AP Statistics · Unit 4 · 10-20% of the exam · ~18 class periods

Inference for Quantitative Data: Means

AP Statistics Unit 4 covers inference for means: sampling distributions of sample means, the t-distribution, and one-sample, matched-pairs, and two-sample t confidence intervals and tests. It is 10 to 20 percent of the multiple-choice section and about 18 class periods.

AP Statistics: Unit 4 (topics 4.1 Sampling Distributions for Sample Means, 4.2 Constructing a Confidence Interval for a Population Mean or Population Mean Difference, 4.3 Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference, 4.4 Setting Up a Test for a Population Mean or Population Mean Difference, 4.5 Carrying Out a Test for a Population Mean or Population Mean Difference, 4.6 Sampling Distributions for the Difference Between Two Sample Means, 4.7 Constructing a Confidence Interval for the Difference Between Two Population Means, 4.8 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means, 4.9 Setting Up a Test for the Difference Between Two Population Means, 4.10 Carrying Out a Test for the Difference Between Two Population Means). Unit 4 (Inference for Quantitative Data: Means) is 10 to 20 percent of the multiple-choice section and about 18 class periods in the Fall 2026 redesigned course, covering one-sample, matched-pairs, and two-sample t procedures.

What Unit 4 covers

Unit 4 is where you do inference for quantitative data, meaning numerical measurements whose center is a mean. It builds directly on the ideas you used for proportions in Unit 3, but the parameter is now a population mean, written μ\mu (the Greek letter mu), instead of a proportion.

The College Board weights Unit 4 at 10 to 20 percent of the multiple-choice section and allots about 18 class periods. There are 10 topics, split into two halves: one-sample and matched-pairs procedures (topics 4.1 to 4.5), then two-sample procedures (topics 4.6 to 4.10).

The big change from Unit 3 is that you almost never know the population standard deviation σ\sigma (sigma) for a quantitative variable. You estimate it with the sample standard deviation ss, and that forces you to use the t-distribution instead of the normal z-distribution. See t-test vs z-test for the full contrast.

The 10 topics in Unit 4

  • 4.1 Sampling distributions for sample means. Find the mean and standard deviation of the sampling distribution of the sample mean xˉ\bar{x} (x-bar), and check when it is approximately normal.
  • 4.2 Constructing a confidence interval for a population mean or mean difference. Describe t-distributions and build a one-sample t-interval, including for matched pairs.
  • 4.3 Justifying a claim based on that interval. Interpret the interval and the confidence level, and relate sample size, width, and margin of error.
  • 4.4 Setting up a test for a population mean or mean difference. Choose the one-sample t-test, state hypotheses about μ\mu or the population mean difference μd\mu_d, and check conditions.
  • 4.5 Carrying out that test. Compute the t-statistic and p-value, then decide by comparing the p-value to the significance level α\alpha (alpha).
  • 4.6 Sampling distributions for the difference between two sample means. Find the mean and standard deviation of xˉ1xˉ2\bar{x}_1 - \bar{x}_2.
  • 4.7 Constructing a confidence interval for the difference between two population means. Build a two-sample t-interval.
  • 4.8 Justifying a claim based on that interval. Interpret it, and use whether the interval contains 0.
  • 4.9 Setting up a test for the difference between two population means. Choose the two-sample t-test and state the hypotheses.
  • 4.10 Carrying out that test. Compute the two-sample t-statistic and p-value and conclude in context.

Sampling distributions for sample means

For a population with mean μ\mu and standard deviation σ\sigma, when the sampled values are independent, the sampling distribution of the sample mean xˉ\bar{x} has mean μxˉ=μ\mu_{\bar{x}} = \mu and standard deviation σxˉ=σ/n\sigma_{\bar{x}} = \sigma/\sqrt{n}, where nn is the sample size.

That standard deviation shrinks as nn grows, which is why larger samples give more precise estimates. The central limit theorem tells you the shape. If the population is normal, xˉ\bar{x} is normal for any nn. If the population is not normal, xˉ\bar{x} is approximately normal once n30n \ge 30, and a strongly skewed population may need a sample much larger than 30.

You can watch this happen in the sampling distribution interactive.

