91Ó°ÊÓ

Sample size is $n=1029$.

Number of success is $x=607$.

Sample proportion would be $\hat{p}=\frac{x}{n}$

$=\frac{607}{1029}$

$=0.590$

For men:

A significance level of $\mathrm{Î\pm }$,

To find $z_{\mathrm{Î\pm }/2}$using table II.

The significance level is $1\%$. i.e. $\mathrm{Î\pm }=0.01$.

Using table II,

$z_{\mathrm{Î\pm }/2}=z_{0.01/2}$

$=z_{0.005}$

$=2.58.$

The confidence interval for $p$is

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

$⇒\mathrm{CI}=0.590Â\pm 2.58\sqrt{\frac{0.590(1-0.590)}{1029}}$

$⇒\mathrm{CI}=0.590Â\pm 0.04$

$⇒\mathrm{CI}=0.55\text{to}0.63$

$⇒\mathrm{CI}=55\%\text{to}63\%$

For women:

Sample size is $n=1223$.

Number of success is $x=648$.

Sample proportion would be $\hat{p}=\frac{x}{n}$

$=\frac{648}{1223}$

$=0.5298$

A significance level of $\mathrm{Î\pm }$,

To find $z_{\mathrm{Î\pm }/2}$using table II.

Significance level is $1\%$.

i.e. $\mathrm{Î\pm }=0.01$.

Using table II,

$z_{\mathrm{Î\pm }/2}=z_{0.01/2}$

$=z_{0.005}$

$=2.58.$

The confidence interval for $p$is given as-

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

$⇒\mathrm{CI}=0.5298Â\pm 2.58\sqrt{\frac{0.5298(1-0.5298)}{1223}}$

$⇒\mathrm{CI}=0.4930\text{to}56.66$

$⇒\mathrm{CI}=59.30\%\text{to}56.66\%$

One could assume that the data does not provide adequate information to indicate that men and women have different percentages of smartphone owners.