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# Advanced Placement Statistics Practice Test

This is a practice test for the multiple-choice section of the Advanced Placement Statistics Exam. Like the actual AP Statistics test, this test includes 40 questions covering a variety of topics.

## Test Instructions

Each multiple-choice question has five response options. Choose the best answer. Then, click the "Next" button or the "Back" button to see another question.

The multiple-choice portion of the Advanced Placement (AP) Statistics Exam is a timed test, lasting 90 minutes. Use the Timer (top left) to simulate an actual test, or pause the Timer to work at your own pace. Use the Test Performance monitor (bottom left) to check your test results. Test questions are color-coded.

• Blue. A question that has not been answered.
• Green. A question that has been answered correctly.
• Red. A question that has been answered incorrectly.

Choose the best answer to the question. To see another question, click the "Next" button or the "Back" button. To see the solution, click the "Show solution" button.

## Problem 1

Which of the following statements are true? (Check one)

I. Categorical variables are the same as qualitative variables.
II. Categorical variables are the same as quantitative variables.
III. Quantitative variables can be continuous variables.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

## Solution

The correct answer is (E). Categorical variables are just another name for qualitative variables. And quantitative variables are numeric variables, so they can be continuous variables. Categorical variables, however, are not quantitative variables.

## Problem 2

A coin is tossed three times. What is the probability that it lands on heads exactly one time?

(A) 0.125
(B) 0.250
(C) 0.333
(D) 0.375
(E) 0.500

## Solution

The correct answer is (D). If you toss a coin three times, there are a total of eight possible outcomes. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Of the eight possible outcomes, three have exactly one head. They are: HTT, THT, and TTH. Therefore, the probability that three flips of a coin will produce exactly one head is 3/8 or 0.375.

## Problem 3

An auto analyst is conducting a satisfaction survey, sampling from a list of 10,000 new car buyers. The list includes 2,500 Ford buyers, 2,500 GM buyers, 2,500 Honda buyers, and 2,500 Toyota buyers. The analyst selects a sample of 400 car buyers, by randomly sampling 100 buyers of each brand.

Is this an example of a simple random sample?

(A) Yes, because each buyer in the sample was randomly sampled.
(B) Yes, because each buyer in the sample had an equal chance of being sampled.
(C) Yes, because car buyers of every brand were equally represented in the sample.
(D) No, because every possible 400-buyer sample did not have an equal chance of being chosen.
(E) No, because the population consisted of purchasers of four different brands of car.

## Solution

The correct answer is (D). A simple random sample requires that every sample of size n (in this problem, n is equal to 400) have an equal chance of being selected. In this problem, there was a 100 percent chance that the sample would include 100 purchasers of each brand of car. There was zero percent chance that the sample would include, for example, 99 Ford buyers, 101 Honda buyers, 100 Toyota buyers, and 100 GM buyers. Thus, all possible samples of size 400 did not have an equal chance of being selected; so this cannot be a simple random sample.

The fact that each buyer in the sample was randomly sampled is a necessary condition for a simple random sample, but it is not sufficient. Similarly, the fact that each buyer in the sample had an equal chance of being selected is characteristic of a simple random sample, but it is not sufficient. The sampling method in this problem used random sampling and gave each buyer an equal chance of being selected; but the sampling method was actually stratified random sampling.

The fact that car buyers of every brand were equally represented in the sample is irrelevant to whether the sampling method was simple random sampling. Similarly, the fact that population consisted of buyers of different car brands is irrelevant.

## Problem 4

Which of the following statements is true?

I. The center of a confidence interval is a population parameter.
II. The bigger the margin of error, the smaller the confidence interval.
III. The confidence interval is a type of point estimate.
IV. A population mean is an example of a point estimate.

(A) I only
(B) II only
(C) III only
(D) IV only
(E) None of the above.

## Solution

The correct answer is (E). The center of a confidence interval is a sample statistic, not a population parameter. The confidence interval is equal to the sample statistic plus or minus the margin of error; so the confidence interval gets bigger as the margin of error gets bigger. A confidence interval is a type of interval estimate, not a type of point estimate. A population mean is not an example of a point estimate; a sample mean is an example of a point estimate.

## Problem 5

A sample consists of four observations: {1, 3, 5, 7}. What is the standard deviation?

(A) 2
(B) 2.58
(C) 6
(D) 6.67
(E) None of the above

## Solution

The correct answer is (B). First, we need to compute the sample mean.

x = ( 1 + 3 + 5 + 7 ) / 4 = 4

Then, we plug all of the known values into the formula for the standard deviation of a sample, as shown below:

s = sqrt [ Σ ( xi - x )2 / ( n - 1 ) ]
s = sqrt { [ ( 1 - 4 )2 + ( 3 - 4 )2 + ( 5 - 4 )2 + ( 7 - 4 )2 ] / ( 4 - 1 ) }
s = sqrt { [ ( -3 )2 + ( -1 )2 + ( 1 )2 + ( 3 )2 ] / 3 }
s = sqrt { [ 9 + 1 + 1 + 9 ] / 3 } = sqrt (20 / 3) = sqrt ( 6.67 ) = 2.58

## Problem 6

A card is drawn randomly from a deck of ordinary playing cards. You win \$10 if the card is a spade or an ace. What is the probability that you will win the game?

(A) 1/13
(B) 13/52
(C) 4/13
(D) 17/52
(E) None of the above.

