91Ó°ÊÓ

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:

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

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

$≈-0.0191$

and $\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

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

$≈0.1017$

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.