A political candidate has asked you to to conduct a poll to determine what percentage of people support her. If the candidate only wants a 1.5% margin of error at a 97.5% confidence level, what size

Question

A political candidate has asked you to to conduct a poll to determine what percentage of people support her. If the candidate only wants a 1.5% margin of error at a 97.5% confidence level, what size of sample is needed?

Answer 1

1st thing, I'd argue w the candidate that 97.5% confidence is anal retentive and that 80% wd be sufficient, given how opinions shift over time and it's better to focus resources on quality of data over quantity. But if E=z(alpha/2)*Sqrt(p0*(1-p0)/n) then (E/z)^2=p0*(1-p0)/n and (z/E)^2*p0*(1-p0)=n. If E=.015, and to be conservative p0=.5, and z=the excel output of Norm.s.inv(.9875) or 2.241 then one must calculate (2.241/.015)^2*.5*.5=5581, after rounding to the nearest integer. If an 80% conf interval were used instead and one conservatively presumed that p0=.35, it would be (1.282/.015)^2*.4*.6=1,662, or less than a third of the sample size.