## Solution

The correct answer is C. Let S = the event that the card is a spade; and let A = the event that the card is an ace. We know the following:

• There are 52 cards in the deck.
• There are 13 spades, so P(S) = 13/52.
• There are 4 aces, so P(A) = 4/52.
• There is 1 ace that is also a spade, so P(S A) = 1/52.

Therefore, based on the rule of addition:

P(S A) = P(S) + P(A) - P(S A)
P(S A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13

## Problem 7

Which of the following statements is true.

I. The standard error is computed solely from sample attributes.
II. The standard deviation is computed solely from sample attributes.
III. The standard error is a measure of central tendency.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

## Solution

The correct answer is (A). The standard error can be computed from a knowledge of sample attributes - sample size and sample statistics. The standard deviation cannot be computed solely from sample attributes; it requires a knowledge of one or more population parameters. The standard error is a measure of variability, not a measure of central tendency.

## Problem 8

Nine hundred (900) high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 2.7, and the standard deviation was 0.4. What is the margin of error, assuming a 95% confidence level?

(A) 0.013
(B) 0.025
(C) 0.500
(D) 1.960
(E) None of the above.

## Solution

The correct answer is (B). To compute the margin of error, we need to find the critical value and the standard error of the mean. To find the critical value, we take the following steps.

• Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05
• Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 = 0.975
• Find the critical z score. Since the sample size is large, the sampling distribution will be roughly normal in shape. Therefore, we can express the critical value as a z score. For this problem, it will be the z score having a cumulative probability equal to 0.975. Then, using an online calculator (e.g., Stat Trek's free normal distribution calculator), a handheld graphing calculator, or the standard normal distribution table, we find the cumulative probability associated with the z-score. Using the Normal Distribution Calculator, we find that the critical value is 1.96. (On the actual AP Statistics exam, you may need to use a graphing calculator or a normal table, since Stat Trek's analytical tools will not be available.)

Next, we find the standard error of the mean, using the following equation:

SEx = s / sqrt( n ) = 0.4 / sqrt( 900 ) = 0.4 / 30 = 0.013

And finally, we compute the margin of error (ME).

ME = Critical value x Standard error = 1.96 * 0.013 = 0.025

## Problem 9

A national achievement test is administered annually to 3rd graders. The test has a mean score of 100 and a standard deviation of 15. If Jane's z-score is 1.20, what was her score on the test?

(A) 82
(B) 88
(C) 100
(D) 112
(E) 118

## Solution

The correct answer is (E). From the z-score equation, we know

z = (X - μ) / σ

where z is the z-score, X is the value of the element, μ is the mean of the population, and σ is the standard deviation.

Solving for Jane's test score (X), we get

X = ( z * σ) + 100 = ( 1.20 * 15) + 100 = 18 + 100 = 118

## Problem 10

Which of the following is a discrete random variable?

I. The average height of a randomly selected group of boys.
II. The annual number of sweepstakes winners from New York City.
III. The number of presidential elections in the 20th century.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

## Solution

The correct answer is B. The annual number of sweepstakes winners is an integer value and it results from a random process; so it is a discrete random variable. The average height of a group of boys could be a non-integer, so it is not a discrete variable. And the number of presidential elections in the 20th century is an integer, but it does not vary and it does not result from a random process; so it is not a random variable.

## Problem 11

Which of the following statements are true? (Check one)

I. A sample survey is an example of an experimental study.
II. An observational study requires fewer resources than an experiment.
III. The best method for investigating causal relationships is an observational study.

(A) I only
(B) II only
(C) III only
(D) All of the above.
(E) None of the above.

## Solution

The correct answer is (E). In a sample survey, the researcher does not assign treatments to survey respondents. Therefore, a sample survey is not an experimental study; rather, it is an observational study. An observational study may or may not require fewer resources (time, money, manpower) than an experiment. The best method for investigating causal relationships is an experiment - not an observational study - because an experiment features randomized assignment of subjects to treatment groups. Randomization "evens out" the effects of extraneous variables, which makes it easier for the researcher to identify causal effects of treatment variables.

## Problem 12

Suppose we want to estimate the average weight of an adult male in Dekalb County, Georgia. We draw a random sample of 1,000 men from a population of 1,000,000 men and weigh them. We find that the average man in our sample weighs 180 pounds, and the standard deviation of the sample is 30 pounds. What is the 95% confidence interval.

(A) 180 + 1.86
(B) 180 + 3.0
(C) 180 + 5.88
(D) 180 + 30
(E) None of the above.

## Solution

The answer is (A). To specify the confidence interval, we work through the four steps below.

• Identify a sample statistic. Since we are trying to estimate the mean weight in the population, we choose the mean weight in our sample (180) as the sample statistic.

• Select a confidence level. In this case, the confidence level is defined for us in the problem. We are working with a 95% confidence level.

• Find the margin of error. Previously, we described how to compute the margin of error. The key steps are shown below.

• Find standard error. The standard error (SE) of the mean is:

SE = s / sqrt( n ) = 30 / sqrt(1000) = 30/31.62 = 0.95

• Find critical value. The critical value is a factor used to compute the margin of error. To express the critical value as a t score (t*), follow these steps.

