In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows: $n _ { 1 } = 20$ , $\bar { x } _ { 1 } = 10.8$ , $s _ { 1 } = 0.90$ . $n _ { 2 } = 18$ , $\bar { x } _ { 2 } = 9.6$ , $s _ { 2 } = 1.10$ . Assume that the population variances are unequal. The standard error of the sampling distribution of the sample mean difference $\bar { x } _ { 1 } - \bar { x } _ { 2 }$ is equal to:
A) 0.3247.
B) 0.3282.
C) 0.1054.
D) 0.1125.

Question

Quizplus · Accepted Answer

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

$$
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{(0.9)^2}{20} + \frac{(1.1)^2}{18}} \approx 0.3282
$$

Therefore, the correct answer is (B).

In Testing Whether the Means of Two Normal Populations Are n1=20n _ { 1 } = 20n1​=20

In Testing Whether the Means of Two Normal Populations Are $n _ { 1 } = 20$