91Ó°ÊÓ

Given in the question that,$x_1=18,n_1=30,x_2=10,n_2=20$

we need to use the two-proportions plus-four z-interval procedure to find the required confidence interval for the difference between the no population proportions.

The given values are, $x_1=18,n_1=30,x_2=10,n_2=20$, and $95\%$ confidence interval.

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

The value of ${\tilde{p}}_1$is calculated as,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

$=\frac{18+1}{30+2}$

$=0.59$

The formula for ${\tilde{p}}_2$is given by,

${\tilde{p}}_2$

The value of ${\tilde{p}}_2$is calculated as,

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

$=\frac{10+1}{20+2}$

$=0.5$

The value of $z$at $\mathrm{Î\pm }/2$from the z-score table is $1.96$.

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

$\left({\tilde{p}}_1−{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}â‹\dots \sqrt{\frac{{\tilde{p}}_1\left(1−{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1−{\tilde{p}}_2\right)}{n_2+2}}=(0.59−0.5)Â\pm 1.96$

role="math" localid="1651309362069" $.\sqrt{\frac{0.59(1−0.59)}{30+2}+\frac{0.5(1−0.5)}{20+2}}$

$=0.09Â\pm 0.270$

role="math" localid="1651309497132" $=−0.018$to$0.36$

As a result, the $-0.018$ to $0.36$ confidence interval for the difference in two-population proportion is necessary.

Given in the question that,$x_1=18,n_1=30,x_2=10,n_2=20$

we need to compare result with the corresponding confidence interval found in parts$\left(d\right)$of Exercises 11.100-11.105

The given values are, $x_1=18,n_1=30,x_2=10,n_2=20$, and $95\%$confidence interval.

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

The formula for ${\tilde{p}}_2$is given by,

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

Using the two-proportions plus-four z-interval approach, the needed confidence interval for the difference between the two-population proportion is $-0.018$ to $0.36$. The results are in line with the stated exercise outcomes, which have a $95\%$ confidence level.