2019 AP Statistics Practice Exam: Ethical Study Guide

AP Statistics Practice

2019 International Practice Exam AP Statistics: Ethical Study Guide

2019 International Practice Exam AP Statistics: Ethical Study Guide practice bank: Solve the visible multiple-choice and free-response questions, then compare every step with the worked answers.

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2019 International Practice Exam AP Statistics: Ethical Study Guide: formulas and targets

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Mixed current-course AP Statistics multiple-choice practice

Question 1. Binomial Distribution

For 18 independent students at a school district in Riverbend during a community outreach cycle, each meets the lunch-program participation criterion with probability 0.237. Find P(X=3) and give the binomial mean and SD.

A. P=0.0133 because combinations and failures are omitted.
B. P(X=3)=0.1878, μ=4.266, σ=1.804.
C. μ=0.237, σ=0.181; ignore n.
D. P=0.3519 because exactly means at most.

Answer: B

X~Bin(18,0.237). P(X=3)=C(18,3)(0.237)³(0.763)¹⁵=0.1878. μ=np=4.266 and σ=√np(1−p)=1.804.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: P(X=3)=0.1878, μ=4.266, σ=1.804. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).
Choice C: It omits a required part of the calculation. The correct comparison or result is: P(X=3)=0.1878, μ=4.266, σ=1.804. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).
Choice D: It does not match the requested calculation. The correct comparison or result is: P(X=3)=0.1878, μ=4.266, σ=1.804. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).

Question 2. Sampling Distributions

A population at a school district in Capital Region during a monthly quality review has mean 96 and SD 23. For random samples of size 64, find the mean and SD of x̄ and P(x̄>99.59) under a normal/CLT approximation.

A. Mean 96, SD 2.875, probability 0.1059.
B. Mean 1.5, SD 23, probability 0.1059.
C. Mean 96, SD 23, because averaging does not change spread.
D. Mean 96, SD 0.359, probability 0.8941.

Answer: A

μ_x̄=μ=96. σ_x̄=σ√n=23√64=2.875. The cutoff has z=(99.59−96)2.875=1.249, so P(x̄>cutoff)=0.1059.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 96, SD 2.875, probability 0.1059. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 96, SD 2.875, probability 0.1059. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.
Choice D: It does not match the requested calculation. The correct comparison or result is: Mean 96, SD 2.875, probability 0.1059. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.

Question 3. Sampling Distribution of a Sample Mean

A population at a public health department in Central County during a regional benchmarking study has mean 46 and SD 14. For random samples of size 64, find the mean and SD of x̄ and P(x̄>46.88) under a normal/CLT approximation.

A. Mean 46, SD 1.75, probability 0.3075.
B. Mean 46, SD 14, because averaging does not change spread.
C. Mean 46, SD 0.219, probability 0.6925.
D. Mean 0.719, SD 14, probability 0.3075.

Answer: A

μ_x̄=μ=46. σ_x̄=σ√n=14√64=1.75. The cutoff has z=(46.88−46)1.75=0.503, so P(x̄>cutoff)=0.3075.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 46, SD 1.75, probability 0.3075. Key check: Larger n reduces standard error and the CLT improves normal approximation.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 46, SD 1.75, probability 0.3075. Key check: Larger n reduces standard error and the CLT improves normal approximation.
Choice D: It does not match the requested calculation. The correct comparison or result is: Mean 46, SD 1.75, probability 0.3075. Key check: Larger n reduces standard error and the CLT improves normal approximation.

Question 4. Sampling Distribution of a Sample Proportion

For a population proportion p=0.513, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.55).

A. Mean 0.513, SD 0.0017 without the square root.
B. Mean 0.004, SD 0.0417.
C. Mean 0.513, SD 0.0417, probability 0.8128.
D. Mean 0.513, SD 0.0417, probability 0.1872.

Answer: D

μ_p̂=p=0.513. σ_p̂=√p(1−p)n=√0.513(0.487)144=0.0417. z=(0.55−0.513)0.0417≈0.888, so the upper-tail probability is 0.1872.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: Mean 0.513, SD 0.0417, probability 0.1872. Key check: The standard deviation depends on the population proportion and sample size.
Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 0.513, SD 0.0417, probability 0.1872. Key check: The standard deviation depends on the population proportion and sample size.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 0.513, SD 0.0417, probability 0.1872. Key check: The standard deviation depends on the population proportion and sample size.

Question 5. One-Proportion z Interval

In an SRS of 240 residents from a public health department in Westview during a service-improvement study, 156 meet the vaccination appointment completion criterion. Construct a 90% one-proportion z interval and interpret it.

A. (0.599, 0.701) with a population-proportion interpretation.
B. (0.599, 0.701); 90% of sampled observations lie inside this interval.
C. (0.619, 0.681); omit the critical value.
D. (0.625, 0.675); halve the margin of error.

Answer: A

p̂=156240=0.65. Conditions include randomization/independence and 156 successes and 84 failures, both at least 10. SE=√p̂(1−p̂)n=0.0308; z*=1.645; ME=0.051. The interval is (0.599, 0.701). We are 90% confident that the true population proportion lies in this interval.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: (0.599, 0.701) with a population-proportion interpretation. Key check: The interval estimates one population proportion.
Choice C: It omits a required part of the calculation. The correct comparison or result is: (0.599, 0.701) with a population-proportion interpretation. Key check: The interval estimates one population proportion.
Choice D: It does not match the requested calculation. The correct comparison or result is: (0.599, 0.701) with a population-proportion interpretation. Key check: The interval estimates one population proportion.

Question 6. Two-Proportion z Interval

At a grocery cooperative in Westview during a pre-exam training cycle, Group 1 has 113230 meeting a criterion and Group 2 has 61130. Construct a 95% interval for p₁−p₂.

A. (-0.085, 0.129) for p₁−p₂.
B. (-0.129, 0.085) while still labeling it p₁−p₂.
C. Use a pooled standard error because all two-proportion procedures pool.
D. (-0.033, 0.077); omit 1.96.

