91Ó°ÊÓ

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40;80\mathrm{\%}$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:

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1+1}{n_1+2} \\ =\frac{18+1}{40+2} \\ =0.45 \end{gathered}$$

Consider the formula for ${\tilde{p}}_2$as follow:

${\tilde{p}}_2=\frac{x_2+1}{n_2+2}$

So, the value of ${\tilde{p}}_2$is:

localid="1651426030441" $$\begin{gathered} {\tilde{p}}_2=\frac{x_2+1}{n_2+2} \\ =\frac{30+1}{40+2} \\ =0.74 \end{gathered}$$

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

$$\begin{gathered} 80=100(1−\mathrm{Î\pm }) \\ \mathrm{Î\pm }=0.2 \end{gathered}$$

We observed that the values of $z$at $a/2$from the table of $z$score is $1.282$

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

$\left({\tilde{p}}_1−{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{{\tilde{p}}_1\left(1−{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1−{\tilde{p}}_2\right)}{n_2+2}}=(0.45−0.74)Â\pm 1.282×\sqrt{\frac{0.45(1−0.45)}{40+2}+\frac{0.74(1−0.74)}{40+2}}$

$$\begin{gathered} =−0.29Â\pm 0.131 \\ =−0.421\text{to}−0.159 \end{gathered}$$

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40;80\mathrm{\%}\text{confidence interval}$

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 technique, the needed confidence interval for the difference between the two-population proportion is $-0.421$to $-0.159$.