Chapter 10.1, Problem 13E

a.

To determine

To construct: The 99% confidence interval for difference in population proportion

Answer to Problem 13E

The confidence interval is $\left(0.060,0.114\right)$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{Number of men}\left(n_1\right)=2253\\ \text{Number of women}\left(n_2\right)=2629\\ \text{Number of men that live with their parents}\left(x_1\right)=986\\ \text{Number of women that live with their parents}\left(x_2\right)=923\end{array}$

Formula used:

The formula to construct the confidence interval for difference in population proportion is:

  $\begin{array}{l}CI=\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}\\ Where,\\ {\widehat{p}}_1=\frac{x_1}{n_1}\\ {\widehat{p}}_2=\frac{x_2}{n_2}\end{array}$

Calculation:

The obtained Excel output is:

From the above output, the confidence interval is $\left(0.060,0.114\right)$ .

b.

To determine

To explain: Whether using the above computed confidence interval it could be concluded that the difference exits between the two population proportions.

Explanation of Solution

The above constructed confidence interval does not contain 0. Thus, it could be said that theredifference exits between the two population proportions of men and women that live with their parents.