In each of Exercises 11.122-11.127, we have given the numbers of successes and the sample sizes for simple random samples for independent random samples from two populations. In each case,

a. use the rwo-proportions plus-four $z$-interval procedure to find the required confidence interval for the difference between the two population proportions.

b. compare your result with the corresponding confidence interval found in parts (d) of Exercises 11.100-11.105, if finding such a confidence interval was appropriate.

$x_1=14,n_1=20,x_2=8,n_2=20;90\mathrm{\%}$confidence interval

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 $\alpha$first :

$$\begin{gathered} 90=100(1-\alpha ) \\ \alpha =0.1 \end{gathered}$$

We observed that the values of $z$at $\alpha /2$ from the table of $z$score is $1.645$

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}\cdot \sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.68-0.41)\pm 1.645\times \sqrt{\frac{0.68(1-0.68)}{20+2}+\frac{0.41(1-0.41)}{20+2}}$

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

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 $0.508$.