Use your result from Exercise $11.132$to show that a $(1-\alpha )$level confidence interval for the difference between two population proportions that has a margin of error of at most $E$can be obtained by choosing

$n_1=n_2=0.5{\left(\frac{z_{\alpha /2}}{E}\right)}^2$

rounded up to the nearest whole number.

Given in the question that, we need to to show that a $(1-\alpha )$level confidence interval for the difference between two population proportions that has a margin of error of at most $E$can be obtained by choosing

$n_1=n_2=0.5{\left(\frac{z_{\alpha /2}}{E}\right)}^2$

by using the result from exercise 11.132

The expressions given are$n_1=n_2=0.50{\left(\frac{z_{u/2}}{E}\right)}^2$

When estimating the proportional differences between two populations, the margin of error is,

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

The margin of error in estimating the proportional differences between two populations is,

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

Substitute $n_1=n_2=n$and

${\tilde{p}}_1\left(1-{\tilde{p}}_1\right)=0.25$

${\tilde{p}}_2\left(1-{\tilde{p}}_2\right)=0.25$

$E=z_{\alpha /2}\sqrt{\frac{0.25}{n}+\frac{0.25}{n}}$

${\left(\frac{E}{z_{\alpha /2}}\right)}^2=\frac{0.25}{n}+\frac{0.25}{n}$

${\left(\frac{E}{z_{\alpha /2}}\right)}^2=\frac{0.25+0.25}{n}$

${\left(\frac{E}{z_{\alpha /2}}\right)}^2=\frac{0.50}{n}$

Simplify even more,

$\frac{n}{0.50}={\left(\frac{z_{\alpha /2}}{E}\right)}^2$

$n=0.50{\left(\frac{z_{\alpha /2}}{E}\right)}^2$

$n_1=n_2=0.50{\left(\frac{z_{\alpha /2}}{E}\right)}^2$

By selecting $n_1=n_2=0.50{\left(\frac{z_{a/2}}{E}\right)}^2$, a $(1-\alpha )$level confidence interval for the difference between two population proportions with a margin of error of at most $E$can be generated.