Sampling distribution of the difference between two sample means $ \bar{y}_1 - \bar{y}_2$:

When we draw a sample of size $ n_1$ from population 1, and a sample of size $ n_2$ from population 2, we can compute the mean of a variable $ y$ in sample 1 and in sample 2, and then compute the difference between the two sample means: $ \bar{y}_1 - \bar{y}_2$. Now suppose that we repeated these steps many times. Specifically, suppose that we drew an infinite number of of group 1 and group 2 samples, each time of size $ n_1$ and $ n_2$. Each time we have a group 1 and group 2 sample, we could compute the difference between the two sample means: $ \bar{y}_1 - \bar{y}_2$. Different samples would give different sample means and differences. The distribution of all these differences $ \bar{y}_1 - \bar{y}_2$ is the sampling distribution of $ \bar{y}_1 - \bar{y}_2$. Note that this sampling distribution is purely hypothetical. We would never really draw an infinite number of group 1 and group 2 samples, but hypothetically, we could.

Standard deviation:

Suppose that the assumptions of the two sample $ z$ test hold:

• Within population 1, the variable $ y$ is normally distributed with mean $\mu_1$ and standard deviation $\sigma_1$; within population 2, the variable $ y$ is normally distributed with mean $\mu_2$ and standard deviation $\sigma_2$
• The population standard deviations $\sigma_1$ and $\sigma_2$ are known
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another

Note that the $ z$ statistic $ z = \frac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}}$ thus indicates how many standard deviations $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ the observed difference $\bar{y}_1 - \bar{y}_2$ is removed from 0: the difference $\mu_1 - \mu_2$ according to the null hypothesis.