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When we calculate Fisher's LSD test, why do we use Mean Square of Error which is the variance of all groups (pooled variance), as a variance of each individual group mean? In the denominator it's: sqrt(MSE/n1+ MSE/n2). 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?

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Testing this way is one of the strengths of analysis of variance and related techniques. The underlying model specifies that the residual errors around all the group means have the same distribution. Any one estimate (from a single group or from 2 particular groups being compared) of the residual-error variance will be based on fewer cases and thus be less precise than an estimate obtained by pooling among all groups. Pooling provides a single best estimate of the error variance which is then used for all subsequent tests. It thus helps to avoid faulty comparisons due to the vagaries of error estimates based on particular samplings of small numbers of cases in individual groups.

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