Where’s Egypt? In a Pew Research poll, $287$out of $522$randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When$520$ randomly selected U.S. women were asked, $233$ were able to do so.

a. Construct and interpret a $95\%$ confidence interval for the difference in the true

proportion of U.S. men and U.S. women who can identify Egypt on a map.

b. Based on your interval, is there convincing evidence of a difference in the true

proportions of U.S. men and women who can identify Egypt on a map? Justify your

answer.

It is given that $x_1=287$

$x_2=233$

$n_1=522$

$n_2=520$

$c=95\%=0.95$

The three conditions are:

Random: Samples are independent random samples.

Independent: $522$US men are less than $10\%$of population. Same is true for women.

Normal: Success in two samples are $287,233$and failures are $235,287$. All are greater than ten.

All conditions are satisfied.

Sample Proportion: ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{287}{522}=0.5498$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{233}{520}=0.4481$

Confidence Interval:

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

$=(0.5498-0.4481)-1.96\times \sqrt{\frac{0.5498(1-0.5498)}{522}+\frac{0.4481(1-0.4481)}{520}}$

$\approx -0.0191$

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

$=(0.5498-0.4481)+1.96\times \sqrt{\frac{0.5498(1-0.5498)}{522}+\frac{0.4481(1-0.4481)}{520}}$

$\approx 0.1017$

The confidence interval is$(-0.0191,0.1017)$

As the confidence interval $(-0.0191,0.1017)$does not contain zero. It is unlikely that population proportions are equal.

So, there is evidence of difference in true proportion of US men and women who can identify Egypt on map or not.