Chapter 10.2, Problem 76E

(a)

To determine

To construct the 95% confidence interval that would estimate the proportion of all adults who use internet is:

Explanation of Solution

Given information:

  $\begin{array}{l}n=2092x=1318\\ \end{array}$

Where $n$ is number of adults.

And $x$ is the number users who used the internet.

Formula used:

The formula for 95% confidence interval for true difference in the population proportion $P$ is as follows:

  $\widehat{p}\pm Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Where $\widehat{p}$ is sample proportion of users use internet.

  $\widehat{p}=\frac{x}{n}$

Calculation:

For the given problem $p$ are obtained as below:

  $\widehat{p}=\frac{1318}{2092}=0.6300$

The critical value of Z at 5% level of significance is $Z_{\alpha /2}=Z_{0.025}=1.96$

Therefore, from the formula the confidence interval will be,

  $\begin{array}{l}(0.6300)\pm 1.96\sqrt{\frac{(0.6300)(1-0.6300)}{2092}}\\ 0.63\pm 0.02069\end{array}$

Hence 95% confidence interval to estimate the true differences in the proportion of users and non-users who expect business to have Web sites is $(0.60931,0.65069)$

(b)

To determine

To construct the 95% confidence interval that would estimate the true difference in the proportion of users and non-users who expect business to have Web sites:

Explanation of Solution

Given information:

  $\begin{array}{l}{n}_1=1318x_1=1041\\ n_2=774x_2=294\end{array}$

Where $n_1\&n_2$ are number of users and non-users.

Also $x_1\&x_2$ are number users who expect business to have websites and the number of non-users who expect business to have website respectively.

Formula used:

The formula for 95% confidence interval for true difference in the population proportion $(P_1-P_2)$ is as follows:

  $(\widehat{p_1}-\widehat{p_2})\pm Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\frac{(\widehat{p_1})(1-\widehat{p_1})}{n_1}+\frac{(\widehat{p_2})(1-\widehat{p_2})}{n_2}}$

Where $\widehat{p_1}\&\widehat{p_2}$ are sample proportions.

  $\widehat{p_1}=\frac{x_1}{n_1}$ and $\widehat{p_2}=\frac{x_2}{n_2}$

Calculation:

For the given problem $\widehat{p_1}\&\widehat{p_2}$ are obtained as below:

  $\begin{array}{l}\widehat{p_1}=\frac{1041}{1318}=0.7898\\ \widehat{p_2}=\frac{294}{774}=0.3798\end{array}$

The critical value of Z at 5% level of significance is $Z_{\alpha /2}=Z_{0.025}=1.96$

Therefore, from the formula the confidence interval will be,

  $\begin{array}{l}(0.7898-0.3798)\pm 1.96\sqrt{\frac{(0.7898)(1-0.7898)}{1318}+\frac{(0.3798)(1-0.3798)}{774}}\\ 0.41\pm 0.0406\end{array}$

Hence 95% confidence interval to estimate the true differences in the proportion of users and non-users who expect business to have Web sites is $(0.3694,0.4506)$