Calculate the sample size for a proportion and a mean, with finite population, design effect and response rate, the margin of error for a given n and the confidence interval, with the formula and the values filled in.
To estimate a proportion, the sample size is n₀ = z² · p · (1 − p) / E², where z is the critical value of the confidence level (1.96 for 95%), p the expected proportion and E the margin of error, all as fractions. With 95% confidence, a 5% margin and p = 50% (the worst case, which gives the largest sample) the result is 384.15, rounded up: 385 respondents. At 99% confidence it is 664; with a 3% margin, 1,068; with a 10% margin, 97. If the population is small and known, Cochran's finite population correction, n = n₀ / (1 + (n₀ − 1) / N), lowers the number: for N = 1,000 it drops to 278, for N = 500 to 218 and for N = 10,000 to 370; above roughly 100,000 the effect is negligible (383). To estimate a mean, n₀ = (z · σ / E)², as in the NIST/SEMATECH e-Handbook: with σ = 15 and a margin of 2, that is 217 observations; if σ was estimated, Student's t requires the smallest sample whose real margin fits the target, here 219. Real surveys adjust two more things: the design effect (deff), which multiplies n when sampling by clusters, and the response rate, which sets how many to invite (385 respondents at a 30% response rate need 1,284 invitations). In the other direction, the margin of error for a given n is E = z · √(p(1 − p) / n): with 1,000 respondents it is ±3.1 percentage points and with 2,000, ±2.2. For a proportion's confidence interval, the tool shows four methods side by side; the Wald interval, taught in many textbooks, fails with small samples or extreme proportions, and Wilson or Clopper-Pearson are more reliable. A mean's interval uses Student's t with n − 1 degrees of freedom. The comparison tab computes the size per group to detect a difference between two means or two proportions at the desired power, using the normal approximation. All formulas assume probability samples; with convenience samples, the margin of error does not have its usual statistical meaning.
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For a proportion, use n₀ = z² · p · (1 − p) / E². At 95% confidence (z = 1.96), a 5% margin and p = 50%, n₀ = 384.15, so 385 people. If the population is finite, apply n = n₀ / (1 + (n₀ − 1) / N). Then multiply by the design effect, if any, and divide by the expected response rate to learn how many to invite.
When unsure, use 50%: it is the case that needs the largest sample.
1 means simple random sampling. Cluster surveys use values above 1.
How many of those invited should answer. 100% means everyone answers.
With 385 respondents, the estimate of a proportion around 50% will have a margin of error of ±5 percentage points, at 95% confidence.
z = 1.96 (95%)n₀ = z² · p · (1 − p) / E² = 1.96² · 0.5 · 0.5 / 0.05² = 384.15⌈384.15⌉ = 385| Margin | 90% | 95% | 99% |
|---|---|---|---|
| ±1% | 6,764 | 9,604 | 16,588 |
| ±2% | 1,691 | 2,401 | 4,147 |
| ±3% | 752 | 1,068 | 1,844 |
| ±4% | 423 | 601 | 1,037 |
| ±5% | 271 | 385 | 664 |
| ±7% | 139 | 196 | 339 |
| ±10% | 68 | 97 | 166 |