AP Statistics FRQ Practice 2027: 12 Sets, 48 FRQs

Worked answer

1. A suitable question is: Among students enrolled in the district’s three high schools, what proportion eat breakfast on a typical school day, and does access to grab-and-go breakfast increase that proportion? The population is all 1,800 district high-school students; the response is whether a student eats breakfast on the measured school day.
2. Separate the roster by high school, randomly sample students within each school, and choose sample sizes proportional to each school’s enrollment. That preserves representation while keeping selection random within strata.
3. Use students (or naturally intact classrooms if the treatment must be delivered by classroom) as experimental units. Randomly assign comparable units to grab-and-go access or usual breakfast access, keep measurement rules the same, and compare breakfast participation.
4. Random sampling supports generalization to the population from which the sample was randomly selected. Random assignment supports a cause-and-effect conclusion for the experimental units, assuming implementation is otherwise comparable. Having both gives the strongest combination of generalizability and causal inference.

Worked answer

The mean is 17.56 minutes, the median is 16, and the sample standard deviation is about 5.48. With the median excluded when splitting the ordered list, Q1 = 14, Q3 = 20, so IQR = 6.

The fences are 14 − 1.5(6) = 5 and 20 + 1.5(6) = 29; therefore 30 is a high outlier by this rule. Median and IQR are the more resistant summaries for a distribution with this unusually large value. Replacing 30 with 42 increases the mean and standard deviation substantially, while the median remains 16 and the quartile positions—and therefore IQR—remain unchanged for this ordered sample.

Worked answer

Let p be the true proportion of all students at the university who used the library during the previous week. The estimate is p̂ = 84/140 = 0.600.

The sample was random; if the university has at least 1,400 students, the 10% condition is satisfied. The sample contains 84 successes and 56 failures, both at least 10. The standard error is about 0.0414, so the 95% interval is 0.600 ± 1.96(0.0414) = (0.519, 0.681).

Interpretation: We are 95% confident that the true proportion of all students at this university who used the library during the previous week is between about 51.9% and 68.1%.

Worked answer

The least-squares line is ŷ = 57.933 + 3.829x, with r ≈ 0.995 and r² ≈ 0.989. Within this observed range, each additional study hour is associated with an increase of about 3.83 predicted quiz points. About 98.9% of the observed variation in quiz scores is explained by the linear relationship with study hours.

At x = 4.5, the predicted score is about 75.16. At x = 4, ŷ ≈ 73.25, so the residual is 74 − 73.25 ≈ +0.75. Because this is observational, students who study more may differ in prior preparation, motivation, course attendance, or other variables; association does not establish causation.

Worked answer

A suitable question asks whether receiving nightly reminders changes the mean proportion of assigned review problems completed. Create separate ninth-grade and twelfth-grade blocks, then randomly assign students within each block to reminders or no reminders. Use the same assignment length, scoring rule, and observation period for both treatment groups so treatment is not confounded with measurement.

Random assignment can support a causal comparison for the participating students. Because volunteers were not randomly sampled from the whole school, broad generalization to every student is not automatically justified.

Worked answer

Participants: 54/72 = 0.750. Nonparticipants: 45/78 ≈ 0.577. Overall: 99/150 = 0.660. The participant completion rate is about 17.3 percentage points higher.

This is an association, but participation was voluntary. Motivation, prior achievement, available time, and other variables can differ between those who chose the program and those who did not. Without random assignment, the observed difference does not by itself establish causation.

Worked answer

Let μ be the population mean response time. A one-sample t interval is appropriate because σ is unknown and is estimated by s. The random sample supports independence; if sampling without replacement, the population should be at least 360 requests. With n = 36 and no extreme outliers, the t procedure is reasonable.

SE = 10.8/√36 = 1.8. With df = 35, t* ≈ 2.030. The interval is 72.4 ± 2.030(1.8) = (68.75, 76.05) minutes. We are 95% confident that the population mean response time lies between about 68.8 and 76.1 minutes.

Worked answer

For X ~ Binomial(20, 0.08), P(X ≤ 1) ≈ 0.5169. The binomial model requires binary outcomes, independent trials, fixed n, and constant p. E(X) = np = 1.6 and SD(X) = √np(1−p) ≈ 1.213.

For n = 100, μ_p̂ = 0.08 and SD_p̂ = √[0.08(0.92)/100] ≈ 0.02713. Using a normal approximation gives z ≈ 1.474 for p̂ = 0.12, so P(p̂ > 0.12) ≈ 0.0702. The exact binomial model does not require a normal shape; the p̂ normal approximation does, so np and n(1−p) must be sufficiently large.

