit Theorem is incredibly important, and it provides a foundation for much of statistics. As we begin applying the Central Limit Theorem, be mindful of the two technical conditions: the observations must be independent, and the sample size must be suﬃciently large such that np ≥ 10
and n(1 − p) ≥ 10. HOW TO VERIFY SAMPLE OBSERVATIONS ARE INDEPENDENT If the observations are from a random process such as tossing a coin, then they are independent. If the observations are from a random sample with replacement, then they are independent. If the observations are from a simple random sample (without replacement), we can treat them as independent if the sample size is less than 10% of the population size. If a sample is from a seemingly random process, e.g. an occasional error on an assembly line, checking independence is more diﬃcult. In this case, use your best judgement. When the sample exceeds 10% of the population size, the methods we discuss tend to overes- timate the sampling error slightly versus what we would get using more advanced methods.1 An interesting question to answer is, what happens when np < 10 or n(1 − p) < 10? We can simulate drawing samples of diﬀerent sizes where, say, the true proportion is p = 0.25. Here’s a sample of size 10: no, no, yes, yes, no, no, no, no, no, no In this sample, we observe a sample proportion of yeses of ˆp = 2 10 = 0.2. We can simulate many such proportions to understand the sampling distribution for ˆp when n = 10 and p = 0.25, which we’ve plotted in Figure 4.2 alongside a normal distribution with the same mean and variability. These distributions have a number of imprrtant diﬀerences. Normal: N (0.25, 0.14) n = 10, p = 0.25 Unimodal? Yes Yes Smooth? Yes No Symmetric? Yes No Notice that the success-failure condition was not satisﬁed when n = 10 and p = 0.25: np = 10 × 0.25 = 2.5 n(1 − p) = 10 × 0.75 = 7.5 This single sampling distribution does not show that the success-failure condition is the perfect guideline, but we have found that the guideline did correctly identify that a normal distribution might not be appropriate. 1For example, we could use what’s called the ﬁnite population correction factor: if the sample is of size n N −1 to obtain a and the population size is
N, then we can multiple the typical standard deviation formula by smaller, more precise estimate of the actual standard deviation. When n < 0.1 × N, this correction factor is relatively close to 1. N −n 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 223 Figure 4.2: Left: simulations of ˆp when the sample size is n = 10 and the population proportion is p = 0.25. Right: a normal distribution with the same mean (0.25) and standard deviation (0.137). We can complete several additional simulations, shown in Figures 4.3 and 4.4, and we can see some trends: 1. When either np or n(1 − p) is small, the distribution is more discrete, i.e. not continuous. 2. When np or n(1 − p) is smaller than 10, the skew in the distribution is more noteworthy. 3. The larger both np and n(1 − p), the more normal the distribution. This may be a little harder to see for the larger sample size in these plots as the variability also becomes much smaller. 4. When np and n(1 − p) are both very large, the distribution’s discreteness is hardly evident, and the distribution looks much more like a normal distribution. So far we’ve only focused on the skew and discreteness of the distributions. We haven’t considered how the mean and standard error of the distributions change. Take a moment to look back at the graphs, and pay attention to three things: 1. The centers of the distribution are always at the population proportion, p, that was used to generate the simulation. Because the sampling distribution for ˆp is always centered at the population parameter p, it means the sample proportion ˆp is unbiased when the data are independent and drawn from such a population. 2. For a particular population proportion p, the variability in the sampling distribution decreases as the sample size n becomes larger. This will likely align with your intuition: an estimate based on a larger sample size will tend to be more accurate. 3. For a particular sample size, the variability will be largest when p = 0.5. The diﬀerences may be a little subtle, so take a close look. This reﬂects the role of the proportion p in the standard error formula: SE =. The standard error is largest when p = 0
.5. p(1−p) n At no point will the distribution of ˆp look perfectly normal, since ˆp will always be take discrete values (x/n). It is always a matter of degree, and we will use the standard success-failure condition with minimums of 10 for np and n(1 − p) as our guideline within this book. THREE IMPORTANT FACTS ABOUT THE DISTRIBUTION OF A SAMPLE PROPORTION ˆpˆpˆp When the observations can be considered independent, such as from a random sample of less than 10% of the population, the distribution of the sample proportion can be described as follows. 1. CENTER: The mean of a sample proportion is p. 2. SPREAD: The SD of a sample proportion is p(1−p) n. 3. SHAPE: When np ≥ 10 and n(1 − p) ≥ 10, the sample proportion closely follows a normal distribution. Using these facts, we can now answer the question posed at the beginning of this section. Sample Proportions0.00.10.20.30.40.50.60.70.8010002000Frequency−0.20.00.20.40.6 224 CHAPTER 4. SAMPLING DISTRIBUTIONS Figure 4.3: Sampling distributions for several scenarios of p and n. Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90. Columns: n = 10 and n = 25. n = 10n = 25p = 0.10.00.20.40.60.81.00.00.20.40.60.81.0p = 0.20.00.20.40.60.81.00.00.20.40.60.81.0p = 0.50.00.20.40.60.81.00.00.20.40.60.81.0p = 0.80.00.20.40.60.81.00.00.20.40.60.81.0p = 0.90.00.20.40.60.81.00.00.20.40.60.81.0 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 225 Figure 4.
4: Sampling distributions for several scenarios of p and n. Rows: p = 0.10, p = 0.20, p = 0.50, p = 0.80, and p = 0.90. Columns: n = 50, n = 100, and n = 250. n = 50n = 100n = 2500.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.00.00.20.40.60.81.0 226 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.1.3 Normal approximation for the sampling distribution for ˆpˆpˆp EXAMPLE 4.2 Find the probability that less than 30% of a random sample of 400 people will be blood type O+ if the population proportion is 35%. In the previous section we veriﬁed that the observations can be treated as independent and that np and n(1 − p) are at least 10. The mean of the sample proportion is 0.35 and the standard deviation for the sample proportion is given by approximation to estimate the probability: 0.35(1−0.35) 400 = 0.024. We can ﬁnd a Z-score and use normal Z = ˆp − µ ˆp σ ˆp = 0.30 − 0.35 0.024 = −2.1 P (Z < −2.1) = 0.0179 We leave it to the reader to construct a ﬁgure for this example. EXAMPLE 4.3 The probability 0.0179 is the same probability we calculated when we found the probability of getting fewer than 120 with blood type O+ out of
400! Why is this? Notice that 120/400 = 0.30. Using the binomial distribution to ﬁnd the probability of fewer than 120 with blood type O+ in the sample is equivalent to using the distribution of ˆp to ﬁnd the probability of a sample proportion less than 0.30. GUIDED PRACTICE 4.4 Given a population that is 50% male, what is the probability that a random sample of size 200 would have greater than 55% males? Remember to verify that conditions for normal approximation are met.2 2First, verify the conditions: There is a random sample, and the sample size is much smaller than the population size, so observations can be considered independent. Also, np = 200(0.5) = 100 ≥ 10 and n(1 − p) = 200(0.5) = 100 ≥ 10, so the normal approximation is reasonable. Next we ﬁnd the mean and standard deviation of ˆp: µ ˆp = p = 0.50 σ ˆp = p(1 − p) n = 0.5(0.5) 200 = 0.0354 Then we ﬁnd a Z-score and ﬁnd the upper tail of the normal distribution: Z = ˆp − µ ˆp σ ˆp = 0.55 − 0.5 0.0354 = 1.412 → P (Z > 1.412) = 0.07 The probability of getting a sample proportion greater than 55% is about 0.07. 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 227 Section summary • A Z-score represents the number of standard deviations a value in a data set is above or below the mean. To calculate a Z-score use: Z = x−mean SD. • The standard deviation of ˆp describes the typical error or distance of the sample proportion from the population proportion. It also tells us how much the sample proportion is likely to vary from one random sample to another. • The sampling distribution for the sample proportion ˆp for a random sample of size n is identical to the binomial distribution with parameters n and p, but with a change of scale, i.e. diﬀerent mean and diﬀerent SD, but same shape. • The same success-failure condition for the binomial distribution
holds for a sample propor- tion ˆp. • Three important facts about the sampling distribution for the sample proportion ˆp, where the observations can be considered independent: – The mean of a sample proportion is denoted by µ ˆp, and it is equal to p. (center ) – The SD of a sample proportion is denoted by σ ˆp, and it is equal to p(1−p) n. (spread ) – When np ≥ 10 and n(1 − p) ≥ 10, the distribution of the sample proportion will be approximately normal. (shape) • We use these properties when solving the following type of normal approximation problems involving a sample proportion. Find the probability of getting more / less than % yeses in a sample of size n. 1. Identify n and p. Verify than observations can be treated as independent and that np ≥ 10 and n(1 − p) ≥ 10, which implies that normal approximation is reasonable. 2. Calculate the Z-score. Use µ ˆp = p and σ ˆp = tion. p(1−p) n to standardize the sample propor- 3. Find the appropriate area under the normal curve. 228 CHAPTER 4. SAMPLING DISTRIBUTIONS Exercises 4.1 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.95. The ﬁgure below shows what the distribution of a sample proportion looks like when the sample size is n = 20, n = 100, and n = 500. (a) What does each point (observation) in each of the samples represent? (b) Describe the distribution of the sample proportion, ˆp. How does the distribution of the sample proportion change as n becomes larger? 4.2 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.5. The ﬁgure below shows what the distribution of a sample proportion looks like when the sample size is n = 20, n = 100, and n = 500. What does each point (observation) in each of the samples represent? Describe how the distribution of the sample proportion, ˆp, changes as n becomes larger. 0.700.750.800.850.900.951.000.700.750.800.850.900
.951.000.700.750.800.850.900.951.000.20.40.60.80.20.40.60.80.20.40.60.8 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 229 4.3 Vegetarian college students. Suppose that 8% of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since n ≥ 30. (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where 12% are vegetarians would be considered unusual. (d) A random sample of 250 college students where 12% are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250. 4.4 Young Americans, Part I. About 77% of young adults think they can achieve the American dream. Determine if the following statements are true or false, and explain your reasoning.3 (a) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 20 is left skewed. (b) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since n ≥ 30. (c) A random sample of 60 young Americans where 85% think they can achieve the American dream would be considered unusual. (d) A random sample of 120 young Americans where 85% think they can achieve the American dream would be considered unusual. 4.5 Distribution of ˆpˆpˆp. simple random sample of size n = 50. Suppose the true population proportion were p = 0.5 and a researcher takes a (a) Find and interpret the standard deviation of the sample proportion ˆp. (b) Calculate the probability that the sample proportion will be larger than 0.55 for a random sample of size 50. 4.6 Distribution of ˆpˆpˆp. Suppose the true population proportion were p = 0.6 and a researcher takes a simple random sample of size n = 50. (a) Find and interpret the standard deviation of the sample proportion ˆp. (b) Calcul
