Question

Show that, for two populations, $MSE=s_{\mathrm{p}}^2$, where $s_{\mathrm{p}}^2$is the pooled variance defined in Section 10.2 on page 407 . Conclude that$\sqrt{MSE}$ is the pooled sample standard deviation, $s_{\mathrm{p}}$.

Short answer

The standard deviation of the pool$s_p=\sqrt{MSE}$.

Step-by-step solution

Given Information

$s_p^2$ is the pooled variance.

Explanation

We have two populations based on the data.

So, $k=2,$

$n=n_1+n_2$

Error sum of squares is defined as

$SSE=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2$

Mean square error is defined as

$MSE=\frac{SSE}{n-k}$

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

$=s_p^2$

$\Rightarrow MSE=s_p^2$

The standard deviation of the pool is $s_p=\sqrt{MSE}$.

Hence, proved.