• Compute alpha (α): α = 1 - (confidence level / 100) = 0.05
• Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 = 0.975
• Find the degrees of freedom (df): df = n - 1 = 1000 - 1 = 999
• The critical value is the t score having 999 degrees of freedom and a cumulative probability equal to 0.975. Using an online calculator (e.g., Stat Trek's free t Distribution Calculator), a handheld graphing calculator, or a t distribution table, we find that the t score associated with a cumulative probability of 0.975 is 1.96. (On the actual AP Statistics exam, you may need to use a graphing calculator or a t table, since Stat Trek's analytical tools will not be available.)

Note: We might also have expressed the critical value as a z score. Because the sample size is large, a z score analysis produces the same result - a critical value equal to 1.96.

• Compute margin of error (ME): ME = critical value * standard error = 1.96 * 0.95 = 1.86

• Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level. Therefore, we can be 95% confident that the population means falls within the interval 180 + 1.86.

## Problem 13

The stemplot below shows the number of hot dogs eaten by contestants in a recent hot dog eating contest.

 80 70 60 50 40 30 20 10 1 4 7 2 2 6 0 2 5 7 9 9 5 7 9 7 9 1

Which of the following statements are true?

I. The range is 70.
II. The median is 46.
III. The mean is 47.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III

## Solution

The correct answer is (D). The range is equal to the biggest value minus the smallest value. The biggest value is 81, and the smallest value is 11; so the range is equal to 81 -11 or 70. The median is equal to the middle value in the data set. Here, we have an even number of values - 45 and 47 - in the middle of the data set. Their average is (45 + 47)/2 or 46, so the median is equal to 46. The mean is equal to the average of all the values - 45.56.

## Problem 14

The number of adults living in homes on a randomly selected city block is described by the following probability distribution.

 Number of adults, x 0 1 2 3 4 or more Probability, P(x) 0.05 0.2 0.5 0.15 ???

What is the probability that 4 or more adults reside at a randomly selected home?

(A) 0.10
(B) 0.15
(C) 0.25
(D) 0.50
(E) There is not enough information to answer this question.

## Solution

The correct answer is A. The sum of all the probabilities is equal to 1. Therefore, the probability that four or more adults reside in a home is equal to 1 - (0.05 + 0.20 + 0.50 + 0.15) or 0.10.

## Problem 15

A major metropolitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. They asked whether the paper should increase its coverage of local news. Forty percent of the sample wanted more local news. What is the 99% confidence interval for the proportion of readers who would like more coverage of local news?

(A) 0.30 to 0.50
(B) 0.32 to 0.48
(C) 0.35 to 0.45
(D) 0.37 to 0.43
(E) 0.39 to 0.41

## Solution

The answer is (D). The approach that we used to solve this problem is valid when the following conditions are met.

• The sampling method must be simple random sampling. This condition is satisfied; the problem statement says that we used simple random sampling.
• The sample should include at least 10 successes and 10 failures. Suppose we classify a "more local news" response as a success, and any other response as a failure. Then, we have 0.40 * 1600 = 640 successes, and 0.60 * 1600 = 960 failures - plenty of successes and failures.
• If the population size is much larger than the sample size, we can use an "approximate" formula for the standard deviation or the standard error. This condition is satisfied, so we will use a simple "approximate" formula for the standard error.

Since the above requirements are satisfied, we can use the following four-step approach to construct a confidence interval.

• Identify a sample statistic. Since we are trying to estimate a population proportion, we choose the sample proportion (0.40) as the sample statistic.

• Select a confidence level. The confidence level is defined for us in the problem statement. We are working with a 99% confidence level.

• Find the margin of error. Elsewhere on this site, we show how to compute the margin of error when the sampling distribution is approximately normal. The key steps are shown below.

• Find standard deviation or standard error. Since we do not know the population proportion, we cannot compute the standard deviation; instead, we compute the standard error. And since the population is more than 10 times larger than the sample, we can use the following formula to compute the standard error (SE) of the proportion:

SE = sqrt [ p(1 - p) / n ] = sqrt [ (0.4)*(0.6) / 1600 ] = sqrt [ 0.24/1600 ] = 0.012

• Find critical value. The critical value is a factor used to compute the margin of error. Because the sampling distribution is approximately normal and the sample size is large, we can express the critical value as a z score by following these steps.

• Compute alpha (α): α = 1 - (confidence level / 100) = 1 - (99/100) = 0.01
• Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.01/2 = 0.995
• The critical value is the z score having a cumulative probability equal to 0.995. Using an online calculator (e.g., Stat Trek's free Normal Distribution Calculator), a handheld graphing calculator, or a normal distribution table, we find that the z score associated with a cumulative probability of 0.995 is 2.58. (On the actual AP Statistics exam, you may need to use a graphing calculator or a normal distribution table, since Stat Trek's analytical tools will not be available.)

• Compute margin of error (ME): ME = critical value * standard error = 2.58 * 0.012 = 0.03

• Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level.

Therefore, the 99% confidence interval is 0.37 to 0.43. That is, we are 99% confident that the true population proportion is in the range defined by 0.4 + 0.03.

## Problem 16

Suppose a simple random sample of 150 students is drawn from a population of 3000 college students. Among sampled students, the average IQ score is 115 with a standard deviation of 10. What is the 99% confidence interval for the students' IQ score?

(A) 115 + 0.01
(B) 115 + 0.82
(C) 115 + 2.1
(D) 115 + 2.6
(E) None of the above.

