Answer:
The minimum sample size is N=1537.
Step-by-step explanation:
We want to know the sample size to estimate the proportion of voters with a margin of error of 0.05, with a 95% of confidence.
The margin of error can be defined as:
[tex]e=UL-LL=(p+z*\sqrt{\frac{p(1-p)}{N}}) -(p-z*\sqrt{\frac{p(1-p)}{N}})=2z\sqrt{\frac{p(1-p)}{N}}[/tex]
We can calculate N from this
[tex]e=2z\sqrt{\frac{p(1-p)}{N}}\\\\\frac{p(1-p)}{N}=(\frac{e}{2z})^2\\\\N= (\frac{2z}{e})^2*p(1-p)[/tex]
The sample size will be calculated for p=0.5, which is the proportion that requires the larger sample size (maximum variance).
The z-value for a 95% CI is z=1.96.
The minimum sample size is then
[tex]N= (\frac{2z}{e})^2*p(1-p)=(\frac{2*1.96}{0.05})^2*0.5(1-0.5)=6146.56*0.5*0.5=1536.64[/tex]
The minimum sample size is N=1537.