Answer: A

p̂₁=0.491, p̂₂=0.469, difference=0.022. The unpooled SE is √0.491(0.509)230+0.469(0.531)130=0.0548. The 95% interval is 0.022±1.96(0.0548)=(-0.085, 0.129).

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: (-0.085, 0.129) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.
Choice C: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: (-0.085, 0.129) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.
Choice D: It omits a required part of the calculation. The correct comparison or result is: (-0.085, 0.129) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.

Question 7. One-Sample t Interval

A random sample of 34 learners at a digital learning platform in Mountain Region during a service-improvement study has mean 87.41 and SD 8.36 for lesson completion. Construct a 95% t interval for the population mean.

A. Use a z interval because the sample SD is known.
B. (84.49, 90.33) for the population mean.
C. (85.98, 88.84); omit t*.
D. (71.02, 103.8); use raw SD instead of SE.

Answer: B

df=33, t*=2.035, SE=s√n=8.36√34=1.434, so the interval is 87.41±2.917=(84.49, 90.33). The interpretation concerns the population mean, conditional on randomization/independence and an acceptable population shape or robust sample size.

Why the other choices fail

Choice A: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: (84.49, 90.33) for the population mean. Key check: Use t because population σ is unknown.
Choice C: It omits a required part of the calculation. The correct comparison or result is: (84.49, 90.33) for the population mean. Key check: Use t because population σ is unknown.
Choice D: It does not match the requested calculation. The correct comparison or result is: (84.49, 90.33) for the population mean. Key check: Use t because population σ is unknown.

Question 8. Null and Alternative Hypotheses

A school district in Prairie District during a semester-long cohort study tests H₀:p=0.4 against H_a:p≠0.4 using 93 successes in n=290. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. z=-2.757, p=0.0058; reject H₀.
B. Interpret the p-value as the probability that H₀ is true.
C. Use p̂ in the null SE and report p=0.321.
D. Always reject H₀ when the sample proportion differs numerically from p₀.

Answer: A

p̂=93290=0.321. Under H₀, SE=√0.4(0.6)290=0.0288 and z=(0.321−0.4)0.0288=-2.757. The two-sided p-value is 0.0058. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Why the other choices fail

Choice B: It misinterprets evidence or exceeds the scope supported by the study design. The correct comparison or result is: z=-2.757, p=0.0058; reject H₀. Key check: The alternative direction must match the research question before seeing data.
Choice C: It does not match the requested calculation. The correct comparison or result is: z=-2.757, p=0.0058; reject H₀. Key check: The alternative direction must match the research question before seeing data.
Choice D: It makes a conclusion that the data do not support. The correct comparison or result is: z=-2.757, p=0.0058; reject H₀. Key check: The alternative direction must match the research question before seeing data.

Question 9. Type I Error, Type II Error, and Power

A municipal emergency dispatch center in New England network during a community outreach cycle tests H₀:p=0.4 against H_a:p>0.4 for the response time criterion. Describe Type I and Type II errors, then state one change that generally increases power against p=0.52.

A. Type I and Type II errors are both calculation mistakes.
B. Decreasing sample size always raises power by making results more variable.
C. Type I rejects a true H₀; Type II fails to reject a false H₀; increasing n generally raises power.
D. Power is the probability that H₀ is true.

Answer: C

For the municipal emergency dispatch center in New England network during a community outreach cycle, a Type I error is concluding that the population proportion meeting the response time criterion exceeds 0.4 when it actually equals 0.4. A Type II error is failing to conclude it exceeds 0.4 when the true proportion is 0.52. Increasing sample size or α generally increases power against that alternative; improving measurement can also help.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Type I rejects a true H₀; Type II fails to reject a false H₀; increasing n generally raises power. Key check: Power is the probability of rejecting H₀ when a specified alternative is true.
Choice B: It makes a conclusion that the data do not support. The correct comparison or result is: Type I rejects a true H₀; Type II fails to reject a false H₀; increasing n generally raises power. Key check: Power is the probability of rejecting H₀ when a specified alternative is true.
Choice D: It misinterprets evidence or exceeds the scope supported by the study design. The correct comparison or result is: Type I rejects a true H₀; Type II fails to reject a false H₀; increasing n generally raises power. Key check: Power is the probability of rejecting H₀ when a specified alternative is true.

Question 10. One-Proportion z Test

A regional airport authority in Mountain Region during a summer implementation review tests H₀:p=0.6 against H_a:p>0.6 using 119 successes in n=220. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

A. z=-1.789, p=0.9632; fail to reject H₀.
B. Interpret the p-value as the probability that H₀ is true.
C. Always reject H₀ when the sample proportion differs numerically from p₀.
D. Use p̂ in the null SE and report p=0.541.

Answer: A

p̂=119220=0.541. Under H₀, SE=√0.6(0.4)220=0.033 and z=(0.541−0.6)0.033=-1.789. The right-tailed p-value is 0.9632. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

Why the other choices fail

Choice B: It misinterprets evidence or exceeds the scope supported by the study design. The correct comparison or result is: z=-1.789, p=0.9632; fail to reject H₀. Key check: The test statistic is z=(p̂−p₀)√p₀(1−p₀)n.
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: z=-1.789, p=0.9632; fail to reject H₀. Key check: The test statistic is z=(p̂−p₀)√p₀(1−p₀)n.
Choice D: It does not match the requested calculation. The correct comparison or result is: z=-1.789, p=0.9632; fail to reject H₀. Key check: The test statistic is z=(p̂−p₀)√p₀(1−p₀)n.

Question 11. Two-Proportion z Test

For independent samples at a university advising center in North Valley during a randomized pilot period, Group 1 has 132180 successes and Group 2 has 127210. Test H₀:p₁=p₂ against H_a:p₁≠p₂.

A. z=2.68, two-sided p=0.0074.
B. Pool by averaging p̂1 and p̂2 without weighting sample sizes.
C. Use a one-sample t test because two sample proportions are means of zeros and ones.
D. Use the unpooled confidence-interval standard error for the equality test.

