Sample size for a margin of error

Fix the margin of error you want, set the margin of error formula equal to it, and solve for $n$: $n = \left(\frac{z^*}{ME}\right)^2 p^*(1-p^*)$, where $p^*$ (p-star) is a planning value for the proportion chosen before any data exist. With no prior estimate use $p^* = 0.5$, since $p^*(1-p^*)$ peaks at 0.25 there and so returns the largest, safest $n$. Nothing on the right comes from a sample, which is what makes this a planning calculation rather than inference.

For a margin of error of 0.05 at 95% confidence with $p^* = 0.5$: $n = (1.96/0.05)^2(0.25) = (39.2)^2(0.25) = 384.16$.

The answer is 385, and the round-up is a rule you can check rather than accept. Put both candidates back into the margin of error formula. At $n = 384$, $1.96\sqrt{0.25/384} = 0.050010$, which is above the target. At $n = 385$, $1.96\sqrt{0.25/385} = 0.049945$, which meets it. Rounding 384.16 to the nearest whole number gives 384 and misses, by a hair, but it misses. Since $n$ sits under a square root in the denominator, the margin falls as $n$ rises, so the first whole number at or past 384.16 is the smallest sample that works.

"So 385 people guarantees a margin of error of 0.05" reads the formula backwards. The planning value $p^*$ is a guess; the margin you finally report is rebuilt from the $\hat{p}$ (p-hat) your sample actually delivers. Planning with $p^* = 0.5$ can only overshoot: if those 385 people split 40 to 60, the realized margin is $1.96\sqrt{0.40(0.60)/385} = 0.048936$, narrower than promised. Plan with a smaller $p^*$ taken from an earlier study and that protection is gone, because a sample proportion landing nearer 0.5 than your planning value gives a wider interval than you budgeted for.

One edge case: when the formula lands exactly on a whole number, nothing rounds. At $ME = 0.02$ and $p^* = 0.5$, $n = (98)^2(0.25) = 2401$ exactly, and 2401 returns a margin of 0.020000.