Chapter 10.1, Problem 25E

To determine

To construct: and interpret a 95 percent interval for $p_1-p_2$ .

Answer to Problem 25E

(0.100, 0.382)

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=44\\ n_1=88\\ x_2=21\\ n_2=81\end{array}$

Formula used:

Sample proportion

  $\widehat{p}=\frac{x}{n}$

Confidence interval for $p_1-p_2$

Lower part is

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)-Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\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}}$

Upper part is

  $\left({\widehat{p}}_1-{\widehat{p}}_2\right)+Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\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}}$

Calculation:

Sample proportion

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}=\frac{44}{88}=0.5\\ {\widehat{p}}_2=\frac{x_2}{n_2}=\frac{21}{81}=0.259\end{array}$

For confidence interval $1-\alpha =0.95$

  $\Rightarrow \alpha =0.05$

  $Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=Z_{0.025}=1.96$

Confidence interval for $p_1-p_2$

  $\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)-Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\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}}\\ =\left(0.5-0.259\right)-1.96\sqrt{\frac{0.5\left(1-0.5\right)}{88}+\frac{0.259\left(1-0.259\right)}{81}}=0.100\end{array}$

  $\begin{array}{l}\left({\widehat{p}}_1-{\widehat{p}}_2\right)+Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\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}}\\ =\left(0.5-0.259\right)+1.96\sqrt{\frac{0.5\left(1-0.5\right)}{88}+\frac{0.259\left(1-0.259\right)}{81}}=0.382\end{array}$

By calculating the proportion difference, it is 95 percent confident that the proportion difference between 0.100 and 0.382