ate the probability that the sample proportion will be larger than 0.65 for a random sample of size 50. It is believed that nearsightedness aﬀects about 8% of all children. We 4.7 Nearsighted children. are interested in ﬁnding the probability that fewer than 12 out of 200 randomly sampled children will be nearsighted. (a) Estimate this probability using the normal approximation to the binomial distribution. (b) Estimate this probability using the distribution of the sample proportion. (c) How do your answers from parts (a) and (b) compare? 4.8 Social network use. The Pew Research Center estimates that as of January 2014, 89% of 18-29 year olds in the United States use social networking sites.4 Calculate the probability that at least 95% of 500 randomly sampled 18-29 year olds use social networking sites. 4.9 CLT for proportions, Part 1. Deﬁne the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.1. 4.10 CLT for proportions, Part 2. Deﬁne the term “sampling distribution” of the sample proportion, and describe how the shape, center, and spread of the sampling distribution change as the sample size increases when p = 0.8. 3A. Vaughn. “Poll ﬁnds young adults optimistic, but not about money”. In: Los Angeles Times (2011). 4Pew Research Center, Washington, D.C. Social Networking Fact Sheet, accessed on May 9, 2015. 230 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.2 Sampling distribution for a sample mean If bags of chips are produced with an average weight of 15 oz and a standard deviation of 0.1 oz, what is the probability that the average weight of 30 bags will be within 0.1 oz of the mean? The answer is not 68%! To answer this question we must visualize and understand what is called the sampling distribution for a sample mean. Learning objectives 1. Describe the center, spread, and shape of the sampling distribution for a sample mean. 2. Distinguish between the standard deviation of a population and the standard deviation of a sampling distribution. 3. Explain the content and importance of the Central Limit Theorem. 4. Ident
ify and explain the conditions for using normal approximation involving a sample mean. 5. Check the appropriate conditions and, when met, carry out normal approximation involving a sample mean or sample sum. 4.2.1 The mean and standard deviation of ¯x¯x¯x In this section we consider a data set called run17, which represents all 19,961 runners who ﬁnished the 2017 Cherry Blossom 10 mile run in Washington, DC. Part of this data set is shown in Figure 4.5, and the variables are described in Figure 4.6. ID 1 2... 16923 16924 time 92.25 106.35... 122.87 93.30 age 38.00 33.00... 37.00 27.00 gender state M MD M DC...... VA F DC F Figure 4.5: Four observations from the run17 data set. variable time age gender state description Ten mile run time, in minutes Age, in years Gender (M for male, F for female) Home state (or country if not from the US) Figure 4.6: Variables and their descriptions for the run17 data set. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 231 These data are special because they include the results for the entire population of runners who ﬁnished the 2017 Cherry Blossom Run. We took a simple random sample of this population, which is represented in Figure 4.7. A histogram summarizing the time variable in the run17samp data set is shown in Figure 4.8. ID 1983 8192... 1287 time 88.31 100.67... 89.49 age 59 32... 26 gender state M MD VA M...... M DC Figure 4.7: Three observations for the run17samp data set, which represents a simple random sample of 100 runners from the 2017 Cherry Blossom Run. Figure 4.8: Histogram of time for a single sample of size 100. The average of the sample is in the mid-90s and the standard deviation of the sample s ≈ 17 minutes. From the random sample represented in run17samp, we guessed the average time it takes to run 10 miles is 95.61 minutes. Suppose we take another random sample of 100 individuals and take its mean: 95.30 minutes. Suppose we took another (93.43 minutes) and another (94.16 minutes), and so on. If we do this many many times – which we can do only because
we have the entire population data set – we can build up a sampling distribution for the sample mean when the sample size is 100, shown in Figure 4.9. SAMPLING DISTRIBUTION The sampling distribution represents the distribution of the point estimates based on samples of a ﬁxed size from a certain population. It is useful to think of a point estimate as being drawn from such a distribution. Understanding the concept of a sampling distribution is central to understanding statistical inference. The sampling distribution shown in Figure 4.9 is unimodal and approximately symmetric. It is also centered exactly at the true population mean: µ = 94.52. Intuitively, this makes sense. The sample mean should be an unbiased estimator of the population mean. Because we are considering the distribution of the sample mean, we will use µ¯x = 94.52 to describe the true mean of this distribution. Time (Minutes)Frequency608010012014005101520 232 CHAPTER 4. SAMPLING DISTRIBUTIONS Figure 4.9: A histogram of 1000 sample means for run time, where the samples are of size n = 100. This histogram approximates the true sampling distribution for the sample mean, with mean µ¯x and standard deviation σ¯x. We can see that the sample mean has some variability around the population mean, which can be quantiﬁed using the standard deviation of this distribution of sample means. The standard deviation of the sample mean tells us how far the typical estimate is away from the actual population mean, 94.52 minutes. It also describes the typical error of a single estimate, and is denoted by the symbol σ¯x. STANDARD DEVIATION OF AN ESTIMATE The standard deviation associated with an estimate describes the typical error or uncertainty associated with the estimate. EXAMPLE 4.5 Looking at Figures 4.8 and 4.9, we see that the standard deviation of the sample mean with n = 100 is much smaller than the standard deviation of a single sample. Interpret this statement and explain why it is true. The variation from one sample mean to another sample mean is much smaller than the variation from one individual to another individual. This makes sense because when we average over 100 values, the large and small values tend to cancel each other out. While many individuals have a time under 90 minutes, it would be unlikely for the average of 100 runners to be less than 90 minutes. Sample meanFrequency7080901001101
20050100The distribution of sample means,shown here, is much narrower thanthe distribution of raw observations. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 233 GUIDED PRACTICE 4.6 (a) Would you rather use a small sample or a large sample when estimating a parameter? Why? (b) Using your reasoning from (a), would you expect a point estimate based on a small sample to have smaller or larger standard deviation than a point estimate based on a larger sample?5 When considering how to calculate the standard deviation of a sample mean, there is one problem: there is no obvious way to estimate this from a single sample. However, statistical theory provides a helpful tool to address this issue. In the sample of 100 runners, the standard deviation of the sample mean is equal to one-tenth of the population standard deviation: 15.93/10 = 1.59. In other words, the standard deviation of the sample mean based on 100 observations is equal to SD¯x = σ¯x = σx√ n = 15.93 √ 100 = 1.59 where σx is the standard deviation of the individual observations. This is no coincidence. We can show mathematically that this equation is correct when the observations are independent using the probability tools of Section 3.4. COMPUTING SD FOR THE SAMPLE MEAN Given n independent observations from a population with standard deviation σ, the standard deviation of the sample mean is equal to SD¯x = σ¯x = σ √ n (4.7) A reliable method to ensure sample observations are independent is to conduct a simple random sample consisting of less than 10% of the population. GUIDED PRACTICE 4.8 The average of the runners’ ages is 35.05 years with a standard deviation of σ = 8.97. A simple random sample of 100 runners is taken. (a) What is the standard deviation of the sample mean? (b) Would you be surprised to get a sample of size 100 with an average of 36 years?6 GUIDED PRACTICE 4.9 (a) Would you be more trusting of a sample that has 100 observations or 400 observations? (b) We want to show mathematically that our estimate tends to be better when the sample size is larger. If the standard deviation of the individual observations is 10, what is our estimate of the standard deviation of the mean when the sample size is
100? What about when it is 400? (c) Explain how your answer to (b) mathematically justiﬁes your intuition in part (a).7 5(a) Consider two random samples: one of size 10 and one of size 1000. Individual observations in the small sample are highly inﬂuential on the estimate while in larger samples these individual observations would more often average each other out. The larger sample would tend to provide a more accurate estimate. (b) If we think an estimate is better, we probably mean it typically has less error. Based on (a), our intuition suggests that a larger sample size corresponds to a smaller standard deviation. 6(a) Use Equation (4.7) with the population standard deviation to compute the standard deviation of the sample mean: SD¯y = 8.97/ 100 = 0.90 years. (b) It would not be surprising. 36 years is about 1 standard deviation from the true mean of 35.05. Based on the 68, 95 rule, we would get a sample mean at least this far away from the true mean approximately 100% − 68% = 32% of the time. √ √ 7(a) Extra observations are usually helpful in understanding the population, so a point estimate with 400 observations seems more trustworthy. (b) The standard deviation of the mean when the sample size is 100 is given by 400 = 0.5. The larger sample has a smaller standard deviation of SD100 = 10/ the mean. (c) The standard deviation of the mean of the sample with 400 observations is lower than that of the sample with 100 observations. The standard deviation of ¯x describes the typical error, and since it is lower for the larger sample, this mathematically shows the estimate from the larger sample tends to be better – though it does not guarantee that every large sample will provide a better estimate than a particular small sample. 100 = 1. For 400: SD400 = 10/ √ 234 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.2.2 The Central Limit Theorem revisited In Figure 4.9, the sampling distribution for the sample mean looks approximately normally distributed. Will the sampling distribution for a mean always be nearly normal? To address this question, we will investigate three cases to see roughly when the approximation is reasonable. We consider three data sets: one from a uniform distribution, one from an exponential distribution, and the other from a normal distribution. These distributions are shown
in the top panels of Figure 4.10. The uniform distribution is symmetric, and the exponential distribution may be considered as having moderate skew since its right tail is relatively short (few outliers). The left panel in the n = 2 row represents the sampling distribution for ¯x if it is the sample mean of two observations from the uniform distribution shown. The dashed line represents the closest approximation of the normal distribution. Similarly, the center and right panels of the n = 2 row represent the respective distributions of ¯x for data from exponential and log-normal distributions. GUIDED PRACTICE 4.10 Examine the distributions in each row of Figure 4.10. What do you notice about the sampling distribution for the mean as the sample size, n, becomes larger?8 EXAMPLE 4.11 In general, would normal approximation for a sample mean be appropriate when the sample size is at least 30? Yes, the sampling distributions when n = 30 all look very much like the normal distribution. However, the more non-normal a population distribution, the larger a sample size is necessary for the sampling distribution to look nearly normal. DETERMINING IF THE SAMPLE MEAN IS NORMALLY DISTRIBUTED If the population is normal, the sampling distribution for ¯x will be normal for any sample size. The less normal the population, the larger n needs to be for the sampling distribution for ¯x to be nearly normal. However, a good rule of thumb is that for almost all populations, the sampling distribution for ¯x will be approximately normal if n ≥ 30. This brings us to the Central Limit Theorem, the most fundamental theorem in Statistics. CENTRAL LIMIT THEOREM When taking a random sample of independent observations from a population with a ﬁxed mean and standard deviation, the distribution of ¯x approaches the normal distribution as n increases. 8The normal approximation becomes better as larger samples are used. However, in the case when the population is normally distributed, the normal distribution of the sample mean is normal for all sample sizes. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 235 Figure 4.10: Sampling distributions for the mean at diﬀerent sample sizes and for three diﬀerent distributions. The dashed red lines show normal distributions. 0:1populationdistributionsuniform−20246exponential−20246normal−202460:1n = 2−202402020:1n = 50120120120
