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 calculated using the formula $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$, where $\sigma_1^2$ and $\sigma_2^2$ are the population variances and $n_1$ and $n_2$ are the sample sizes. Plugging in the given values, $\sqrt{\frac{42}{36} + \frac{50}{36}} = \sqrt{\frac{92}{36}} = \sqrt{2.5556}$, which equals approximately 1.598, not 2.5556.

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