91Ó°ÊÓ

Given in the question that, $x_1=15,n_1=20,x_2=18,n_2=30;90\mathrm{\%}$confidence interval

We have to determine the needed confidence interval for the difference in the proportions of the two populations.

Let's compute the value of ${\tilde{p}}_1$as follow:

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1+1}{n_1+2} \\ =\frac{15+1}{20+2} \\ =0.73 \end{gathered}$$

Then calculate the value of ${\tilde{p}}_2$

role="math" localid="1651428777770" $$\begin{gathered} {\tilde{p}}_2=\frac{x_2+1}{n_2+2} \\ =\frac{18+1}{30+2} \\ =0.59 \end{gathered}$$

Let's find the value of $\mathrm{Î\pm }$first

$$\begin{gathered} 90=100(1−\mathrm{Î\pm }) \\ \mathrm{Î\pm }=0.1 \end{gathered}$$

We observed that the values of $z$at $\mathrm{Î\pm }/2$from the table of $z$score is $1.645$

For the difference between the two-population proportion, the needed confidence interval is determined as:

$\left({\tilde{p}}_1−{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{{\tilde{p}}_1\left(1−{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1−{\tilde{p}}_2\right)}{n_2+2}}=(0.73−0.59)Â\pm 1.645×\sqrt{\frac{0.73(1−0.73)}{20+2}+\frac{0.59(1−0.59)}{30+2}}$

$$\begin{gathered} =0.14Â\pm 0.129 \\ =0.011\text{to}0.269 \end{gathered}$$

We have to determine the needed confidence interval for the difference in the proportions of the two populations.

Using the two-proportions plus-four $z$-interval approach, the needed confidence interval for the difference between the two-population proportion is $0.011$to$0.269$.