AP Statistics · Topic 4.2 · Unit 4

AP Stats 4.2: Confidence Interval for a Mean

By Jude Wallis · Published July 28, 2026

Topic 4.2 builds a one-sample t-interval for a population mean: x-bar plus or minus t-star times s over the square root of n, with degrees of freedom n minus 1. Use the t-distribution because sigma is unknown and s replaces it. Paired differences use the same interval.

AP Statistics: Unit 4 (topics 4.2). Topic 4.2 (Constructing a Confidence Interval for a Population Mean or Population Mean Difference) in the Fall 2026 AP Statistics course. Unit 4 is 10 to 20 percent of the multiple-choice section.

What topic 4.2 covers

Topic 4.2 is where you first estimate a population mean $\mu$ (mu) with an interval. It introduces the t-distribution, has you identify the correct procedure, and asks you to justify it by checking conditions before you calculate. The College Board name for the procedure is a one-sample t-interval for a population mean.

Because the population standard deviation $\sigma$ (sigma) is almost never known for a quantitative variable, you estimate it with the sample standard deviation $s$ . That single fact is why means use $t$ rather than the $z$ you used for proportions in Unit 3.

t-distributions

t-distributions, also called Student's t-distributions, are a family of symmetric, bell-shaped curves that are standardized like the normal curve but have wider tails. Each one is set by a parameter called the degrees of freedom (df), which comes from the sample size. For a one-sample interval, $df = n - 1$ .

When df is small, a t-distribution has a lower peak and fatter tails than the standard normal, which widens the interval to account for the extra uncertainty in estimating $\sigma$ . As df grows, the t-distribution gets closer to the standard normal (mean 0, standard deviation 1). You read the critical value $t^*$ from the t-table using your confidence level and $df = n - 1$ .

The interval, standard error, and margin of error

The point estimate for $\mu$ is the sample mean $\bar{x}$ (x-bar). The interval is

\bar{x} \pm t^* \frac{s}{\sqrt{n}},

where $t^*$ is the critical value for the central C% of a t-distribution with $n - 1$ degrees of freedom. The standard error of the mean is $SE = \dfrac{s}{\sqrt{n}}$ , and the margin of error is $t^* \cdot SE$ . For a matched pairs design you first subtract within each pair to get one sample of differences, then run the identical interval on those differences to estimate the population mean difference $\mu_d$ ; always state the order of subtraction. The confidence interval calculator and one-sample t-test calculator check your arithmetic.

Conditions to justify the interval

A one-sample t-interval requires three conditions, each stated in context:

Randomization: the data come from a random sample or a randomized experiment.
10% condition: when sampling without replacement, $n \le 0.10N$ , where $N$ is the population size.
Sample data (normality): the population is stated to be approximately normal, or $n \ge 30$ ; if $n < 30$ , the sample should be free from strong skewness and outliers. For matched pairs, apply this to the differences, using 30 or more differences or a difference sample with no strong skew or outliers.

For the full recipe shared with proportions, see how to calculate a confidence interval, and for the deeper contrast with z see t-test vs z-test.

95% one-sample t-interval for a mean

A produce inspector weighs a random sample of 9 tomatoes from a large shipment. The weights, in grams, are 110, 112, 114, 116, 118, 120, 122, 124, and 126. The weights show no strong skew or outliers. Construct a 95% confidence interval for the mean weight of tomatoes in the shipment.

Sample mean: $\bar{x} = \frac{110 + 112 + 114 + 116 + 118 + 120 + 122 + 124 + 126}{9} = \frac{1062}{9} = 118$ grams.
Deviations from 118 are -8, -6, -4, -2, 0, 2, 4, 6, 8; their squares are 64, 36, 16, 4, 0, 4, 16, 36, 64, which sum to 240.
Sample standard deviation: $s = \sqrt{\frac{240}{9 - 1}} = \sqrt{\frac{240}{8}} = \sqrt{30} \approx 5.4772$ grams.
Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{5.4772}{\sqrt{9}} = \frac{5.4772}{3} \approx 1.8257$ grams.
Degrees of freedom: $df = 9 - 1 = 8$ . For 95% confidence, $t^* = 2.306$ .
Margin of error: $t^* \cdot SE = 2.306 \times 1.8257 \approx 4.210$ grams.
Interval: $118 \pm 4.210 = (113.79,\ 122.21)$ grams.

You are 95% confident that the mean weight of tomatoes in the shipment is between about 113.79 and 122.21 grams. Because $n = 9 < 30$ , this relies on the sample being free from strong skew and outliers.

Frequently asked questions

Why use t instead of z for a mean?

You use t because the population standard deviation sigma is unknown and you replace it with the sample standard deviation s. That substitution adds uncertainty, which the wider tails of the t-distribution account for. If you somehow knew sigma exactly, a z-interval would apply, but that almost never happens with quantitative data.

What degrees of freedom do I use?

For a one-sample t-interval or a matched-pairs interval on the differences, the degrees of freedom are n minus 1, where n is the number of observations (or the number of paired differences). Use that df to read the critical value t-star from the t-table.