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

Short Answer

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: $\begin{array}{|l|l|}\hline A&\text { 0.3247}\\\hline B&\text { 0.3282}\\\hline C&\text {0.1054 }\\\hline D&\text { 0.1125}\\\hline\end{array}$