Two independent samples of sizes 20 and 30 are randomly selected from two normally distributed populations. Assume that the population variances are unknown but equal. In order to test the difference between the population means, $\mu _ { 1 } - \mu _ { 2 }$ , the sampling distribution of the sample mean difference, $\bar { x } _ { 1 } - \bar { x } _ { 2 }$ , is:
A)normal.
B)Student-t with 50 degrees of freedom.
C)Student-t with 48 degrees of freedom.
D)None of these choices.

Question

Quizplus · Accepted Answer

When the population variances are unknown but assumed equal, and the samples are independent, the sampling distribution of the difference between the sample means follows a Student-t distribution with degrees of freedom calculated using the formula: $df = n_1 + n_2 - 2$. Here, $n_1 = 20$ and $n_2 = 30$, so $df = 20 + 30 - 2 = 48$.

Two Independent Samples of Sizes 20 and 30 Are Randomly μ1−μ2\mu _ { 1 } - \mu _ { 2 }μ1​−μ2​

Two Independent Samples of Sizes 20 and 30 Are Randomly $\mu _ { 1 } - \mu _ { 2 }$