AP Statistics · Topic 4.6 · Unit 4

AP Stats 4.6: Sampling Distribution of a Difference

By Jude Wallis · Published July 28, 2026

Topic 4.6 describes the sampling distribution of x-bar-1 minus x-bar-2. Its mean is mu-1 minus mu-2, and its standard deviation is the square root of sigma-1 squared over n-1 plus sigma-2 squared over n-2. It is approximately normal when both populations are normal or both sample sizes reach 30.

AP Statistics: Unit 4 (topics 4.6). Topic 4.6 (Sampling Distributions for the Difference Between Two Sample Means) in the Fall 2026 AP Statistics course.

What topic 4.6 covers

Topic 4.6 opens the two-sample half of Unit 4. It describes the sampling distribution of the difference between two sample means, $\bar{x}_1 - \bar{x}_2$ , drawn from two independent populations. This is the foundation for the two-sample confidence interval in topic 4.7 and the two-sample t-test in topics 4.9 and 4.10.

The parameter of interest is the difference between population means, $\mu_1 - \mu_2$ . As with one sample, you need the center, spread, and shape of the sampling distribution before you can do inference.

Mean and standard deviation of the difference

For two independent populations with means $\mu_1, \mu_2$ and standard deviations $\sigma_1, \sigma_2$ , when the sampled values are independent, the sampling distribution of $\bar{x}_1 - \bar{x}_2$ has

\mu_{\bar{x}_1 - \bar{x}_2} = \mu_1 - \mu_2 \qquad \sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}.

The mean of the difference is just the difference of the means. The standard deviation adds the two variances of the individual sample means, then takes the square root; you add variances, not standard deviations, because the samples are independent. Later, when the population standard deviations are unknown, replacing each $\sigma$ with a sample $s$ gives the standard error used for inference.

Conditions and shape

For two samples drawn without replacement, two conditions support independence:

Randomization: the data come from two independent random samples, or from a randomized experiment where treatments are randomly assigned.
10% condition: each sample is at most 10% of its population, $n_1 \le 0.10N_1$ and $n_2 \le 0.10N_2$ . Data from a randomized experiment do not need this check.

The difference $\bar{x}_1 - \bar{x}_2$ can be modeled by a normal distribution if both populations are normal. If neither population is normal, the central limit theorem still makes the difference approximately normal when both $n_1 \ge 30$ and $n_2 \ge 30$ . You can build intuition for both sample means at once in the sampling distribution interactive.

Interpret each result within the context of the two specific populations. The mean of the difference describes the average gap you expect between the two sample means, the standard deviation describes how much that gap varies from one pair of samples to the next, and a probability answers how surprising a particular observed gap would be if the population means were as stated.

Mean, standard deviation, and a probability for a difference of means

Scores on a standardized test are approximately normal at two schools. School A has mean 500 and standard deviation 40; school B has mean 480 and standard deviation 30. Independent random samples of $n_1 = 64$ students from A and $n_2 = 36$ from B are taken. Find the mean and standard deviation of $\bar{x}_A - \bar{x}_B$ , then find the probability that this difference exceeds 27.07.

Mean of the difference: $\mu_{\bar{x}_A - \bar{x}_B} = 500 - 480 = 20$ points.
First variance term: $\frac{\sigma_A^2}{n_A} = \frac{40^2}{64} = \frac{1600}{64} = 25$ .
Second variance term: $\frac{\sigma_B^2}{n_B} = \frac{30^2}{36} = \frac{900}{36} = 25$ .
Standard deviation: $\sigma_{\bar{x}_A - \bar{x}_B} = \sqrt{25 + 25} = \sqrt{50} \approx 7.0711$ points.
Both populations are normal, so the difference is normal. Standardize 27.07: $z = \frac{27.07 - 20}{7.0711} \approx \frac{7.07}{7.0711} \approx 1.00$ .
Upper-tail area: $P(Z > 1.00) = 1 - 0.8413 = 0.1587$ .

The sampling distribution of $\bar{x}_A - \bar{x}_B$ is normal with mean 20 and standard deviation about 7.07 points. There is about a 0.1587 probability that the sample from A averages more than 27.07 points above the sample from B.

Frequently asked questions

Why do I add the variances instead of the standard deviations?

Because the two samples are independent, their variances add when you form the difference, and variance is the square of standard deviation. So you compute sigma-1 squared over n-1 plus sigma-2 squared over n-2, then take the square root at the end. Adding the standard deviations directly would overstate the spread of the difference.

Does subtracting the means change the spread?

No. Whether you look at the sum or the difference of two independent sample means, the variances still add, so the standard deviation is the same square-root expression either way. Only the center changes: the mean of the difference is mu-1 minus mu-2.