"How many people do I need to survey?" has a real answer, and it isn't "as many as possible." It falls out of three inputs: how confident you want to be, how tight your margin of error should be, and what you already know about the answer. Here's the formula, the correction for small populations, and worked examples you can check by hand.
A survey estimates a number: the share of customers who'd cancel, the average time on hold. Two things can be wrong with an estimate: it's off by some amount (that's the margin of error), and the true value might sit outside your interval some fraction of the time (that's what confidence level measures). Sample size is the dial that improves both at once. Bigger sample, tighter margin at the same confidence.
The catch is the price curve. Margin of error shrinks with the square root of n, so cutting the margin in half quadruples the sample. This one fact explains most survey design decisions, including why national polls interview about a thousand people and stop.
For estimating a proportion (a percentage), the standard tool is Cochran's formula:
n0 = z² x p(1-p) / e²
Run the classic case: z=1.96, p=0.5, e=0.05. Then n0 = (1.96)² x 0.25 / (0.05)² = 3.8416 x 0.25 / 0.0025 = 0.9604 / 0.0025 = 384.16. Round up (always up, since 384.16 people don't exist): 385 responses. That's the famous number behind "margin of error ±5 points at 95% confidence."
Cochran's formula assumes the population is effectively infinite. When you're surveying a defined small group, the finite population correction applies:
n = n0 / (1 + (n0 − 1)/N)
where N is the population size. Say your company has 1,000 employees. Take the 384.16 from above: n = 384.16 / (1 + 383.16/1000) = 384.16 / 1.38316 = 277.7, round up to 278. Survey 278 randomly chosen employees and you know the company-wide answer to ±5% at 95% confidence.
| Population N | 95% / 5% sample | Sampling fraction |
|---|---|---|
| 100 | 80 | 80% |
| 500 | 218 | 44% |
| 1,000 | 278 | 28% |
| 5,000 | 357 | 7.1% |
| 10,000 | 370 | 3.7% |
| 1,000,000 | 385 | 0.04% |
Read that last row again. A million-person population needs 385 people — 0.04% of it. Precision comes from the sample's absolute size, not the fraction sampled. Skip the hand arithmetic entirely with the sample size calculator, which applies the correction automatically.
Percentages aren't the only thing surveys measure. When your statistic is an average, the formula becomes:
n = (z x s / e)²
s is the standard deviation, which you take from a pilot run, last year's data, or a comparable study. e is the margin in the same units as s. Worked example: you want average delivery time to within ±3 minutes at 95% confidence, and last quarter's times had a standard deviation of 15 minutes. n = (1.96 x 15 / 3)² = 9.8² = 96.04, so 97 timed deliveries. Need only ±5 minutes? (1.96 x 15/5)² = 34.6, so 35.
The mean formula hides the same lesson as the proportion one: sample size squares with the z you demand. Dropping from 95% to 99% confidence multiplies the workload by (2.576/1.96)² = 1.73 with zero change in the margin.
The formulas return completed responses you can use, not emails sent. Bridge the gap with your response rate: need 385 completes at a 20% response rate? Invite 385 / 0.20 = 1,925 people. In experiments and clinical-style studies, inflate for expected attrition the same way, plus 5-10% buffer for unusable data.
And remember what the math assumes: a simple random sample. A convenience sample of 2,000 Twitter replies doesn't have a ±2.2% margin of error; it has an unknown bias that no n fixes. If you're rusty on the probability underneath, the probability calculator is good warm-up, and once data lands, the average calculator handles the descriptive side.
Confidence level, margin, population size. Get the required sample for both proportions and means.
Sample Size Calculator →One more habit worth building: sanity-check any published poll's claimed margin with the back-solve. At n=1,000, the margin at 95% is 1.96 x √(0.25/1000) = ±3.1%, which is exactly what serious pollsters report. If a press release claims ±2% from 600 respondents, something else is going on.
At the standard settings, 95% confidence with a 5% margin of error and p=0.5, you need 385 completed responses for a large or unknown population. If you can name your population and it's small, apply the finite population correction: 278 for 1,000 people, 218 for 500, 80 for 100. Completed responses, not invitations; if your response rate is 20%, invite five times as many people.
Because a random sample's precision depends on its own size, not on what fraction of the population it represents. Going from 10,000 to 10 million people moves the requirement from 370 to 385 at 95%/5%. Sample size scales with 1/e², the squared margin of error, and not with N. That's the counterintuitive core of survey sampling.
Halve the margin and you quadruple the sample. At 95% with p=0.5, a 10% margin needs 97 responses, 5% needs 385, 3% needs 1,068, and 1% needs 9,605. The margin enters the formula squared, so precision gets expensive fast, which is why national pollsters live around n=1,000 and a ±3.1% margin.