AP Statistics 2026 FRQ | TopEduPrep Study Guide

Question 1 · 15 points

Comparing Two Distributions (Goats)

Unit 1 · Descriptive Stats

A goat farmer raises two breeds, Breed H and Breed J. He takes independent random samples of 14 goats from each breed and records weight (pounds).

Weights of Breed H goats: 48, 48, 55, 56, 56, 57, 62, 66, 72, 72, 72, 73, 80, 80

Breed J is shown only as a boxplot with whiskers from about 50 to 87 lb, a box roughly from 57 to 80 lb, and a median line near 64 lb.

Part A

Use the data list to find the five-number summary for Breed H.

Step 1 — Sort & count. The list is already sorted, n = 14 (even).

Step 2 — Median. With n = 14, the median is the average of the 7th and 8th values: (62 + 66) / 2 = 64.

Step 3 — Q1. Take the lower half (first 7 values: 48, 48, 55, 56, 56, 57, 62). Its median (the 4th value) is 56.

Step 4 — Q3. Take the upper half (last 7 values: 66, 72, 72, 72, 73, 80, 80). Its median (the 4th value) is 72.

Step 5 — Min & Max. Min = 48, Max = 80.

Five-number summary: Min = 48, Q1 = 56, Median = 64, Q3 = 72, Max = 80

Beginner tip: When n is even, split the data exactly in half (don't include the overall median in either half) before finding Q1/Q3 as the "median of each half."

Part B

Compare center and variability of Breed H vs. Breed J, in context.

Center. Breed H's median (64 lb) is about the same as Breed J's median (≈ 64 lb read from the boxplot) — so the two breeds have similar typical weight.

Variability. Breed H's IQR = 72 − 56 = 16 lb. Breed J's IQR (read from the boxplot) ≈ 80 − 57 = 23 lb. Breed J's range (≈ 87 − 50 = 37 lb) is also larger than Breed H's range (80 − 48 = 32 lb).

Breed J goats have roughly the same median weight as Breed H, but Breed J's weights are more spread out (larger IQR and range) — Breed J shows more variability.

Beginner tip: "Compare in context" means don't just say "the IQR is bigger" — say whose weight is more spread out and use the units (pounds). Always name both groups.

Part C

i. What shape feature is visible in the stem-and-leaf plot but not in a boxplot built from the five-number summary? ii. Why can't a boxplot show it?

i. The stem-and-leaf plot (4|88, 5|5667, 6|26, 7|2223, 8|00) shows gaps and clustering — the data bunch up in the 40s–50s and again in the 70s–80s, with a relative gap around the low 60s. A boxplot would look smooth and wouldn't reveal this clustered/gapped shape.

ii. A boxplot is built from only five summary numbers (min, Q1, median, Q3, max). It doesn't plot individual data values, so any gaps, clusters, or multiple peaks between those five numbers are invisible — the boxplot just draws a straight box and whiskers regardless of how the values inside are actually distributed.

Beginner tip: This is a classic AP Stats concept: boxplots are great for comparing center/spread quickly, but they always hide shape details like gaps, clusters, and multiple peaks. Only dotplots, histograms, and stem-and-leaf plots show individual values.

Key concepts — Q1

Five-number summary = Min, Q1, Median, Q3, Max
IQR = Q3 − Q1 measures spread (resistant to outliers)
Boxplots summarize but hide shape detail (gaps, clusters, modality)

Question 2 · 10 points

Designing an Experiment (Rosebushes)

Unit 3 · Collecting Data

Holly, a botanist, grows 30 rosebushes in a controlled greenhouse. She randomly assigns 15 to get ½ cup of coffee grounds weekly; the other 15 get none. After 3 months she counts the roses on each bush.

Part A

Identify: i. Treatments ii. Experimental units iii. Response variable

i. Treatments (2): (1) coffee grounds added weekly, (2) no coffee grounds added.

ii. Experimental units: the 30 individual rosebushes.

iii. Response variable: the number of roses produced per rosebush after three months.

Beginner tip: "Treatment" = the condition being tested; "experimental unit" = the thing the treatment is applied to; "response variable" = what you measure afterward. Keep these three roles separate — mixing them up is the most common point loss on experiment-design FRQs.

