Sample Size Calculator
How many responses a survey needs to hit the margin of error you want, with the finite population correction applied when the group is small enough for it to matter. Every count is rounded up, because you cannot survey a fraction of a person.
How many responses the survey needs
How often a survey run this way should land inside its stated margin. 95% is the reporting convention almost everywhere.
How far the result may sit from the truth, in percentage points. It is squared in the formula: halving it costs four times the responses.
The share you expect to answer one way. Leave it at 50 unless you have a prior estimate — that is the split that demands the most responses, so it cannot catch you short.
Everyone the survey could have asked. Only worth filling in for a small, countable group — above about a million it changes nothing.
384.16 rounded up, because nobody can survey a fraction of a person. No population size is needed: sampling error depends on how many people you ask, not on how many you could have asked.
- Responses needed
- 385
- Before rounding up
- 384.16
- Ignoring population size
- 385
- Finite population correction
- Not applied — population treated as unlimited
- z multiplier
- 1.96
| Margin of error | Responses needed |
|---|---|
| ±1% | 9,604 |
| ±2% | 2,401 |
| ±3% | 1,068 |
| ±5% | 385 |
| ±10% | 97 |
- A 50% split is the conservative assumption: p(1 − p) peaks at 0.25 there, so no other expected proportion needs a larger sample. If the true split turns out to be 70/30, this survey simply beats its promised margin.
- Rounded up, always. The formula returns a fraction of a person, and rounding down would quietly deliver a wider margin of error than the one promised alongside the result.
- This is completed responses, not invitations. Divide by the response rate you expect to get the number of people to contact.
- Sampling error only. A margin of error says nothing about who declined to answer or who your list never reached — in 1936 the Literary Digest called the US presidential election for the wrong candidate on 2.4 million replies.
The requirement, and the correction that shrinks it
The first formula is the confidence-interval formula for a proportion, turned inside out: instead of asking what margin a given sample produces, it asks what sample a given margin requires.
n = z² × p(1 − p) ∕ e² • n ′ = n ∕ (1 + (n − 1) ∕ N)- z
- The multiplier for your confidence level — 1.960 at 95%
- p
- The proportion you expect to answer one way, as a decimal
- e
- The margin of error you are willing to live with, as a decimal
- n
- Responses needed, ignoring how large the population is
- N
- The population — everyone the survey could have reached
- n ′
- Responses needed after the finite population correction
Both figures are always rounded up. The formula returns 384.16 responses, and there is no such thing as 0.16 of a respondent. Rounding down would deliver a margin of error slightly wider than the one printed alongside the result, which is the direction of error that makes the published number false; rounding up costs one extra interview.
Why the population size barely matters
Look at the main formula again: N does not appear in it. Sampling error depends on how many people you asked, not on how many you could have asked. A well-stirred pot and a well-stirred lake both give up their secret to the same spoonful.
The population only enters through the correction, and the correction only bites when the sample is a real fraction of the whole. Here is the requirement at 95% confidence and a ±5-point margin, as the population grows:
| Population | Responses needed | Share of the population |
|---|---|---|
| 500 | 218 | 44% |
| 1,000 | 278 | 28% |
| 5,000 | 357 | 7% |
| 10,000 | 370 | 3.7% |
| 100,000 | 383 | 0.4% |
| 1,000,000 | 385 | 0.04% |
| 10,000,000 | 385 | 0.004% |
| Unlimited | 385 | — |
Read the last four rows slowly. Going from a population of ten thousand to a population of ten million — a thousandfold increase — raises the requirement from 370 responses to 385. Fifteen more interviews. And past a million, nothing changes at all: the correction has become invisible against the uncorrected 384.16.
This is the arithmetic that makes a national poll of about a thousand people legitimate, and it is the single most counterintuitive fact in survey design. The instinct that a bigger country needs a proportionally bigger poll is wrong, and it is wrong by orders of magnitude. What a bigger sample buys is a narrower margin — a thousand respondents gets you to about ±3.1 points — and that is the only thing it buys.
