Question

Suppose that the variable under consideration is normally distributed in each of two populations and that the population standard deviations are equal. Further, suppose that you want to perform a hypothesis test to decide whether the populations have different means, that is, whether ${\mu }_1\ne {\mu }_2$. If independent simple random samples are used, identify two hypothesis-testing procedures that you can use to carry out the hypothesis test.

Short answer

The $t-$ test and the $z-$ test are two different tests for comparing population means.

Step-by-step solution

Given Information

The variable being studied has a normal distribution.

The standard deviations of the population are also equal.

Explanation

Assume that the population standard deviations are the same based on the information provided.

Consider the following hypothesis:

Null hypothesis $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis $H_a:{\mu }_1\ne {\mu }_2$

Under Null hypothesis, the $t-$test is defined as

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)}{s_p\sqrt{\left(1/n_1\right)+\left(1/n_2\right)}}$

$\approx t_{n_1+n_2-2}$

The pooled standard deviation $s_p=\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}$

Under Null hypothesis, the $z-$test is defined as

$z=\frac{\left(\overline{x_1}-\overline{x_2}\right)}{\sigma \sqrt{\left(1/n_1\right)+\left(1/n_2\right)}}$

$z\approx N(0,1)$

The population standard deviation $\sigma ={\sigma }_1$,

$\sigma ={\sigma }_2$

As a result, both tests can be used.

$t$- test and $z$ - test.