## Solution

The correct answer is (C). The approach that we used to solve this problem is valid when the following conditions are met.

Since the above requirements are satisfied, we can use the following four-step approach to construct a confidence interval.

• Identify a sample statistic. Since we are trying to estimate a population mean, we choose the sample mean (115) as the sample statistic.

• Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 99% confidence level.

• Find the margin of error. Elsewhere on this site, we show how to compute the margin of error when the sampling distribution is approximately normal. The key steps are shown below.

• Find standard deviation or standard error. Since we do not know the standard deviation of the population, we cannot compute the standard deviation of the sample mean; instead, we compute the standard error (SE). Because the sample size is much smaller than the population size, we can use the "approximate" formula for the standard error.

SE = s / sqrt( n ) = 10 / sqrt(150) = 10 / 12.25 = 0.82

• Find critical value. The critical value is a factor used to compute the margin of error. Because the standard deviation of the population is unknown, we express the critical value as a t score rather than a z score. To find the critical value, we take these steps.

• Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 99/100 = 0.01
• Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.01/2 = 0.995
• Find the degrees of freedom (df): df = n - 1 = 150 - 1 = 149
• The critical value is the t score having 149 degrees of freedom and a cumulative probability equal to 0.995. Using an online calculator (e.g., Stat Trek's free t Distribution Calculator), a handheld graphing calculator, or a t distribution table, we find that the t score associated with a cumulative probability of 0.995 is 2.61. (On the actual AP Statistics exam, you may need to use a graphing calculator or a t table, since Stat Trek's analytical tools will not be available.)

Note: We might also have expressed the critical value as a z score. Because the sample size is fairly large, a z score analysis produces a similar result - a critical value equal to 2.58.

• Compute margin of error (ME): ME = critical value * standard error = 2.61 * 0.82 = 2.1

• Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level.

Therefore, the 99% confidence interval is 112.9 to 117.1. That is, we are 99% confident that the true population mean is in the range defined by 115 + 2.1.

## Problem 17

Consider the boxplot below.

 2 4 6 8 10 12 14 16 18

Which of the following statements are true?

I. The distribution is skewed right.
II. The interquartile range is about 8.
III. The median is about 10.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

## Solution

The correct answer is (B). Most of the observations are on the high end of the scale, so the distribution is skewed left. The interquartile range is indicated by the length of the box, which is 18 minus 10 or 8. And the median is indicated by the vertical line running through the middle of the box, which is roughly centered over 15. So the median is about 15.

## Problem 18

The number of adults living in homes on a randomly selected city block is described by the following probability distribution.

 Number of adults, x 1 2 3 4 Probability, P(x) 0.25 0.5 0.15 0.1

What is the standard deviation of the probability distribution?

(A) 0.50
(B) 0.62
(C) 0.79
(D) 0.89
(E) 2.10

## Solution

The correct answer is D. The solution has three parts. First, find the expected value; then, find the variance; then, find the standard deviation. Computations are shown below, beginning with the expected value.

E(X) = Σ [ xi * P(xi) ]
E(X) = 1*0.25 + 2*0.50 + 3*0.15 + 4*0.10 = 2.10

Now that we know the expected value, we find the variance.

σ2 = Σ [ xi - E(x) ]2 * P(xi)
σ2 = (1 - 2.1)2 * 0.25 + (2 - 2.1)2 * 0.50 + (3 - 2.1)2 * 0.15 + (4 - 2.1)2 * 0.10
σ2 = (1.21 * 0.25) + (0.01 * 0.50) + (0.81) * 0.15) + (3.61 * 0.10) = 0.3025 + 0.0050 + 0.1215 + 0.3610 = 0.79

And finally, the standard deviation is equal to the square root of the variance; so the standard deviation is sqrt(0.79) or 0.89.

## Problem 19

Which of the following statements are true?

I. Random sampling is a good way to reduce response bias.
II. To guard against bias from undercoverage, use a convenience sample.
III. Increasing the sample size tends to reduce survey bias.
IV. To guard against nonresponse bias, use a mail-in survey.

(A) I only
(B) II only
(C) III only
(D) IV only
(E) None of the above.

## Solution

The correct answer is (E). None of the statements is true. Random sampling provides strong protection against bias from undercoverage bias and voluntary response bias; but it is not effective against response bias. A convenience sample does not protect against undercoverage bias; in fact, it sometimes causes undercoverage bias. Increasing sample size does not affect survey bias. And finally, using a mail-in survey does not prevent nonresponse bias. In fact, mail-in surveys are quite vulnerable to nonresponse bias.

## Problem 20

Twenty-two students were randomly selected from a population of 1000 students. The sampling method was simple random sampling. All of the students were given a standardized English test and a standardized math test. Test results are summarized below.

 Student English Math Difference, d (d - d)2 1 95 90 5 16 2 89 85 4 9 3 76 73 3 4 4 92 90 2 1 5 91 90 1 0 6 53 53 0 1 7 67 68 -1 4 8 88 90 -2 9 9 75 78 -3 16 10 85 89 -4 25 11 90 95 -5 36
 Student English Math Difference, d (d - d)2 12 85 83 2 1 13 87 83 4 9 14 85 83 2 1 15 85 82 3 4 16 68 65 3 4 17 81 79 2 1 18 84 83 1 0 19 71 60 11 100 20 46 47 -1 4 21 75 77 -2 9 22 80 83 -3 16

Σ(d - d)2 = 270
d = 1

What is the 90% confidence interval for the mean difference between student scores on the math and English tests? Assume that the mean differences are approximately normally distributed.

