Sample size calculator
A 95% confidence level with a ±5% margin and no assumption about the answer needs 385 responses, whatever the population. 384 is the figure most often quoted, from rounding the 384.15 rather than rounding it up, and it is why so many surveys land at "about 400 respondents". Assuming 50% is deliberately the worst case: any other expected proportion needs fewer.
Normal-approximation sample size for a single proportion. It is unreliable when the expected proportion is close to 0 or 100 per cent, where the interval it is sized for can run past the ends of the scale.
Sample size for a proportion is z² × p(1−p) ÷ margin², rounded up. At 95% confidence with a ±5% margin and no prior expectation, that is 385 responses, and the population size barely changes it unless the population is small.
How to work out sample size
The formula sizes one proportion at one point in time, using the normal approximation to the binomial. Read it as the number of responses that makes a 95 per cent interval about the answer no wider than the margin you asked for, on the assumption that the responses are a random draw from the group you care about. Each of those clauses is doing work.
The expected proportion is not a cosmetic setting
Fifty per cent maximises p(1−p), so leaving it there gives the largest sample any answer could require and the number is safe whatever comes back. Move it and the requirement drops fast: 10 per cent needs 139 responses for the same ±5 points, and 1 per cent needs 16.
Sixteen is where the approximation quietly stops being arithmetic anyone should act on. An interval of 1 per cent plus or minus 5 points runs from −4 to 6, which is not a range a proportion can occupy, and with sixteen people at a one-in-a-hundred rate the expected number of positive answers is 0.16. The usual guard is to require both n × p and n × (1 − p) to be at least about 5 before trusting a normal approximation for a proportion; at the extremes here they are nowhere near it. If you are measuring something rare, size the study on the count of rare events you need to observe, not on a percentage margin.
The population correction, in real numbers
The second size row applies the finite population correction, n ÷ (1 + (n − 1) ÷ N), which matters only when your sample would be a meaningful slice of the whole group. At ±5 per cent and 95 per cent confidence the unadjusted 385 becomes 80 in a population of 100, 218 in 500, 278 in 1,000, 370 in 10,000 and 383 in 100,000. Past a few tens of thousands the correction has nothing left to do.
This is the source of the standard surprise that a national poll and a town poll need roughly the same number of people. Precision comes from how many you asked, not from what fraction of the group they were. It also carries a warning in the other direction: when the correction bites, you are sampling a large share of a small group, and the people who decline are then a large share too.
Sampling error is the smaller problem
The number here prices exactly one thing, the randomness of who happened to land in the sample. It says nothing about non-response bias, a leading question, an unrepresentative sample frame, or a panel that has learned what researchers want to hear. Ten thousand self-selected website visitors are worse evidence than four hundred properly randomised responses, and no arithmetic on this page will reveal that. Treat it as the floor below which the sampling error alone makes the result useless, rather than as the size at which the result becomes trustworthy.
What people use it for
- Planning a survey to a stated margin of error
- Sizing one cell of an A/B test
- Checking whether a published poll had the sample to say what it said
- Deciding when to stop collecting responses
- Costing fieldwork against the precision it buys
Questions
385 for ±5% at 95% confidence, 1,068 for ±3%, and 9,604 for ±1%. Each is the worst case, at an expected proportion of 50%.
Only when the sample is a real fraction of it. Against a population of 1,000 the 385 falls to 278; against 100,000 it falls to 383.
Because p(1−p) is largest there, so it needs the biggest sample. Any other expected answer needs fewer responses.
No. A large biased sample is confidently wrong, and adding respondents from the same skewed frame makes it more confident rather than less wrong.
Five points is common for general research and three for political polling. The honest answer is whatever margin still lets you act differently on the two ends of the interval.
Because part of a respondent cannot be collected. The ±5% case is 384.15 before rounding, so 385 is the first whole number that meets the target and 384 is the same figure rounded the other way.
Not well. With an expected proportion near 0 or 100 per cent the normal approximation breaks down; size the work on how many events you need to see instead.
It sizes an estimate, not a comparison. A test needs a power calculation on the difference you want to detect, which will normally ask for more than this.