Chapter 10, Problem T10.2SPT

To determine

To select the correct option for, which of the following is a $95\%$ confidence interval for the difference in the proportion of workers at the two companies who would admit to using sick leave when they weren’t ill.

Answer to Problem T10.2SPT

Option (b) is correct.

Explanation of Solution

It is given that,

  $\begin{array}{l}{x}_1=35\\ x_2=17\\ n_1=125\\ n_2=68\\ c=95\%\end{array}$

The sample proportion is the number of successes divided by the sample size:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{35}{125}\\ =0.28\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{17}{68}\\ =0.25\end{array}$

For confidence level $1-\alpha =0.95$ , determine $z_{\alpha /2}=z_{0.025}$ using table II:

  $z_{\alpha /2}=1.96$

The endpoints of the confidence interval for $p_1-p_2$ are then:

  $\begin{array}{l}({\widehat{p}}_1-{\widehat{p}}_2)\pm z_{\alpha /2}\sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.28-0.25)\pm 1.96\sqrt{\frac{0.28(1-0.28)}{125}+\frac{0.25(1-0.25)}{68}}\\ =0.03\pm 1.96\sqrt{\frac{0.28(0.72)}{125}+\frac{0.25(0.75)}{68}}\end{array}$

Thus, the correct option is (b) option.