Sample size for a margin of error

Set the margin of error formula equal to your target and solve for n, which gives $n = \left(\frac{z^*}{ME}\right)^2 p^*(1-p^*)$, where $p^*$ (p-star) is a planning value for the proportion. With no prior estimate, use $p^* = 0.5$, because $p^*(1-p^*)$ is largest there and so gives the safe upper bound on the sample size. For example, for a margin of error of 0.03 at 95% confidence, $n = (1.96/0.03)^2(0.25) = (65.3333)^2(0.25) = 1067.11$, so you survey 1068 people. Always round up, never to the nearest whole number, since rounding down would leave the margin of error above your target.