Chapter 10, Problem R10.2RE

To determine

To calculate and interpret a $99\%$ confidence interval for the difference in the true proportion of U.S. teens and adults who use Facebook.

Answer to Problem R10.2RE

The $99\%$ confidence interval is $(-0.0427,0.3027)$ .

Explanation of Solution

It is given in the question that:

Let $p_T$ be the true proportion of all U.S. teens who use Facebook.

And $p_A$ be the true proportion of all U.S. adults who use Facebook.

Now, it is also given that:

  $\begin{array}{l}{x}_1=71\\ x_2=58\\ n_1=n_2=100\end{array}$

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

  $\begin{array}{l}{\widehat{p}}_T=\frac{x_1}{n_1}\\ =\frac{71}{100}\\ =0.71\\ {\widehat{p}}_A=\frac{x_2}{n_2}\\ =\frac{58}{100}\\ =0.58\end{array}$

For the confidence interval $1-\alpha =0.99$ , determine $z_{\alpha /2}=z_{0.005}$ ,

  $z_{\alpha /2}=2.5758$

The endpoints of the confidence interval for $p_T-p_A$ are then:

  $\begin{array}{l}({\widehat{p}}_T-{\widehat{p}}_A)+z_{\alpha /2}\sqrt{\frac{{\widehat{p}}_T(1-{\widehat{p}}_T)}{n_1}+\frac{{\widehat{p}}_A(1-{\widehat{p}}_A)}{n_2}}\\ =(0.71-0.58)+2.5758\sqrt{\frac{0.71(1-0.71)}{100}+\frac{0.58(1-0.58)}{100}}\\ =0.3027\\ ({\widehat{p}}_T-{\widehat{p}}_A)-z_{\alpha /2}\sqrt{\frac{{\widehat{p}}_T(1-{\widehat{p}}_T)}{n_1}+\frac{{\widehat{p}}_A(1-{\widehat{p}}_A)}{n_2}}\\ =(0.71-0.58)-2.5758\sqrt{\frac{0.71(1-0.71)}{100}+\frac{0.58(1-0.58)}{100}}\\ =-0.0427\end{array}$

Thus, the $99\%$ confidence interval is $(-0.0427,0.3027)$ .

This interprets that the true value of statistics lies within this confidence interval if there is hypothesis testing.