91Ó°ÊÓ

$n=2092$,

$x=1318$

Where $n$is number of adults.

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

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

$\hat{p}Â\pm Z_{\mathrm{Î\pm }/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

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

$\hat{p}=\frac{x}{n}$

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

localid="1650370085599" $\hat{p}=\frac{1318}{2092}$

$=0.6300$

The critical value of $Z$at $5\%$level of significance is localid="1650370192109" $Z_{\mathrm{Î\pm }/2}=Z_{0.025}$

$=1.96$

$(0.6300)Â\pm 1.96\sqrt{\frac{(0.6300)(1-0.6300)}{2092}}$

$0.63Â\pm 0.02069$

$n_1=1318,$

$x_1=1041$

$n_2=774,$

$x_2=294$

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 respectivelv.

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

$\left(\widehat{p_1}-\widehat{p_2}\right)Â\pm Z_a/2\sqrt{\frac{\left(\widehat{p_1}\right)\left(1-\widehat{p_1}\right)}{n_1}+\frac{\left(\widehat{p_2}\right)\left(1-\widehat{p_2}\right)}{n_2}}$

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

localid="1650370283641" $\widehat{p_1}=\frac{x_1}{n_1}\text{and}$

$\widehat{p_2}=\frac{x_2}{n_2}$

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

localid="1650370304988" $\widehat{p_1}=\frac{1041}{1318}$

$=0.7898$

localid="1650370333928" $\widehat{p_2}=\frac{294}{774}$

$=0.3798$

The critical value of $Z$at $5\%$level of significance is localid="1650370356144" $Z_{\mathrm{Î\pm }/2}=Z_{0.025}$

$=1.96$

Therefore, from the formula the confidence interval will be,$(0.7898-0.3798)Â\pm 1.96\sqrt{\frac{(0.7898)(1-0.7898)}{1318}+\frac{(0.3798)(1-0.3798)}{774}}$

$0.41Â\pm 0.0406$.