Answer: A

p̂_c=(132+127)(180+210)=0.664. SE₀=√p̂_c(1−p̂_c)(1180+1210)=0.048. z=(0.733−0.605)0.048=2.68 and the two-sided p-value is 0.0074. Reject H₀ at .05; evidence suggests different population proportions.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: z=2.68, two-sided p=0.0074. Key check: The pooled estimate is (x₁+x₂)(n₁+n₂).
Choice C: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: z=2.68, two-sided p=0.0074. Key check: The pooled estimate is (x₁+x₂)(n₁+n₂).
Choice D: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: z=2.68, two-sided p=0.0074. Key check: The pooled estimate is (x₁+x₂)(n₁+n₂).

Question 12. One-Sample t Test

A random sample of 23 bus trips at a city transit agency in South Harbor during a service-improvement study has x̄=75.25, s=17.11. Test H₀:μ=73 against H_a:μ≠73.

A. Use z with population SD equal to the sample mean.
B. The p-value is the probability the population mean equals the sample mean.
C. t=0.631, df=22, p=0.5348.
D. Use df=n because one parameter is estimated.

Answer: C

t=(75.25−73)(17.11√23)=0.631, df=22. The two-sided p-value is 0.5348. Fail to reject H₀ at .05; the data do not provide sufficiently strong evidence that the population mean differs.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: t=0.631, df=22, p=0.5348. Key check: Check randomization/independence and population shape or sample size.
Choice B: It does not match the requested calculation. The correct comparison or result is: t=0.631, df=22, p=0.5348. Key check: Check randomization/independence and population shape or sample size.
Choice D: It does not match the requested calculation. The correct comparison or result is: t=0.631, df=22, p=0.5348. Key check: Check randomization/independence and population shape or sample size.

Question 13. Chi-Square Homogeneity and Independence

At a food safety laboratory in Midwest consortium during a school-year data collection, observed category counts are [[39, 57, 48], [48, 54, 19]]. Test the null claim of same categorical distribution across two groups; calculate expected counts, χ², df, and p-value.

A. Degrees of freedom equal the grand total minus one.
B. χ²=11.656, df=2, p=0.0029.
C. Expected counts equal observed counts, so χ² is always zero.
D. Use a two-proportion z test even though there are more than two categories.

Answer: B

Expected counts are [[47.28, 60.32, 36.41], [39.72, 50.68, 30.59]]. χ²=Σ(O−E)²E=11.656, df=2, p=0.0029. Reject the null at .05; the distributions differ.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: χ²=11.656, df=2, p=0.0029. Key check: Expected count=(row total)(column total)(grand total).
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: χ²=11.656, df=2, p=0.0029. Key check: Expected count=(row total)(column total)(grand total).
Choice D: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: χ²=11.656, df=2, p=0.0029. Key check: Expected count=(row total)(column total)(grand total).

Question 14. Scatterplots and Correlation

A scatterplot for 54 travelers at a regional airport authority in Coastal Plains during a regional benchmarking study is roughly linear with correlation r=0.81. Interpret direction and strength, and state one limitation of r.

A. r is the slope, so y changes by r units for every one-unit x increase.
B. A strong positive linear association; r does not establish causation.
C. r proves that x causes y because its magnitude is nonzero.
D. r describes any curved pattern as long as both variables are quantitative.

Answer: B

2019 International Practice Exam AP Statistics: Ethical Study Guide — Question 14. Scatterplots and Correlation: The association is strong and positive: larger x values tend to occur with larger y values. Correlation describes linear association only, is sensitive to outliers, has no units, and does not establish causation.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: A strong positive linear association; r does not establish causation. Key check: Correlation measures linear association and is not resistant to outliers.
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: A strong positive linear association; r does not establish causation. Key check: Correlation measures linear association and is not resistant to outliers.
Choice D: It does not match the requested calculation. The correct comparison or result is: A strong positive linear association; r does not establish causation. Key check: Correlation measures linear association and is not resistant to outliers.

Question 15. Least-Squares Regression Line

For appointment wait time at a university advising center in Central County during a weekday operations study, x̄=44.191, ȳ=76.054, sx=6.123, sy=15.962, and r=0.82. Find the least-squares line and predict y at x=46.89.

A. ŷ=76.054+0.82x because r is the slope.
B. The line cannot be calculated from summary statistics.
C. ŷ=2.138+-18.411x; interchange slope and intercept.
D. ŷ=-18.411+2.138x; prediction 81.824.

Answer: D

b₁=r(sy/sx)=0.82(15.9626.123)=2.138. b₀=ȳ−b1x̄=76.054−2.138(44.191)=-18.411. Thus ŷ=-18.411+2.138x, and at x=46.89, ŷ=81.824. The slope predicts a 2.138-unit change in y for each one-unit increase in x, within the observed range.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: ŷ=-18.411+2.138x; prediction 81.824. Key check: The least-squares line minimizes the sum of squared vertical residuals.
Choice B: It makes a conclusion that the data do not support. The correct comparison or result is: ŷ=-18.411+2.138x; prediction 81.824. Key check: The least-squares line minimizes the sum of squared vertical residuals.
Choice C: It reverses the stated order or sign. The correct comparison or result is: ŷ=-18.411+2.138x; prediction 81.824. Key check: The least-squares line minimizes the sum of squared vertical residuals.

Question 16. Residuals and Regression Diagnostics

A regression model at a community bank in Metro East during a regional benchmarking study is ŷ=19.625+2.056x. For x=14.015, the observed y is 46.893. Find and interpret the residual.

A. Residual 46.893 because the observed value is the error.
B. Residual -1.547; the model overpredicted by 1.547.
C. Residual 1.547; compute predicted minus observed.
D. Residual 48.44 because the prediction is the unexplained component.