:1n = 12012012012n = 300.51.01.50.51.01.50.51.01.5 236 CHAPTER 4. SAMPLING DISTRIBUTIONS EXAMPLE 4.12 Sometimes we do not know what the population distribution looks like. We have to infer it based on the distribution of a single sample. Figure 4.11 shows a histogram of 20 observations. These represent winnings and losses from 20 consecutive days of a professional poker player. Based on this sample data, can the normal approximation be applied to the distribution of the sample mean? We should consider each of the required conditions. (1) These are referred to as time series data, because the data arrived in a particular sequence. If the player wins on one day, it may inﬂuence how she plays the next. To make the assumption of independence we should perform careful checks on such data. (2) The sample size is 20, which is smaller than 30. (3) There are two outliers in the data, both quite extreme, which suggests the population may not be normal and instead may be very strongly skewed or have distant outliers. Outliers can play an important role and aﬀect the distribution of the sample mean and the estimate of the standard deviation of the sample mean. Since we should be skeptical of the independence of observations and the extreme upper outliers pose a challenge, we should not use the normal model for the sample mean of these 20 observations. If we can obtain a much larger sample, then the concerns about skew and outliers would no longer apply. Figure 4.11: Sample distribution of poker winnings. These data include two very clear outliers. These are problematic when considering the normality of the sample mean. For example, outliers are often an indicator of very strong skew. EXAMINE DATA STRUCTURE WHEN CONSIDERING INDEPENDENCE Some data sets are collected in such a way that they have a natural underlying structure between observations, e.g. when observations occur consecutively. Be especially cautious about independence assumptions regarding such data sets. WATCH OUT FOR STRONG SKEW AND OUTLIERS Strong skew in the population is often identiﬁed by the presence of clear outliers in the data. If a data set has prominent outliers, then a larger sample size will be needed for the sampling distribution for ¯x to be normal. There are no simple guidelines for what sample size is big enough for each situation
. However, we can use the rule of thumb that, in general, an n of at least 30 is suﬃcient for most cases. Poker Winnings and Losses (US$)Frequency−200−1000100200300400500024681012 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 237 4.2.3 Normal approximation for the sampling distribution for ¯x¯x¯x At the beginning of this chapter, we used normal approximation for populations or for data that had an approximately normal distribution. When appropriate conditions are met, we can also use the normal approximation to estimate probabilities about a sample average. We must remember to verify that the conditions are met and use the mean µ¯x and standard deviation σ¯x for the sampling distribution for the sample average. THREE IMPORTANT FACTS ABOUT THE DISTRIBUTION OF A SAMPLE MEAN ¯x¯x¯x When the observations can be considered independent, such as from a random sample of less than 10% of the population, the distribution of the sample mean can be described as follows. 1. The mean of a sample mean is denoted by µ¯x, and it is equal to µ. 2. The SD of a sample mean is denoted by σ¯x, and it is equal to σ√ n. 3. When the population is normal or when n ≥ 30, the sample mean closely follows a normal distribution. EXAMPLE 4.13 In the 2017 Cherry Blossom 10 mile run, the average time for all of the runners is 94.52 minutes with a standard deviation of 8.97 minutes. The distribution of run times is approximately normal. Find the probabiliy that a randomly selected runner completes the run in less than 90 minutes. Because the distribution of run times is approximately normal, we can use normal approximation. Z = x − µ σ = 90 − 94.52 8.97 P (Z < −0.504) = 0.3072 = −0.504 There is a 30.72% probability that a randomly selected runner will complete the run in less than 90 minutes. EXAMPLE 4.14 Find the probabiliy that the average of 20 runners is less than 90 minutes. Here, n = 20 < 30, but the distribution of the population, that is, the distribution of run times is stated to be approximately normal. Because of this, the sampling distribution will be normal for any sample size. = �
�¯x = = 2.01 σ √ n ¯x − µ¯x σ¯x 8.97 √ 20 90 − 94.52 2.01 P (Z < −2.25) = 0.0123 Z = = = −2.25 There is a 1.23% probability that the average run time of 20 randomly selected runners will be less than 90 minutes. 238 CHAPTER 4. SAMPLING DISTRIBUTIONS EXAMPLE 4.15 The average of all the runners’ ages is 35.05 years with a standard deviation of σ = 8.97. The distribution of age is somewhat skewed. What is the probability that a randomly selected runner is older than 37 years? Because the distribution of age is skewed and is not normal, we cannot use normal approximation for this problem. In order to answer this question, we would need to look at all of the data. GUIDED PRACTICE 4.16 What is the probability that the average of 50 randomly selected runners is greater than 37 years?9 REMEMBER TO DIVIDE BY When ﬁnding the probability that an average or mean is greater or less than a particular value, n to calculate the correct SD. remember to divide the standard deviation of the population by √ √ √ √ n n n 9Because n = 50 ≥ 30, the sampling distribution for the mean is approximately normal, so we can use normal approximation for this problem. The mean is given as 35.05 years. σ¯x = σ √ n = 8.97 √ 50 = 1.27 z = ¯x − µ¯x σ¯x = 37 − 35.05 1.27 = 1.535 P (Z > 1.535) = 0.062 There is a 6.2% chance that the average age of 50 runners will be greater than 37. 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 239 Section summary • The symbol ¯x denotes the sample average. ¯x for any particular sample is a number. However, ¯x can vary from sample to sample. The distribution of all possible values of ¯x for repeated samples of a ﬁxed size from a certain population is called the sampling distribution for ¯x. • The standard deviation of ¯x describes the typical error or distance of the sample mean from the population mean. It also tells us how much the sample mean is likely to vary
from one random sample to another. • The standard deviation of ¯x will be smaller than the standard deviation of the population by a factor of n. The larger the sample, the better the estimate tends to be. √ • Consider taking a simple random sample from a population with a ﬁxed mean and standard deviation. The Central Limit Theorem ensures that regardless of the shape of the original population, as the sample size increases, the distribution of the sample average ¯x becomes more normal. • Three important facts about the sampling distribution for the sample average ¯x where the observations can be treated as independent: – The mean of a sample mean is denoted by µ¯x, and it is equal to µ. (center ) – The SD of a sample mean is denoted by σ¯x, and it is equal to σ√ n. (spread ) – When the population is normal or when n ≥ 30, the sample mean closely follows a normal distribution. (shape) • These facts are used when solving the following two types of normal approximation prob- lems involving a sample mean or a sample sum. A: Find the probability that a sample average will be greater/less than a certain value. 1. Verify that the observations can be treated as independent and that either the pop- ulation is approximately normal or n ≥ 30. 2. Calculate the Z-score. Use µ¯x = µ and σ¯x = σ√ 3. Find the appropriate area under the normal curve. n to standardize the sample average. B: Find the probability that a sample sum/total will be greater/less than a certain value. 1. Convert the sample sum into a sample average, using ¯x = sum n. 2. Do steps 1-3 from Part A above. 240 CHAPTER 4. SAMPLING DISTRIBUTIONS Exercises 4.11 Ages of pennies, Part I. The histogram below shows the distribution of ages of pennies at a bank. (a) Describe the distribution. (b) Sampling distributions for means from simple random samples of 5, 30, and 100 pennies is shown in the histograms below. Describe the shapes of these distributions and comment on whether they look like what you would expect to see based on the Central Limit Theorem. 4.12 Ages of pennies, Part II. The mean age of the pennies from Exercise 4.11 is 10.44 years with a standard deviation of
9.2 years. Using the Central Limit Theorem, calculate the means and standard deviations of the distribution of the mean from random samples of size 5, 30, and 100. Comment on whether the sampling distributions shown in Exercise 4.11 agree with the values you compute. A housing survey was conducted to determine the price of a typical home in 4.13 Housing prices. Topanga, CA. The mean price of a house was roughly $1.3 million with a standard deviation of $300,000. There were no houses listed below $600,000 but a few houses above $3 million. (a) Is the distribution of housing prices in Topanga symmetric, right skewed, or left skewed? Hint: Sketch the distribution. (b) Would you expect most houses in Topanga to cost more or less than $1.3 million? (c) Can we estimate the probability that a randomly chosen house in Topanga costs more than $1.4 million using the normal distribution? (d) What is the probability that the mean of 60 randomly chosen houses in Topanga is more than $1.4 million? (e) How would doubling the sample size aﬀect the standard deviation of the mean? 4.14 Stats ﬁnal scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the ﬁnal exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the ﬁnal) but a few students scored below 20 points. (a) Is the distribution of scores on this ﬁnal exam symmetric, right skewed, or left skewed? (b) Would you expect most students to have scored above or below 70 points? (c) Can we calculate the probability that a randomly chosen student scored above 75 using the normal distribution? (d) What is the probability that the average score for a random sample of 40 students is above 75? (e) How would cutting the sample size in half aﬀect the standard deviation of the mean? Penny ages01020304050x n = 5051015202530x n = 30681012141618x n = 100891011121314 4.2. SAMPLING DISTRIBUTION FOR A SAMPLE MEAN 241 4.15 Identify distributions, Part
I. Four plots are presented below. The plot at the top is a distribution for a population. The mean is 10 and the standard deviation is 3. Also shown below is a distribution of (1) a single random sample of 100 values from this population, (2) a distribution of 100 sample means from random samples with size 5, and (3) a distribution of 100 sample means from random samples with size 25. Determine which plot (A, B, or C) is which and explain your reasoning. 4.16 Identify distributions, Part II. Four plots are presented below. The plot at the top is a distribution for a population. The mean is 60 and the standard deviation is 18. Also shown below is a distribution of (1) a single random sample of 500 values from this population, (2) a distribution of 500 sample means from random samples of each size 18, and (3) a distribution of 500 sample means from random samples of each size 81. Determine which plot (A, B, or C) is which and explain your reasoning. 4.17 Weights of pennies. The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation of 0.03 grams. (a) What is the probability that a randomly chosen penny weighs less than 2.4 grams? (b) Describe the sampling distribution for the mean weight of 10 randomly chosen pennies. (c) What is the probability that the mean weight of 10 pennies is less than 2.4 grams? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from (a) and (c) if the weights of pennies had a skewed distribution? 05101520Populationm = 10s = 3Plot A67891011121301020Plot B5101501020Plot C0102091011020406080100Populationm = 60s = 18Plot A54565860626466050100Plot B020406080100050100Plot C5055606570050100 242 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.18 CFLBs. A manufacturer of compact ﬂuorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours. (a) What is the probability that a randomly chosen light bulb
