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.
更新于 已验证的示例：6
A random sample of 385 gives a 95% confidence interval no wider than ±5 percentage points whatever the true proportion turns out to be (p = 50% is the worst case). Halving the margin roughly quadruples the sample; the formula assumes simple random sampling and says nothing about 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.
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.
核验来源：Cochran's formula 1.959964² × 0.25 / 0.05² = 384.146 (Python NormalDist); the widely published figure is 385
核验来源：Finite population correction n₀/(1 + (n₀ − 1)/N) in Python
核验来源：Finite population correction in Python: 384.146/(1 + 383.146/100)
核验来源：(1.959964 × 15 / 3)² = 96.04 in Python
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% is the worst case, 385 is enough for any true proportion in 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 that sample is big enough whatever the true value turns out to be. 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.
准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例：6。 例如，“Survey at 95%, ±5%, p = 50% (defaults)”根据Cochran's formula 1.959964² × 0.25 / 0.05² = 384.146 (Python NormalDist); the widely published figure is 385进行核验。
NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.2 Sample sizes required; Cochran, W. G. (1977). Sampling Techniques, 3rd ed., §4.4 (sample size for proportions and the finite population correction); Cohen, J. (1988). Statistical Power Analysis for the Behavioral Sciences, 2nd ed., Table 2.4.1.
此计算器包含 6 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。
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