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: $$S _ { pooled} ^ { 2 } = \frac { 1 } { n _ { m } + n _ { w } - 2 } \left[ \sum _ { i = 1 } ^ { n _ { m } } \left( Y _ { i } - \overline { Y _ { m } } \right) ^ { 2 } + \sum _ { i = 1 } ^ { n _ { w } } \left( Y _ { i } - \bar { Y } _ { w } \right) ^ { 2 } \right]$$ 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: $$S _ { pooled} ^ { 2 } = \frac { 1 } { n _ { m } + n _ { w } - 2 } \left[ \sum _ { i = 1 } ^ { n _ { m } } \left( Y _ { i } - \overline { Y _ { m } } \right) ^ { 2 } + \sum _ { i = 1 } ^ { n _ { w } } \left( Y _ { i } - \bar { Y } _ { w } \right) ^ { 2 } \right]$$ 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