Your textbook states that when you test for differences in means and you assume that
the two population variances are equal, then an estimator of the population variance is
the following "pooled" estimator: $$s _ { \text {pooled } } ^ { 2 } = \frac { 1 } { n _ { m } + n _ { w } - 2 } \left[ \sum _ { i = 1 } ^ { n _ { m } } \left( Y _ { i } - \bar { Y } _ { m } \right) ^ { 2 } + \sum _ { i = 1 } ^ { n _ { w } } \left( Y _ { i } - \bar { Y } _ { w } \right) ^ { 2 } \right]$$ 2
Explain why this pooled estimator can be looked at as the weighted average of the two
variances.

Question

Quizplus · Accepted Answer

The Answer of Your textbook states that when you test for differences in means and you assume that
the two population variances are equal, then an estimator of the population variance is
the following "pooled" estimator: $$s _ { \text {pooled } } ^ { 2 } = \frac { 1 } { n _ { m } + n _ { w } - 2 } \left[ \sum _ { i = 1 } ^ { n _ { m } } \left( Y _ { i } - \bar { Y } _ { m } \right) ^ { 2 } + \sum _ { i = 1 } ^ { n _ { w } } \left( Y _ { i } - \bar { Y } _ { w } \right) ^ { 2 } \right]$$ 2
Explain why this pooled estimator can be looked at as the weighted average of the two
variances.

Your Textbook States That When You Test for Differences in Means