Chapter 10.1, Problem 33E

To determine

To select from the option that which of the following is the correct margin of error for a $99\%$ confidence interval for the difference in the proportion of male and female college students who worked for pay last summer.

Answer to Problem 33E

The option (c) is the correct option.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=410\\ x_2=484\\ n_1=500\\ n_2=550\\ c=0.99\end{array}$

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{410}{500}\\ =0.82\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{484}{550}\\ =0.88\end{array}$

Now, for the confidence interval $1-\alpha =0.99$ , we have to find out the value of $z_{\alpha /2}=z_{0.005}$ using the table. Thus the z score will be:

  $z_{\alpha /2}=2.576$

Thus, the margin of error will be:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =2.576\times \sqrt{\frac{0.82(1-0.82)}{500}+\frac{0.88(1-0.88)}{550}}\\ =2.576\times \sqrt{\frac{0.88(1-0.88)}{550}+\frac{0.82(1-0.82)}{500}}\\ =2.576\times \sqrt{\frac{0.88(0.120)}{550}+\frac{0.82(0.180)}{500}}\end{array}$

Thus, option (c) is the correct option.