Chapter 10.1, Problem 26E

To determine

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

Answer to Problem 26E

(0.0175, 0.3075)

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=34\\ n_1=80\\ x_2=21\\ n_2=80\end{array}$

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

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}=\frac{34}{88}=0.425\\ {\widehat{p}}_2=\frac{x_2}{n_2}=\frac{21}{80}=0.2625\end{array}$

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(0.425-0.2625\right)-1.96\sqrt{\frac{0.425\left(1-0.425\right)}{80}+\frac{0.2625\left(1-0.2625\right)}{80}}=0.0175\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(0.425-0.2625\right)+1.96\sqrt{\frac{0.425\left(1-0.425\right)}{80}+\frac{0.2625\left(1-0.2625\right)}{80}}=0.3075\end{array}$

By calculating the proportion difference, it is 95 percent confident that the proportion difference is between 0.0175 and 0.3075