# Estimating error bounds for proportion estimation for a given sample size

Given an error bound, I can calculate the sample size to estimate proportion. E.g. for 10% the sample size is 100. But this relies on maximum variance being $\sqrt{\frac{0.25}{n}}$ which is achieved when p=0.5. Now say the result on the sample is 1%. So the actual p can be in the range (0%, 11%). But for this range, the variance is lower than the above and is $0.11*0.89/n=0.0979$. This yields that the error bound is in fact lower, leading to a lower maximum variance etc.

What is the formula to determine the actual error bound given a sample size and estimated proportion?

$1.96 \times \sqrt{1/4n}$ is the correct upper bound of the margin of error with 0.95 confidence. In some ways your search is futile because if $p$ is far from 0.5, no one would accept a large absolute margin of error, but would instead insist that you achieve a reasonable relative margin of error. The net result is that the needed sample size doesn't change very much as you lower $p$ and in fact can increase drastically. BBR discusses both the absolute and relative (odds) scales.