The Sample Size Needed to Estimate the Difference Between Two $\mathrm { E }$

The sample size needed to estimate the difference between two population proportions to within a margin of error $\mathrm { E }$ with a confidence level of $1 - \alpha$ can be found as follows: in the expression
$$\mathrm { E } = \mathrm { z } _ { \alpha } / 2 \sqrt { \frac { \mathrm { p } _ { 1 } \mathrm { q } _ { 1 } } { \mathrm { n } _ { 1 } } + \frac { \mathrm { p } _ { 2 } \mathrm { q } _ { 2 } } { \mathrm { n } _ { 2 } } }$$
replace $\mathrm { n } _ { 1 }$ and $\mathrm { n } _ { 2 }$ by $\mathrm { n }$ (assuming both samples have the same size) and replace each of $\mathrm { p } _ { 1 } , \mathrm { q } _ { 1 } , \mathrm { p } _ { 2 }$ , and $\mathrm { q } _ { 2 }$ by $0.5$ (because their values are not known) . Then solve for $\mathrm { n }$ .

Use this approach to find the size of each sample if you want to estimate the difference between the proportions of men and women who plan to vote in the next presidential election. Assume that you want $99 \%$ confidence that your error is no more than $0.05$ .

A) 1087
B) 769
C) 516
D) 1327