Hypothesis Tests for Two Means: Known Population Variances

When two independent samples are drawn from populations with known standard deviations $\sigma_1$ and $\sigma_2,$ and we want to test a claim about the difference of the population means $\mu_1 - \mu_2,$ we use the two-sample $z$-test.

The null hypothesis has the form

$$H_0:\ \mu_1 - \mu_2 = \delta_0,$$

where $\delta_0$ is the hypothesized value of the difference. Most commonly $\delta_0 = 0,$ which corresponds to testing whether the two population means are equal. The alternative is one of

$$H_a:\ \mu_1 - \mu_2 \ne \delta_0,\qquad H_a:\ \mu_1 - \mu_2 > \delta_0,\qquad H_a:\ \mu_1 - \mu_2 < \delta_0.$$

Given sample means $\bar{x}_1,\bar{x}_2$ from samples of sizes $n_1,n_2,$ the test statistic is

$$z = \dfrac{(\bar{x}_1 - \bar{x}_2) - \delta_0}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}.$$

Under $H_0,$ this statistic follows the standard normal distribution $N(0,1).$

For example, suppose $n_1 = 25,\ \bar{x}_1 = 30,\ \sigma_1 = 4$ and $n_2 = 25,\ \bar{x}_2 = 28,\ \sigma_2 = 3,$ and we test $H_0: \mu_1 - \mu_2 = 0.$ Then

$$z = \dfrac{30 - 28}{\sqrt{\dfrac{16}{25} + \dfrac{9}{25}}} = \dfrac{2}{\sqrt{1}} = 2.$$