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AP Statistics

College Board · 5 units · exam-weighted · high school

AP Statistics: every unit with its share of the exam, as College Board publishes it, with practice questions per unit and a countdown to the exam.

Set up AP Statistics → One set per unit, 5 of them pre-written and answer-checked. Free, no account needed.

Units and exam weighting

Unit 1: Exploring One-Variable Data and Collecting Data 20–30% of the exam

Categorical and quantitative variables, graphical and numerical summaries, normal distributions, sampling methods, bias, experimental design.

3 of 14 questions in this unit, with the worked answer

A city collects data on residents' zip codes to determine service districts. Even though it is recorded as a number, zip code is best classified as what type of variable?

Answer. Categorical variable

Zip codes are labels that identify a location or group; averaging them would be meaningless, which is the defining test for a categorical variable regardless of its numeric appearance.

The number of pets owned by each student in a class is recorded. What type of variable is this, and why?

Answer. Quantitative discrete, because it is a countable numeric value (0, 1, 2, ...) with meaningful arithmetic, and it can only take whole-number values.

Quantitative variables measure or count something; since the number of pets is a whole-number count rather than a measurement on a continuous scale, it is discrete rather than continuous.

Exam scores are approximately normal with mean 75 and standard deviation 5. According to the Empirical Rule, what percent of scores fall between 65 and 85?

Answer. 95%

65 and 85 are each exactly 2 standard deviations from the mean (75 ± 2×5), and the Empirical Rule states about 95% of values in a normal distribution fall within 2 standard deviations of the mean.

All 14 questions for Unit 1, with the working →

What people get wrong

  1. Dividing by n instead of n-1 when computing a sample standard deviation by hand. Instead, use $n-1$ (or trust the calculator's Sx, not σx, for sample data) since dividing by n underestimates spread.
  2. Concluding causation from an observational study (e.g., 'vitamin use causes lower heart disease rates'). Instead, state only that the variables are associated and note a plausible confounding variable, since causal claims require random assignment.
  3. Confusing stratified and cluster sampling. Instead, remember that stratified sampling takes a random sample from every group (each stratum is represented), while cluster sampling randomly selects a few whole groups and surveys everyone within them.
  4. Reporting the mean and standard deviation for a skewed distribution or one with outliers. Instead, check the shape first (from a graph or the mean-vs-median comparison) and report the median and IQR when the data are not roughly symmetric.

Unit 2: Probability, Random Variables, and Probability Distributions 15–25% of the exam

Probability rules, conditional probability and independence, random variables, binomial and geometric distributions, sampling distributions.

3 of 14 questions in this unit, with the worked answer

At a school, 40% of students play a sport, 30% play an instrument, and 15% do both. What is the probability a randomly chosen student plays a sport or an instrument?

Answer. 0.55

Use the general addition rule since the events overlap: P(sport or instrument) = P(sport) + P(instrument) − P(both) = 0.40 + 0.30 − 0.15 = 0.55. Subtracting the overlap avoids double-counting students who do both.

In a survey of 200 students, 120 are girls and 80 are boys. Of the girls, 90 passed an exam; of the boys, 50 passed. What is P(passed | girl)?

Answer. 0.75

Conditional probability restricts attention to the given group. P(pass | girl) = (number of girls who passed) / (total girls) = 90/120 = 0.75.

Using the same survey (200 students, 120 girls with 90 passing, 80 boys with 50 passing), are 'being a girl' and 'passing the exam' independent? Justify with numbers.

Answer. No, they are not independent

Overall P(pass) = 140/200 = 0.70. But P(pass | girl) = 90/120 = 0.75. Since 0.75 ≠ 0.70, knowing a student is a girl changes the probability of passing, so the events are dependent.

All 14 questions for Unit 2, with the working →

What people get wrong

  1. Writing P(A|B) when the problem gives P(B|A), or vice versa — conditional probability is not symmetric. Always identify which event is the 'given' condition before dividing.
  2. Assuming two events are independent because they seem unrelated in context. Instead, compute P(A), P(B), and P(A and B) (or P(B|A)) and check the independence equation numerically.
  3. Using the simple addition rule P(A)+P(B) for events that overlap. Instead, always subtract P(A and B) unless the problem explicitly states or a Venn diagram confirms the events are mutually exclusive.
  4. Applying the binomial formula to a without-replacement scenario drawn from a small population, where trials are not actually independent. Instead, check that the population is much larger than the sample (a common rule of thumb is at least 10 times the sample size) before treating it as binomial.

Unit 3: Inference for Categorical Data: Proportions 15–25% of the exam

Confidence intervals and significance tests for one and two proportions, errors and power, chi-square tests.

3 of 13 questions in this unit, with the worked answer

What three conditions must be verified before constructing a confidence interval for a single proportion?

Answer. Random, 10%, and Large Counts.

Random ensures no bias in how data were collected; the 10% condition (n ≤ 0.10N) keeps trials approximately independent when sampling without replacement; Large Counts requires $n\hat p \ge 10$ and $n(1-\hat p)\ge 10$ so the normal approximation is valid.

In a class of 30 students, 18 support extending the lunch period. Construct a 95% confidence interval for the true proportion of all students who would support this.

Answer. Approximately (0.425, 0.775).

$\hat p = 18/30 = 0.6$. SE $=\sqrt{0.6(0.4)/30}=0.0894$. Margin of error $=1.96(0.0894)=0.175$. Interval: $0.6\pm0.175 = (0.425, 0.775)$.

A 95% confidence interval for the proportion of adults who favor a new policy is (0.42, 0.58). Interpret this interval.

Answer. We are 95% confident that the true proportion of all adults who favor the policy is between 0.42 and 0.58.

The interpretation must reference the population parameter and the specific context, not just repeat the numbers.

