Sample size calculator: determine how many survey respondents you need for a target confidence level and margin of error. Free, no signup, runs in your browser.
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Every figure above is calculated locally in your browser from the assumptions shown. No inputs are sent anywhere. See the methodology section below for the formulas used.
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Sample size calculator: determine how many survey respondents you need for a target confidence level and margin of error. Free, no signup, runs in your browser.
Transparent methodology
How this calculator works
Reviewed August 2026 · BoringToolsKit Editorial Team
Formula
n = z^2 x p x (1 - p) / margin^2, with z = 1.96 at 95% confidence and p = 0.5 the conservative proportion.
Worked example
A survey with +/-5% margin at 95% confidence needs 385 respondents.
BoringToolsKit. “Sample Size Calculator.” boringtoolskit.com/sample-size-calculator/ (reviewed August 2026). Free to reference in articles, syllabi, and answer posts with a link.
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Worked example: 5 percent margin on a population of 1,000
Take the defaults: 95 percent confidence, 5 percent margin, and a 50/50 split for p. The z value for 95 percent is 1.96, so the uncorrected size is 1.96 x 1.96 x 0.5 x 0.5, all divided by 0.05 x 0.05. The numerator is 3.8416 x 0.25 = 0.9604, and the denominator is 0.0025, giving 384.16. Rounding up gives 385 people to survey.
Now suppose the whole group you care about is 1,000 people, such as all employees at a company. The finite population correction divides by 1 + (384.16 - 1) / 1,000 = 1.38316, so 384.16 / 1.38316 = 277.7, which rounds up to 278. The correction trimmed 107 people from the requirement because sampling a large share of a small population already reduces uncertainty. With a population of 10,000 the same inputs give 369, and above roughly 100,000 the correction stops mattering much: you get 384, essentially the open-population answer.
Where the 50 percent assumption costs and saves you
The 50 percent default is the worst case for p(1-p), which peaks at 0.25 there. If you genuinely expect 30 percent of respondents to answer yes, p(1-p) = 0.21, and a 5 percent margin at 95 percent confidence needs only 1.96 x 1.96 x 0.21 / 0.0025 = 323.9, so 324 people instead of 385. Estimating a rare attribute like a 2 percent defect rate needs just 77 people at the same precision. Guessing p badly in the safe direction (using 0.5) only costs extra surveying, never precision.
Tightening the margin is the expensive lever: sample size scales with 1 divided by the margin squared. Cutting the margin from 5 percent to 3 percent at p = 0.5 raises the requirement from 385 to 0.9604 / 0.0009 = 1,067.1, so 1,068 people, nearly triple. Going to 2 percent needs 2,401. Also note what this tool does not cover: it assumes simple random sampling, so stratified, cluster, or quota designs need a design effect multiplier, and it says nothing about nonresponse. If you expect 20 percent of invitees to ignore the survey, divide the requirement by 0.8, turning 385 invites into a panel of about 482.