With an expected 20/80 split on survey answers, error 4% & 99% confidence interval what sample size do I need?

Confidence interval for the sample proportion is





phat +- z*sqrt(phat(1-phat)/n))





The error is determined by the plus or minus part of this expression. If you want a 99% confidence interval, then z = 2.576. Since you expect a 20/80 split, phat = 0.2. And you want this plus or minus number to be less than or equal to 0.04 (4%). So...





2.576*sqrt(0.2(1-0.2)/n) = 0.04





Then solve for n





sqrt(0.2(0.8)/n) = 0.04/2.576 = 0.01553


0.16/n = 0.01553^2 = 0.00024112


n/0.16 = 1/0.00024112 = 4147.36


n = 4147.36*0.16 = 663.5776,





which you should always round up, so you would need a sample of 664 or more to make the error less than or equal to 0.04.