Worked answer

The QR-code plan undercovers residents who do not visit that park and creates voluntary-response bias because people who choose to scan may have unusually strong opinions. A stronger plan uses a resident sampling frame, divides residents into geographic districts, randomly samples within every district, and weights only if the sampling plan requires it.

A neutral question could ask, “On a scale from 1 to 5, how satisfied are you with the condition and accessibility of city public parks during the past six months?” Nonresponse can still bias results if responders and nonresponders systematically differ in satisfaction.

Worked answer

For x = 80, z = (80−68)/8 = 1.50, and Φ(1.50) ≈ 0.9332, so the score is around the 93.3rd percentile. For 60 to 76, the z-scores are −1 and +1, giving probability ≈ 0.6827.

The correct statement is that about 93.3% of values in the Normal model are at or below 80—not that 80 is “93.3% above the mean.” Standardization is always arithmetic, but a Normal probability statement depends on a defensible Normal model for the distribution.

Worked answer

p̂A = 0.600, p̂B = 0.450, and the pooled estimate under H0 is 126/240 = 0.525. The random/independent samples and large expected success/failure counts support the two-proportion z test.

SE under H0 ≈ 0.06447, z ≈ 2.327, and the two-sided p-value ≈ 0.0200. Reject H0 at α = 0.05. The data provide convincing evidence that the population positive-response proportions differ between message A and message B.

Worked answer

The line is ŷ = 11.8 + 3.2x, r ≈ 0.9995, and r² ≈ 0.9990. At x = 7, ŷ = 34.2. At x = 8, ŷ = 37.4, so the residual is −0.4.

Within the observed range, each additional training hour is associated with about 3.2 more predicted tasks completed. Roughly 99.9% of observed variation in y is explained by the linear relationship with x. Predicting at 25 hours is extreme extrapolation far beyond the observed 2–10 hour range; the relationship need not continue there.

Worked answer

Each participant serves as their own control, which removes much person-to-person variation in baseline typing ability. Randomly assign each participant to seated-first or standing-first, using a fair random mechanism. Define d = accuracy while standing − accuracy while seated, so d > 0 means better accuracy while standing.

A carryover effect could occur if fatigue or practice from the first task influences the second. Use comparable but different tasks, allow a washout/rest period, and randomize order to balance order effects.

Worked answer

μ_x̄ = 120 and σ_x̄ = 24/√64 = 3. For x̄ = 126, z = 2, so P(x̄ > 126) ≈ 0.0228.

Averages vary less from sample to sample than individual observations vary in the population; averaging n independent observations reduces standard deviation by √n. At n = 144, the standard error becomes 24/12 = 2.

Worked answer

μd is the population mean time saved by the new system under the chosen difference definition. The data must be matched, the pairs should be independent, and the distribution of differences should have no severe skew/outliers for n = 12.

t = 3.4/(2.1/√12) ≈ 5.609, df = 11, and the one-sided p-value is about 0.00008. Reject H0. There is very strong evidence that the new system reduces mean completion time. A 95% CI is approximately (2.07, 4.73) minutes saved, which lies entirely above 0 and agrees with the test.

Worked answer

Row totals are both 70; column totals are 70, 43, and 27, with N = 140. Expected counts for each row are 35, 21.5, 13.5. The chi-square statistic is χ² ≈ 5.754 with df = 2, giving p ≈ 0.0563.

At α = 0.05, fail to reject independence; the sample does not provide quite enough evidence of an association between device type and preferred study format. The approximation is based on the null-model expected cell counts, so those—not the observed counts—control the large-count check. Random sampling can support generalization to the sampled population, but the variables were not randomly assigned, so causation is not justified.

Worked answer

The email list excludes riders who pay by cash, use anonymous cards, or buy tickets through channels not linked to email, so it does not represent all weekday riders. A stronger plan samples trips, stops, or rider records across routes and times, then randomly selects riders within the chosen frame so each major rider type has a chance to be selected.

A neutral question might ask, “During your weekday transit trips in the past 30 days, how satisfied were you with on-time performance on a 1–5 scale?” Compare response rates across sampled route/time/rider strata and follow up with a random subset of nonresponders if feasible; systematic differences in response can bias the result.