Why means use the t-distribution

Because you rarely know σ\sigma, you replace it with the sample standard deviation ss. The quantity s/ns/\sqrt{n} is the standard error of the mean, an estimate of the true standard deviation σ/n\sigma/\sqrt{n}. Those two ideas trip up many students, so read standard error vs standard deviation.

Using ss adds extra uncertainty, so the standardized statistic follows a t-distribution rather than the standard normal. t-distributions are symmetric and bell-shaped with heavier tails than the normal curve. Each one is set by its degrees of freedom (df); for one sample, df=n1df = n - 1. As df grows, the t-distribution approaches the standard normal, and you look up critical values on the t-table.

One-sample and matched-pairs t procedures

A one-sample t-interval estimates μ\mu as xˉ±tsn\bar{x} \pm t^* \dfrac{s}{\sqrt{n}}, where tt^* is the critical value for the central C percent of a t-distribution with n1n - 1 degrees of freedom. The one-sample t-test uses the statistic

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}

where μ0\mu_0 is the value of the mean claimed by the null hypothesis.

Matched pairs, for example a pre-test and post-test on the same students, are handled by taking the difference within each pair to get one sample of differences. You then run the same one-sample procedures on those differences, estimating the population mean difference μd\mu_d. The one-sample t-test calculator runs both the interval and the test.

Two-sample t procedures

When you compare two independent groups, the parameter is the difference between population means μ1μ2\mu_1 - \mu_2. The sampling distribution of xˉ1xˉ2\bar{x}_1 - \bar{x}_2 has standard deviation σ12n1+σ22n2\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}, and its standard error replaces each σ\sigma with the matching sample standard deviation ss.

The two-sample t-interval is (xˉ1xˉ2)±ts12n1+s22n2(\bar{x}_1 - \bar{x}_2) \pm t^* \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}, and the two-sample t-test uses

t=(xˉ1xˉ2)0s12n1+s22n2.t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}.

The degrees of freedom fall between n1+n22n_1 + n_2 - 2 and the smaller of n11n_1 - 1 and n21n_2 - 1, and on the exam you find them with technology. If a two-sample interval contains 0, you do not have convincing evidence of a difference. The two-sample t-test calculator handles the arithmetic.

Conditions to check before any t procedure

Every t procedure in Unit 4 asks you to verify conditions before you trust the result, and you should state them explicitly on the free-response section.

  • Randomization: the data come from a random sample or a randomized experiment, using two independent random samples for two-sample procedures.
  • 10 percent: when sampling without replacement, each sample is at most 10 percent of its population, written n0.10Nn \le 0.10N, where NN is the population size.
  • Sample data (normality): the population is stated to be approximately normal, or n30n \ge 30. If n<30n < 30, the sample should be free of strong skewness and outliers. For matched pairs, apply this to the differences.

State your hypotheses in terms of population parameters, not sample statistics, and write the conclusion in context by comparing the p-value to α\alpha.

How Unit 4 fits the exam

Unit 4 mirrors Unit 3 step for step, so the College Board expects you to keep the procedures straight and pick the right one from the wording of a problem. A question asking whether data give convincing evidence of something is asking for a significance test, not a description.

Free-response question 3 is always a full inference problem, a hypothesis test or a confidence interval, and Unit 4 procedures are fair game there. A graphing calculator with statistical capabilities is expected, and a formula sheet and tables are provided. For the wider picture see the AP Statistics hub and the official AP Statistics course page.

95% confidence interval for a mean (one sample)

A random sample of 6 phones of one model runs for 8, 10, 12, 14, 9, and 13 hours on a full charge. Assume battery life is approximately normal. Construct a 95% confidence interval for the mean battery life μ\mu.

  1. Find the sample mean: xˉ=8+10+12+14+9+136=666=11\bar{x} = \frac{8 + 10 + 12 + 14 + 9 + 13}{6} = \frac{66}{6} = 11 hours.

  2. Find the sample standard deviation. The deviations from 11 are -3, -1, 1, 3, -2, 2; their squares are 9, 1, 1, 9, 4, 4, which sum to 28. So s=2861=5.62.3664s = \sqrt{\frac{28}{6 - 1}} = \sqrt{5.6} \approx 2.3664 hours.