(A) 1 + 0.8
(B) 1 + 1.0
(C) 1 + 1.3
(D) 1 + 2.0
(E) 1 + 3.6

## Solution

The answer is (C). The approach that we used to solve this problem is valid when the following conditions are met.

• The sampling method must be simple random sampling. This condition is satisfied; the problem statement says that we used simple random sampling.
• The sampling distribution should be approximately normally distributed. The problem statement says that the differences were normally distributed; so this condition is satisfied.

Since the above requirements are satisfied, we can use the following four-step approach to construct a confidence interval.

• Identify a sample statistic. Since we are trying to estimate a population mean difference in math and English test scores, we use the sample mean difference (d = 1) as the sample statistic.

• Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 90% confidence level.

• Find the margin of error. Elsewhere on this site, we show how to compute the margin of error when the sampling distribution is approximately normal. The key steps are shown below.

• Find standard deviation or standard error. Since we do not know the standard deviation of the population, we cannot compute the standard deviation of the sample mean; instead, we compute the standard error (SE). Since the sample size is much smaller than the population size, we can use the approximation equation for the standard error.

sd = sqrt [ (Σ(di - d)2 / (n - 1) ] = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586

SE = sd / sqrt( n ) = 3.586 / [ sqrt(22) ] = 3.586/4.69 = 0.765

• Find critical value. The critical value is a factor used to compute the margin of error. Because the standard deviation of the population is unknown, we express the critical value as a t score rather than a z score. Since the sample size is fairly small, we choose the t score. To find the critical value, we take these steps.

• Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 90/100 = 0.10
• Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.10/2 = 0.95
• Find the degrees of freedom (df): df = n - 1 = 22 - 1 = 21
• The critical value is the t score having 21 degrees of freedom and a cumulative probability equal to 0.95. Using an online calculator (e.g., Stat Trek's free t Distribution Calculator), a handheld graphing calculator, or a t distribution table, we find that the t score associated with a cumulative probability of 0.95 is 1.72. (On the actual AP Statistics exam, you may need to use a graphing calculator or a t table, since Stat Trek's analytical tools will not be available.)
• Compute margin of error (ME): ME = critical value * standard error = 1.72 * 0.765 = 1.3

• Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level.

Therefore, the 90% confidence interval is -0.3 to 2.3. That is, we are 90% confident that the true population proportion is in the range defined by 1 + 1.3.

## Problem 21

Below, the cumulative frequency plot shows height (in inches) of college basketball players.

What is the interquartile range?

(A) 3 inches
(B) 6 inches
(C) 25 inches
(D) 50 inches
(E) None of the above

## Solution

The correct answer is (B). The interquartile range is the middle range of the distribution, defined by Q3 minus Q1.

Q1 is the height for which the cumulative percentage is 25%. To find Q1 from the cumulative frequency plot, follow the grid line to the right from the Y axis at 25%. This line intersects the curve over the X axis at a height of about 71 inches. This means that 25% of the basketball players are at most 71 inches tall, so Q1 is 71.

To find Q3, follow the grid line to the right from the Y axis at 75%. This line intersects the curve over the X axis at a height of about 77 inches. This means that 75% of the basketball players are at most 77 inches tall, so Q1 is 77.

Since the interquartile range is Q3 minus Q1, the interquartile range is 77 - 71 or 6 inches.

## Problem 22

Suppose X and Y are independent random variables. The variance of X is equal to 16; and the variance of Y is equal to 9. Let Z = X - Y.

What is the standard deviation of Z?

(A) 2.65
(B) 5.00
(C) 7.00
(D) 25.0
(E) It is not possible to answer this question, based on the information given.

## Solution

The correct answer is B. The solution requires us to recognize that Variable Z is a combination of two independent random variables. As such, the variance of Z is equal to the variance of X plus the variance of Y.

Var(Z) = Var(X - Y) = Var(X) + Var(Y) = 16 + 9 = 25

The standard deviation of Z is equal to the square root of the variance. Therefore, the standard deviation is equal to the square root of 25, which is 5.

## Problem 23

Acme Toy Company sells baseball cards in packages of 100. Three types of players are represented in each package -- rookies, veterans, and All-Stars. The company claims that 30% of the cards are rookies, 60% are veterans, and 10% are All-Stars. Cards from each group are randomly assigned to packages.

Suppose you bought a package of cards and counted the players from each group. What method would you use to test Acme's claim that 30% of the production run are rookies; 60%, veterans; and 10%, All-Stars.

(A) Chi-square goodness of fit test
(B) Chi-square test for homogeneity
(C) Chi-square test for independence
(D) One-sample t test
(E) Matched pairs t-test

## Solution

The answer is (A). The chi-square goodness of fit test is used to find out whether an observed pattern of categorical data is consistent with a specified distribution. In this problem, we are dealing with a categorical variable -- the type of baseball card. And we want to determine whether the distribution in our package is consistent with the production distribution claimed by Acme Toy Company. So the chi-square goodness of fit test is appropropriate.

The other chi-square options (the independence test and the homogeneity test) involve comparing data from two samples. Since we only have one sample in this baseball card problem, they are not appropriate. The t-tests are not appropriate, since they are used with quantitative data; and this problem involves categorical data.