Part B

Describe how to randomly assign treatments so each gets the same number of units.

Number the 30 rosebushes 1 through 30. Using a random number generator (or a random digit table), generate random numbers to select 15 distinct bush numbers without replacement. Assign the coffee-grounds treatment to those 15 bushes; assign the remaining 15 bushes to the no-coffee-grounds group.

Key phrase graders look for: "without replacement" + "assign the selected 15 to one treatment, the rest to the other."

Beginner tip: Always mention how you'd generate the random selection (e.g., random number generator) and that it's done without replacement — otherwise a bush could get picked twice.

Part C

Explain the meaning of "statistically significant" here.

It means: if there really were no difference in mean number of roses between the two treatments (i.e., coffee grounds have no effect), then the probability of seeing a difference in sample means at least as large as the one Holly observed — just from the random assignment of bushes to groups — would be less than 0.05.

Because that probability is so small, the observed difference is unlikely to be due to chance alone, giving convincing evidence that coffee grounds really do affect the number of roses.

Beginner tip: "Statistically significant" is always a statement about a p-value being small — always frame it as "if the null hypothesis were true, this result would be unlikely to happen by chance."

Key concepts — Q2

Treatment vs. experimental unit vs. response variable
Random assignment (not random sampling) creates comparable groups
"Statistically significant" = small p-value = unlikely due to chance

Question 3 · 15 points

Normal, Binomial & Geometric (Team Song)

Unit 4–5 · Random Variables

Performance time of a team song ~ Normal(μ = 109 sec, σ = 16 sec). All performances are independent.

Part A

P(a performance takes longer than 120 seconds)?

Standardize. z = (120 − 109) / 16 = 0.6875

Use normal table/calculator. P(Z > 0.6875) = 1 − Φ(0.6875)

P(X > 120) ≈ 0.246 (about 24.6%)

Beginner tip: Draw the normal curve, shade the area to the right of 120, then convert to a z-score before reading the table — sketching first prevents sign errors.

Part B

10 performances selected; X = number that take longer than 120 sec. Find P(X ≥ 3).

Identify the model. X ~ Binomial(n = 10, p = 0.246) — fixed trials, two outcomes ("long" / "not long"), independent, constant p from Part A.

Complement rule. P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [P(0) + P(1) + P(2)]

P(X ≥ 3) ≈ 0.462 (about 46.2%)

Beginner tip: "At least 3" almost always means "use the complement": 1 − P(0, 1, or 2) is much less work than adding up P(3) through P(10).

Part C

Y = number of games Ben attends until a performance exceeds 120 sec. i. Mean of Y ii. SD of Y

Identify the model. Y ~ Geometric(p = 0.246) — counts trials up to and including the first "success" (a long performance).

i. Mean. μ_Y = 1/p = 1 / 0.246 ≈ 4.07 games

ii. SD. σ_Y = √(1 − p) / p ≈ 3.53 games

Mean ≈ 4.07 games, SD ≈ 3.53 games

Beginner tip: Binomial counts "successes out of a fixed n." Geometric counts "trials needed to get the first success" — the phrase "until" is the trigger word for geometric.

Part D

Interpret the standard deviation from Part C(ii) in context.

Typically, the number of games Ben attends until he hears a performance longer than 120 seconds will vary from the mean of 4.07 games by about 3.53 games.

Beginner tip: Every SD interpretation follows this template: "[Values] typically vary from the mean of [mean] by about [SD] [units]." Fill in the blanks and you're done.

Key concepts — Q3

Normal → z-score → table/calculator for probability
Binomial: fixed n trials, count of successes; "at least" → use complement
Geometric: trials until first success; mean = 1/p, sd = √(1−p)/p

Question 4 · 8 points

Two-Sample t-Test (Fertilizers)

Unit 6 · Inference for Means

58 trees get Brand C, 58 trees get Brand N, randomly assigned. Number of oranges recorded per tree.

	n	Mean	SD
Brand C	58	141	15
Brand N	58	148	19

At α = 0.05, is there convincing evidence the mean number of oranges differs between the two fertilizers?

Full Solution

Carry out a complete two-sample t-test.

