AP Statistics · Topic 3.10 · Unit 3

AP Stats 3.10: CI for Two Proportions

By Jude Wallis · Published July 28, 2026

A two-sample z-interval for the difference between two proportions is (p-hat-1 minus p-hat-2) plus or minus z* times sqrt(p-hat-1(1-p-hat-1)/n1 + p-hat-2(1-p-hat-2)/n2). Check randomization, the 10% condition, and at least 10 successes and 10 failures in each group.

AP Statistics: Unit 3 (topics 3.10). CED topic 3.10 (Constructing a Confidence Interval for the Difference Between Two Population Proportions), skills 2.C, 3.E, 4.E.

What topic 3.10 covers

Topic 3.10 builds an interval estimate for how two population proportions differ. The correct procedure is the two-sample z-interval for a difference between two population proportions. The point estimate is $\hat{p}_1 - \hat{p}_2$ , and the interval reports a range of plausible values for the true difference $p_1 - p_2$ in context. This is the two-group version of the interval you built in topic 3.3, and it is the standard way to compare two groups without running a formal test.

The formula and its pieces

The interval is

(\hat{p}_1 - \hat{p}_2) \pm z^{*}\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}.

The standard error is the square root term, an estimate of the standard deviation of $\hat{p}_1 - \hat{p}_2$ . Because this is an interval and not a test, each group keeps its own $\hat{p}$ ; the standard errors are not pooled. The margin of error is $z^{*}$ times that standard error, and the critical value $z^{*}$ comes from the confidence level, such as 1.96 for 95%.

Conditions

A two-sample z-interval requires three conditions:

Randomization: two independent random samples or a randomized experiment.
10% condition: when sampling without replacement, each sample is at most 10% of its population.
Normality: the observed successes and failures in both groups, $n_1\hat{p}_1$ , $n_1(1-\hat{p}_1)$ , $n_2\hat{p}_2$ , and $n_2(1-\hat{p}_2)$ , are all at least 10.

The 10% condition is unnecessary when the data come from a randomized experiment, because random assignment, not random sampling, supplies the independence.

Reading the interval

The order of subtraction sets the meaning of the interval, so decide which group is group 1 before computing and state it clearly. A positive value for $p_1 - p_2$ means group 1 has the larger proportion, and a negative value means group 2 does. Whether the interval includes 0 is the key question that topic 3.11 uses to justify a claim, so keep track of the sign of both endpoints.

A wider interval signals less precision, usually from small samples or a high confidence level. Because the interval already answers whether a difference is plausible, many free-response questions ask you to build it and then interpret it, rather than to run a separate test on the same data. Stating the two groups and the direction of subtraction before you interpret keeps the final sentence clear and easy to score.

95% interval for a difference in proportions

In group 1, 90 of 200 people recommend a product; in group 2, 80 of 250 do. Build a 95% confidence interval for $p_1 - p_2$ .

Sample proportions: $\hat{p}_1 = \dfrac{90}{200} = 0.45$ and $\hat{p}_2 = \dfrac{80}{250} = 0.32$ .
Point estimate: $0.45 - 0.32 = 0.13$ .
Standard error: $\sqrt{\dfrac{0.45 \times 0.55}{200} + \dfrac{0.32 \times 0.68}{250}} = \sqrt{0.0012375 + 0.0008704} = \sqrt{0.0021079} \approx 0.04591$ .
Critical value for 95%: $z^{*} = 1.96$ .
Margin of error: $1.96 \times 0.04591 \approx 0.0900$ .
Interval: $0.13 \pm 0.0900$ .

The 95% interval for $p_1 - p_2$ is about $(0.040, 0.220)$ . You are 95% confident the true difference in recommendation proportions lies in this range.

Frequently asked questions

Why is the standard error not pooled here like in a two-proportion test?

Pooling assumes the two proportions are equal, which is the null hypothesis in a test. A confidence interval makes no such assumption, so each group uses its own $\hat{p}$ in the standard error.