Find the margin of error for a survey from sample size and confidence level
Enter your sample size, confidence level and the response proportion to find the margin of error for a survey or poll. The result updates as you change the inputs.
e = z·√(p·(1−p)/n), with a finite-population correction when a population size is given.
The margin of error is the radius of the confidence interval around a survey result. If a poll reports 52% support with a ±3% margin at 95% confidence, the true value is most likely between 49% and 55%. It captures sampling error — the random variation from surveying a sample rather than the whole population — not bias from how the survey was designed or conducted.
For a proportion, the margin of error is e = z · √(p · (1 − p) ÷ n), where z is the critical value for the confidence level, p is the sample proportion, and n is the sample size. When the sample is a large fraction of a known population, the finite-population correction √((N − n) ÷ (N − 1)) shrinks the margin.
At 95% confidence (z ≈ 1.96), with p = 50% and n = 1,000: e = 1.96 × √(0.25 ÷ 1000) ≈ 3.1%. So a result from 1,000 respondents carries roughly a ±3% margin — typical for a national poll.
Because the sample size is under a square root, the margin shrinks slowly: quadrupling the sample only halves the margin. That is why polls rarely exceed a few thousand respondents — the extra precision is not worth the cost. To plan the sample size for a target margin instead, use the sample size calculator.
Multiply the z critical value for your confidence level by the square root of p·(1−p)/n, where p is the proportion and n is the sample size.
If you have a result, use it. If not, 50% gives the largest (most conservative) margin of error.
It helps, but with diminishing returns — because n is under a square root, you must quadruple the sample to halve the margin of error.