If two random samples of sizes $n _ { 1 }$ and $n _ { 2 }$ are selected independently from two populations with means $\mu _ { 1 }$ and $\mu _ { 2 }$ , then the mean of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , equals:

A)$\mu_{1}+\mu_{2}$
B) $\mu_{1}-\mu_{2}$
C) $\mu_{1}/\mu_{2}$
D) $\mu_{1}\mu_{2}$ :

Question

If two random samples of sizes

n _ { 1 }

and

n _ { 2 }

are selected independently from two populations with means

\mu _ { 1 }

and

\mu _ { 2 }

, then the mean of the sampling distribution of the sample mean difference,

\bar { X } _ { 1 } - \bar { X } _ { 2 }

, equals:

A)$\mu_{1}+\mu_{2}$
B) $\mu_{1}-\mu_{2}$
C) $\mu_{1}/\mu_{2}$
D) $\mu_{1}\mu_{2}$ :

Quizplus · Accepted Answer

The mean of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$, is equal to the difference between the population means, $\mu _ { 1 } - \mu _ { 2 }$, regardless of the sample sizes. Therefore, the best choice is B.

If Two Random Samples of Sizes n1n _ { 1 }n1​

If Two Random Samples of Sizes $n _ { 1 }$