Given the following information about a hypothesis test of the difference between two means based on independent random samples,what is the standard deviation of the difference between the two means? Assume that the samples are obtained from normally distributed populations having equal variances.
H_A: ?_A > ?_B, $\bar { X } _ { 1 } = 12$
$\bar { X } _ { 2 } = 9$
S₁ = 5,s₂ = 3,n₁ = 13,n₂ = 10.

Question

Given the following information about a hypothesis test of the difference between two means based on independent random samples,what is the standard deviation of the difference between the two means? Assume that the samples are obtained from normally distributed populations having equal variances.
H_A: ?_A > ?_B,

\bar { X } _ { 1 } = 12

\bar { X } _ { 2 } = 9

S₁ = 5,s₂ = 3,n₁ = 13,n₂ = 10.

Quizplus · Accepted Answer

The standard deviation of the difference between the two means is:$$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$$Since the samples are obtained from normally distributed populations having equal variances, we can use the pooled variance to estimate the common variance:$$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$$$$s_p^2 = \frac{(13-1)(5)^2 + (10-1)(3)^2}{13 + 10 - 2} = 16.2$$Therefore, the standard deviation of the difference between the two means is:$$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{16.2}{13} + \frac{16.2}{10}} = 1.792$$

Given the Following Information About a Hypothesis Test of the Difference