Fisher's LSD test. When we calculate Fisher's LSD test, why do we use Mean Square of Error, $\sigma_e$, which is the variance of all groups (pooled variance), as a variance of each individual group mean? In the denominator it's: 
$$\sqrt{{{\sigma_e} \over {n_1}}+ {{\sigma_e} \over {n_2}}}$$
Why don't we use group variance instead of MSE? Why don't we use a t-test just like the procedure of testing differences betweeen two sample means for independent samples?
 A: If you are assuming (as usual in a one-way ANOVA) that all
$g$ treatment groups have the same variance, then Fisher's
LSD has greater power if you use
data from all $g$ groups to get the standard error for the
difference. 
Recall that Fisher's LSD is to be used only if the F-test
in the ANOVA has found that there are significant differences
among the group means.
This method of finding the standard error is what distinguishes
Fisher's LSD from doing ${g \choose 2}$ tw0-sample tests to try to
find the pattern of differences among the group means.
Note: Of course, if group means truly have quite
different population means, the F-test in the ANOVA does
not have the anticipated distribution, and Fisher's LSD
does not work properly either. In that case, there is a
separate-variances Welch-style ANOVA procedure that is
appropriate, and LSD would not be used as the method
of multiple comparisons among group means. Bonferroni seems
to be the multiple comparison method of choice.
Some software implements the separate-variances ANOVA
(for example, R) and other software does not (for example, Minitab).
