91影视

Given

$x_1=46$

$n_1=63$

$x_2=27$

$n_2=56$

The sample proportion is calculated by dividing the sample size by the number of successes:

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{46}{63} \\ 鈮�0.730 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{27}{56} \\ 鈮�0.482 \end{gathered}$$

For confidence level $1-伪=0.90$, determine $z_{伪/2}=z_{0.05}$using table II (look up $0.05$in the table, the $z$-score is then the found $z$-score with the opposite sign):

$z_{伪/2}=1.645$

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

localid="1650449367380" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-z_{伪/2}路\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.730-0.482)-1.645\sqrt{\frac{0.730(1-0.730)}{63}+\frac{0.482(1-0.482)}{56}} \\ 鈮�0.105 \end{gathered}$$

localid="1650449382868" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)+z_{伪/2}路\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.730-0.482)+1.645\sqrt{\frac{0.730(1-0.730)}{63}+\frac{0.482(1-0.482)}{56}} \\ 鈮�0.391 \end{gathered}$$

Given

$x_1=46$

$n_1=63$

$x_2=27$

$n_2=56$

Result part (a)

$(0.105,0.391)$

The confidence interval does not contain $0$, then it is very unlikely that the population proportions are equal and thus there is convincing evidence of a difference between the population proportions.