AP Statistics · Topic 3.9 · Unit 3

AP Stats 3.9: Difference of Two Proportions

By Jude Wallis · Published July 28, 2026

The sampling distribution of p-hat-1 minus p-hat-2 has mean p1 minus p2 and standard deviation sqrt(p1(1-p1)/n1 + p2(1-p2)/n2). It is approximately normal when all four expected counts are at least 10 and the samples are independent.

AP Statistics: Unit 3 (topics 3.9). CED topic 3.9 (Sampling Distributions for the Difference Between Sample Proportions), skills 3.D, 4.D, 4.E.

What topic 3.9 covers

Topic 3.9 moves from one proportion to comparing two. It describes the sampling distribution of the difference $\hat{p}_1 - \hat{p}_2$ for two independent populations with proportions $p_1$ and $p_2$ . You learn its mean, its standard deviation, and the conditions that make it approximately normal, which sets up the two-sample interval and test in later topics. The structure mirrors topic 3.2, with a second group added and its variability folded in.

Mean and standard deviation

When the two samples are independent, the difference $\hat{p}_1 - \hat{p}_2$ has mean

\mu_{\hat{p}_1 - \hat{p}_2} = p_1 - p_2

and standard deviation

\sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}.

The variances of the two independent sample proportions add, then you take the square root. You do not subtract under the root, and you never add the standard deviations directly.

Conditions

Two conditions justify the standard deviation, and one more justifies the normal shape.

Randomization: two independent random samples, or a randomized experiment.
10% condition: when sampling without replacement, each sample is at most 10% of its population, so $n_1 \le 0.10N_1$ and $n_2 \le 0.10N_2$ .
Normality: the expected successes and failures in both groups, $n_1p_1$ , $n_1(1-p_1)$ , $n_2p_2$ , and $n_2(1-p_2)$ , are all at least 10.

For an experiment, only the randomization condition is required, met by randomly assigning treatments to units.

Interpreting the distribution

Read the mean, standard deviation, and any probability in the context of the two specific populations being compared. The mean $p_1 - p_2$ is what you are trying to estimate, and a value of 0 means the two proportions are equal. The standard deviation is the typical distance the observed difference $\hat{p}_1 - \hat{p}_2$ falls from the true difference across repeated pairs of samples. Because this spread depends on both sample sizes, a small group can inflate the standard deviation even when the other group is large, so balanced sample sizes tend to give more precise estimates.

When you later run a two-sample interval or test, this standard deviation, estimated from the data, becomes the standard error in the denominator, so the reasoning here carries straight into topics 3.10 through 3.13. Keeping the mean, the standard deviation, and the shape distinct in your notes now makes those later procedures much easier to set up correctly.

Center and spread of a difference in proportions

Population 1 has $p_1 = 0.50$ with $n_1 = 100$ ; population 2 has $p_2 = 0.40$ with $n_2 = 120$ . Find the mean and standard deviation of $\hat{p}_1 - \hat{p}_2$ .

Mean: $p_1 - p_2 = 0.50 - 0.40 = 0.10$ .
First variance term: $\dfrac{0.50 \times 0.50}{100} = \dfrac{0.25}{100} = 0.0025$ .
Second variance term: $\dfrac{0.40 \times 0.60}{120} = \dfrac{0.24}{120} = 0.0020$ .
Add: $0.0025 + 0.0020 = 0.0045$ .
Square root: $\sqrt{0.0045} \approx 0.0671$ .

The mean is 0.10 and the standard deviation is about 0.0671. All four expected counts (50, 50, 48, 72) exceed 10, so the distribution is approximately normal.

Frequently asked questions

Do I add or subtract the two variances?

You add them. The two sample proportions are independent, and variances of independent quantities add whether you are looking at a sum or a difference. You take the square root of that sum to get the standard deviation.