Sample size for a margin of error

Choose n so that a confidence interval for a mean or proportion has at most a desired margin of error E, rounding up and optionally applying the finite population correction.

n_0 = \Big(\frac{z_{1-\alpha/2}\,\sigma}{E}\Big)^2

Mean

$\sigma$: planning value for the population SD (pilot study, past data)
$E$: desired margin of error (half-width)
$z_{1-\alpha/2}$: normal quantile for the confidence level

n_0 = \frac{z_{1-\alpha/2}^2\; p(1-p)}{E^2}

Proportion (p = 0.5 is the conservative choice)

$p$: planning value for the proportion; 0.5 gives the largest $n$

n = \frac{n_0}{1 + (n_0 - 1)/N}

Finite population adjustment; always round UP

$N$: population size

When to use

Planning a survey or study whose goal is a precise estimate. For detecting an effect with a test, plan for power instead (see errors and power and A/B sample size).

Assumptions

Simple random sampling; a sensible planning value for σ or $p$ ; the margin applies to the full sample (subgroups need their own $n$ ).

How to interpret

Halving the margin quadruples $n$ .
Round up to the next whole number; rounding down misses the target.
Plan for nonresponse: divide by the expected response rate.

Common mistakes

Rounding to the nearest integer.
Using an optimistic σ, which yields too small a sample.
Forgetting that the margin only covers sampling error, not bias.

Example

Mean with $\sigma = 15$ , $E = 3$ , 95%: $n_0 = (1.96 \times 15/3)^2 = 96.04$ , so $n = 97$ . Proportion with $E = 0.03$ and $p = 0.5$ : $n_0 = 1067.07$ , so $n = 1068$ ; from a population of $N = 5000$ the FPC reduces this to $n = 880$ .