$x_1=30,n_1=80,x_2=15,n_2=20;95\%$confidence interval

Given in the question that,$x_1=30,n_1=80,x_2=15,n_2=20$

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

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

substitute the values of $x_1,n_1$

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

$=\frac{30+1}{80+2}$

$=0.23$

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

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

Substitute the values of$x_2,n_2$

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

$=\frac{15+1}{20+2}$

$=0.73$

Calculate the value of $\alpha$,

$95=100(1-\alpha )$

$\alpha =0.05$

The value of $z$ at $\alpha /2$ from the $z$-score table is $1.96$.

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.23-0.73)\pm 1.96$

$.\sqrt{\frac{0.23(1-0.23)}{80+2}+\frac{0.73(1-0.73)}{20+2}}$

$=-0.5\pm 0.270$

$=-0.707$to $-0.293$

As a result, the needed confidence interval for the proportional difference between the two populations is

$-0.707$to$-0.293\text{.}$

Given in the question that $x_1=30,n_1=80,x_2=15,n_2=20$

we need to compare result with the corresponding confidence interval found in parts of Exercises 11.100-11.105.

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 technique, the needed confidence interval for the difference between the two-population proportion is $-0.707$ to$-0.293$. The results are in line with the stated exercise outcomes, which have a $95\%$ confidence level.