The correction earns its keep at the other end. Surveying a department of 200 people needs 132 responses rather than 385, because at that size you are sampling a large slice of the entire population and running out of people to be wrong about. As a rule of thumb, fill in the population when your sample would exceed about 5% of it, and leave it blank otherwise — blank is the conservative choice.
Why the expected proportion is 50%
The requirement scales with p(1 − p), the variance of a yes/no answer. That product is a parabola with its peak at exactly p = 0.5:
| Expected proportion | p(1 − p) | Responses needed |
|---|---|---|
| 10% or 90% | 0.09 | 139 |
| 20% or 80% | 0.16 | 246 |
| 30% or 70% | 0.21 | 323 |
| 40% or 60% | 0.24 | 369 |
| 50% | 0.25 | 385 |
Because 0.25 is the maximum, assuming a 50/50 split asks for the largest sample any split could possibly need. That makes it the only assumption you cannot be caught out by. Guess 20% when the truth turns out to be 50% and the survey comes back with a margin of error wider than the one you promised, which is a broken published number. Guess 50% when the truth is 20% and you have collected 139 responses more than you strictly needed, which is a bill. Only lower it below 50% when you have a genuine prior estimate — a previous wave of the same survey, a known base rate — and are prepared to defend it.
A ±5-point survey at 95% confidence
Take the defaults above. At 95% confidence z is 1.960, so z² is 3.8416. With no prior estimate p is 0.5, so p(1 − p) is 0.25. A margin of 5 percentage points is e = 0.05, so e² is 0.0025. Put them together: 3.8416 × 0.25 ÷ 0.0025 = 384.16 responses, which rounds up to 385.
Now suppose that survey is going to the 10,000 members of a professional association, so the population is finite and worth declaring. The correction is 384.16 ÷ (1 + 383.16 ÷ 10,000) = 384.16 ÷ 1.038316 = 369.98, which rounds up to 370. Fifteen fewer responses, because the sample is 3.7% of everyone there is.
Suppose instead you want ±2.5 points from the same population. Halving the margin quadruples the uncorrected requirement to 1,536.64, or 1,537 responses — and against a population of 10,000 the correction brings that back to 1,333. Note what happened: a modest improvement in precision cost four times the fieldwork, while a thousandfold increase in population cost fifteen responses. Precision is expensive; population is nearly free.
A margin of error is not a measure of accuracy
Everything on this page describes sampling error: the spread you would see if you drew repeated random samples from the same population and asked them the same question. It is a statement about randomness, and it assumes the sample really was drawn at random from the population you care about.
The errors that sink real surveys are not random. Coverage error is the people your list never included. Non-response bias is the people who saw it and did not reply — and they are almost never a random subset of those who did. Question wording moves answers by more than most margins of error. None of these shrink as n grows. A larger sample measures a skewed population more precisely.
The canonical demonstration is the 1936 Literary Digest poll. It collected about 2.4 million replies — a sample whose theoretical margin of error is a few hundredths of a point — and predicted that Alf Landon would beat Franklin Roosevelt. Roosevelt took 46 of the 48 states. The magazine had mailed its ballots to subscribers, car owners and telephone subscribers, and only a quarter of those mailed replied. Its sample was enormous, precise, and about the wrong population. Sample size fixes noise. Nothing in this calculator fixes bias.
What this calculator assumes
- Simple random sampling. Cluster and stratified designs carry a design effect that multiplies the requirement — commonly 1.5 to 2 for a clustered household survey — and this tool does not apply one.
- The margin of error is for a single proportion, expressed in percentage points. Sizing for a mean, or for the difference between two groups, uses a different formula.
- The figure is completed responses, not invitations. Divide by your expected response rate to get the number of people to contact.
- Both counts are rounded up, always — including every row of the comparison table in the results panel. Two populations whose exact requirements differ by a fraction of a response can therefore show the same whole number.
- The normal approximation behind the formula assumes a reasonable number of responses on both sides. For very rare outcomes — a 1% response rate to a yes/no question — the requirement here is a floor rather than a reliable answer.