lasts more than 10,500 hours? (b) Describe the distribution of the mean lifespan of 15 light bulbs. (c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? 4.19 Songs on an iPod. Suppose an iPod has 3,000 songs. The histogram below shows the distribution of the lengths of these songs. We also know that, for this iPod, the mean length is 3.45 minutes and the standard deviation is 1.63 minutes. (a) Calculate the probability that a randomly selected song lasts more than 5 minutes. (b) You are about to go for an hour run and you make a random playlist of 15 songs. What is the probability that your playlist lasts for the entire duration of your run? Hint: If you want the playlist to last 60 minutes, what should be the minimum average length of a song? (c) You are about to take a trip to visit your parents and the drive is 6 hours. You make a random playlist of 100 songs. What is the probability that your playlist lasts the entire drive? 4.20 Spray paint, Part II. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. (a) What is the probability that the area covered by a can of spray paint is more than 27 square feet? (b) Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square feet must each can be able to cover to spray paint all 540 square feet? (c) What is the probability that you can cover a 540 square feet area using 20 cans of spray paint? (d) If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution? 4.21 Wireless routers. John is shopping for wireless routers and is overwhelmed by the number of available options. In order to get a feel for the average price, he takes a random sample of 75 routers and ﬁnds that the average price for this sample is $75
and the standard deviation is $25. (a) Based on this information, how much variability should he expect to see in the mean prices of repeated samples, each containing 75 randomly selected wireless routers? (b) A consumer website claims that the average price of routers is $80. Is a true average of $80 consistent with John’s sample? 4.22 Betting on dinner, Part II. Exercise 3.42 introduces a promotion at a restaurant where prices of menu items are determined randomly following some underlying distribution. We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all ﬁve baskets of fries? Length of song02468100200400600800 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 243 4.3 Sampling distribution for a difference In this section, we consider the case where instead of having one random sample from a population of interest, we have two independent random samples and we are interested in how far the sample values might be from one another. We describe the sampling distribution for the diﬀerence of sample proportions and the diﬀerence of sample means, and we ﬁnd the probability that the diﬀerence would be greater than or less than a certain amount. Learning objectives 1. Describe the center, spread, and shape of the sampling distribution for a diﬀerence of sample proportions. 2. Describe the center, spread, and shape of the sampling distribution for a diﬀerence of sample means. 3. Verify appropriate conditions and, if met, carry out the normal approximation using a diﬀer- ence of sample proportions or a diﬀerence of sample means. 4.3.1 The mean and SD for a difference of two random variables (review) In Section 3.4, we saw how to ﬁnd the mean and standard deviation of a diﬀerence of two random variables X and Y. We found that: E(X − Y ) = E(X) − E(Y ) That is, if we have the mean of each random variable, µX and µY, then the mean of the diﬀerence of the random variables is simply
the diﬀerences of the individual means. This should seem very straightforward. The situation is a little more complex when looking at the variability of the diﬀerence of X and Y. Here we’re going to require a condition be met, speciﬁcally that X and Y are independent random variables. When that independence condition is met, then the following formula for the standard deviation of the diﬀerence, X − Y, holds: SD(X − Y ) = (SD(X))2 + (SD(Y ))2 These two formulas from probability play an important role in applications where we look at the diﬀerence of two sample proportions (ˆp1 − ˆp2) or the diﬀerence of sample means (¯x1 − ¯x2). 244 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.3.2 Difference of sample proportions In Section 4.1 we considered a county where the proportion of people with blood type O+ is known to be 35%, and we described the distribution of the sample statistic ˆp from a random sample of size n from this population. Let us now consider two counties. In County 1, it is known that 35% of people have blood type O+. In County 2, it is known that 30% of people have blood type O+. If we take a random sample of size 50 from County 1 and a random sample of size 50 from County 2, what is the probability that we will get a higher proportion with blood type O+ in the County 2 sample? We know that the expected proportion with blood type O+ is lower in County 2. However, due to random variability, there is still some chance that the sample from County 2 will have a higher proportion with blood type O+. First, we need to ﬁnd the mean (expected value) and the standard deviation for the diﬀerence of sample proportions ˆp1 − ˆp2. For this, we use the two formulas from Section 4.3.1 in the context of the diﬀerence of sample proportions by substituting X = ˆp1 and Y = ˆp2: E(ˆp1 − ˆp2) = E(ˆp1) − E(ˆp2) = p1 − p2 SD(ˆp1 − ˆp2
) = (SD(ˆp1))2 + (SD(ˆp2))2 = p1(1 − p1) n1 + p2(1 − p2) n2. That is, given two independent random samples, the distribution of all possible values of ˆp1 − ˆp2 is centered on the true diﬀerence p1 − p2 and the typical distance or error of ˆp1 − ˆp2 from p1 − p2 is given by. Using µ for mean and σ for SD, we summarize this as follows. p1(1−p1) n1 + p2(1−p2) n2 MEAN AND STANDARD DEVIATION OF A DIFFERENCE OF SAMPLE PROPORTIONS The mean and standard deviation of the diﬀerence of sample proportions describe the center and spread of the distribution of all possible diﬀerences ˆp1 − ˆp2. Given population proportions p1 and p2 and random samples of size n1 and n2 where the independence condition is satisﬁed, we have the following. µ ˆp1− ˆp2 = p1 − p2 σ ˆp1− ˆp2 = p1(1 − p1) n1 + p2(1 − p2) n2 As we saw previously, the independence condition is satisﬁed if the data is collected from an experiment with two randomly assigned treatments or collected from 2 independent random samples, where each sample size is less than 10% of the population size if done without replacement. Having described the center and spread of the distribution of the diﬀerence of sample proportions, we need to understand the shape of the distribution. When looking at the sum or diﬀerence of random variables, if each variable is nearly normal, then the sum and diﬀerence are also nearly normal. This property will be very useful, as it says that when each sample proportion has a nearly normal distribution, then the diﬀerence of sample proportions will also be nearly normal, and we can use normal approximation to estimate probabilities that the diﬀerence will be greater or less than some value. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 245 EX
AMPLE 4.17 Let’s return to the blood type example. In County 1, it is known that 35% of people have blood type O+. In County 2, it is known that 30% of people have blood type O+. If we take a random sample of size 50 from County 1 and a random sample of size 50 from County 2, what is the probability that we will get a higher proportion with blood type O+ in the County 2 sample? We want to ﬁnd P (ˆp1 < ˆp2). This is equivalent to P (ˆp1 − ˆp2 < 0). Let us ﬁrst determine whether the independence condition is satisﬁed and whether normal approximation will be appropriate. We have two random samples, and the samples are independent of each other as they are from distinct populations. It is reasonable to assume that each sample size is less than 10% of the county population size. We now check the success-failure condition for each group. n1p1 = 50 × 0.35 = 17.5 ≥ 10 n2p2 = 50 × 0.30 = 15.0 ≥ 10 n1(1 − p1) = 50 × (1 − 0.35) = 32.5 ≥ 10 n2(1 − p2) = 50 × (1 − 0.30) = 35.0 ≥ 10 The independence condition is satisﬁed and the success-failure condition is met for both groups, so the distribution of ˆp1 − ˆp2 can be said to be nearly normal. We now ﬁnd the mean and the standard deviation of ˆp1 − ˆp2. µ ˆp1− ˆp2 = p1 − p2 = 0.35 − 0.30 = 0.05 σ ˆp1− ˆp2 = p1(1 − p1) n1 + p2(1 − p2) n2 = 0.35(1 − 0.35) 50 + 0.30(1 − 0.30) 50 = 0.0935 Recall, we want to ﬁnd P (ˆp1 − ˆp2 < 0). We use the calculated mean and standard deviation of ˆp1 − ˆp2 to ﬁnd a Z
-score for the value of interest, which is 0.05 0.0935 = −0.535 Using technology to ﬁnd the area to the left of -0.535 under the standard normal curve, we obtain P (ˆp1 − ˆp2) ≈ P (Z < −0.535) = 0.296. Even though County 2 has a lower proportion of people with blood type O+, with these small sample sizes, there is still a 29.6% chance that the sample from County 2 ends up with a higher proportion with blood type O+ than the sample from County 1. −0.140.050.24x = 0 246 CHAPTER 4. SAMPLING DISTRIBUTIONS 4.3.3 Difference of sample means In Section 4.2 we started with all of the data from the 2017 Cherry Blossom Run, and we considered what the sampling distribution for a mean would look like for random samples of size n. The population mean for all the runners is 94.52 minutes and the population standard deviation is 8.97 minutes. Imagine taking two independent random samples of size 50 from this population. We know that the sampling distribution for the mean would have the same center and spread in each sample, but what about the sampling distribution for the diﬀerence of the sample means? What is the likelihood that the sample means from these two independent random samples would diﬀer by more than 3 minutes? To answer this, we consider two scenarios. First, consider the possibility that the mean of sample 1 is more than 3 minutes greater than the mean of sample 2. That is, we want to ﬁnd P(¯x1 − ¯x2 > 3). Second, it is also possible that the mean of sample 2 is more than 3 minutes greater than the mean of sample 1, so we also need to ﬁnd P(¯x2 − ¯x1 > 3), which we could also write as P(¯x1 − ¯x2 < −3). We now see that we are interested in the upper and lower tail of the distribution of ¯x1 − ¯x2. As we will soon see, the distribution of ¯x1 − ¯x2 follows a normal distribution when certain conditions are met, and in those cases we can ﬁnd one tail and then double it thanks to the symmetry of the normal distribution. Let’s ﬁnd the mean (expected value)
and the standard deviation for the diﬀerence of sample means, ¯x1 − ¯x2. We again use the formulas discussed in Section 4.3.1 for the diﬀerence in two independent random normal variables, X − Y, but this time we substitute in X = ¯x1 and Y = ¯x2: E(¯x1 − ¯x2) = E(¯x1) − E(¯x2) = µ1 − µ2 SD(¯x1 − ¯x2) = (SD(¯x1))2 + (SD(¯x2))2 = σ2 1 n1 + σ2 2 n2. That is, given two independent random samples, the distribution of all possible values of ¯x1 − ¯x2 is centered on the true diﬀerence µ1 − µ2 and the typical distance or error of ¯x1 − ¯x2 from µ1 − µ2 is given by. σ2 1 n1 + σ2 2 n2 MEAN AND STANDARD DEVIATION OF A DIFFERENCE OF SAMPLE MEANS The mean and standard deviation of the diﬀerence of sample means describe the center and spread of the distribution of all possible diﬀerences ¯x1 − ¯x2. Given population means µ1 and µ2, individual population standard deviations σ1 and σ2, and two random samples of size n1 and n2, then if the independence condition is satisﬁed, we have the following: µ¯x1−¯x2 = µ1 − µ2 σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 Recall that in Section 4.3.2, we also noted that the sum or diﬀerence of random variables will also nearly normal, if each variable is itself nearly normal and the two random variables are also independent of each other. We will frequently use this principle to determine when the diﬀerence of sample means can be modeled using a normal distribution. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 247 EXAMPLE 4.18 Let’s return to the Cherry Blossom Run application. We have that the population mean for all the runners in the 2017 Cherry Blossom Run is 94.52 minutes and the population standard deviation is