Worked answer

Section B has the higher mean by 1.8 points, but the medians are almost the same (14.1 vs 13.9). Section B also has much larger SD (5.4 vs 3.1), while the IQRs are similar (4.4 vs 4.2). That pattern is consistent with a high-end tail or influential high values in Section B, because nonresistant measures changed much more than resistant measures.

However, numerical summaries do not uniquely determine shape. A graph or the raw data would be needed to identify skewness, multimodality, gaps, or specific outliers conclusively.

Worked answer

The observed difference is 82.1 − 77.4 = 4.7. The standard error is about 2.045; Welch df is about 72.8, giving t* ≈ 1.993. The 95% CI is approximately (0.62, 8.78).

Assuming independent random samples and no severe distribution problems for these sample sizes, we are 95% confident that Group 1’s population mean exceeds Group 2’s by about 0.6 to 8.8 units. Because 0 is not in the interval, equal population means are not supported by this interval at the corresponding two-sided 5% level.

Worked answer

The probabilities are all between 0 and 1 and sum to 1. E(X) = 1.8 alerts/day. The standard deviation is about 0.927 alerts/day.

Over many days under the same process, the long-run average number of alerts per day approaches 1.8. Eighteen alerts is a 10-day total, while SD(X) describes one-day variability; they are different random quantities on different scales. A 10-day total would need its own distributional analysis rather than being compared directly with the one-day standard deviation.

Worked answer

Within each field, randomly assign approximately half the plots to fertilizer A and half to B, then compare yields within blocks before combining evidence. Blocking removes predictable field-to-field soil variation from the treatment comparison, improving precision.

Replication provides repeated observations under each treatment and lets random variation be assessed. Random assignment balances lurking variables in expectation and supports a causal interpretation of treatment differences; replication alone does not create causal validity.

Worked answer

The line is ŷ = 19.6 − 2.171x, r ≈ −0.991, and r² ≈ 0.982. Each additional practice session is associated with about 2.17 fewer predicted errors. About 98.2% of observed variation in error count is explained by the linear relationship with practice sessions.

At x = 4, ŷ ≈ 10.914, so the residual is 10 − 10.914 ≈ −0.914. A negative slope describes the direction of association between x and y; it does not mean y must take negative values within the observed range.

Worked answer

H0: the Yes/No population distribution is the same across the three regions. Ha: at least one region has a different distribution. Since 80 Yes and 70 No responses occur among 150 total, each 50-person row has expected counts 26.67 Yes and 23.33 No, all well above 5.

χ² = 6.000, df = 2, p ≈ 0.0498. At α = 0.05, reject H0; the data provide evidence that the Yes/No population distribution is not the same across all three regions.

Worked answer

μ_p̂ = 0.62 and SD_p̂ = √[0.62(0.38)/150] ≈ 0.03963. np = 93 and n(1−p) = 57, so the Normal approximation is strong.

For p̂ = 0.55, z ≈ −1.766, giving P(p̂ < 0.55) ≈ 0.0387. That is uncommon under p = 0.62, but random samples can produce tail outcomes. The calculation assumes p = 0.62; observing one unusual sample is evidence to investigate, not proof by itself that the parameter changed.

Worked answer

A suitable question asks whether app use is associated with mean weekly exercise time in a defined population. Baseline motivation, health status, access to exercise facilities, income, and prior activity are plausible confounders. Matching only on age balances one variable, not every systematic difference between self-selected users and nonusers.

For a causal feature test, recruit eligible participants who all have access to the study platform, then randomly assign one group to receive the feature (for example, goal reminders) and the other to a control version. Compare a prespecified exercise outcome using the same measurement window.

Worked answer

E(X) = 6.5 points and SD(X) = 4.5 points. Over many plays, the average award approaches 6.5 points; the SD describes the typical scale of one-play deviation around that mean.

An expected value is a long-run weighted average and need not be an attainable single-play outcome. Shifting probability mass from 0 to 15 increases the expected value because more probability is placed on a larger outcome.

Worked answer

Under H0, np0 = 82.5 and n(1−p0) = 67.5; with random sampling and the 10% condition if relevant, the one-proportion z test is appropriate. p̂ = 92/150 ≈ 0.6133, z ≈ 1.559, and the one-sided p-value ≈ 0.0595.

At α = 0.05, fail to reject H0; the sample does not provide quite enough evidence that the population preference proportion exceeds 0.55. A Type I error would be concluding p > 0.55 when in fact p = 0.55. Increasing sample size generally increases power for a fixed true alternative.

Worked answer

For n = 36, μ_x̄ = 500 and SE = 60/6 = 10. The bounds 485 and 515 correspond to z = −1.5 and +1.5, so the probability is about 0.8664.