Answer: B

Predicted y=19.625+2.056(14.015)=48.44. Residual=observed−predicted=46.893−48.44=-1.547. The model overpredicted the observed response by 1.547 response units.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Residual -1.547; the model overpredicted by 1.547. Key check: A useful linear model has residuals randomly scattered around zero.
Choice C: It reverses the stated order or sign. The correct comparison or result is: Residual -1.547; the model overpredicted by 1.547. Key check: A useful linear model has residuals randomly scattered around zero.
Choice D: It does not match the requested calculation. The correct comparison or result is: Residual -1.547; the model overpredicted by 1.547. Key check: A useful linear model has residuals randomly scattered around zero.

Question 17. Categorical and Quantitative Variables

The wildlife clinic in Central County during a randomized pilot period records recovery time as “0=no, 1=yes.” Classify the variable and explain whether arithmetic on the recorded values is meaningful.

A. The variable is categorical, and the classification follows what the values represent rather than whether digits appear.
B. Its type cannot be determined unless the sample mean is known.
C. It is always categorical because the organization created the variable.
D. It is always quantitative because the computer stores numbers.

Answer: A

2019 International Practice Exam AP Statistics: Ethical Study Guide — Question 17. Categorical and Quantitative Variables: The variable is categorical. The numbers are labels, so means or numerical distances between codes are not meaningful.

Why the other choices fail

Choice B: It makes a conclusion that the data do not support. The correct comparison or result is: The variable is categorical, and the classification follows what the values represent rather than whether digits appear. Key check: Units belong to quantitative variables; category codes remain categorical.
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: The variable is categorical, and the classification follows what the values represent rather than whether digits appear. Key check: Units belong to quantitative variables; category codes remain categorical.
Choice D: It makes a conclusion that the data do not support. The correct comparison or result is: The variable is categorical, and the classification follows what the values represent rather than whether digits appear. Key check: Units belong to quantitative variables; category codes remain categorical.

Question 18. Frequency and Relative Frequency Tables

For four categories of algebra benchmark completion at a public high school in Desert County during a yearly program evaluation, observed counts are [58, 21, 58, 21]. Find the relative frequency for category 2 and state its meaning.

A. 7.524; divide the total by the category count.
B. 21; relative frequency is reported as the original count.
C. 0.21; always divide a count by 100.
D. 0.133; about 13.3% of the observed units are in category 2.

Answer: D

The total is 158. Category 2 has relative frequency 21158=0.133, so 13.3% of the observed students were in category 2.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: 0.133; about 13.3% of the observed units are in category 2. Key check: Relative frequency equals category frequency divided by the total count.
Choice B: It does not match the requested calculation. The correct comparison or result is: 0.133; about 13.3% of the observed units are in category 2. Key check: Relative frequency equals category frequency divided by the total count.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: 0.133; about 13.3% of the observed units are in category 2. Key check: Relative frequency equals category frequency divided by the total count.

Question 19. Two-Way Tables and Conditional Relative Frequency

A two-way table for a wildlife clinic in Riverbend during a spring 2027 pilot has Group A counts 30 meeting and 55 not meeting the recovery time criterion; Group B counts are 64 and 21. Compare the conditional proportions that meet the criterion and comment on association.

A. Use row denominators: 0.353 for A and 0.753 for B; compare those conditional proportions.
B. Use the grand total for both groups and compare 0.176 with 0.376.
C. Use column totals because “given group” always means condition on the outcome.
D. Compare the raw meeting counts only; denominators are irrelevant.

Answer: A

Within Group A, p̂_A=30(30+55)=0.353. Within Group B, p̂_B=64(64+21)=0.753. The difference is -0.4 (40.0% percentage points in magnitude). Because the conditional proportions differ, the table suggests an association between group and outcome.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: Use row denominators: 0.353 for A and 0.753 for B; compare those conditional proportions. Key check: Compare conditional distributions to assess association between categorical variables.
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: Use row denominators: 0.353 for A and 0.753 for B; compare those conditional proportions. Key check: Compare conditional distributions to assess association between categorical variables.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: Use row denominators: 0.353 for A and 0.753 for B; compare those conditional proportions. Key check: Compare conditional distributions to assess association between categorical variables.

Question 20. Exploring One-Variable Data

The grocery cooperative in Mountain Region during a randomized pilot period records nine checkout time values: [12, 19, 35, 41, 52, 53, 65, 67, 81]. Calculate the mean, median, and IQR, then describe what each contributes to a distribution summary.

A. Mean 47.222, median 52, IQR 93.
B. Mean 52, median 47.222, IQR 69.
C. Mean 425, median 52, IQR 19.5.
D. Mean 47.222, median 52, IQR 39.

Answer: D

The mean is 4259=47.222 and the median is the fifth ordered value, 52. Q1=27, Q3=66, so IQR=39. Mean uses every value and is sensitive to extremes; median and IQR are resistant summaries of center and middle-half spread.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Mean 47.222, median 52, IQR 39. Key check: Mean and standard deviation are nonresistant; median and IQR are resistant.
Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 47.222, median 52, IQR 39. Key check: Mean and standard deviation are nonresistant; median and IQR are resistant.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 47.222, median 52, IQR 39. Key check: Mean and standard deviation are nonresistant; median and IQR are resistant.

Question 21. Percentiles and z-Scores

At a county library in Capital Region during a service-improvement study, weekly program attendance has mean 88 and standard deviation 8. A value is 98.4. Find and interpret its z-score.

A. z=10.4; do not divide by the standard deviation.
B. z=1.3; the value is 1.3 standard deviations above the mean.
C. z=-1.3; reverse the numerator.
D. z=12.3; ignore the mean.

Answer: B

z=(98.4−88)8=1.3. The value is 1.3 standard deviations above the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 22. Normal Distribution

Model sample concentration at a food safety laboratory in Cedar Grove during a pre-exam training cycle as Normal(μ=76, σ=13). Find P(69.5≤X≤96.8) and interpret the area.

A. 0.637; about 63.7% of the modeled population lies between the cutoffs.
B. 0.363; subtract the central area from 1.
C. 0.309; use only the lower cumulative area.
D. 0.945; use only the upper cumulative area.

