91Ó°ÊÓ

Given in the question that, $x_1=14,n_1=20,x_2=8,n_2=20;90\mathrm{\%}$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:

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1+1}{n_1+2} \\ =\frac{14+1}{20+2} \\ =0.68 \end{gathered}$$

Then determine the value of ${\tilde{p}}_2$:

$$\begin{gathered} {\tilde{p}}_2=\frac{x_2+1}{n_2+2} \\ =\frac{8+1}{20+2} \\ =0.41 \end{gathered}$$

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

$\left({\tilde{p}}_1−{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{{\tilde{p}}_1\left(1−{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1−{\tilde{p}}_2\right)}{n_2+2}}=(0.68−0.41)Â\pm 1.645×\sqrt{\frac{0.68(1−0.68)}{20+2}+\frac{0.41(1−0.41)}{20+2}}$

$$\begin{gathered} =0.27Â\pm 0.238 \\ =0.032\text{to}0.508 \end{gathered}$$

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 $0.032$to $0.508$.