- Nothing here evaluates whether your sampling frame reaches the population you mean, which is usually the larger risk by a wide margin.
Sample size FAQ
Why is 385 enough for a country of 300 million?
Because the formula for sampling error does not contain the population size at all. Precision depends on how many people you asked, not on how many you could have asked — a spoonful tells you as much about a well-stirred pot as about a well-stirred lake. The population only enters through the finite population correction, and that correction is meaningful just when your sample is a large fraction of the whole. At a 5-point margin you need 370 responses from a population of 10,000, 383 from 100,000, and 385 from ten million or from everyone alive. Fifteen extra responses buys a thousandfold increase in population.
Why is 50% the default expected proportion?
Because it is the assumption that cannot catch you short. The requirement scales with p(1 − p), which peaks at exactly 0.25 when p is 0.5 and falls away on both sides — 0.24 at 40%, 0.21 at 30%, 0.09 at 10%. Assuming a 50/50 split therefore asks for the largest sample any split could need. Guess 20% when the truth is 50% and your survey comes back with a margin wider than the one you promised; guess 50% when the truth is 20% and you have merely collected 139 responses more than you needed.
Does a bigger sample fix a biased survey?
No, and this is the failure that actually sinks surveys. A margin of error describes sampling error only — the spread you would see across repeated random samples from the same population. It says nothing about who never saw your survey, who declined to answer, or how the question was worded. Those biases do not shrink with n; a larger sample simply measures a skewed picture more precisely. In 1936 the Literary Digest collected about 2.4 million replies, a sample so large its sampling error was a rounding error, and called the US presidential election for the wrong candidate by a wide margin. The people it could reach and the people who bothered to reply were not the electorate.
How many people do I need to invite to get this many responses?
Divide by the response rate you expect. The figure this tool returns is completed responses, so at a 30% response rate, 385 completions means inviting about 1,300 people. Response rates for cold email surveys are often well under 10%, which is usually the binding constraint on a study rather than the arithmetic here. It is also worth remembering that a low response rate is itself a bias risk, not just a volume problem — the people who reply are rarely a random subset of the people who did not.
Why does halving the margin of error quadruple the sample?
Because the margin is squared in the denominator. Going from ±5 points to ±2.5 multiplies the requirement by four, from 385 to 1,537; getting to ±1 point takes 9,604. This is the same square-root wall that governs every estimate built from a sample: precision improves with √n, so buying twice the precision costs four times the data. It is why national polls cluster around a thousand respondents — that is roughly where ±3 points sits, and the next meaningful step up in precision costs several times as much.
Should I use the finite population correction?
Only when your sample is a meaningful share of the population — as a rough rule, more than about 5% of it. Surveying 400 people out of 2,000 employees is a large slice of the whole, and the correction legitimately cuts the requirement (to 323 in that case). Surveying 400 out of a city of 500,000 is not, and the correction changes the answer by a response or two. When in doubt, leave the population blank: the uncorrected figure is the conservative one.
Is this the right calculator for measuring an average rather than a percentage?
No. This formula is for a proportion — the share of people who answer one way. Sizing a sample for a mean uses n = (z × σ ÷ e)², which needs an estimate of the standard deviation of the thing you are measuring and expresses the margin in the unit of the measurement rather than in percentage points. The structure is the same: the margin is squared in the denominator, so the same quadrupling rule applies.
Sources and review notes
- NIST/SEMATECH e-Handbook of Statistical Methods §7.2.2.2 — Sample sizes required, the standard derivation of n from a target margin of error
- NIST/SEMATECH e-Handbook §7.2.4.1 — Confidence intervals for a proportion, the interval this requirement is derived from
The z multipliers are the published two-sided values from the standard normal distribution, rounded to three decimals as they appear in every printed table. The finite population correction is the standard n ∕ (1 + (n − 1) ∕ N). Both are constants and formulas rather than data, so there is nothing here to go stale.