Answer: A

Standardize the bounds: zL=(69.5−76)13=-0.5 and zU=(96.8−76)13=1.6. P=0.945−0.309=0.637, so about 63.7% of modeled samples fall in the interval.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 23. Sampling Methods

A county library in Metro East during a weekday operations study plans to choose a random start from the first 20 records and then select every 20th record. Identify the sampling method and explain one defining feature.

A. Systematic sampling; A random start is followed by a fixed selection interval.
B. Random assignment, because sampling methods assign treatments.
C. Voluntary response, because selected individuals may answer questions.
D. Convenience sampling, because the researcher uses a list.

Answer: A

2019 International Practice Exam AP Statistics: Ethical Study Guide — Question 23. Sampling Methods: This is a systematic sample. A random start is followed by a fixed selection interval.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: Systematic sampling; A random start is followed by a fixed selection interval. Key check: Random sampling supports generalization; it does not by itself establish causation.
Choice C: It does not match the requested calculation. The correct comparison or result is: Systematic sampling; A random start is followed by a fixed selection interval. Key check: Random sampling supports generalization; it does not by itself establish causation.
Choice D: It does not match the requested calculation. The correct comparison or result is: Systematic sampling; A random start is followed by a fixed selection interval. Key check: Random sampling supports generalization; it does not by itself establish causation.

Question 24. Sampling Bias

A food safety laboratory in Mountain Region during a pre-exam training cycle asks a leading question that praises one response before asking for an opinion. Identify the main source of bias and explain why increasing the sample size would not necessarily remove it.

A. Confounding from random assignment; the study has too many treatments.
B. Sampling variability only; a larger sample always eliminates it.
C. No bias is possible because a survey produces numerical data.
D. Response Bias; the mechanism is systematically flawed, so a larger n does not guarantee less bias.

Answer: D

The main problem at the food safety laboratory in Mountain Region during a pre-exam training cycle is response bias: the plan asks a leading question that praises one response before asking for an opinion. This mechanism systematically distorts who is represented or what is reported about sample concentration. A larger sample produced by the same flawed mechanism can estimate the wrong target more precisely; it does not eliminate systematic bias.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Response Bias; the mechanism is systematically flawed, so a larger n does not guarantee less bias. Key check: Undercoverage, nonresponse, response bias, convenience, and voluntary response differ.
Choice B: It makes a conclusion that the data do not support. The correct comparison or result is: Response Bias; the mechanism is systematically flawed, so a larger n does not guarantee less bias. Key check: Undercoverage, nonresponse, response bias, convenience, and voluntary response differ.
Choice C: It does not match the requested calculation. The correct comparison or result is: Response Bias; the mechanism is systematically flawed, so a larger n does not guarantee less bias. Key check: Undercoverage, nonresponse, response bias, convenience, and voluntary response differ.

Question 25. Experimental Design

A community bank in Westview during a six-week field trial will compare two programs affecting mobile-deposit adoption. Describe a randomized design that uses previous experience as a block and explain the benefit.

A. Measure only one treatment because blocking removes the need for a control group.
B. Randomly sample a population but let participants choose treatments.
C. Block by previous experience, randomize treatments within each block, and compare responses; this controls block-to-block variation.
D. Assign the better-performing block to one treatment and the other block to the second treatment.

Answer: C

At the community bank in Westview during a six-week field trial, group the customers by previous experience, then randomly assign units within each block to the two programs affecting mobile-deposit adoption. Keep other conditions as similar as practical and compare responses. Blocking controls variation associated with previous experience, which can improve precision; random assignment supports a causal comparison for these experimental units.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Block by previous experience, randomize treatments within each block, and compare responses; this controls block-to-block variation. Key check: Random assignment supports cause-and-effect conclusions for the experimental units.
Choice B: It does not match the requested calculation. The correct comparison or result is: Block by previous experience, randomize treatments within each block, and compare responses; this controls block-to-block variation. Key check: Random assignment supports cause-and-effect conclusions for the experimental units.
Choice D: It does not match the requested calculation. The correct comparison or result is: Block by previous experience, randomize treatments within each block, and compare responses; this controls block-to-block variation. Key check: Random assignment supports cause-and-effect conclusions for the experimental units.

Question 26. Probability Rules

For two events A and B in a grocery cooperative in North Valley during a regional benchmarking study, P(A)=0.65, P(B)=0.364, and P(A∩B)=0.063. Find P(A∪B) and determine whether the events are mutually exclusive.

A. 1.014, and they are mutually exclusive.
B. 0.049, and mutual exclusivity cannot be assessed.
C. 0.063, because union and intersection are the same.
D. 0.951, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.65+0.364−0.063=0.951. Because P(A∩B) is not zero, the events are not mutually exclusive.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.951, and the events are not mutually exclusive. Key check: Use complement, addition, and multiplication rules with overlap and dependence handled correctly.
Choice B: It makes a conclusion that the data do not support. The correct comparison or result is: 0.951, and the events are not mutually exclusive. Key check: Use complement, addition, and multiplication rules with overlap and dependence handled correctly.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.951, and the events are not mutually exclusive. Key check: Use complement, addition, and multiplication rules with overlap and dependence handled correctly.

Question 27. Conditional Probability and Independence

In a county library in Cedar Grove during a yearly program evaluation, among Group A, 53 meet and 51 do not meet a criterion; among Group B, 33 meet and 22 do not. Find P(meet|A) and compare it with P(meet) to assess independence.

A. Independence is established whenever both groups contain at least 10 observations.
B. P(meet|A)=0.51; compare it with the overall probability 0.541.
C. P(meet|A)=0.541 by definition.
D. P(meet|A)=0.616; condition on those who meet instead of Group A.

Answer: B

P(meet|A)=53(53+51)=0.51. Overall P(meet)=(53+33)(53+51+33+22)=0.541. The probabilities differ, so meeting the criterion and group membership are not independent.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: P(meet|A)=0.51; compare it with the overall probability 0.541. Key check: A and B are independent when P(A|B)=P(A), equivalently P(A∩B)=P(A)P(B).
Choice C: It does not match the requested calculation. The correct comparison or result is: P(meet|A)=0.51; compare it with the overall probability 0.541. Key check: A and B are independent when P(A|B)=P(A), equivalently P(A∩B)=P(A)P(B).
Choice D: It does not match the requested calculation. The correct comparison or result is: P(meet|A)=0.51; compare it with the overall probability 0.541. Key check: A and B are independent when P(A|B)=P(A), equivalently P(A∩B)=P(A)P(B).

