AP Statistics — comparing two means with t procedures

This material trains the second half of Unit 4 of the AP Statistics course framework for 2026–27: comparing two population means with two-sample t procedures. It is written for students preparing for the May 2027 exam at introductory college level, with a graphing calculator at hand.

The quiz takes each skill in turn. You find the mean and standard deviation of the sampling distribution of a difference of sample means (variances add, standard deviations do not) and use it for a probability; choose a two-sample t-interval for a study with two independent samples; see which condition is not needed when the data come from a randomized experiment and when a small sample with an outlier breaks the sample data condition; compute a two-sample t-interval with the degrees of freedom given by technology; read an interval that contains zero; write hypotheses for a one-sided comparison; compute a two-sample t statistic; recognize which reported degrees of freedom are possible; state a conclusion for a large p-value without claiming the means are equal; and decide what a randomized experiment with volunteers allows you to conclude. Explanations show the working and say why each wrong answer is wrong.

The flashcards summarize the formulas, the degrees-of-freedom bounds, the three conditions for samples and for experiments, the difference between two-sample and paired designs, and the scope of conclusions.

The written work is a printable sheet of eight free-response tasks modeled on the exam's inference question and on the multi-part questions that combine data collection with inference: a complete two-sample t-test for an experiment, a complete interval for two regions, choosing between paired and two-sample procedures for three designs, a probability for a difference of sample means, conditions for experiments versus samples, conservative and technology degrees of freedom, a p-value with a possible Type I error, and the scope of conclusions for four designs. Handwritten answers are photographed and checked against model answers and key points.

The oral exam has an examiner ask one question at a time about a comparison of two groups, from recognizing the design to computing, interpreting and stating the scope of the conclusion, and it ends with short, precise feedback.

One-sample and paired t procedures are covered by a separate material of this category. All scenarios are invented, and all questions are original practice items written by Zestly, not released exam questions. The material follows the published course framework but is independent practice and does not predict an exam score.

  • Describe the sampling distribution of a difference between two sample means and use it for probabilities
  • Choose between two-sample and paired t procedures from the design of a study
  • Check the randomization, 10% and sample data conditions for two samples and for randomized experiments
  • Construct and interpret a two-sample t-interval and use it to judge whether population means differ
  • Set up and carry out a two-sample t-test, with degrees of freedom from technology or the conservative rule
  • State conclusions with the correct scope: cause and effect from random assignment, generalization from random sampling

Practice material written by Zestly, based on the College Board AP Statistics course framework effective fall 2026 (Unit 4, Inference for Quantitative Data: Means, topics 4.6–4.10). Original questions, not released exam items.

Sample question

In a large city, the time residents spend commuting each day has mean 68 minutes and standard deviation 10 minutes in the northern district, and mean 62 minutes and standard deviation 8 minutes in the southern district. Independent random samples of 50 northern and 40 southern residents are taken. What are the mean and standard deviation of the sampling distribution of $\bar{x}_N - \bar{x}_S$?

See the answer

Mean 6 minutes, standard deviation about 1.90 minutes

The mean of $\bar{x}_N - \bar{x}_S$ is $\mu_N - \mu_S = 68 - 62 = 6$. For independent samples the VARIANCES add: $\sqrt{\frac{10^2}{50} + \frac{8^2}{40}} = \sqrt{2 + 1.6} = \sqrt{3.6} \approx 1.90$. The value 3.60 is the variance, 2.68 adds the two standard deviations $10/\sqrt{50} + 8/\sqrt{40}$ instead of the variances, and 18 adds the population standard deviations.

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