In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows: n1 = 25, $\bar { X } _ { 1 }$ = 7.30, s1 = 1.05.
N2 = 30, $\bar { X } _ { 2 }$ = 6.80, s2 = 1.20.
Assume that the population variances are equal. Then the standard error of the sampling distribution of the sample mean difference $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ is equal to:
A)0.3072.
B)0.0917.
C)0.3028.
D)0.0944.

Question

Quizplus · Accepted Answer

The standard error of the sampling distribution of the sample mean difference is:$$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}} = \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2}}$$where s is the pooled standard deviation, which is:$$s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$$Plugging in the given values, we get:$$s = \sqrt{\frac{(25 - 1)1.05^2 + (30 - 1)1.20^2}{25 + 30 - 2}} = 1.125$$Therefore, the standard error of the sampling distribution of the sample mean difference is:$$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{1.125^2}{25} + \frac{1.125^2}{30}} = 0.3072$$So the correct answer is A.

In Testing Whether the Means of Two Normal Populations Are Xˉ1\bar { X } _ { 1 }Xˉ1​

In Testing Whether the Means of Two Normal Populations Are $\bar { X } _ { 1 }$