## Problem 24

Suppose a researcher conducts an experiment to test a hypothesis. If she doubles her sample size, which of the following will increase?

I. The power of the hypothesis test.
II. The effect size of the hypothesis test.
III. The probability of making a Type II error.

(A) I only
(B) II only
(C) III only
(D) All of the above
(E) None of the above

## Solution

The answer is (A). Increasing sample size makes the hypothesis test more sensitive - more likely to reject the null hypothesis when it is, in fact, false. Thus, it increases the power of the test. The effect size is not affected by sample size. And the probability of making a Type II error gets smaller, not bigger, as sample size increases.

## Problem 25

College

High school
7
3 6 6
1 2 3 4
6 8 8 9
2 8

3
0
1
2
3
4
5
6
7

0 0 3 5
1 2 4 4 6
1 8 9
0 1

The back-to-back stemplot on the right shows the number of books read in a year by a random sample of college and high school students. Which of the following statements are true?

I. One college student read seven books.
II. The college median is equal to the high school median.
III. The mean is greater than the median in both groups.

(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III

## Solution

The correct answer is (E). All of the college students read books during the year; the fewest books read by a college student was seven. In both groups, the median is equal to 24. And the mean number of books read per year is 25.3 for high school students and 30.4 for college students; so the mean is greater than the median in both groups.

## Problem 26

Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?

(A) 0.028
(B) 0.161
(C) 0.167
(D) 0.333
(E) There is not enough information to answer this question.

## Solution

The answer is (B). This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is:

b(2; 5, 0.167) = nCx * Px * (1 - P)n - x = 5C2 * (0.167)2 * (0.833)3 = 10 * (0.167)2 * (0.833)3
b(2; 5, 0.167) = 0.161

## Problem 27

With respect to experimental design, which of the following statements are true?

I. Blinding controls for the effects of confounding.
II. Randomization controls for effects of lurking variables.
III. Each experimental factor has one treatment level.

(A) I only
(B) II only
(C) III only
(D) All of the above.
(E) None of the above.

## Solution

The correct answer is (B). By randomly assigning subjects to treatment levels, randomization spreads potential effects of lurking variables roughly evenly across treatment levels. Blinding ensures that subjects in control and treatment conditions experience the placebo effect equally, but it does not guard against confounding. And finally, each factor has two or more treatment levels. If a factor had only one treatment level, each subject in the experiment would get the same treatment on that factor. As a result, that factor would be confounded with every other factor in the experiment.

## Problem 28

In hypothesis testing, which of the following statements is always true?

I. The P-value is greater than the significance level.
II. The P-value is computed from the significance level.
III. The P-value is the parameter in the null hypothesis.
IV. The P-value is a test statistic.
V. The P-value is a probability.

(A) I only
(B) II only
(C) III only
(D) IV only
(E) V only

## Solution

The answer is (E). The P-value is the probability of observing a sample statistic as extreme as the test statistic. It can be greater than the significance level, but it can also be smaller than the significance level. It is not computed from the significance level, it is not the parameter in the null hypothesis, and it is not a test statistic.

## Problem 29

A national consumer magazine reported the following correlations.

• The correlation between car weight and car reliability is -0.30.
• The correlation between car weight and annual maintenance cost is 0.20.

Which of the following statements are true?

I. Heavier cars tend to be less reliable.
II. Heavier cars tend to cost more to maintain.
III. Car weight is related more strongly to reliability than to maintenance cost.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III

## Solution

The correct answer is (E). The correlation between car weight and reliability is negative. This means that reliability tends to decrease as car weight increases. The correlation between car weight and maintenance cost is positive. This means that maintenance costs tend to increase as car weight increases.

The strength of a relationship between two variables is indicated by the absolute value of the correlation coefficient. The correlation between car weight and reliability has an absolute value of 0.30. The correlation between car weight and maintenance cost has an absolute value of 0.20. Therefore, the relationship between car weight and reliability is stronger than the relationship between car weight and maintenance cost.

## Problem 30

Bob is a high school basketball player. He is a 70% free throw shooter. That means his probability of making a free throw is 0.70. What is the probability that Bob makes his first free throw on his fifth shot?

(A) 0.0024
(B) 0.0057
(C) 0.0081
(D) 0.0720
(E) 0.1681

## Solution

The answer is (B). This is an example of a geometric distribution, which is a special case of a negative binomial distribution. Therefore, this problem can be solved using the negative binomial formula or the geometric formula. We demonstrate each approach below, beginning with the negative binomial formula.

The probability of success (P) is 0.70, the number of trials (x) is 5, and the number of successes (r) is 1. We enter these values into the negative binomial formula.

b*(x; r, P) = x-1Cr-1 * Pr * Qx - r
b*(5; 1, 0.7) = 4C0 * 0.71 * 0.34
b*(5; 3, 0.7) = 0.00567

Now, we demonstate a solution based on the geometric formula.

g(x; P) = P * Qx - 1
g(5; 0.7) = 0.7 * 0.34 = 0.00567

Notice that each approach yields the same answer.

## Problem 31

An archer claims that 25% of her shots will be in the center of the target (i.e., a bulls-eye). A sports writer plans to test this claim by sampling 300 shots. If the 300 shots result in 60 or fewer bulls-eyes (i.e., 20% bulls-eyes), the writer will reject the archer's claim.

