91影视

It is given that $x_1=986$

$x_2=923$

$n_1=2253$

$n_2=2629$

$c=99\%=0.99$

The conditions are:

Random: Samples are independent random samples.

Independent: $2252$men are $<10\%$of all mean and it applies to women also.

Normal: In first sample, there are $983$success and $2253-983=1270$failures.

In second sample, there are $923$success and $2629-923=1706$failures. All are greater than ten.

All conditions are satisfied.

Sample proportion is ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{986}{2253}=0.438$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{923}{2629}=0.351$

For $1-伪=0.99$,using table, $z_{a/2}=2.575$

Confidence interval is $\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{a/2}脳\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.438-0.351)-2.575脳\sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}鈮�0.051$

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

$=(0.438-0.351)+2.575脳\sqrt{\frac{0.438(1-0.438)}{2253}+\frac{0.351(1-0.351)}{2629}}鈮�0.123$

Hence, confidence interval is$\left(0.051,0.123\right)$

As Confidence Interval is $(0.051,0.123)$. it does not contain zero. It is unlikely that population proportion are equal.

Hence, it gives convincing evidence for difference between population proportions.