All 13 questions for Unit 3, with the working →

What people get wrong

  1. Using the pooled proportion when constructing a two-proportion confidence interval. The pooled proportion is only for the significance test of H0: p1 = p2; confidence intervals must use each sample's own $\hat p_1$ and $\hat p_2$ in the standard error.
  2. Checking the Large Counts condition using n alone instead of the actual expected successes and failures. Always compute $n\hat p$, $n(1-\hat p)$ (or expected counts in a table) and confirm each is at least 10 for proportions or 5 for chi-square before proceeding.
  3. Interpreting a confidence interval as 'there is a 95% probability the true proportion is in this interval.' The parameter is fixed, not random — the correct interpretation is about the long-run percentage of intervals from repeated sampling that would capture the true value.
  4. Mixing up degrees of freedom formulas — using (r−1)(c−1) for a goodness-of-fit test or k−1 for a two-way table. Match the formula to the test: k−1 for goodness of fit, (r−1)(c−1) for independence or homogeneity.

Unit 4: Inference for Quantitative Data: Means 10–20% of the exam

t-distributions, confidence intervals and tests for one mean, paired data and two means, conditions and interpretation.

3 of 13 questions in this unit, with the worked answer

A random sample of 25 college students has a mean study time of 15.2 hours/week with sample standard deviation 3.4 hours. Construct a 95% confidence interval for the true mean study time (t* = 2.064, df = 24).

Answer. (13.80, 16.60) hours

SE = 3.4/√25 = 0.68. Margin of error = 2.064 × 0.68 ≈ 1.40. CI = 15.2 ± 1.40, giving approximately (13.80, 16.60) hours.

Interpret the confidence interval (13.80, 16.60) hours from the study-time problem.

Answer. We are 95% confident that the true mean weekly study time of all college students (in the population sampled) is between 13.80 and 16.60 hours.

A correct CI interpretation always refers to the population parameter, stated in context, not to individual students or to a probability about this specific interval.

A cereal box claims a mean weight of 16 oz. A quality control manager weighs a random sample of 20 boxes and finds mean 15.7 oz, s = 0.6 oz. Test at α = 0.05 whether the true mean weight differs from 16 oz. Give the test statistic and conclusion (df = 19, two-tailed critical t ≈ 2.093).

Answer. t ≈ −2.24, p ≈ 0.037 < 0.05, so reject H0: there is evidence the true mean weight differs from 16 oz.

SE = 0.6/√20 ≈ 0.1342. t = (15.7−16)/0.1342 ≈ −2.236. Since |t| = 2.236 > 2.093 (or p ≈ 0.037 < 0.05), reject the null hypothesis.

All 13 questions for Unit 4, with the working →

What people get wrong

  1. Using z* instead of t* whenever σ is unknown (which is almost always). Instead, always use the t-distribution with the correct degrees of freedom when estimating the standard deviation from the sample.
  2. Treating two independent samples as if they should use a pooled-variance formula. AP Statistics convention is to always use the unpooled two-sample t procedure for means, not the pooled t-test taught in some other courses.
  3. Confusing paired and independent designs — using the two-sample formula on data that are actually paired (same subjects measured twice) inflates the standard error and gives wrong conclusions. Instead, recognize paired designs and reduce to differences first.
  4. Interpreting a confidence interval as 'there is a 95% probability the true mean is in this specific interval' or '95% of the data lies in this interval.' Instead, state that in repeated sampling, about 95% of intervals constructed this way would capture the true population mean.

Unit 5: Regression Analysis 10–20% of the exam

Scatterplots, correlation, least-squares regression, residuals, transformations, inference for the slope.

3 of 14 questions in this unit, with the worked answer

A scatterplot of engine size (liters) vs highway mpg shows a consistent downward linear pattern with little scatter, except one hybrid car with a small engine and unusually high mpg. Describe the association and this car's role in the data.

Answer. The association is negative, linear, and fairly strong; the hybrid car is an outlier because it deviates sharply from the overall pattern (and could be influential if it pulls the slope toward zero).

Describing a scatterplot requires stating form, direction, strength, and outliers. A point that departs from the pattern in y for its x-value is an outlier; whether it's influential depends on its effect on the fitted line.

The correlation between height in inches and weight in pounds for a group of adults is r = 0.82. If height is converted to centimeters, what is the new correlation?

Answer. 0.82 (unchanged)

Correlation is invariant under linear transformations of either variable, including unit conversions, because r is a standardized measure of linear association.

In simple linear regression, r² = 0.72 for a model predicting plant height from monthly rainfall. Which is the correct interpretation?

Answer. 72% of the variation in plant height is explained by the linear relationship with rainfall.

r² is always interpreted as a percentage of variation in the response explained by the explanatory variable, not as the correlation itself or as prediction accuracy for individual points.

All 14 questions for Unit 5, with the working →

What people get wrong

  1. Interpreting r² as if it were r (e.g. saying 'r² = 0.72 means a strong correlation of 0.72'). Instead, always state r² as a percentage of variation explained, and square-root it (keeping the sign of the slope) if you actually need r.
  2. Extrapolating the regression line beyond the range of observed x-values and treating the prediction as reliable. Instead, explicitly flag predictions outside the data range as extrapolation and note the relationship may not hold there.
  3. Interpreting slope or intercept without units or context, e.g. 'y increases by 3.85' instead of 'exam score increases by 3.85 points, on average, for each additional hour studied.' The AP rubric requires context in every interpretation.
  4. Running a t-test or CI for the slope without first checking the residual plot for linearity and constant spread. Instead, state explicitly that the residual plot shows no pattern and roughly constant spread before proceeding to inference.

How it works

Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Readiness is weighted the way the exam is, so a weak unit that carries 18% of the score pulls harder than one that carries 6%. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.

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