8.97 minutes. If we take two independent random samples of 50 runners, what is the probability that the sample means from these two samples will diﬀer by more than 3 minutes? We want to ﬁnd P (¯x1 − ¯x2 > 3) + P (¯x1 − ¯x2 < −3). Let us ﬁrst determine whether the independence condition is satisﬁed and whether normal approximation will be appropriate. We have two random samples, and the samples are distinct and independent of each other. The sample sizes are each less than 10% of total population of runners. Also, because the sample sizes n1 and n2 are both 50, and 50 ≥ 30, we don’t need the distribution of all run times to be nearly normal. With these conditions met, we have shown that the distribution of ¯x1 − ¯x2 is nearly normal. Next, we ﬁnd the mean and the standard deviation of ¯x1 − ¯x2: µ¯x1−¯x2 = µ1 − µ2 = 94.52 − 94.52 = 0 σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2 = 8.972 50 + 8.972 50 = 1.79 To ﬁnd P (¯x1 − ¯x2 > 3), we use the calculated mean and standard deviation of ¯x1 − ¯x2 to ﬁrst ﬁnd a Z-score for the diﬀerence of interest, which is 3.79 = 1.676 Using technology, to ﬁnd the area to the right of 1.676 under the standard normal curve, we have: P (¯x1 − ¯x2 > 3) ≈ P (Z > 1.676) = 0.047. Because the normal distribution of interest is centered at zero, the P (¯x1 − ¯x2 < −3) tail area will have the same size. So to get the total of the two tail areas, we double 0.047. Even though the samples are from the same population of runners, there is still a 2 × 0.047 = 0.094 probability, or a 9.4% chance, that the sample means will diﬀer by more than 3 minutes. −5.37−3.58−1.7901
.793.585.37x = −3x = 3 248 CHAPTER 4. SAMPLING DISTRIBUTIONS Section summary • When two random variables each follow a nearly normal distribution, the distribution of their diﬀerence also follows a nearly normal distribution. • Both ˆp1 − ˆp2 and ¯x1 − ¯x2 are statistics that can take on diﬀerent values from one random sample to the next. As such, they have sampling distributions that can be described by their center, spread, and shape. • Three important facts about the sampling distribution for the diﬀerence of sample proportions ˆp1 − ˆp2 where the observations can be treated as independent: – The mean of the diﬀerence of sample proportions, denoted by µ ˆp1− ˆp2, is equal to p1 − p2. (center) – The SD of the diﬀerence of sample proportions, denoted by σ ˆp1− ˆp2, is equal to p1(1−p1) n1 + p2(1−p2) n2. (spread) – When both groups meet the success-failure condition, the diﬀerence of sample proportions can be modeled using a normal distribution. (shape) • Three important facts about the sampling distribution for the diﬀerence of sample means ¯x1 − ¯x2 where the observations can be treated as independent: – The mean of the diﬀerence of sample means, denoted by µ¯x1−¯x2, is equal to µ1 − µ2. (center) – The SD of the diﬀerence of sample means, denoted by σ¯x1−¯x2, is equal to σ2 1 n1 + σ2 2 n2. (spread) – When both populations are nearly normal or when n1 ≥ 30 and n2 ≥ 30, the diﬀerence of sample means can be modeled using a normal distribution. (shape) • When the diﬀerence of sample proportions ˆp1 − ˆp2 or the diﬀerence of sample means ¯x1 − ¯x2 follow a nearly normal distribution, we can ﬁnd the probability that the diﬀerence is greater than or
less than a certain amount by ﬁnding a Z-score and using the normal approximation. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 249 Exercises 4.23 Difference of proportions, Part 1. The fraction of workers who are considered “supercommuters”, because they commute more than 90 minutes to get to work, varies by state. Suppose the following were the exact values for Nebraska and New York: State Nebraska New York Proportion Supercommuters 0.01 0.06 Now suppose that we plan a study to survey 1000 people from each state, and we will compute the sample proportions ˆpN E for Nebraska and ˆpN Y for New York. (a) What is the associated mean and standard deviation of ˆpN E? (b) What is the associated mean and standard deviation of ˆpN Y? (c) Calculate and interpret the mean and standard deviation associated with the diﬀerence in sample pro- portions for the two groups, ˆpN Y − ˆpN E. (d) How are the standard deviations from parts (a), (b), and (c) related? 4.24 Difference of proportions, Part 2. The fraction of workers who are considered “supercommuters”, because they commute more than 90 minutes to get to work, varies by state. Suppose the following were the exact values for Nebraska and New York: State Nebraska New York Proportion Supercommuters 0.01 0.06 Now suppose that we plan a study to survey 1000 people from each state, and we will compute the sample proportions ˆpN E for Nebraska and ˆpN Y for New York. (a) What distribution is associated with the diﬀerence ˆpN Y − ˆpN E? Justify your answer. (b) Determine the probability that ˆpN Y − ˆpN E will be larger than 0.055. (c) Determine the probability that ˆpN Y − ˆpN E will be smaller than 0.4. 4.25 Difference of means, Part 1. Suppose we will collect two random samples from the following distributions: Sample 1 Sample 2 Mean 15 20 Standard Deviation 20 10 Sample Size 50 30 In each of the parts below, consider the
sample means ¯x1 and ¯x2 that we might observe from these two samples. (a) What is the associated mean and standard deviation of ¯x1? (b) What is the associated mean and standard deviation of ¯x2? (c) Calculate and interpret the mean and standard deviation associated with the diﬀerence in sample means for the two groups, ¯x2 − ¯x1. (d) How are the standard deviations from parts (a), (b), and (c) related? 4.26 Difference of means, Part 2. Suppose we will collect two random samples from the following distributions: Sample 1 Sample 2 Mean 15 20 Standard Deviation 20 10 Sample Size 50 30 In each of the parts below, consider the sample means ¯x1 and ¯x2 that we might observe from these two samples. (a) What distribution is associated with the diﬀerence ¯x2 − ¯x1? Justify your answer. (b) Determine the probability that ¯x2 − ¯x1 will be larger than 7. (c) Determine the probability that ¯x2 − ¯x1 will be smaller than 3. (d) Determine the probability that ¯x2 − ¯x1 will be smaller than 0. 250 CHAPTER 4. SAMPLING DISTRIBUTIONS Chapter highlights This chapter began by introducing the idea of a sampling distribution. As with any distribution, we can summarize a sampling distribution with regard to its center, spread, and shape. A common thread that ran through this chapter is the application of normal approximation (introduced in Section 2.3) to diﬀerent sampling distributions. The key steps are included for each of the normal approximation scenarios below. To verify that observations can be considered independent, verify that you have one of the following: a random process, a random sample with replacement, or a random sample without replacement of less than 10% of the population. To satisfy the independence condition when working with two groups, we require 2 independent random samples with replacement, 2 independent samples without replacement of less than 10% of their populations, or an experiment with 2 randomly assigned treatments. For completion and comparison purposes, we include cases introduced in earlier chapters as well in the overview below. 1. Normal approximation for numerical data: (introduced in Section 2.3) • Verify that observations can be treated as independent and that population is approxi- mately normal. • Use a normal model with mean µ and SD �
�. 2. Normal approximation for a sample proportion (with categorical data): • Verify that observations can be treated as independent and that np ≥ 10 and n(1−p) ≥ 10. • Use a normal model with mean µ ˆp = p and SD σ ˆp = p(1−p) n. 3. Normal approximation for a sample mean (with numerical data): • Verify that observations can be treated as independent and that population is approxi- mately normal or that n ≥ 30. • Use a normal model with mean µ¯x = µ and SD σ¯x = σ√ n. 4. Normal approximation for a diﬀerence of sample proportions: • Verify that observations can be treated as independent and that n1p1 ≥ 10, n1(1 − p1) ≥ 10, n2(1 − p2) ≥ 10, and n2(1 − p2) ≥ 10. • Use a normal model with mean µ ˆp1− ˆp2 = p1 − p2 and SD σ ˆp1− ˆp2 = p1(1−p1) n1 + p2(1−p2) n2. 5. Normal approximation for a diﬀerence of sample means: • Verify that observations can be treated as independent and that both populations are nearly normal or both n1 and n2 are ≥ 30. • Use a normal model with mean: µ¯x1−¯x2 = µ1 − µ2 and SD: σ¯x1−¯x2 = σ2 1 n1 + σ2 2 n2. Cases 1, 3 and 5 are for numerical variables, while cases 2 and 4 are for categorical yes/no variables. In the case of proportions and counts, we never look to see if the population is normal. That would not make sense because a yes/no variable cannot have a normal distribution. The Central Limit Theorem is the mathematical rule that ensures that when the sample size is suﬃciently large, the sample mean/sum and sample proportion/count will be approximately normal. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 251 Chapter exercises 4.27 University admissions. Suppose a university announced that it admitted 2,500 students for the following year’s freshman class. However
, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the oﬀer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class? 4.28 SAT scores. SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certiﬁcate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probability this student’s score will be at least 1500? (The material covered in Section 3.2 would be useful for this question.) 4.29 Overweight baggage. Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. 4.30 Survey response rate. Pew Research reported that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond?10 Suppose weights of the checked baggage of airline passengers 4.31 Overweight baggage, Part II. follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. What is the probability that the total weight of 10 bags is greater than 460 lbs? 4.32 Chocolate chip cookies. Students are asked to count the number of chocolate chips in 22 cookies for a class activity. The packaging for these cookies claims that there are an average of 20 chocolate chips per cookie with a standard deviation of 4.37 chocolate chips. (a) Based on this information, about how much variability should they expect to see in the mean number of chocolate chips in random samples of 22 chocolate chip cookies? (b) What is the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips if the company’s claim on the packaging is true? Assume that the distribution of chocolate chips in these cookies is approximately normal. (c) Assume the students got 14.77 as the average in their sample of 22 cookies. Do you have conﬁdence or not in the company’s claim that the true average is 20? Explain your
reasoning. 4.33 Young Hispanics in the US. The 2019 Current Population Survey (CPS) estimates that 36.0% of the people of Hispanic origin in the Unites States are under 21 years old.11 Calculate the probability that at least 35 people among a random sample of 100 Hispanic people living in the United States are under 21 years old. 4.34 Poverty in the US. The 2019 Current Population Survey (CPS) estimates that 19.2% of Mississippians live in poverty, which makes Mississippi the state with the highest poverty rate in the United States.12 We are interested in ﬁnding out the probability that at least 200 people among a random sample of 1,000 Mississippians live in poverty. (a) Estimate this probability using the normal approximation to the binomial distribution. (b) Estimate this probability using the distribution of the sample proportion. (c) How do your answers from parts (a) and (b) compare? 10Pew Research Center, Assessing the Representativeness of Public Opinion Surveys, May 15, 2012. 11United States Census Bureau. https://www.census.gov/data/tables/2019/demo/hispanic-origin/2019-cps.html. The Hispanic Population in the United States: 2019. Web. Census 12United Bureau. States https://www.census.gov/data/tables/time-series/demo/income- poverty/historical-poverty-people.html. Historical Poverty Tables: People and Families - 1959 to 2020. Web. 252 Chapter 5 Foundations for inference 5.1 Estimating unknown parameters 5.2 Conﬁdence intervals 5.3 Introducing hypothesis testing 253 Statistical inference is primarily concerned with understanding and quantifying the uncertainty of parameter estimates. While the equations and details change depending on the setting, the foundations for inference are the same throughout all of statistics. We start with a familiar topic: the idea of using a sample proportion to estimate a population proportion. Next, we create what’s called a conﬁdence interval, which is a range of values where the true population value is likely to lie. Finally, we introduce a hypothesis testing framework, which allows us use data to formally evaluate claims about the population, such as whether a survey provides strong evidence that a candidate has the support of a majority of the voting population. For videos, slides, and other resources, please visit www.open