1. Hypotheses. H₀: μ_C = μ_N Hₐ: μ_C ≠ μ_N (μ_C, μ_N = true mean number of oranges for each fertilizer)

2. Conditions. Random assignment ✓ (given). Independence: 58 + 58 < 10% of all possible trees isn't needed here since it's an experiment with random assignment — treat observations within each group as independent. Normal/large sample: n = 58 in each group is large (≥ 30), so the sampling distribution of the difference in means is approximately Normal by the Central Limit Theorem. ✓

3. Test statistic. SE = √(15²/58 + 19²/58) ≈ 3.18
t = (141 − 148) / 3.18 ≈ −2.20

4. p-value. Using the (Welch) t-distribution, df ≈ 108 (or the conservative df = 57 if computing by hand): p-value ≈ 0.030

5. Conclusion. Since p ≈ 0.030 < α = 0.05, we reject H₀.

There is convincing statistical evidence that the true mean number of oranges differs between trees fertilized with Brand C and Brand N.

Beginner tip: A full inference FRQ always has the same 4-part skeleton: Hypotheses → Conditions → Test statistic & p-value → Conclusion in context. Missing any one of these four costs points even if your math is correct.

On your calculator (2-SampTTest with stats), leave "Pooled" set to No since we're not told the population variances are equal.

Key concepts — Q4

Two-sample t-test compares two population means using sample data
Always write hypotheses in terms of parameters, not statistics
Conclusion must reference the context and compare p-value to α

Question 5 · 15 points

Two-Way Tables & Mosaic Plots (Athletes)

Unit 4 · Probability

Age-group by sport for all 4,193 professional athletes (population data, not a sample):

Age	Basketball	Football	Baseball	Total
< 25	232	807	259	1,298
25–<30	175	1,326	620	2,121
30–<35	90	287	276	653
35+	19	41	61	121
Total	516	2,461	1,216	4,193

Part A

i. P(football player)? ii. P(25 ≤ Age < 30 | football player)?

i. P(football) = 2,461 / 4,193 ≈ 0.587 (about 58.7%)

ii. Restrict to the football column only: P(25–30 | football) = 1,326 / 2,461 ≈ 0.539 (about 53.9%)

Beginner tip: The word "given" always means: narrow your denominator down to only the group after "given." Here that's the 2,461 football players, not all 4,193 athletes.

Part B

i. Does b match part A(i) or A(ii)? ii. What does x = b·h represent?

i. b is the width of the football column, which represents the overall proportion of all athletes who play football — that's the unconditional probability from Part A(i).

ii. h is the height of the 25–30 segment within the football column — the conditional probability from Part A(ii). So x = b·h is the joint probability: P(football AND 25 ≤ Age < 30) = 1,326 / 4,193 ≈ 0.316 (31.6%).

Beginner tip: In a mosaic plot, area = probability. Multiplying "width × height" is really just applying the general multiplication rule: P(A and B) = P(A) · P(B | A).

Part C

i. Are "Baseball" and "35 ≤ Age" mutually exclusive? ii. Are they independent?

i. Mutually exclusive? No. There are 61 athletes who are baseball players and 35 or older (the table cell isn't 0), so both events can happen to the same athlete.

ii. Independent? Check whether P(Baseball) × P(35+) = P(Baseball and 35+).
P(Baseball) = 1,216/4,193 ≈ 0.290, P(35+) = 121/4,193 ≈ 0.0289
Product ≈ 0.290 × 0.0289 ≈ 0.00837
Actual P(Baseball and 35+) = 61/4,193 ≈ 0.01455

Since 0.00837 ≠ 0.01455, the events are NOT independent.

Beginner tip: Mutually exclusive and independent are different ideas! Mutually exclusive asks "can both happen at once?" Independent asks "does knowing one change the probability of the other?" Always show the actual multiplication check for independence — don't just eyeball it.

Part D

Is a chi-square test for independence appropriate here?

No. A chi-square test for independence is an inference procedure — its purpose is to generalize a pattern seen in a sample to a larger population. Here the table already contains data on all 4,193 professional athletes (the entire population), not a sample. Since we already know the true relationship for the whole population, there's nothing left to infer — we can just compute the exact association directly from the table, with no need for a significance test.

Beginner tip: Whenever an AP Stats question says "for all ___" or gives you the full population, that's your cue to say inference isn't appropriate — significance tests exist to handle the uncertainty of using a sample, and population data has no such uncertainty.

