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AP Statistics mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 25 questions and 4 free-response questions · about 140 minutes

A full-length practice test built from our exam-style questions on every unit of AP Statistics. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page
  1. Before you start
  2. Section I
  3. Section II
  4. After the mock

Before you start

Section I: 25 exam-style questions

Q1Exam-style
Test scores have mean 70 and standard deviation 8. Find the z-score of a score of 82.

Hint

z = (x − mean)/SD.

Worked solution
  1. z = (82 − 70)/8 = 12/8.
  2. z = 1.5: the score is 1.5 standard deviations above the mean.
Q2Exam-style
Adult heights are approximately normal with mean 170 cm and standard deviation 6 cm. What proportion are taller than 182 cm?

Hint

Standardise, then use the normal table or normalcdf.

Worked solution
  1. z = (182 − 170)/6 = 2.
  2. P(Z > 2) = 1 − 0.9772.
  3. Proportion = 0.0228, about 2.3%.
Q3Exam-style
A researcher divides a city into neighbourhoods by income level, then takes a simple random sample from each income level. What kind of sample is this?

Hint

Is every group sampled, or are whole groups chosen?

Worked solution
  1. The population is split into groups (strata) of similar individuals, and a random sample is taken from every group.
  2. That is a stratified random sample: B.
Q4Exam-style
SAT-style scores are approximately normal with mean 500 and standard deviation 100. What score is at the 90th percentile?

Hint

Find the z-score with 0.90 of the area below it, then unstandardise.

Worked solution
  1. invNorm(0.90) gives z = 1.2816.
  2. x = 500 + 1.2816 × 100 = 628.
Q5Challenge
An observational study finds that people who drink more coffee live longer. Why can't we conclude that coffee causes a longer life?

Hint

What does an experiment have that an observational study does not?

Worked solution
  1. Without random assignment of coffee drinking, coffee drinkers may differ in other ways that affect lifespan.
  2. Those confounding variables mean we cannot conclude cause and effect: B.
Q6Exam-style
In a school, 30% of students play a sport, and 12% play a sport and play an instrument. What is the probability that a student who plays a sport also plays an instrument?

Hint

This is a conditional probability: given that they play a sport.

Worked solution
  1. P(instrument | sport) = P(both)/P(sport).
  2. = 0.12/0.30 = 0.4.
Q7Exam-style
A basketball player makes 30% of her three-point attempts, independently. In 10 attempts, what is the probability she makes exactly 3?

Hint

Binomial with n = 10 and p = 0.3.

Worked solution
  1. P(X = 3) = 10C3(0.3)3(0.7)7 = 120 × 0.027 × 0.0824.
  2. P(X = 3) = 0.267.
Q8Exam-style
Each of 5 independent light bulbs is faulty with probability 0.2. What is the probability that at least one is faulty?

Hint

Use the complement: 1 − P(none faulty).

Worked solution
  1. P(none faulty) = 0.85 = 0.328.
  2. P(at least one) = 1 − 0.328 = 0.672.
Q9Exam-style
X is binomial with n = 100 and p = 0.4. Find the standard deviation of X.

Hint

σ = √(np(1 − p)).

Worked solution
  1. σ = √(100 × 0.4 × 0.6) = √24.
  2. σ = 4.90.
Q10Challenge
A population has mean 50 and standard deviation 12. For random samples of size 36, find the probability that the sample mean is greater than 53.

Hint

Find the standard deviation of the sample mean first. With n = 36 the central limit theorem lets you use a normal model.

Worked solution
  1. σx̄ = 12/√36 = 2.
  2. z = (53 − 50)/2 = 1.5.
  3. P(x̄ > 53) = P(Z > 1.5) = 0.0668.
Q11Exam-style
Using the sample of 150 students with p̂ = 0.3, find the margin of error for a 95% confidence interval.

Hint

Margin = z* × SE, with z* = 1.96.

Worked solution
  1. SE = 0.0374.
  2. Margin = 1.96 × 0.0374 = 0.0733.
Q12Exam-style
A coin is tossed 200 times and lands heads 112 times. Find the z test statistic for H0: p = 0.5.

Hint

Use p0 = 0.5 in the standard error, not p̂.

Worked solution
  1. p̂ = 0.56. SE = √(0.5 × 0.5/200) = 0.0354.
  2. z = (0.56 − 0.5)/0.0354 = 1.70.
Q13Exam-style
For that test, with Ha: p > 0.5, find the p-value.

