Case 1
- Z-score (confidence level)
- 1.96
- Expected proportion (p)
- 0.5
- Margin of error (E)
- 0.05
384.16
Open with these valuesSample Size Calculator
384.16
Result: 384.16At 95 % confidence and a margin of five points you need 384.16 responses — 385, because you always round a sample size up. The margin is the expensive part: tightening it to three points takes 1068, nearly three times as many.
Worked examples
Case 2
- Z-score (confidence level)
- 1.96
- Expected proportion (p)
- 0.5
- Margin of error (E)
- 0.03
1,067.11
Open with these valuesCase 3
- Z-score (confidence level)
- 2.576
- Expected proportion (p)
- 0.2
- Margin of error (E)
- 0.02
2,654.31
Open with these valuesHow it's calculated
n = z² × p × (1 − p) ÷ E²
- StepEnter the z-score of your confidence level: 1.645, 1.96 or 2.576.
- StepEnter the proportion you expect, or 0.5 if you have no estimate.
- StepEnter the margin of error you can live with, as a decimal.
- ResultRound the result up to the next whole respondent.
What this number means
The number on this page is a budget: how many responses a survey needs before its result is precise enough to be worth quoting. It comes from turning the margin-of-error formula around. A proportion measured on n people carries an uncertainty of √(p × (1 − p) ÷ n), and the confidence level multiplies that by its z-score, which gives the margin E = z × √(p × (1 − p) ÷ n). Solving for n moves the z and the square root to the other side and squares both: n = z² × p × (1 − p) ÷ E². At the values set here that is 1.96² = 3.8416, times 0.5 × 0.5 = 0.25 for 0.9604, divided by 0.05² = 0.0025, leaving 384.16 — 385 respondents once rounded up. What the figure does not promise is a good survey. It assumes every respondent is drawn at random from exactly the group you want to describe, and the margin of error covers only the luck of that draw. Non-response, a skewed sampling frame, a leading question — none of them appear in the formula, and none of them shrink as n grows. 385 answers from whoever felt like replying are not ±5 points from the truth; they carry an unmeasured bias on top.
The population size is not in the formula
n = z² × p × (1 − p) ÷ E² has no term for how many people the group holds. It is the standard formula for an infinite or very large population, which fits most surveys.
The margin is the expensive lever
It sits in the denominator, squared. At 95 % going from ±5 to ±3 points lifts the requirement from 384.16 to 1067.11, and ±2.5 points takes 1536.64.
Use p = 0.5 when you have no estimate
The term p × (1 − p) is largest at 0.5, so that choice gives the most conservative sample. However the answers split, it stays big enough.
Always round up
Nobody can survey a fraction of a person, and rounding down would leave the margin slightly too wide. 384.16 becomes 385 respondents.
Commonly misread
A survey of a million people needs a far bigger sample than one of ten thousand.
The formula holds no population size at all; at 95 % and ±5 points both need 385. Only once the sample passes roughly 5 % of the group does a finite population correction lower the requirement.
384.16 respondents is the answer.
A sample size is always rounded up, so it is 385. Rounding down would leave the margin of error slightly too wide.
Halving the margin of error costs twice the sample.
E is squared, so halving it quadruples the sample. At 95 % with p = 0.5, ±5 points needs 384.16 and ±2.5 points needs 1536.64.
Reference table
| z, p, E | Confidence and margin | Respondents |
|---|---|---|
| 1.645, 0.5, 0.05 | 90 %, ±5 points | 270.60 |
| 1.96, 0.5, 0.05 | 95 %, ±5 points | 384.16 |
| 2.576, 0.5, 0.05 | 99 %, ±5 points | 663.58 |
| 1.96, 0.5, 0.03 | 95 %, ±3 points | 1067.11 |
| 2.576, 0.2, 0.02 | 99 %, ±2 points | 2654.31 |
Questions
How do I calculate sample size?
Use n = z² × p × (1 − p) ÷ E², with z the z-score of the confidence level, p the expected proportion and E the margin of error. At 95 % confidence, p = 0.5 and a margin of 0.05 that comes to 384.16, which you round up to 385 respondents.
Which z-score should I use?
It comes from the confidence level: 1.645 for 90 %, 1.96 for 95 % and 2.576 for 99 %. A higher level means a larger z and a larger sample. For surveys and polls 95 % is the usual choice.
Why is the proportion usually set to 0.5?
The term p × (1 − p) is largest at p = 0.5, so that value gives the most conservative — the biggest — sample. With no reliable estimate of the true proportion, 0.5 keeps the sample big enough however the answers split.
Why do I round the sample size up?
You cannot survey a fraction of a person, and rounding down would leave the margin of error slightly too wide. The calculator shows the exact figure, so 384.16 becomes 385 respondents.
Does this account for the size of the population?
No. This is the standard formula for an infinite or very large population, which fits most surveys. For a small, finite population you can apply a finite population correction afterwards, which lowers the requirement once your sample passes roughly 5 % of the whole group.
Sources and last check
Information, not professional advice.