intro.org/ahss 254 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1 Estimating unknown parameters Companies such as the Gallup and Pew Research frequently conduct polls as a way to understand the state of public opinion or knowledge on many topics, including politics, scientiﬁc understanding, brand recognition, and more. How well do these polls estimate the opinion or knowledge of the broader population? Why is a larger sample generally preferable to a smaller sample? And what role does the concept of a sampling distribution, introduced in the previous chapter, play in answering these questions? Learning objectives 1. Explain the diﬀerence between probability and inference and identify when to use which one. 2. Understand the purpose and use of a point estimate. 3. Understand how to measure the variability/error in a point estimate. 4. Recognize the relationship between the standard error of a point estimate and the standard deviation of a sample statistic. 5. Understand how changing the sample size aﬀects the variability/error in a point estimate. 5.1.1 Point estimates With this chapter, we move from the world of probability to the world of inference. Whereas probability involves using a known population value (parameter) to make a prediction about the likelihood of a particular sample value (statistic), inference involves using a calculated sample value (statistic) to estimate or better understand an unknown population value (parameter). For both of these, the concept of the sampling distribution is fundamental. Suppose a poll suggested the US President’s approval rating is 45%. We would consider 45% to be a point estimate of the approval rating we might see if we collected responses from the entire population. This entire-population response proportion is generally referred to as the parameter of interest, and when the parameter is a proportion, we denote it with the letter p. We typically estimate the parameter by collecting information from a sample of the population; we compute the observed proportion in the sample and we denote this sample proportion as ˆp. Unless we collect responses from every individual in the sample, p remains unknown, and we use ˆp as our point estimate for p. 5.1. ESTIMATING UNKNOWN PARAMETERS 255 The diﬀerence we observe from the poll versus the parameter is called the error in the estimate. Generally, the error consists of two aspects: sampling error and bias. Bias describes a systematic tendency to over- or under-estimate the true population value
. For instance, if we took a political poll but our sample didn’t include a roughly representative distribution of the political parties, the sample would likely skew in a particular direction and be biased. Taking a truly random sample helps avoid bias. However, as we saw in Chapter 1, even with a random sample, various types of response bias can still be present. For example, if we were taking a student poll asking about support for a new college stadium, we’d probably get a biased estimate of the stadium’s level of student support by wording the question as, Do you support your school by supporting funding for the new stadium? We try to minimize bias through thoughtful data collection procedures, but bias can creep into our estimates without us even be aware. Sampling error is uncertainty in a point estimate that happens naturally from one sample to the next. Much of statistics, including much of this book, is focused on understanding and quantifying sampling error. Remember though, that sampling error does not account for the possible eﬀects of leading questions or other types of response bias. When we measure sampling error, we are measuring the expected variability in a point estimate that arises from randomly sampling only a subset of the population. EXAMPLE 5.1 In Chapter 2, we found the summary statistics for the number of characters in a set of 50 email data. These values are summarized below. ¯x median sx 11,160 6,890 13,130 Estimate the population mean based on the sample. The best estimate for the population mean is the sample mean. That is, ¯x = 11, 160 is our best estimate for µ. GUIDED PRACTICE 5.2 Using the email data, what quantity should we use as a point estimate for the population standard deviation σ?1 1Again, intuitively we would use the sample standard deviation s = 13, 130 as our best estimate for σ. 256 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1.2 Understanding the variability of a point estimate Suppose the proportion of American adults who support the expansion of solar energy is p = 0.88, which is our parameter of interest.2 If we were to take a poll of 1000 American adults on this topic, the estimate would not be perfect, but how close might we expect the sample proportion in the poll would be to 88%? We want to understand, how does the sample proportion ˆp behave when the true population proportion is 0.88.
3 Let’s ﬁnd out! We can simulate responses we would get from a simple random sample of 1000 American adults, which is only possible because we know the actual support expanding solar energy to be 0.88. Here’s how we might go about constructing such a simulation: 1. There were about 250 million American adults in 2018. On 250 million pieces of paper, write “support” on 88% of them and “not” on the other 12%. 2. Mix up the pieces of paper and pull out 1000 pieces to represent our sample of 1000 American adults. 3. Compute the fraction of the sample that say “support”. Any volunteers to conduct this simulation? Probably not. While this physical simulation is totally impractical, we can simulate it thousands, even millions, of times using computer code. We’ve written a short computer simulation and run it 10,000 times. The results are show in Figure 5.1 in case you are curious what the computer code looks like. In this simulation, the sample gave a point estimate of ˆp1 = 0.894. We know the population proportion for the simulation was p = 0.88, so we know the estimate had an error of 0.894 − 0.88 = +0.014. and 12% are "not". # 1. Create a set of 250 million entries, where 88% of them are "support" # pop size <- 250000000 possible entries <- c(rep("support", 0.88 * pop size), rep("not", 0.12 * pop size)) # 2. Sample 1000 entries without replacement. sampled entries <- sample(possible entries, size = 1000) # 3. Compute p-hat: count the number that are "support", then divide by # sum(sampled entries == "support") / 1000 the sample size. Figure 5.1: For those curious, this is code for a single ˆp simulation using the statistical software called R. Each line that starts with # is a code comment, which is used to describe in regular language what the code is doing. We’ve provided software labs in R at openintro.org/stat/labs for anyone interested in learning more. One simulation isn’t enough to get a great sense of the distribution of estimates we might expect in the simulation, so we should run more simulations. In a second simulation, we get ˆp
2 = 0.885, which has an error of +0.005. In another, ˆp3 = 0.878 for an error of -0.002. And in another, an estimate of ˆp4 = 0.859 with an error of -0.021. With the help of a computer, we’ve run the simulation 10,000 times and created a histogram of the results from all 10,000 simulations in Figure 5.2. This distribution of sample proportions is called a sampling distribution. We can characterize this sampling distribution as follows: 2We haven’t actually conducted a census to measure this value perfectly. However, a very large sample has suggested the actual level of support is about 88%. 3Note: 88% written as a proportion would be 0.88. It is common to switch between proportion and percent. 5.1. ESTIMATING UNKNOWN PARAMETERS 257 Center. The center of the distribution is µ ˆp = 0.880, which is the same as the parameter. Notice that the simulation mimicked a simple random sample of the population, which is a straightforward sampling strategy that helps avoid sampling bias. Spread. The standard deviation of the distribution is σ ˆp = 0.010. Shape. The distribution is symmetric and bell-shaped, and it resembles a normal distribution. These ﬁndings are encouraging! When the population proportion is p = 0.88 and the sample size is n = 1000, the sample proportion ˆp tends to give a pretty good estimate of the population proportion. We also have the interesting observation that the histogram resembles a normal distribution. Figure 5.2: A histogram of 10,000 sample proportions sampled from a population where the population proportion is 0.88 and the sample size is n = 1000. SAMPLING DISTRIBUTIONS ARE NEVER OBSERVED, BUT WE KEEP THEM IN MIND In real-world applications, we never actually observe the sampling distribution, yet it is useful to always think of a point estimate as coming from such a hypothetical distribution. Understanding the sampling distribution will help us characterize and make sense of the point estimates that we do observe. EXAMPLE 5.3 If we used a much smaller sample size of n = 50, would you guess that the standard error for ˆp would be larger or smaller than when we used n = 1000? Intuitively, it seems like more data
is better than less data, and generally that is correct! The typical error when p = 0.88 and n = 50 would be larger than the error we would expect when n = 1000. Example 5.3 highlights an important property we will see again and again: a bigger sample tends to provide a more precise point estimate than a smaller sample. Remember though, that this is only true for random samples. Additionally, a bigger sample cannot correct for response bias or other types of bias that may be present. Sample Proportions0.840.860.880.900.920250500750Frequency 258 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.1.3 Introducing the standard error Point estimates only approximate the population parameter. How can we quantify the expected variability in a point estimate ˆp? The discussion in Section 4.1 tells us how. The variability in the distribution of ˆp is given by its standard deviation. If we know the population proportion, we can calculate the standard deviation of the point estimate ˆp. In our simulation we knew p was 0.88. Thus we can calculate the standard deviation as SD ˆp = p(1 − p) n = (.088)(1 − 0.88) n = 0.01 If we now look at the sampling distribution, we see that the typical distance sample proportions are from the true value of 0.88 is about 0.01. EXAMPLE 5.4 Consider a random sample of size 80 from a county population. We ﬁnd that 15% of the sample support a controversial new ballot measure at the state level. How far is our estimate likely to be from the true percent that support the measure? We would like to calculate the standard deviation of ˆp, but we run into a serious problem: p is unknown. In fact, when doing inference, p must be unknown, otherwise it is illogical to try to estimate it. We cannot calculate the SD, but we can estimate it using, you might have guessed, the sample proportion ˆp. This estimate of the standard deviation is known as the standard error, or SESESE for short. SE ˆp = ˆp(1 − ˆp) n EXAMPLE 5.5 Calculate and interpret the SE of ˆp for the previous example. SE ˆp = ˆp(1 − ˆp) n = 0