Question 28. Random Variables

A random variable X for a public high school in Desert County during a regional benchmarking study takes values [0, 1, 2, 3] with probabilities [0.117, 0.151, 0.089, 0.643]. Verify validity and find E(X).

A. The distribution is valid and E(X)=2.258.
B. E(X)=1.5; average the outcomes without probabilities.
C. The distribution is invalid because an expected value must be an integer.
D. The distribution is invalid unless all outcomes have equal probability.

Answer: A

Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.117)+1(0.151)+2(0.089)+3(0.643)=2.258.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: The distribution is valid and E(X)=2.258. Key check: A random variable assigns a numerical value to each outcome.
Choice C: It makes a conclusion that the data do not support. The correct comparison or result is: The distribution is valid and E(X)=2.258. Key check: A random variable assigns a numerical value to each outcome.
Choice D: It does not match the requested calculation. The correct comparison or result is: The distribution is valid and E(X)=2.258. Key check: A random variable assigns a numerical value to each outcome.

Question 29. Expected Value and Standard Deviation of a Random Variable

A municipal water office in Sunbelt district during a two-month observation window outcome pays $41 with probability 0.191 and loses $11 otherwise. Find and interpret the expected net value.

A. $-1.07 per repetition in the long run.
B. $7.83 because losses are ignored.
C. $15 because outcomes are averaged equally.
D. $30 because probability does not affect expected value.

Answer: A

E(X)=0.191($41)+0.809(−$11)=$-1.07. Over many independent repetitions under the same conditions, the average net result approaches about $-1.07 per repetition; it is not a guaranteed result on one trial.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: $-1.07 per repetition in the long run. Key check: Expected value is a long-run mean, not necessarily a possible single outcome.
Choice C: It does not match the requested calculation. The correct comparison or result is: $-1.07 per repetition in the long run. Key check: Expected value is a long-run mean, not necessarily a possible single outcome.
Choice D: It does not match the requested calculation. The correct comparison or result is: $-1.07 per repetition in the long run. Key check: Expected value is a long-run mean, not necessarily a possible single outcome.

Question 30. Binomial Distribution

For 23 independent households at a recycling program in Atlantic Corridor during a randomized pilot period, each meets the weekly material weight criterion with probability 0.272. Find P(X=10) and give the binomial mean and SD.

A. μ=0.272, σ=0.198; ignore n.
B. P=0.9723 because exactly means at most.
C. P(X=10)=0.0409, μ=6.256, σ=2.134.
D. P=0 because combinations and failures are omitted.

Answer: C

X~Bin(23,0.272). P(X=10)=C(23,10)(0.272)¹⁰(0.728)¹³=0.0409. μ=np=6.256 and σ=√np(1−p)=2.134.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: P(X=10)=0.0409, μ=6.256, σ=2.134. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).
Choice B: It does not match the requested calculation. The correct comparison or result is: P(X=10)=0.0409, μ=6.256, σ=2.134. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).
Choice D: It omits a required part of the calculation. The correct comparison or result is: P(X=10)=0.0409, μ=6.256, σ=2.134. Key check: For X~Bin(n,p), μ=np and σ=√np(1−p).

Question 31. Sampling Distributions

A population at a solar installer in Coastal Plains during a randomized pilot period has mean 101 and SD 16. For random samples of size 36, find the mean and SD of x̄ and P(x̄>103.7) under a normal/CLT approximation.

A. Mean 101, SD 0.444, probability 0.8416.
B. Mean 101, SD 2.667, probability 0.1584.
C. Mean 101, SD 16, because averaging does not change spread.
D. Mean 2.806, SD 16, probability 0.1584.

Answer: B

μ_x̄=μ=101. σ_x̄=σ√n=16√36=2.667. The cutoff has z=(103.7−101)2.667=1.001, so P(x̄>cutoff)=0.1584.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Mean 101, SD 2.667, probability 0.1584. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 101, SD 2.667, probability 0.1584. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.
Choice D: It does not match the requested calculation. The correct comparison or result is: Mean 101, SD 2.667, probability 0.1584. Key check: A sampling distribution describes a statistic over all possible samples of a fixed size.

Question 32. Sampling Distribution of a Sample Mean

A population at a school district in Pine Ridge during a baseline measurement week has mean 64 and SD 13. For random samples of size 64, find the mean and SD of x̄ and P(x̄>66.03) under a normal/CLT approximation.

A. Mean 64, SD 0.203, probability 0.8942.
B. Mean 64, SD 13, because averaging does not change spread.
C. Mean 64, SD 1.625, probability 0.1058.
D. Mean 1, SD 13, probability 0.1058.

Answer: C

μ_x̄=μ=64. σ_x̄=σ√n=13√64=1.625. The cutoff has z=(66.03−64)1.625=1.249, so P(x̄>cutoff)=0.1058.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: Mean 64, SD 1.625, probability 0.1058. Key check: Larger n reduces standard error and the CLT improves normal approximation.
Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 64, SD 1.625, probability 0.1058. Key check: Larger n reduces standard error and the CLT improves normal approximation.
Choice D: It does not match the requested calculation. The correct comparison or result is: Mean 64, SD 1.625, probability 0.1058. Key check: Larger n reduces standard error and the CLT improves normal approximation.

Question 33. Sampling Distribution of a Sample Proportion

For a population proportion p=0.564, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.604).

A. Mean 0.564, SD 0.0331, probability 0.1131.
B. Mean 0.564, SD 0.0331, probability 0.8869.
C. Mean 0.003, SD 0.0331.
D. Mean 0.564, SD 0.0011 without the square root.