Hint

The p-value is the area to the right of the test statistic.

Worked solution
  1. P(Z > 1.70).
  2. p-value = 0.045.
Q14Exam-style
A test gives a p-value of 0.03. Which is the correct interpretation?

Hint

A p-value is calculated assuming H0 is true.

Worked solution
  1. The p-value is the probability, assuming H0, of a statistic at least as extreme as the one observed.
  2. The answer is C.
Q15Challenge
In two random samples of 100, 60 of the first group and 45 of the second support a proposal. Find the z statistic for H0: p1 = p2.

Hint

Use the pooled proportion in the standard error.

Worked solution
  1. Pooled p̂ = (60 + 45)/200 = 0.525.
  2. SE = √(0.525 × 0.475 × (1/100 + 1/100)) = 0.0706.
  3. z = (0.60 − 0.45)/0.0706 = 2.12.
Q16Exam-style
A random sample of 36 bags of rice has mean mass 52 g and standard deviation 6 g. Find the t statistic for H0: μ = 50.

Hint

t = (x̄ − μ0)/(s/√n).

Worked solution
  1. SE = 6/√36 = 1.
  2. t = (52 − 50)/1 = 2.
Q17Exam-style
A random sample of 16 has standard deviation 8. Find the margin of error of a 95% t interval for the mean.

Hint

t* has 15 degrees of freedom.

Worked solution
  1. SE = 8/√16 = 2. With df = 15, t* = 2.131.
  2. Margin = 2.131 × 2 = 4.26.
Q18Exam-style
A two-sided t test has t = 2.0 with 35 degrees of freedom. Find the p-value.

Hint

Two-sided: double the tail area.

Worked solution
  1. P(t > 2.0) with df = 35 is about 0.0267.
  2. p-value = 2 × 0.0267 = 0.053.
Q19Exam-style
Twenty students each take a test before and after a revision course. Which procedure compares the scores?

Hint

Are the two sets of scores independent, or linked student by student?

Worked solution
  1. Each student gives two linked measurements, so analyse the 20 differences.
  2. That is a paired t test: B.
Q20Challenge
Group 1: mean 78, SD 10, n = 25. Group 2: mean 72, SD 12, n = 30. Find the two-sample t statistic for the difference in means.

Hint

SE = √(s12/n1 + s22/n2).

Worked solution
  1. SE = √(100/25 + 144/30) = √(4 + 4.8) = 2.966.
  2. t = (78 − 72)/2.966 = 2.02.
Q21Exam-style
For two variables, r = 0.8, sx = 4 and sy = 10. Find the slope of the least-squares regression line.

Hint

b = r(sy/sx).

Worked solution
  1. b = 0.8 × 10/4.
  2. b = 2.
Q22Exam-style
The same data have x̄ = 20 and ȳ = 50. Find the intercept of the regression line.

Hint

The line passes through (x̄, ȳ).

Worked solution
  1. a = ȳ − b x̄ = 50 − 2 × 20.
  2. a = 10.
Q23Exam-style
The correlation between two variables is −0.6. What fraction of the variation in y is explained by the regression on x?

Hint

Square r.

Worked solution
  1. r2 = (−0.6)2 = 0.36, so 36%.
Q24Exam-style
A regression of height on age uses children aged 2 to 10. Why is it unwise to use it to predict the height of a 30-year-old?

Hint

Is 30 inside the range of ages used to build the line?

Worked solution
  1. The line was fitted for ages 2 to 10; growth stops in adulthood, so the pattern does not continue.
  2. Predicting at age 30 is extrapolation: B.
Q25Challenge
Find the correlation r for the five points (1, 2), (2, 4), (3, 5), (4, 4), (5, 5).

Hint

Use your calculator's linear regression, or r = Σ(zxzy)/(n − 1).

Worked solution
  1. x̄ = 3, ȳ = 4. Σ(x − x̄)(y − ȳ) = 6, Σ(x − x̄)2 = 10, Σ(y − ȳ)2 = 6.
  2. r = 6/√(10 × 6) = 6/7.746.
  3. r = 0.775.

Section II: 4 free-response questions

Free-response question 1: Exploring one-variable data and collecting data

Experimental design style · 10 points · about 22 minutes

A teacher wants to know whether a new revision app improves test scores more than revising from notes. Forty students volunteer. Their scores on last term's test vary widely.