.15(1 − 0.15) 80 = 0.04 The typical or expected error in our estimate is 4%. EXAMPLE 5.6 If we quadruple the sample size from 80 to 320, what will happen to the SE? SE ˆp = ˆp(1 − ˆp) n = 0.15(1 − 0.15) 320 = 0.02 The larger the sample size, the smaller our standard error. This is consistent with intuition: the more data we have, the more reliable an estimate will tend to be. However, quadrupling the sample size does not reduce the error by a factor of 4. Because of the square root, the eﬀect is to reduce the error by a factor 4, or 2. √ 5.1. ESTIMATING UNKNOWN PARAMETERS 259 5.1.4 Basic properties of point estimates We achieved three goals in this section. First, we determined that point estimates from a sample may be used to estimate population parameters. We also determined that these point estimates are not exact: they vary from one sample to another. Lastly, we quantiﬁed the uncertainty of the sample proportion using what we call the standard error. Remember that the standard error only measures sampling error. It does not account for bias that results from leading questions or other types of response bias. When our sampling method produces estimates in an unbiased way, the sampling distribution will be centered on the true value and we call the method accurate. When the sampling method produces estimates that have low variability, the sampling distribution will have a low standard error, and we call the method precise. EXAMPLE 5.7 Using Figure 5.3, which of the distributions were produced by methods that are biased? That are accurate? Order the distributions from most precise to least precise (that is, from lowest variability to highest variability). Distributions (b) and (d) are centered on the parameter (the true value), so those methods are accurate. The methods that produced distributions (a) and (c) are biased, because those distributions are not centered on the parameter. From most precise to least precise, we have (a), (b), (c), (d). Figure 5.3: Four sampling distributions shown, with parameter identiﬁed. Explore these distributions through a Desmos activity at openintro.org/ahss/desmos. EXAMPLE 5.8 Why do we want a point
estimate to be both precise and accurate? If the point estimate is precise, but highly biased, then we will consistently get a bad estimate. On the other hand, if the point estimate is unbiased but not at all precise, then by random chance, we may get an estimate far from the true value. Remember, when taking a sample, we generally get only one chance. sampling distribution that tell us how much conﬁdence we can have in the estimate. It is the properties of the The strategy of using a sample statistic to estimate a parameter is quite common, and it’s a strategy that we can apply to other statistics besides a proportion. For instance, if we want to estimate the average salary for graduates from a particular college, we could survey a random sample of recent graduates; in that example, we’d be using a sample mean ¯x to estimate the population mean µ for all graduates. As another example, if we want to estimate the diﬀerence in product prices for two websites, we might take a random sample of products available on both sites, check the prices on each, and use then compute the average diﬀerence; this strategy certainly wouldn’t give us a perfect measurement of the actual diﬀerence, but it would give us a point estimate. While this chapter emphases a single proportion context, we’ll encounter many diﬀerent contexts throughout this book where these methods will be applied. The principles and general ideas are the same, even if the details change a little. 260 CHAPTER 5. FOUNDATIONS FOR INFERENCE Section summary • In this section we laid the groundwork for our study of inference. Inference involves using known sample values to estimate or better understand unknown population values. • A sample statistic can serve as a point estimate for an unknown parameter. For example, the sample mean is a point estimate for an unknown population mean, and the sample proportion is a point estimate for an unknown population proportion. • It is helpful to imagine a point estimate as being drawn from a particular sampling distribution. • The standard error (SE) of a point estimate tells us the typical error or uncertainty associated with the point estimate. It is also an estimate of the spread of the sampling distribution. • A point estimate is unbiased (accurate) if the sampling distribution (i.e., the distribution of all possible outcomes of the point estimate from repeated samples from the same population) is centered on the true population parameter. • A point
estimate has lower variability (more precise) when the standard deviation of the sampling distribution is smaller. • In a random sample, increasing the sample size n will make the standard error smaller. This is consistent with the intuition that larger samples tend to be more reliable, all other things being equal. • In general, we want a point estimate to be unbiased and to have low variability. Remember: the terms unbiased (accurate) and low variability (precise) are properties of generic point estimates, which are variables that have a sampling distribution. These terms do not apply to individual values of a point estimate, which are numbers. 5.1. ESTIMATING UNKNOWN PARAMETERS 261 Exercises 5.1 Identify the parameter, Part I. For each of the following situations, state whether the parameter of interest is a mean or a proportion. It may be helpful to examine whether individual responses are numerical or categorical. (a) In a survey, one hundred college students are asked how many hours per week they spend on the Internet. (b) In a survey, one hundred college students are asked: “What percentage of the time you spend on the Internet is part of your course work?” (c) In a survey, one hundred college students are asked whether or not they cited information from Wikipedia in their papers. (d) In a survey, one hundred college students are asked what percentage of their total weekly spending is on alcoholic beverages. (e) In a sample of one hundred recent college graduates, it is found that 85 percent expect to get a job within one year of their graduation date. 5.2 Identify the parameter, Part II. For each of the following situations, state whether the parameter of interest is a mean or a proportion. (a) A poll shows that 64% of Americans personally worry a great deal about federal spending and the budget deﬁcit. (b) A survey reports that local TV news has shown a 17% increase in revenue within a two year period while newspaper revenues decreased by 6.4% during this time period. (c) In a survey, high school and college students are asked whether or not they use geolocation services on their smart phones. (d) In a survey, smart phone users are asked whether or not they use a web-based taxi service. (e) In a survey, smart phone users are asked how many times they used a web-based taxi service over the last year. 5.3 Quality control.
As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She ﬁnds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic that we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is 10%. Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be 10%. If we use this proportion to recompute the value in part (e) using p = 0.1 instead of ˆp, does the resulting value change much? 262 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.4 Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic that we can use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data? (g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0.4 instead of ˆp, does the resulting value change much? 5.5 Repeated water samples. A nonproﬁt wants to understand the fraction of households that have elevated levels of lead in their drinking water. They expect at least 5% of homes will have elevated levels of lead, but not more than about 30%. They randomly sample 800 homes and work with the owners to retrieve water samples, and they compute the fraction of these homes with elevated lead levels. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape
of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) If the proportions are distributed around 8%, what is the variability of the distribution? (d) What is the formal name of the value you computed in (c)? (e) Suppose the researchers’ budget is reduced, and they are only able to collect 250 observations per sample, but they can still collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 800 observations? 5.6 Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution. (d) What is the formal name of the value you computed in (c)? (e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations? 5.2. CONFIDENCE INTERVALS 263 5.2 Conﬁdence intervals The site ﬁvethirtyeight.com regularly forecasts support for each candidate in Congressional races, i.e. races in the US House of Representatives and the US Senate. In addition to point estimates, they report conﬁdence intervals.4 What are conﬁdence intervals, and how do we interpret them? Learning objectives 1. Explain the purpose and use of conﬁdence intervals. 2. Construct 95% conﬁdence intervals assuming the point estimate follows a normal distribution. 3. Calculate the critical value for a C% conﬁdence interval when the point estimate follows a normal distribution. 4. Describe how sample size and conﬁdence level aﬀect the width of a conﬁdence interval. 5. Interpret a conﬁdence interval
and the conﬁdence level in context. 6. Draw conclusions with a speciﬁed conﬁdence level about the values of unknown parameters. 7. Calculate and interpret the margin of error for a C% conﬁdence interval. Distinguish between margin of error and standard error. 5.2.1 Capturing the population parameter A point estimate provides a single plausible value for a parameter. However, a point estimate isn’t perfect and will have some standard error associated with it. When estimating a parameter, it is better practice to provide a plausible range of values instead of supplying just the point estimate. A plausible range of values for the population parameter is called a conﬁdence interval. Using only a point estimate is like ﬁshing in a murky lake with a spear, and using a conﬁdence interval is like ﬁshing with a net. We can throw a spear where we saw a ﬁsh, but we will probably miss. On the other hand, if we toss a net in that area, we have a good chance of catching the ﬁsh. If we report a point estimate, we probably will not hit the exact population parameter. On the other hand, if we report a range of plausible values – a conﬁdence interval – we have a good shot at capturing the parameter. 4See: https://projects.fivethirtyeight.com/2018-midterm-election-forecast/senate 264 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.2 Constructing a 95% conﬁdence interval A point estimate is our best guess for the value of the parameter, so it makes sense to build the conﬁdence interval around that value. The standard error is a measure of the uncertainty associated with the point estimate. EXAMPLE 5.9 How many standard errors should we extend above and below the point estimate if we want to be 95% conﬁdent of capturing the true value? First, we observe that the area under the standard normal curve between -1.96 and 1.96 is 95%. When conditions for a normal model are met, the point estimate we observe will be within 1.96 standard deviations of the true value about 95% of the time. Thus, if we want to be 95% conﬁdent of capturing the true
value, we should go 1.96 standard errors on either side of the point estimate. CONSTRUCTING A 95% CONFIDENCE INTERVAL USING A NORMAL MODEL When the sampling distribution for a point estimate can reasonably be modeled as normal, a 95% conﬁdence interval for the unknown parameter can be constructed as: point estimate ± 1.96 × SE of estimate (5.10) We can be 95% conﬁdent that this interval captures the true value. In the next chapters we will determine when we can apply a normal model to a point estimate. For now, we will assume that a normal model is reasonable. EXAMPLE 5.11 The point estimate of the support for a ballot measure from Section 5.1 was 15% from a sample size of 80. The standard error for this point estimate was calculated to be SE = 0.04. Assuming that conditions for a normal model are met, construct and interpret a 95% conﬁdence interval. point estimate ± 1.96 × SE of estimate 0.15 ± 1.96 × 0.04 (0.0716, 0.2284) We are 95% conﬁdent that the true percent of support in this population for the ballot measure is between 7.16% and 22.84%. EXAMPLE 5.12 Suppose the level of support at the state level for the ballot measure has been precisely estimated at 0.20 (20%). Based on the conﬁdence interval above, is there evidence that a smaller proportion support the ballot measure in the speciﬁc county we’re looking at relative to the state as a whole? While the point estimate of 0.15 is lower than 0.20, it appears this deviation may be due to random chance. Because the conﬁdence interval includes the value 0.20, the value of 0.20 is a reasonable value for the proportion of the county population that support the ballot measure. Therefore, based on this conﬁdence interval, we do not have evidence that a smaller proportion support the ballot measure in this county than in the state as a whole. 5.2. CONFIDENCE INTERVALS 265 We can be 95% conﬁdent that a 95% conﬁdence interval captures the true population parameter. However, conﬁdence intervals are imperfect. About 1-in-20