  3. Find the standard error: SE=sn=2.366460.9661SE = \frac{s}{\sqrt{n}} = \frac{2.3664}{\sqrt{6}} \approx 0.9661 hours.

  4. Degrees of freedom are n1=5n - 1 = 5. For 95% confidence, the critical value is t=2.571t^* = 2.571.

  5. Margin of error: tSE=2.571×0.96612.484t^* \cdot SE = 2.571 \times 0.9661 \approx 2.484 hours.

  6. Confidence interval: 11±2.484=(8.516,13.484)11 \pm 2.484 = (8.516, 13.484).

You are 95% confident that the mean battery life for this phone model is between about 8.52 and 13.48 hours. Because n=6<30n = 6 < 30, this relies on the stated assumption that battery life is approximately normal.

One-sample t-test for a population mean

A machine should fill bottles to a mean of 500 mL. A random sample of 5 bottles contains 498, 502, 495, 501, and 499 mL. Assume the fills are approximately normal. Test at α=0.05\alpha = 0.05 whether the mean fill differs from 500 mL.

  1. State hypotheses: H0:μ=500H_0: \mu = 500 and Ha:μ500H_a: \mu \ne 500, where μ\mu is the mean fill volume in mL. The null value is μ0=500\mu_0 = 500.

  2. Sample mean: xˉ=498+502+495+501+4995=24955=499\bar{x} = \frac{498 + 502 + 495 + 501 + 499}{5} = \frac{2495}{5} = 499 mL.

  3. Sample standard deviation. The deviations from 499 are -1, 3, -4, 2, 0; their squares are 1, 9, 16, 4, 0, summing to 30. So s=3051=7.52.7386s = \sqrt{\frac{30}{5 - 1}} = \sqrt{7.5} \approx 2.7386 mL.

  4. Standard error: SE=sn=2.738651.2247SE = \frac{s}{\sqrt{n}} = \frac{2.7386}{\sqrt{5}} \approx 1.2247 mL.

  5. Test statistic: t=xˉμ0SE=4995001.22470.816t = \frac{\bar{x} - \mu_0}{SE} = \frac{499 - 500}{1.2247} \approx -0.816, with df=51=4df = 5 - 1 = 4.

  6. Two-sided p-value from technology: p0.46p \approx 0.46.

Because p0.46>0.05p \approx 0.46 > 0.05, you fail to reject H0H_0. There is not convincing evidence that the machine's mean fill volume differs from 500 mL.

Frequently asked questions

When do you use a t-test instead of a z-test for means?

Use a t procedure whenever the population standard deviation σ\sigma is unknown and you estimate it with the sample standard deviation ss, which is almost always the case for quantitative data. A z procedure for a mean would require you to already know σ\sigma. See t-test vs z-test for the full comparison.

What are the degrees of freedom for a t procedure?

For a one-sample or matched-pairs t procedure, the degrees of freedom are n1n - 1. For a two-sample t procedure, the degrees of freedom fall between n1+n22n_1 + n_2 - 2 and the smaller of n11n_1 - 1 and n21n_2 - 1, and the exam expects you to find them with technology.

How do matched pairs differ from two independent samples?

Matched pairs come from linked measurements, such as a pre-test and post-test on the same students, so you take the difference within each pair and analyze that one sample of differences with a one-sample t procedure. Two independent samples come from separate groups, so you use a two-sample t procedure on the difference μ1μ2\mu_1 - \mu_2.

Every topic in Unit 4

  1. 4.1Sampling Distributions for Sample Means
  2. 4.2Constructing a Confidence Interval for a Population Mean or Population Mean Difference
  3. 4.3Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference
  4. 4.4Setting Up a Test for a Population Mean or Population Mean Difference
  5. 4.5Carrying Out a Test for a Population Mean or Population Mean Difference
  6. 4.6Sampling Distributions for the Difference Between Two Sample Means
  7. 4.7Constructing a Confidence Interval for the Difference Between Two Population Means
  8. 4.8Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means
  9. 4.9Setting Up a Test for the Difference Between Two Population Means
  10. 4.10Carrying Out a Test for the Difference Between Two Population Means