  1. (a) Identify the treatments and the response variable.
  2. (b) Describe how to carry out a completely randomised design with the 40 students.
  3. (c) Explain why blocking by last term's score might be better, and describe how to do it.
  4. (d) Can the results be generalised to all students at the school? Explain.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Treatments: revising with the app, and revising from notes. Response: the score on the next test.

Part (b)

Number the students 1 to 40. Use a random number generator to pick 20 different numbers: those students use the app, the other 20 revise from notes. Compare mean test scores.

Part (c)

Last term's score is likely related to the response, so blocking reduces variability. Pair students with similar scores, and randomly assign one of each pair to the app.

Part (d)

No. The students volunteered, so they may not represent all students; there was no random sample from the school.

Your mark: 0 of 10

Free-response question 2: Probability, random variables and probability distributions

Probability and random variables style · 10 points · about 22 minutes

At a school fair, a game costs $2 to play. A player spins a fair spinner with 10 equal sections: 1 section wins $10, 3 sections win $2, and the rest win nothing. Let W be the player's net gain from one game.

  1. (a) Write down the probability distribution of W.
  2. (b) Find E(W) and interpret it in context.
  3. (c) A player plays 8 games. Find the probability that they win the $10 prize at least once.
  4. (d) Is the number of games until the first $10 win a binomial random variable? Explain.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

W = 8 with probability 0.1; W = 0 with probability 0.3; W = −2 with probability 0.6.

Part (b)

E(W) = 8(0.1) + 0(0.3) − 2(0.6) = −0.4. Over many games, a player loses an average of 40 cents a game.

Part (c)

Binomial, n = 8, p = 0.1. P(at least one) = 1 − 0.98 = 0.570.

Part (d)

No. A binomial variable has a fixed number of trials; here the number of trials is not fixed.

Your mark: 0 of 10

Free-response question 3: Inference for quantitative data: means

Inference for means style · 10 points · about 22 minutes

A coach records the 100 m times of 10 randomly selected sprinters before and after a six-week training programme. The mean of the differences (before minus after) is 0.18 seconds, with standard deviation 0.20 seconds.

  1. (a) State the hypotheses for testing whether the programme reduces times, defining the parameter.
  2. (b) Name the test and check its conditions.
  3. (c) Calculate the test statistic and p-value.
  4. (d) State a conclusion in context at the 5% level.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

μd = the true mean difference (before minus after) for all such sprinters. H0: μd = 0; Ha: μd > 0.

Part (b)

Paired t test. Random: random selection of sprinters. 10%: 10 is less than 10% of all sprinters. Normal: n is small, so check that a graph of the differences shows no strong skew or outliers.

Part (c)

t = 0.18/(0.20/√10) = 0.18/0.0632 = 2.85, df = 9. p-value = P(t > 2.85) = 0.0096.

Part (d)

Since 0.0096 < 0.05, reject H0. There is convincing evidence that the programme reduces 100 m times on average for sprinters like these.

Your mark: 0 of 10

Free-response question 4: Regression analysis

Regression style · 10 points · about 22 minutes

A café records the daily high temperature x (°C) and the number of iced drinks sold y on 20 randomly chosen days. Computer output gives the least-squares line ŷ = −12.5 + 4.2x, with r = 0.91 and s = 6.3.

  1. (a) Interpret the slope in context.
  2. (b) Predict sales on a day with a high of 25°C, and find the residual if 96 iced drinks were sold.
  3. (c) Calculate and interpret r2.
  4. (d) Interpret s = 6.3 in context.
  5. (e) The café manager says hot weather causes more iced drink sales. Can the data support this? Explain.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

For each additional 1°C in daily high temperature, the predicted number of iced drinks sold increases by about 4.2.

Part (b)

ŷ = −12.5 + 4.2(25) = 92.5. Residual = 96 − 92.5 = 3.5.

Part (c)

r2 = 0.912 = 0.828. About 82.8% of the variation in daily iced drink sales is explained by the linear relationship with daily high temperature.

Part (d)

The actual number of iced drinks sold typically differs from the number predicted by the line by about 6.3 drinks.

Part (e)

Not from these data alone: this is an observational study, so other variables (such as weekends or school holidays) may be confounded with temperature.

Your mark: 0 of 10

After the mock

Note every question you missed and its unit, then go back to that unit's practice page: all AP Statistics units. Then try the real thing with official AP Statistics past papers.