(5%) properly constructed 95% conﬁdence intervals will fail to capture the parameter of interest. Figure 5.4 shows 25 conﬁdence intervals for a proportion that were constructed from simulations where the true proportion was p = 0.3. However, 1 of these 25 conﬁdence intervals happened not to include the true value. Figure 5.4: Twenty-ﬁve samples of size n = 300 were simulated when p = 0.30. For each sample, a conﬁdence interval was created to try to capture the true proportion p. However, 1 of these 25 intervals did not capture p = 0.30. GUIDED PRACTICE 5.13 In Figure 5.4, one interval does not capture the true proportion, p = 0.3. Does this imply that there was a problem with the simulations?5 5No. Just as some observations occur more than 1.96 standard deviations from the mean, some point estimates will be more than 1.96 standard errors from the parameter. A conﬁdence interval only provides a plausible range of values for a parameter. While we might say other values are implausible based on the data, this does not mean they are impossible. 0.25p = 0.300.35llllllllllllllllllllllllll 266 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.3 Changing the conﬁdence level Suppose we want to construct a conﬁdence interval with a conﬁdence level somewhat greater than 95%: perhaps we would like a conﬁdence level of 99%. EXAMPLE 5.14 Other things being equal, would a 99% conﬁdence interval be wider or narrower than a 95% conﬁdence interval? Using a previous analogy: if we want to be more conﬁdent that we will catch a ﬁsh, we should use a wider net, not a smaller one. To be 99% conﬁdence of capturing the true value, we must use a wider interval. On the other hand, if we want an interval with lower conﬁdence, such as 90%, we would use a narrower interval. The 95% conﬁdence interval structure provides guidance in how to make intervals with new conﬁdence levels. Below is
a general 95% conﬁdence interval for a point estimate that comes from a nearly normal distribution: point estimate ± 1.96 × SE of estimate (5.15) There are three components to this interval: the point estimate, “1.96”, and the standard error. The choice of 1.96 × SE was based on capturing 95% of the distribution since the estimate is within 1.96 standard deviations of the true value about 95% of the time. The choice of 1.96 corresponds to a 95% conﬁdence level. GUIDED PRACTICE 5.16 If X is a normally distributed random variable, how often will X be within 2.58 standard deviations of the mean?6 Figure 5.5: The area between -z and z increases as \|z\| becomes larger. If the conﬁdence level is 99%, we choose z such that 99% of the normal curve is between -z and z, which corresponds to 0.5% in the lower tail and 0.5% in the upper tail: z = 2.58. 6This is equivalent to asking how often the Z-score will be larger than -2.58 but less than 2.58. (For a picture, see Figure 5.5.) There is ≈ 0.99 probability that the unobserved random variable X will be within 2.58 standard deviations of the mean. standard deviations from the mean−3−2−1012395%, extends −1.96 to 1.9699%, extends −2.58 to 2.58 5.2. CONFIDENCE INTERVALS 267 To create a 99% conﬁdence interval, change 1.96 in the 95% conﬁdence interval formula to be 2.58. Guided Practice 5.16 highlights that 99% of the time a normal random variable will be within 2.58 standard deviations of its mean. Thus, the formula for a 99% conﬁdence interval is point estimate ± 2.58 × SE of estimate (5.17) Figure 5.5 provides a picture of how to identify z based on a conﬁdence level. The number of standard errors we go above and below the point estimate is called the critical value. When the critical value is determined based on a normal model, we call the critical value z. CONFIDENCE INTERVAL FOR ANY CONFIDENCE
LEVEL If the point estimate follows a normal model with standard error SE, then a conﬁdence interval for the population parameter is point estimate ± z × SE of estimate where z depends on the conﬁdence level selected. Finding the value of z that corresponds to a particular conﬁdence level is most easily accomplished by using a new table, called the t-table. For now, what is noteworthy about this table is that the bottom row corresponds to conﬁdence levels. The numbers inside the table are the critical values, but which row should we use? Later in this book, we will see that a t-curve with inﬁnite degrees of freedom corresponds to the normal curve. For this reason, when ﬁnding z, we use the t-table at row ∞. df one tail 1 2 3... 1000 0.100 3.078 1.886 1.638... 1.282 ∞ 1.282 80% 0.050 6.314 2.920 2.353... 1.646 1.645 90% 0.025 12.71 4.303 3.182... 1.962 1.960 95% 0.010 31.82 6.965 4.541... 2.330 2.326 98% 0.005 63.66 9.925 5.841 2.581 2.576 99% Conﬁdence level C Figure 5.6: An abbreviated look at the t-table. The columns correspond to conﬁdence levels. Row ∞ corresponds to the normal curve. FINDING zzz FOR A PARTICULAR CONFIDENCE LEVEL We select z so that the area between -z and z in the normal model corresponds to the conﬁdence level. Use a calculator or use the t-table at row ∞ to ﬁnd the critical value z. GUIDED PRACTICE 5.18 Find the appropriate z value for an 80% conﬁdence interval.7 The normal approximation is crucial to the precision of these conﬁdence intervals. The next two chapters provide detailed discussions about when a normal model can safely be applied to a variety of situations. When a normal model is not a good ﬁt, we will use alternate distributions that better characterize the sampling distribution. 7Using row ∞ on the t-table, we see
that the value that corresponds to an 80% conﬁdence level is 1.282. Therefore, we should use 1.282 as the z value. 268 CHAPTER 5. FOUNDATIONS FOR INFERENCE 5.2.4 Margin of error The conﬁdence intervals we have encountered thus far have taken the form point estimate ± z∗ × SE of estimate Conﬁdence intervals are also often reported as point estimate ± margin of error For example, instead of reporting an interval as 0.09 ± 1.645 × 0.028 or (0.044, 0.136), it could be reported as 0.09 ± 0.046. The margin of error equals the critical value times the standard error of the estimate. The researcher chooses a conﬁdence level, and then calculates the margin of error that corresponds to that conﬁdence level. The margin of error for a 95% conﬁdence interval tells us that we are 95% conﬁdent that the sample statistic is within that margin of error of the true value. Numerically, the margin of error corresponds to the distance between the point estimate and the lower or upper bound of a conﬁdence interval, and thus is half of the total width of the interval. MARGIN OF ERROR When the point estimate follows a normal distribution, margin of error = z × SE of estimate. EXAMPLE 5.19 All other things being equal, will the margin of error be bigger for a 68% conﬁdence interval or a 95% conﬁdence interval? A 95% conﬁdence interval is wider than a 68% conﬁdence interval and has a larger z value, so the 95% conﬁdence interval will have a larger margin of error. In order to be more conﬁdent of capturing the true value, we must create a wider interval. GUIDED PRACTICE 5.20 What is the margin of error for the conﬁdence interval: (0.035, 0.145)?8 8The margin of error is half of the total width of the interval. The margin of error for this interval is 0.145−0.035 = 2 0.055. 5.2. CONFIDENCE INTERVALS 269 5.2.5 Interpreting conﬁ
dence intervals A careful eye might have observed the somewhat awkward language used to describe conﬁdence intervals. Correct interpretation: We are C% conﬁdent that the population parameter is between and. Incorrect language might try to describe the conﬁdence interval as capturing the population parameter with a certain probability.9 Applying the language of probability to a ﬁxed interval or to a ﬁxed parameter is one of the most common errors. As we saw in Figure 5.4, the 95% conﬁdence interval method has a 95% probability of producing an interval that will capture the population parameter. A correct interpretation of the conﬁdence level is that such intervals will capture the population parameter that percent of the time (assuming conditions are met and the probability model is true). However, each individual interval either does or does not capture the population parameter. A correct interpretation of an individual conﬁdence interval cannot involve the vocabulary of probability. Another especially important consideration of conﬁdence intervals is that they only try to capture the population parameter. Our intervals say nothing about the conﬁdence of capturing individual observations, a proportion of the observations, or point estimates. Conﬁdence intervals only attempt to capture population parameters. 5.2.6 Conﬁdence interval procedures: a ﬁve step process Use a conﬁdence interval to estimate a parameter with a particular conﬁdence level, C. (AP EXAM TIP) WHEN CARRYING OUT A CONFIDENCE INTERVAL PROCEDURE, FOLLOW THESE FIVE STEPS: • Identify: Identify the parameter and the conﬁdence level. • Choose: Choose the appropriate interval procedure and identify it by name. • Check: Check that the conditions for the interval procedure are met. • Calculate: Calculate the conﬁdence interval and record it in interval form. CI: point estimate ± critical value × SE of estimate • Conclude: Interpret the interval and, if applicable, draw a conclusion based on whether the interval is entirely above, is entirely below, or contains the value of interest. 9To see that this interpretation is incorrect, imagine taking two random samples and constructing two 95% conﬁdence intervals for an unknown proportion. If these intervals are disjoint, can we say that there is
a 95%+95%=190% chance that the ﬁrst or the second interval captures the true value? 270 CHAPTER 5. FOUNDATIONS FOR INFERENCE Section summary • A point estimate is not perfect; there is almost always some error in the estimate. It is often useful to supply a plausible range of values for the parameter, which we call a conﬁdence interval. • A conﬁdence interval is centered on the point estimate and extends a certain number of standard errors on either side of the estimate, depending upon how conﬁdent one wants to be. For a ﬁxed sample size, to be more conﬁdent of capturing the true value requires a wider interval. • When the sampling distribution for a point estimate can reasonably be modeled as normal, such as with a sample proportion, then the following are true: – A 95% conﬁdence interval is given by: point estimate ± 1.96 × SE of estimate. We can be 95% conﬁdent this interval captures the true value. – A C% conﬁdence interval is given by: point estimate ± z × SE of estimate. We can be C% conﬁdent this interval captures the true value. • For a C% conﬁdence interval described above, we select z such that the area between -z and z under the standard normal curve is C%. Use a t-table at row ∞ to ﬁnd the critical value z. 10 • After interpreting the interval, we can usually draw a conclusion, with C% conﬁdence, about whether a given value X is a reasonable value for the population parameter. When drawing a conclusion based on a conﬁdence interval, there are three possibilities. – We have evidence that the true [parameter]:...is greater than X, because the entire interval is above X....is less than X, because the entire interval is below X. – We do not have evidence that the true [parameter] is not X, because X is in the interval. Interpreting conﬁdence intervals and conﬁdence levels • 68% and 95% are examples of conﬁdence levels. The conﬁdence level tells us the capture rate with repeated sampling. For example, a correct interpretation of a 95%