If two random samples, each of size 36, are selected independently from two populations with variances of 42 and 50, then the standard error of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ , equals 2.5556.

Question

Quizplus · Accepted Answer

The standard error of the sampling distribution of the sample mean difference is equal to:

$$\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

where $s_1^2$ and $s_2^2$ are the variances of the two populations, and $n_1$ and $n_2$ are the sample sizes. Plugging in the given values, we get:

$$\sqrt{\frac{42}{36}+\frac{50}{36}} \approx 3.311$$

Therefore, the statement is false - the standard error of the sampling distribution of the sample mean difference is actually 3.311, not 2.5556.

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If Two Random Samples, Each of Size 36, Are Selected $\bar { X } _ { 1 } - \bar { X } _ { 2 }$