Given the following information about a hypothesis test of the difference between two means based on independent random samples,estimate 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,estimate 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 two means, when the population variances are assumed to be equal, can be estimated using the formula for the pooled standard deviation ($s_p$) and the sample sizes ($n_1$ and $n_2$). The formula for the pooled standard deviation is:$$s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}$$where $s_1$ and $s_2$ are the sample standard deviations, and $n_1$ and $n_2$ are the sample sizes. Given $s_1 = 5$, $s_2 = 3$, $n_1 = 13$, and $n_2 = 10$, we first calculate $s_p$:$$s_p = \sqrt{\frac{(13-1)5^2 + (10-1)3^2}{13 + 10 - 2}} = \sqrt{\frac{12*25 + 9*9}{21}} = \sqrt{\frac{300 + 81}{21}} = \sqrt{\frac{381}{21}} = \sqrt{18.142857}$$$$s_p ≈ 4.26$$Then, the standard deviation of the difference between the two means ($s_{\bar{d}}$) is calculated using the formula:$$s_{\bar{d}} = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$$$$s_{\bar{d}} = 4.26 \sqrt{\frac{1}{13} + \frac{1}{10}} = 4.26 \sqrt{\frac{10+13}{130}} = 4.26 \sqrt{\frac{23}{130}} = 4.26 \sqrt{0.176923}$$$$s_{\bar{d}} ≈ 4.26 \times 0.4205 ≈ 1.79$$Therefore, the closest answer is 1.792 (A).

Given the Following Information About a Hypothesis Test of the Difference