For n = 144, SE = 60/12 = 5; the same ±15 interval is now ±3 SE, giving probability about 0.9973. Larger n concentrates sample means more tightly around μ. It does not shrink the spread of the underlying individual measurements.

Worked answer

Randomly select schools using a probability method, then obtain rosters within selected schools and randomly sample students from each selected school. School selection can use probability proportional to enrollment or the second-stage sample sizes can be adjusted so student inclusion probabilities are known and defensible.

Multistage sampling is operationally cheaper, but students within a school may be more similar than randomly selected students statewide, which can increase sampling variability. A huge convenience sample can still be systematically biased because size does not replace random selection.

Worked answer

Mean(Y) = 1.8(52)+32 = 125.6; SD(Y) = 1.8(8) = 14.4. Median(Y) = 1.8(50)+32 = 122; IQR(Y) = 1.8(10) = 18.

For a positive linear transformation, each value and the center/scale transform together, so z-scores are unchanged. The distribution’s shape and ordering are preserved; only location, scale, and axis labels change.

Worked answer

p̂1 = 110/180 ≈ 0.6111, p̂2 = 84/160 = 0.5250, so the difference is 0.0861. Both samples have at least 10 successes and failures, and the groups/samples must be independent; the 10% condition applies if sampling without replacement.

SE ≈ 0.05366. The 95% CI is 0.0861 ± 1.96(0.05366) = (−0.019, 0.191). The plausible values include 0, so the interval does not establish a nonzero population difference at the corresponding two-sided 5% level.

Worked answer

Zone A support = 45/75 = 0.600; Zone B support = 35/75 ≈ 0.4667; overall support = 80/150 ≈ 0.5333. Support is about 13.3 percentage points higher in Zone A.

If each zone’s residents were randomly sampled, the association can be generalized to the corresponding zone populations, subject to the sampling design. Zone residence was not randomly assigned, so other characteristics can differ between zones; the association is not evidence that changing residence itself causes opinion to change.

Worked answer

Blocking on experience reduces known baseline variation and makes the treatment comparison more precise. Within each experience block, randomly assign employees to the two methods in roughly equal proportions; exact counts can differ by one when a block size is odd.

A primary response might be completion time; error rate could be secondary so a faster method is not judged without accuracy. Random assignment supports causal comparison among participating employees. Because volunteers from one office are not a random sample of all company employees, company-wide generalization is limited.

Worked answer

The estimated p-value is 27/1000 = 0.027. If the null model were true, a result at least this favorable to Ha would occur in about 2.7% of random reassignments under this simulation scheme.

At α = 0.05, the simulation suggests rejecting the null model. Increasing the number of random reassignments reduces Monte Carlo/simulation variability in the estimated p-value; it does not magically remove bias from a poor original study design.

Worked answer

Random sampling supports independence; if sampling without replacement, the population should be at least 250. With n = 25, a roughly symmetric distribution without strong outliers is important. t = (104.6−100)/(9.5/√25) ≈ 2.421, df = 24.

The one-sided p-value is about 0.0117. Reject H0 at α = 0.05. The sample provides convincing statistical evidence that the population mean exceeds 100.

Worked answer

With X ~ Binomial(40, 0.10), P(X ≥ 8) ≈ 0.0419. The expected count is np = 4, so 8 defects are twice the null-model expectation.

A tail probability around 4.2% is relatively unusual if the historical defect probability is still 0.10, so the result provides evidence that the model may no longer fit. The calculation identifies statistical inconsistency, not mechanism; machine wear, material changes, operator differences, or other causes require separate process investigation.

Worked answer

The missing phone-only customers create undercoverage. A better design combines the email and phone/customer-account frames, deduplicates customers, assigns known selection probabilities, and randomly samples from the resulting coverage plan.

A larger sample from the same incomplete email frame reduces random sampling error around the wrong frame-based target but does not remove systematic undercoverage. Duplicates should be identified by customer ID or another defensible matching rule so one customer does not receive unintended extra selection probability.

Worked answer

The line is ŷ = 9.2 + 0.9x, r ≈ 0.990, and r² ≈ 0.980. At x = 35, ŷ = 40.7. At x = 40, ŷ = 45.2, so the residual is +1.8.

Within the observed range, each additional hundred dollars of advertising is associated with about 0.9 more predicted sign-ups. Predicting at x = 100 doubles the largest observed x and assumes the same linear relationship far beyond the data.

