Smartphone Ownership. The Pew Internet \& American Life Project conducted a survey of smartphone ownership. One aspect of the study involved the gender of smartphone owners. Of $1029$ sampled men, $607$ owned a smartphone; and of $1223$ sampled women, $648$ owned a smartphone. At the $1\%$ significance level, do the data provide sufficient evidence to conclude that a difference exists in the percentages of smartphone owners between men and women?

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 $\alpha$,

To find $z_{\alpha /2}$using table II.

The significance level is $1\%$. i.e. $\alpha =0.01$.

Using table II,

$z_{\alpha /2}=z_{0.01/2}$

$=z_{0.005}$

$=2.58.$

The confidence interval for $p$is

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

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

$\Rightarrow \mathrm{CI}=0.590\pm 0.04$

$\Rightarrow \mathrm{CI}=0.55\text{to}0.63$

$\Rightarrow \mathrm{CI}=55\%\text{to}63\%$

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 $\alpha$,

To find $z_{\alpha /2}$using table II.

Significance level is $1\%$.

i.e. $\alpha =0.01$.

Using table II,

$z_{\alpha /2}=z_{0.01/2}$

$=z_{0.005}$

$=2.58.$

The confidence interval for $p$is given as-

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

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

$\Rightarrow \mathrm{CI}=0.4930\text{to}56.66$

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

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