What is the probability that the sports writer will reject the archer's claim, when it is actually true?

(A) 0.01
(B) 0.02
(C) 0.04
(D) 0.08
(E) 0.16

## Solution

The answer is (B). The solution to this problem involves finding the P-value, the probability of making 60 or fewer bulls-eyes, assuming that 25% of the archer's shots are normally bulls-eyes. In other words, the null hypothesis is P = 0.25.

To find the P-value, take the following steps.

• Calculate the standard deviation (σ), assuming that the null hypothesis is true.

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.25 * 0.75) / 300] = sqrt(0.000625) = 0.025

where P is the hypothesized value of population proportion in the null hypothesis, and n is the sample size.

• Compute the z-score test statistic (z).

z = (p - P) / σ = (0.20 - 0.25)/0.025 = -2

where P is the hypothesized value of population proportion in the null hypothesis, and p is the proportion of bulls-eyes observed in the sample.

• The P-value is the probability that a z-score will be less than -2.0. Using an online calculator (e.g., Stat Trek's free Normal Distribution Calculator), a handheld graphing calculator, or a normal distribution table, we find that the z score of -2.0 has a cumulative probability of 0.023. (On the actual AP Statistics exam, you may need to use a graphing calculator or a normal distribution table, since Stat Trek's analytical tools will not be available.)

Note: This problem can also be treated as a binomial experiment. Previously, we showed how to analyze a binomial experiment. The binomial experiment is actually the more exact analysis. It produces a P-value with a cumulative probability of 0.0245. Without a computer, the binomial approach is computationally demanding. Therefore, many statistics texts emphasize the approach presented above, which is a normal approximation of the binomial.

## Problem 32

The Acme Car Company claims that at most 8% of its new cars have a manufacturing defect. A quality control inspector randomly selects 300 new cars and finds that 33 have a defect. Should she reject the 8% claim? Assume that the significance level is 0.05.

(A) Yes, because the P-value is 0.016.
(B) Yes, because the P-value is 0.028.
(C) No, because the P-value is 0.16.
(D) No, because the P-value is 0.28.
(E) There is not enough information to reach a conclusion.

## Solution

The answer is (B). The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P <= 0.08
Alternative hypothesis: P > 0.08

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too big.

• Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

• Analyze sample data. Using sample data, we calculate the standard deviation (σ) and compute the z-score test statistic (z).

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.08 * 0.92) / 300] = sqrt(0.0002453) = 0.0157
z = (p - P) / σ = (.11 - .08)/0.0157 = 1.91

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 1.91. We use the Normal Distribution Calculator to find P(z > 1.91) = 0.028. Thus, the P-value = 0.028.

• Interpret results. Since the P-value (0.028) is less than the significance level (0.05), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the sample included at least 10 successes and 10 failures, and the population size was at least 10 times the sample size.

## Problem 33

In the context of regression analysis, which of the following statements are true?

I. When the sum of the residuals is greater than zero, the model is nonlinear.
II. A random pattern in the residual plot indicates that linear regression is appropriate.
III. Influential points always reduce the correlation coefficient.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III

## Solution

The correct answer is (B). Researchers use residual plots to decide whether data fit a linear or a non-linear model. A random pattern of residuals suggests that a linear model is appropriate; a non-random pattern suggests that a non-linear model may be appropriate. The sum of the residuals is always zero, whether the regression model is linear or nonlinear. And influential points often increase the correlation coefficient.

## Problem 34

Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)

(A) 0.10
(B) 0.18
(C) 0.50
(D) 0.82
(E) 0.90

## Solution

The correct answer is B. As part of the solution to this problem, we assume that test scores are normally distributed. In this way, we use the normal distribution as a model for measurement. Given an assumption of normality, the solution involves three steps.

• First, we transform Molly's test score into a z-score, using the z-score transformation equation.

z = (X - μ) / σ = (940 - 850) / 100 = 0.90

• Then, using an online calculator (e.g., Stat Trek's free normal distribution calculator), a handheld graphing calculator, or the standard normal distribution table, we find the cumulative probability associated with the z-score. In this case, we find P(Z < 0.90) = 0.8159.

• Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.

Thus, we estimate that 18.41 percent of the students tested had a higher score than Molly.

## Problem 35

Which of the following statements are true?

I. A completely randomized design offers no control for lurking variables.
II. A randomized block design controls for the placebo effect.
III. In a matched pairs design, subjects within each pair receive the same treatment.

(A) I only
(B) II only
(C) III only
(D) All of the above.
(E) None of the above.

## Solution

The correct answer is (E). In a completely randomized design, subjects are randomly assigned to treatment conditions. Randomization provides some control for lurking variables. By itself, a randomized block design does not control for the placebo effect. To control for the placebo effect, the experimenter must include a placebo in one of the treatment levels. In a matched pairs design, subjects within each pair are assigned to different treatment levels.

## Problem 36

A sports writer hypothesized that Tiger Woods plays better on par 3 holes than on par 4 holes. He reviewed Woods' performance in a random sample of golf tournaments. On the par 3 holes, Woods made a birdie in 20 out of 80 attempts. On the par 4 holes, he made a birdie in 40 out of 200 attempts. How would you interpret this result?

