Sample size calculator for surveys and studies: n for a margin of error on a proportion or mean, and the per-group n for a target power.
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The planning formula requires 385 observations for a nominal 95% confidence level and margin ±5 percentage points. Using p = 50% gives the largest sample requirement under this approximation. Without population correction, halving the margin quadruples the unrounded sample size. This assumes simple random sampling and the stated variance; it does not account for non-response or bias.
Always rounded up: rounding down would miss the target margin.
For a survey proportion the sample size is n₀ = z²p(1 − p)/E², where z is the critical value for the confidence level and E the margin of error; for a mean it is (zσ/E)². A known population size N reduces this through the finite population correction, n = n₀/(1 + (n₀ − 1)/N). To compare two means, each group needs 2(z₁₋α/₂ + z₁₋β)²/d², where d is the difference to detect divided by σ and n_g is the size per group. This is a normal approximation, not an exact t-test power calculation.
Survey planners and researchers use it before collecting data. The default, 95% confidence and ±5 points with p = 50%, needs 384.1 respondents, rounded up to 385; for a population of 10,000 the correction lowers that to 370.
Results are always rounded up. The formulas assume simple random sampling; cluster samples, weighting and non-response all call for a larger sample.
Cochran's formula gives the sample size for estimating a proportion: n₀ = z² × p(1 − p) ÷ E². At 95% confidence (z = 1.96), an expected 50% and a margin of ±5 points, n₀ = 1.96² × 0.25 ÷ 0.05² ≈ 384.2, which rounds up to 385 respondents.
Worked by hand:
The formula assumes a simple random sample of independent responses and uses a normal approximation. Cluster samples, weighting and non-response all call for a larger sample.
The margin shrinks with the square root of n, so halving it takes four times the sample. At 95% confidence and p = 50%:
The confidence level moves the requirement too. For ±5 points, 90% confidence needs 271 and 99% needs 664.
The ±2.5 figure is 1,537, not 4 × 385 = 1,540, because the factor of four applies before rounding: 4 × 384.146 = 1,536.6.
p(1 − p) is largest at p = 0.5, so 50% gives the biggest requirement for a given margin. Use it unless a reliable earlier estimate exists.
A prior estimate of 20% lowers the requirement to 246 for ±5 points at 95%. If the true share turns out nearer 50%, the achieved margin will be wider than planned. 20% and 80% give the same answer, since p(1 − p) is symmetric.
Only when the sample is a noticeable share of the population. The finite population correction is n = n₀ ÷ [1 + (n₀ − 1) ÷ N], rounded up:
A sample of 1,000 gives about ±3.1 points whether the population is 100,000 or 300 million. When the requirement is over half the population, a census may be simpler.
The correction applies to random sampling without replacement from a defined population of N. It is no discount for having only N volunteers or contacts.
For a mean, the formula is n₀ = (z × σ ÷ E)², with the standard deviation σ and the margin E in the same units. With σ = 15, E = 3 and 95% confidence, n₀ = (1.959964 × 15 ÷ 3)² = 96.04, so n = 97.
σ is the spread of individual observations, usually from a pilot study or earlier data, not the standard error of a mean. Underestimating σ underestimates n, and n grows with σ²: doubling σ quadruples the sample (NIST).
To detect a difference δ between two equal-sized groups with common SD σ, each group needs n = 2(z₁₋α/₂ + z₁₋β)² ÷ d², where d = δ ÷ σ and 1 − β is the power.
Detecting δ = 7.5 when σ = 15 gives d = 0.5. With a two-sided α of 0.05 and 80% power, n = 2 × (1.959964 + 0.841621)² ÷ 0.25 = 62.79, so 63 per group and 126 in total.
Power is the probability of detecting the effect if it is real. Cohen (1988) proposed 80% as a conventional minimum; 90% power raises the example to 85 per group.
A one-sided test at the same α needs 50 per group, but the direction must be fixed before any data are seen.
This is a normal approximation. Cohen's t-based table gives 64 per group for the same case, and very small samples call for an exact t-based calculation.
| Scenario | Required n |
|---|---|
| Default | 385 |
| Half margin | 1537 |
| Population ten thousand | 370 |
| Population hundred | 80 |
| Mean | 97 |
| Power | 63 |
| Power one sided | 50 |
"Default" is the 95%, ±5-point survey at p = 50%, and the next three rows change one input each: margin ±2.5, N = 10,000 and N = 100. "Mean" is σ = 15 with E = 3. The power rows are per group, for d = 0.5, α = 0.05 and 80% power.
For the interval itself once data are in, use the confidence interval calculator. To test the two groups after the study, use the t-test calculator.
Checked against: Cochran's formula 1.959964² × 0.25 / 0.05² = 384.146; the widely published figure is 385
Checked against: Independent calculation
Checked against: Independent calculation
Checked against: Independent calculation
It is what Cochran's formula gives for 95% confidence, a ±5-point margin and p = 50%: 1.96² × 0.25 / 0.05² = 384.1, rounded up to 385. Because p = 50% maximizes p(1 − p), 385 is the conservative requirement under this normal-approximation formula for a large population. A population of 1,000 needs only 278, because each respondent is a larger share of it.
Only when the sample is a noticeable fraction of the population. The finite population correction turns the 385 needed for ±5 points at 95% into 370 for a population of 10,000 and 80 for a population of 100, while any population above 100,000 needs 383 to 385. A sample of 1,000 gives about ±3.1 points whether the population is 100,000 or 300 million.
Use 50% unless you have a reliable prior estimate. p(1 − p) is largest at p = 0.5, so this formula gives its largest sample requirement at that value. A prior estimate lowers the requirement: at 95% confidence and ±5 points, an expected 20% needs 246 respondents instead of 385. If the guess is wrong, the achieved margin will be wider than planned.
The margin shrinks with the square root of n, so halving it needs four times the sample. At 95% confidence and p = 50%, ±5 points needs 385 respondents, ±3 points 1,068, ±2 points 2,401 and ±1 point 9,604. Choosing 90% confidence instead of 95% lowers the ±5-point figure to 271.
Power is the probability that a test detects an effect of a stated size when that effect is real; 1 − power is the Type II error rate β. Cohen (1988) proposed 80% as a conventional minimum. Detecting a difference of 7.5 when σ = 15 (d = 0.5) with a two-sided test at α = 0.05 needs 63 per group for 80% power by the normal approximation; Cohen's t-based table gives 64.
This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.