Chapter 10, Problem T10.8SPT

To determine

To find out the mean and standard deviation of the sampling distribution.

Answer to Problem T10.8SPT

Option (c) is the correct option.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{\widehat{p}}_1=0.8\\ n_1=100\\ {\widehat{p}}_2=0.5\\ n_2=400\end{array}$

The mean of the sampling distribution is the difference in sample proportion, that is as follows:

  $\begin{array}{l}\mu ={\widehat{p}}_1-{\widehat{p}}_2\\ =0.8-0.5\\ 0.3\end{array}$

And the standard deviation of the sampling distribution is calculated as:

  $\begin{array}{l}\sigma =\sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =\sqrt{\frac{0.8(1-0.8)}{100}+\frac{0.5(1-0.5)}{400}}\\ =0.047\end{array}$

Thus, we have, the mean is $0.3$ and the standard deviation is $0.047$ . Therefore, option (c) is the correct option.