Question

The formula for the pooled variance, $s_{\mathrm{p}}^2$, is given on page 407 Show that, if the sample sizes, $n_1$ and $n_2$, are equal, then $s_{\mathrm{p}}^2$ is th mean of $s_1^2$ and $s_2^2$.

Short answer

If the sample sizes, $n_1$ and $n_2$, are equal, then $s_p^2$ is the mean of $s_1^2$ and $s_2^2$.

Step-by-step solution

Given Information

The sample sizes of the two populations, $n_1$ and $n_2$, are the same.

Explanation

A test statistic is used when conducting a hypothesis test based on independent samples to compare the means of two populations with equal and unknown standard deviations. Individual sample variances of two populations, ${s_1}^2$and ${s_2}^2$, are gathered and pooled by weighting them according to their sample size or degree of freedom, $n_1$and $n_2$.

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

Explanation

$s_p^2$incorporates information about both samples' variability into a single estimate of the population variance's common value. As a result, $s_p$is known as a pooled standard variance.

When $n_1=n_2$,

$\Rightarrow s_p^2=\frac{\left(n_1-1\right)s_1^2+\left(n_1-1\right)s_2^2}{n_1+n_1-2}$

$\Rightarrow s_p^2=\frac{\left(n_1-1\right)\left(s_1^2+s_2^2\right)}{2\left(n_1-2\right)}$

$\Rightarrow s_p^2=\frac{\left(s_1^2+s_2^2\right)}{2}$