Answer: A

μ_p̂=p=0.564. σ_p̂=√p(1−p)n=√0.564(0.436)225=0.0331. z=(0.604−0.564)0.0331≈1.21, so the upper-tail probability is 0.1131.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: Mean 0.564, SD 0.0331, probability 0.1131. Key check: The standard deviation depends on the population proportion and sample size.
Choice C: It does not match the requested calculation. The correct comparison or result is: Mean 0.564, SD 0.0331, probability 0.1131. Key check: The standard deviation depends on the population proportion and sample size.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: Mean 0.564, SD 0.0331, probability 0.1131. Key check: The standard deviation depends on the population proportion and sample size.

Question 34. One-Proportion z Interval

In an SRS of 320 parts from a regional manufacturer in Midwest consortium during a school-year data collection, 168 meet the part diameter criterion. Construct a 95% one-proportion z interval and interpret it.

A. (0.47, 0.58) with a population-proportion interpretation.
B. (0.497, 0.553); omit the critical value.
C. (0.498, 0.552); halve the margin of error.
D. (0.47, 0.58); 95% of sampled observations lie inside this interval.

Answer: A

p̂=168320=0.525. Conditions include randomization/independence and 168 successes and 152 failures, both at least 10. SE=√p̂(1−p̂)n=0.0279; z*=1.96; ME=0.055. The interval is (0.47, 0.58). We are 95% confident that the true population proportion lies in this interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: (0.47, 0.58) with a population-proportion interpretation. Key check: The interval estimates one population proportion.
Choice C: It does not match the requested calculation. The correct comparison or result is: (0.47, 0.58) with a population-proportion interpretation. Key check: The interval estimates one population proportion.
Choice D: It does not match the requested calculation. The correct comparison or result is: (0.47, 0.58) with a population-proportion interpretation. Key check: The interval estimates one population proportion.

Question 35. Two-Proportion z Interval

At a municipal water office in Capital Region during a randomized pilot period, Group 1 has 113220 meeting a criterion and Group 2 has 87210. Construct a 95% interval for p₁−p₂.

A. (0.051, 0.147); omit 1.96.
B. (-0.193, -0.006) while still labeling it p₁−p₂.
C. (0.006, 0.193) for p₁−p₂.
D. Use a pooled standard error because all two-proportion procedures pool.

Answer: C

p̂₁=0.514, p̂₂=0.414, difference=0.099. The unpooled SE is √0.514(0.486)220+0.414(0.586)210=0.0479. The 95% interval is 0.099±1.96(0.0479)=(0.006, 0.193).

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: (0.006, 0.193) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.
Choice B: It does not match the requested calculation. The correct comparison or result is: (0.006, 0.193) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.
Choice D: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: (0.006, 0.193) for p₁−p₂. Key check: The interval estimates p₁−p₂; reversing groups reverses signs.

Question 36. One-Sample t Interval

A random sample of 41 learners at a digital learning platform in Coastal Plains during a six-week field trial has mean 44.3 and SD 17.97 for lesson completion. Construct a 90% t interval for the population mean.

A. (41.49, 47.11); omit t*.
B. Use a z interval because the sample SD is known.
C. (39.57, 49.03) for the population mean.
D. (9.08, 79.52); use raw SD instead of SE.

Answer: C

df=40, t*=1.684, SE=s√n=17.97√41=2.806, so the interval is 44.3±4.726=(39.57, 49.03). The interpretation concerns the population mean, conditional on randomization/independence and an acceptable population shape or robust sample size.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: (39.57, 49.03) for the population mean. Key check: Use t because population σ is unknown.
Choice B: It selects a procedure or standard error that does not match the data structure and target. The correct comparison or result is: (39.57, 49.03) for the population mean. Key check: Use t because population σ is unknown.
Choice D: It does not match the requested calculation. The correct comparison or result is: (39.57, 49.03) for the population mean. Key check: Use t because population σ is unknown.

Mixed current-course AP Statistics free-response practice

10-point analytic study rubric used for the sets below
Evidence	Points
Correct target, notation, direction, or group order	2
Correct method/model and defensible conditions	2
Correct setup and execution	3
Contextual interpretation and scope/limitation	2
Clear communication with units and labels	1

FRQ set 1: Expected Value and Standard Deviation of a Random Variable

Scenario. A recycling program in Midwest consortium during a quarterly performance study outcome pays $21 with probability 0.161 and loses $24 otherwise. Find and interpret the expected net value.

Define the event or random variable and state the requested probability or long-run quantity.
Verify the relevant model conditions or probability-distribution requirements.
Show the probability, expected-value, or distribution calculation with labeled terms.
Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

E(X)=0.161($21)+0.839(−$24)=$-16.75. Over many independent repetitions under the same conditions, the average net result approaches about $-16.75 per repetition; it is not a guaranteed result on one trial.

FRQ set 2: Binomial Distribution

Scenario. For 23 independent calls at a municipal emergency dispatch center in North Valley during a six-week field trial, each meets the response time criterion with probability 0.251. Find P(X=12) and give the binomial mean and SD.

Define the event or random variable and state the requested probability or long-run quantity.
Verify the relevant model conditions or probability-distribution requirements.
Show the probability, expected-value, or distribution calculation with labeled terms.
Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

X~Bin(23,0.251). P(X=12)=C(23,12)(0.251)¹²(0.749)¹¹=0.0035. μ=np=5.773 and σ=√np(1−p)=2.079.

FRQ set 3: Sampling Distributions

Scenario. A population at a university advising center in Coastal Plains during a regional benchmarking study has mean 87 and SD 17. For random samples of size 64, find the mean and SD of x̄ and P(x̄>88.7) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=87. σ_x̄=σ√n=17√64=2.125. The cutoff has z=(88.7−87)2.125=0.8, so P(x̄>cutoff)=0.2119.

FRQ set 4: Sampling Distribution of a Sample Mean

Scenario. A population at a regional airport authority in Central County during a spring 2027 pilot has mean 85 and SD 21. For random samples of size 49, find the mean and SD of x̄ and P(x̄>88.75) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=85. σ_x̄=σ√n=21√49=3. The cutoff has z=(88.75−85)3=1.25, so P(x̄>cutoff)=0.1056.