(A) The P-value is < 0.001, very strong evidence that Woods plays better on par 3 holes.
(B) The P-value is between 0.001 and 0.01, strong evidence that Woods plays better on par 3 holes.
(C) The P-value is between 0.01 and 0.05, moderate evidence that Woods plays better on par 3 holes.
(D) The P-value is between 0.05 and 0.10, some evidence that Woods plays better on par 3 holes.
(E) The P-value is > 0.10, little or no support for the notion that Woods plays better on par 3 holes.

## Solution

The answer is (E). The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P3 <= P4
Alternative hypothesis: P3 > P4

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of birdies on par 3 holes (p3) is sufficiently greater than the proportion of birdies on par 4 holes (p4).

• Formulate an analysis plan. For this analysis, the test method is a two-proportion z-test, which is shown below.

• Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p3 * n3 + p4 * n4) / (n3 + n4) = [(0.25 * 80) + (0.20 * 200)] / (80 + 200) = 50/280 = 0.214

SE = sqrt{ p * ( 1 - p ) * [ (1/n3) + (1/n4) ] }
SE = sqrt [ 0.214 * 0.786 * ( 1/80 + 1/200 ) ] = sqrt[ 0.214 * 0.786 * 0.0175 }= sqrt [0.0029548] = 0.0544

z = (p3 - p4) / SE = (0.25 - 0.20)/0.0544 = 0.92

where p3 is the sample proportion of birdies on par 3, where p4 is the sample proportion of birdies on par 4, n3 is the number of par 3 holes, and n4 is the number of par 4 holes.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.92. We use the Normal Distribution Calculator to find P(z > 0.92) = 0.18. Thus, the P-value = 0.18.

• Interpret results. Since the P-value (0.18) is greater than 0.10, we have little support for the notion that Woods plays better on par 3 holes. In short, we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, each population was at least 10 times larger than its sample, and each sample included at least 10 successes and 10 failures.

## Problem 37

In the context of regression analysis, which of the following statements are true?

I. A linear transformation increases the linear relationship between variables.
II. A logarithmic model is the most effective transformation method.
III. A residual plot reveals departures from linearity.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I, II, and III

## Solution

The correct answer is (C). A linear transformation neither increases nor decreases the linear relationship between variables; it preserves the relationship. A nonlinear transformation is used to increase the relationship between variables. The most effective transformation method depends on the data being transformed. In some cases, a logarithmic model may be more effective than other methods; but it other cases it may be less effective. Non-random patterns in a residual plot suggest a departure from linearity in the data being plotted.

## Problem 38

Acme Corporation manufactures light bulbs. The CEO claims that an average Acme light bulb lasts 300 days. A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 50 days. If the CEO's claim were true, what is the probability that 15 randomly selected bulbs would have an average life of no more than 290 days?

(A) 0.100
(B) 0.226
(C) 0.334
(D) 0.443
(E) .775

## Solution

The answer is (B). The first thing we need to do is compute the t score, based on the following equation:

t = [ x - μ ] / [ s / sqrt( n ) ]
t = ( 290 - 300 ) / [ 50 / sqrt( 15) ] = -10 / 12.909945 = - 0.7745966

where x is the sample mean, μ is the population mean, s is the standard deviation of the sample, and n is the sample size.

Then, using an online calculator (e.g., Stat Trek's free T Distribution Calculator), a handheld graphing calculator, or the t distribution table, we find the cumulative probability associated with the t score. For this practice test, we can use the T Distribution Calculator; but on the actual AP Statistics Exam, you may need to use a graphing calculator or a t distribution table.

Since we know the t score, we select "T score" from the Random Variable dropdown box of the T Distribution Calculator. Then, we enter the following data:

• The degrees of freedom are equal to 15 - 1 = 14.
• The t score is equal to - 0.7745966.

The calculator displays the cumulative probability: 0.226. Hence, if the true bulb life were 300 days, there is a 22.6% chance that the average bulb life for 15 randomly selected bulbs would be less than or equal to 290 days.

## Problem 39

Which of the following would be a reason to use a one-sample t-test instead of a one-sample z-test?

I. The standard deviation of the population is unknown.
II. The null hypothesis involves a continuous variable.
III. The sample size is large (greater than 40).

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

## Solution

The answer is (A). When the standard deviation of the population is unknown, the t-test is preferred. Either test can be used when the null hypothesis involves a continuous variable, or when the sample size is large.

## Problem 40

A public opinion poll surveyed a simple random sample of voters. Respondents were classified by gender (male or female) and by voting preference (Republican, Democrat, or Independent). Results are shown below.

 Voting Preferences Republican Democrat Independent Row total Male 200 150 50 400 Female 250 300 50 600 Column total 450 450 100 1000

If you conduct a chi-square test of independence, what is the expected frequency count of male Independents?

(A) 40
(B) 50
(C) 60
(D) 180
(E) 270

## Solution

The answer is (A). To apply the chi-square test for independence, we compute the expected frequency counts for each cell of the table, using the following equation. The computation for males who are classified as Independents is shown below.

Er,c = (nr * nc) / n
E1,3 = (400 * 100) / 1000 = 40000/1000 = 40

where r is the number of levels of gender, c is the number of levels of the voting preference, nr is the number of observations from level r of gender, nc is the number of observations from level c of voting preference, n is the number of observations in the sample, Er,c is the expected frequency count when gender is level r and voting preference is level c, and Or,c is the observed frequency count when gender is level r voting preference is level c.