John wants to compare two means. His sample statistics were $\overline { x _ { 1 } } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9$ and $\bar { x } _ { 2 } = 20.5 , s _ { 2 } ^ { 2 } = 3.6 , n _ { 2 } = 9$ . Assuming equal variances, the pooled variance is:
A)4.5
B)4.9
C)5.1
D)3.8

Question

Quizplus · Accepted Answer

The pooled variance is calculated using the formula $$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$$. Substituting the given values, $$s_p^2 = \frac{(9 - 1)5.4 + (9 - 1)3.6}{9 + 9 - 2} = \frac{8*5.4 + 8*3.6}{16} = \frac{43.2 + 28.8}{16} = \frac{72}{16} = 4.5$$.

John Wants to Compare Two Means x1‾=22.7,s12=5.4,n1=9\overline { x _ { 1 } } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9x1​​=22.7,s12​=5.4,n1​=9

John Wants to Compare Two Means $\overline { x _ { 1 } } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9$