Key concepts — Q5

Marginal probability (row/column totals) vs. conditional probability ("given")
Joint probability = P(A) × P(B|A), shown as area in a mosaic plot
Mutually exclusive ≠ independent — check both separately
Inference (like chi-square tests) requires a sample, not a full population

Question 6 · 15 points

Regression, CI & Prediction Intervals (Baseball)

Unit 2 & 9 · Regression Inference

Scatterplot of Runs vs. Hits for 30 MLB teams. Regression line: Predicted Runs = −372.2 + 0.823(Hits)

Part A

i. Describe the relationship. ii. Predict runs for a team with 1,250 hits.

i. The scatterplot shows a positive, moderately strong to strong, roughly linear relationship between number of hits and number of runs — teams with more hits tend to score more runs.

ii. Predicted Runs = −372.2 + 0.823(1,250) = −372.2 + 1,028.75

Predicted runs ≈ 656.55 runs

Beginner tip: Always describe scatterplots with three things: direction (positive/negative), form (linear/curved), and strength (weak/moderate/strong) — plus mention the context (hits and runs), not just "x and y."

Part B

i. Compare point A to other teams in its salary group. ii. Compare correlation strength between the two salary groups.

i. This part is read directly off Figure 2, so do this visually: find point A, identify its marker (dot = above median salary, square = below median salary), then look at the other points with the same marker that have a similar number of hits. Compare A's number of runs to theirs — is A noticeably higher, lower, or about the same as other teams in its salary group with comparable hits?

ii. For this part, look separately at just the dots (above-median salary) and just the squares (below-median salary). For each group, judge how tightly the points cluster around a straight line: tighter clustering = stronger linear relationship, more scatter = weaker linear relationship. Compare the two patterns to decide which group's relationship is stronger, weaker, or similar.

Because this part depends on precisely reading dot/square positions off the printed scatterplot, work through it using your own copy of Figure 2 — the reasoning method above is exactly what the AP rubric rewards, regardless of which specific answer (stronger/weaker/similar) the actual image supports.

Beginner tip: For "compare a single point to a group" questions, always compare within the same classification only — don't mix dots and squares together.

Part C

i. Find t* for 95% confidence (df = 28). ii. Build the 95% CI for the mean. iii. Build the 95% PI for one team.

i. df = n − 2 = 30 − 2 = 28. From the t-table at 95% confidence: t* = 2.05 (2.048 more precisely).

ii. Confidence interval (mean runs for ALL teams with 1,250 hits). 656.55 ± 2.048(17.48) = 656.55 ± 35.80

95% CI: (620.75, 692.35) runs

iii. Prediction interval (runs for ONE team with 1,250 hits). 656.55 ± 2.048(56.78) = 656.55 ± 116.28

95% PI: (540.27, 772.83) runs

Beginner tip: Same point estimate (656.55) and same t*, but the prediction interval uses a much bigger standard error — that's the whole story of Part D below.

Part D

i. Do sample means vary more or less than individual observations? ii. Why is the PI wider than the CI?

i. A distribution of sample means has less variability than a distribution of individual observations. Averaging values smooths out the extremes — individual observations can be far from the true mean, but averages of many observations tend to cluster closer to it.

ii. The prediction-interval formula has an extra s² added inside the square root, compared with the confidence-interval formula. That extra term accounts for the added variability of a single, individual team's runs around the regression line — on top of the uncertainty in estimating the mean itself. Because the confidence interval only needs to account for uncertainty in the mean (which has less variability, per part i), while the prediction interval must also account for the extra spread of one individual observation, the prediction interval's standard error — and therefore its width — is larger.

The PI (540.27–772.83) is wider than the CI (620.75–692.35) because predicting one team's outcome carries the individual-observation variability on top of the estimation uncertainty that the CI alone captures.

Beginner tip: Think of it as CI = "how sure am I about the average?" vs. PI = "how sure am I about one specific team?" One team is always harder to pin down than the average of many, so PI > CI in width, always.

Key concepts — Q6

Describe scatterplots: direction, form, strength, context
t* depends on confidence level and df = n − 2 for regression
CI estimates a mean; PI predicts a single value — PI is always wider
Averaging reduces variability (this is why CI < PI in width)