Worked answer

Total complete = 100, not complete = 80, and each row total is 60. Under independence, each row has expected counts 33.33 complete and 26.67 not complete, all above 5.

χ² ≈ 14.625, df = 2, p ≈ 0.00067. Reject independence. There is strong evidence of an association between device category and task-completion status in the population represented by the random sample.

Worked answer

The differences have mean 3.5, median 3.5, and sample SD ≈ 1.581. Because d = after − before, positive values indicate improvement; the sample average improvement is 3.5 units.

The same participant contributes both measurements, so the observations are paired and within-person dependence is informative. Treating before and after as independent groups throws away that pairing. With only 10 differences, inspect the difference distribution for strong skewness and outliers before relying on a paired t procedure.

Worked answer

Experimental units are participating users. The treatment is the new focus-mode behavior; a control could show a visually similar neutral interface or standard notification behavior so participants have comparable study interaction. The response might be interruption count or uninterrupted-task time.

Analysts can be blinded to treatment labels during data cleaning/scoring; participants may be partly blinded if both versions look similar, though complete blinding can be difficult if behavior is noticeable. Random assignment is still essential because automatic measurement does not balance motivation, baseline phone use, task difficulty, or other lurking variables.

Worked answer

P(X > 80) = P(Z > 1) ≈ 0.1587. P(X > 60) = P(Z > −1) ≈ 0.8413. Since X > 80 is a subset of X > 60, the conditional probability is 0.1587/0.8413 ≈ 0.1886.

Conditioning restricts attention to scores already above 60, removing the entire lower tail from the reference group. Therefore scores above 80 form a larger fraction of the restricted group than of the whole population.

Worked answer

Assuming independent random samples, the groups are independent; the 10% condition applies if sampled without replacement. With n around 30 in each group, the t procedure is fairly robust unless there is extreme skew or severe outliers.

SE ≈ 1.028, t ≈ 2.335, Welch df ≈ 54.4, and the one-sided p-value ≈ 0.0116. Reject H0 at α = 0.05. The sample provides convincing evidence that Group 1’s population mean exceeds Group 2’s.

Worked answer

The sampling-distribution mean is p1 − p2 = 0.10. The SD is √[0.60(0.40)/100 + 0.50(0.50)/100] = 0.070. All four expected success/failure counts are at least 40, so the Normal approximation is strong.

For a difference of 0.20, z = (0.20−0.10)/0.07 ≈ 1.429, giving tail probability ≈ 0.0766. Random sampling can support generalization to the sampled populations. The calculation compares population proportions; it does not create random assignment or prove that membership in a population causes the outcome.

Worked answer

Plan A is stratified by school because every school contributes sampled students. Plan B is multistage/cluster-like because a random subset of schools is selected first and students are then sampled within selected schools.

Plan B can reduce travel and administration cost, but students within the same school may be more alike, so fewer schools can reduce effective information and increase sampling variability. Choosing convenient schools instead of random schools can systematically overrepresent certain neighborhoods, programs, or school types and undermine district-wide inference.

Worked answer

For X ~ Binomial(10, 0.35), P(X ≥ 5) ≈ 0.2485. E(X) = np = 3.5 and SD(X) = √[10(0.35)(0.65)] ≈ 1.508.

X counts successes in a fixed number of trials. A “trials until first success” variable has a different support and model structure. The revised 2026–27 AP Statistics course removed the geometric distribution from required content, so 2027 core practice should emphasize current probability and binomial reasoning rather than legacy geometric-distribution questions.

Worked answer

μd is the population mean change after − before; negative values mean the response decreased. The matched pairs should be independent of other pairs and the difference distribution should have no severe skew/outliers for n = 16.

t = −2.8/(3.6/√16) ≈ −3.111, df = 15, and the one-sided p-value ≈ 0.0036. Reject H0. The 95% CI is approximately (−4.72, −0.88). The entire interval is below zero, supporting a population mean decrease of roughly 0.9 to 4.7 units under this difference definition.

Worked answer

The regression line is ŷ = 3.2 + 3.0x, r ≈ 0.996, and r² ≈ 0.991. At x = 3.5, ŷ = 13.7. At x = 4, ŷ = 15.2, so the residual is −0.2.

The separate sample proportion is 52/80 = 0.650. The observational regression supports a strong linear association within the observed x-range but not a causal effect of exposure. If the 80 days were randomly sampled from a well-defined population of days, 0.65 is a defensible estimator for that population proportion, subject to sampling error; random sampling still does not create causation.