FRQ set 5: Sampling Distribution of a Sample Proportion

Scenario. For a population proportion p=0.602, an SRS of n=400 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.624).

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_p̂=p=0.602. σ_p̂=√p(1−p)n=√0.602(0.398)400=0.0245. z=(0.624−0.602)0.0245≈0.899, so the upper-tail probability is 0.1844.

FRQ set 6: One-Proportion z Interval

Scenario. In an SRS of 220 animals from a wildlife clinic in Capital Region during a fall 2026 audit, 79 meet the recovery time criterion. Construct a 99% one-proportion z interval and interpret it.

Define the population parameter and preserve the stated group order.
Verify randomization, independence, and procedure-specific success/failure or shape conditions.
Construct the interval from the statistic, critical value, and standard error.
Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.

Model response

p̂=79220=0.359. Conditions include randomization/independence and 79 successes and 141 failures, both at least 10. SE=√p̂(1−p̂)n=0.0323; z*=2.576; ME=0.083. The interval is (0.276, 0.442). We are 99% confident that the true population proportion lies in this interval.

FRQ set 7: Two-Proportion z Interval

Scenario. At a municipal emergency dispatch center in Desert County during a community outreach cycle, Group 1 has 143250 meeting a criterion and Group 2 has 45130. Construct a 95% interval for p₁−p₂.

Define the population parameter and preserve the stated group order.
Verify randomization, independence, and procedure-specific success/failure or shape conditions.
Construct the interval from the statistic, critical value, and standard error.
Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.

Model response

p̂₁=0.572, p̂₂=0.346, difference=0.226. The unpooled SE is √0.572(0.428)250+0.346(0.654)130=0.0522. The 95% interval is 0.226±1.96(0.0522)=(0.124, 0.328).

FRQ set 8: One-Sample t Interval

Scenario. A random sample of 39 parts at a regional manufacturer in South Harbor during a semester-long cohort study has mean 82.14 and SD 14.98 for part diameter. Construct a 95% t interval for the population mean.

Define the population parameter and preserve the stated group order.
Verify randomization, independence, and procedure-specific success/failure or shape conditions.
Construct the interval from the statistic, critical value, and standard error.
Interpret the confidence statement for the population parameter and explain one factor affecting margin of error.

Model response

df=38, t*=2.024, SE=s√n=14.98√39=2.399, so the interval is 82.14±4.856=(77.28, 87). The interpretation concerns the population mean, conditional on randomization/independence and an acceptable population shape or robust sample size.

FRQ set 9: Null and Alternative Hypotheses

Scenario. A housing authority in South Harbor during a regional benchmarking study tests H₀:p=0.5 against H_a:p>0.5 using 226 successes in n=390. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂=226390=0.579. Under H₀, SE=√0.5(0.5)390=0.0253 and z=(0.579−0.5)0.0253=3.139. The right-tailed p-value is 0.0008. At α=.05, reject H₀ and report statistically significant evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 10: Type I Error, Type II Error, and Power

Scenario. A county library in Riverbend during a winter readiness review tests H₀:p=0.5 against H_a:p>0.5 for the weekly program attendance criterion. Describe Type I and Type II errors, then state one change that generally increases power against p=0.58.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

For the county library in Riverbend during a winter readiness review, a Type I error is concluding that the population proportion meeting the weekly program attendance criterion exceeds 0.5 when it actually equals 0.5. A Type II error is failing to conclude it exceeds 0.5 when the true proportion is 0.58. Increasing sample size or α generally increases power against that alternative; improving measurement can also help.

FRQ set 11: One-Proportion z Test

Scenario. A municipal water office in North Valley during a winter readiness review tests H₀:p=0.5 against H_a:p>0.5 using 50 successes in n=120. Compute the one-proportion z statistic and p-value, then conclude at α=.05.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

2019 International Practice Exam AP Statistics: Ethical Study Guide — FRQ set 11: One-Proportion z Test: p̂=50120=0.417. Under H₀, SE=√0.5(0.5)120=0.0456 and z=(0.417−0.5)0.0456=-1.826. The right-tailed p-value is 0.9661. At α=.05, fail to reject H₀ because the sample does not provide sufficiently strong evidence for the alternative. The conclusion is about the population proportion, not proof that H₀ is true or false.

FRQ set 12: Two-Proportion z Test

Scenario. For independent samples at a public health department in Coastal Plains during a fall 2026 audit, Group 1 has 207280 successes and Group 2 has 116190. Test H₀:p₁=p₂ against H_a:p₁≠p₂.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

p̂_c=(207+116)(280+190)=0.687. SE₀=√p̂_c(1−p̂_c)(1280+1190)=0.0436. z=(0.739−0.611)0.0436=2.955 and the two-sided p-value is 0.0031. Reject H₀ at .05; evidence suggests different population proportions.

FRQ set 13: One-Sample t Test

Scenario. A random sample of 47 accounts at a municipal water office in Mountain Region during a baseline measurement week has x̄=60.78, s=10.25. Test H₀:μ=53 against H_a:μ≠53.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

t=(60.78−53)(10.25√47)=5.204, df=46. The two-sided p-value is 0. Reject H₀ at .05; the data provide evidence that the population mean differs from the null value.

FRQ set 14: Chi-Square Homogeneity and Independence

Scenario. At a community college in Prairie District during a winter readiness review, observed category counts are [[25, 63, 55], [29, 39, 19]]. Test the null claim of same categorical distribution across two groups; calculate expected counts, χ², df, and p-value.

State the population parameter and hypotheses, including direction and group order.
Verify the procedure conditions using evidence from the prompt.
Compute the test statistic and p-value from the correct null model or expected counts.
Make a decision at the stated significance level and conclude about the population without treating the p-value as the probability the null is true.

Model response

Expected counts are [[33.57, 63.42, 46.01], [20.43, 38.58, 27.99]]. χ²=Σ(O−E)²E=10.441, df=2, p=0.0054. Reject the null at .05; the distributions differ.