Solve the Problem $\mathrm { E }$ With a Confidence Level Of

Multiple Choice

Solve the problem.
-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
$$E = z _ { \alpha / 2} \sqrt { \frac { p _ { 1 } q _ { 1 } } { \mathrm { n } _ { 1 } } + \frac { \mathrm { p } _ { 2 } \mathrm { q } _ { 2 } } { \mathrm { n } _ { 2 } } }$$
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 $n$ .

Use this approach to find the size of each sample if you want to estimate the difference between the proportions . and women who plan to vote in the next presidential election. Assume that you want $99 \%$ confidence that your no more than $0.03$ .

A) 3017
B) 2135
C) 3685
D) 1432