Question

Smartphone Ownership. Refer to Problem 13 .

a. Find a $99\%$ confidence interval for the difference, $p_1-p_2$, between the proportions of men and women smartphone owners.

b. Interpret your answer in part (a) in terms of the difference in percentages of men and women smartphone owners.

Short answer

a). The gap in the proportion of men and women who own smartphones is somewhere between $0.0062$and $0.1142$, according to $99\%$confidence.

b). One can be $99\%$certain that the gap in smartphone ownership between men and women is between $0.62\%$and $11.42\%$.

Step-by-step solution

Part (a) Step 1: Given Information

Population $1$Men: role="math" localid="1651421541742" $n_1=1029,$

$x_1=607$.

Population $1$Women: $n_2=1223,$

$x_2=648$.

Part (a) Step 2: Explanation

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 endpoints for the confidence interval of $p_1-p_2$are -

$\left({\hat{p}}_1-{\hat{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$$\Rightarrow \mathrm{CI}=(0.590-0.5298)\pm 2.58\sqrt{\frac{0.590(1-0.590)}{1029}+\frac{0.5298(1-0.5298)}{1223}}$$\Rightarrow \mathrm{CI}=0.0602\pm (2.58\times 0.209)$

$\Rightarrow \mathrm{CI}=0.0602\pm 0.0540$

$\Rightarrow \mathrm{CI}=0.0062\text{to}0.1142$

Part (b) Step 2: Given Information

Population $1$Men: $n_1=1029,$

$x_1=607$.

Population $1$Women: $n_2=1223,$

$x_2=648$.

Part (b) Step 2: Explanation

The endpoints for the confidence interval of $p_1-p_2$are -

$\left({\hat{p}}_1-{\hat{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$\Rightarrow \mathrm{CI}=0.0062\text{to}0.1142$

$\Rightarrow \mathrm{CI}=0.62\%\text{to}11.42\%$