How do you calculate survey sample size with Cochran's formula?

Cochran's formula for estimating one proportion is n0 = z² × p(1 - p) / e², where z comes from the confidence level, p is the proportion you expect and e is the margin as a proportion. At 95% confidence z is 1.959964, and with p = 0.5 and e = 0.05 that is 3.841459 × 0.25 / 0.0025 = 384.1459, rounded up to 385 completed responses. For a population of N the page divides n0 by 1 + (n0 - 1) / N and rounds up again: 278 for a population of 1,000 and 370 for 10,000.

How can I calculate survey sample size in Excel?

Use A1 for confidence as a percentage such as 95%, A2 for the expected proportion such as 50%, and A3 for the margin as a proportion such as 5%. In B1 enter =NORM.S.INV((1+A1)/2)^2*A2*(1-A2)/A3^2 for the unrounded n0. =ROUNDUP(B1,0) gives the unbounded completed count. With a population in A4, =ROUNDUP(B1/(1+(B1-1)/A4),0) gives the corrected completed count. Put that count in B2 and a response rate such as 50% in A5; =ROUNDUP(B2/A5,0) gives invitations. The confidence, margin, proportion, population and rate have the same meanings as the fields here.

Why is the answer 385 and not 384?

Because a sample size is rounded up, never to the nearest. n0 at 95% confidence, a 5-point margin and an expected 50% is 384.1459, and 384 responses would give a margin a little wider than 5 points. The page rounds only the final counts: the population correction works on the unrounded 384.1459, which is why a population of 1,000 gives 278 rather than the 279 that correcting 385 would give.

What expected proportion do I enter if I have no estimate?

50% is the conventional entry, because p(1 - p) is largest there, at 0.25, so 50% gives the largest sample of any value. At 95% confidence and a 5-point margin it gives 385, against 246 for an expected 20% and 139 for 10%. The page uses whatever you enter and says how it read it: a bare 0.5 is read as 0.5%, and the note under the result says so.

Does the population size matter?

Much less than people expect once the population is large. At 95% confidence and a 5-point margin, a population of 1,000 gives 278 completed responses, 10,000 gives 370, 1,000,000 gives 384 and an unbounded one 385. Leave the population field blank and the page treats it as unbounded rather than guessing a size.

Why do other calculators give a slightly different number?

Two things move the result: how z was rounded and whether the population correction was applied. Printed tables give 1.96 for 95%; this page computes 1.959964 and also shows the 1.96 version, where n0 is 384.16 instead of 384.1459. Often the rounded-up count is the same, but not always: for a population of 1,000,000 it is 384 with 1.959964 and 385 with 1.96.

How many invitations does a response rate mean?

Enter a response rate and the page divides the completed responses by it and rounds up: 278 completed responses at 50% is 278 / 0.5 = 556 invitations. A rate of 0 is refused, because no number of invitations reaches the sample, and when the invitations come to more than the population you entered, the page says so. The rate only scales the count: it cannot tell whether the people who answer differ from the people who do not.

How do I choose sample size for a confidence interval?

For one survey proportion, the margin is half the planned interval width: a margin of 5 percentage points means a total width of 10 points. The confidence level determines z, and the calculator finds the completed count for that width and expected proportion. This is sample-size planning; an interval calculated from observed responses requires the observed count or proportion as well.

Is this the minimum sample size for a representative survey?

The result is the smallest whole count reaching the chosen margin under this proportion formula and its sampling assumptions. It does not establish representativeness. A large convenience sample can still miss parts of the population; increasing the count does not fix selection or nonresponse bias.

Can I use this for power analysis, A/B tests or a clinical trial?

This page sizes a sample to estimate one proportion using a simple random sample of independent responses and the normal approximation. It applies no design effect for clustered or weighted samples. Power analysis for an A/B test or a comparison of means needs a target difference, a significance threshold and a chosen power; those are different inputs and a different calculation. Clinical-trial sample sizes also depend on the study design and outcome, which this page does not model.

Sample Size Calculator

Estimate how many completed survey responses are needed for one proportion. Choose a confidence level, margin of error and expected proportion; add a population or response rate when known. The working shows each step, and the comparison table covers 90, 95 and 99% and margins of 3 to 6 points.

95% confidenceMargin 5 pointsExpected 50%Population unboundedNo response rate

z at 95% confidence   1.959964, the inverse normal at 0.975p, expected 50%       0.5, so p(1 - p) = 0.25e, margin 5 points    0.05, so e² = 0.0025 n0 = z² × p(1 - p) / e²   = 3.841459 × 0.25 / 0.0025   = 384.1459, rounded up to 385 for an unbounded population population left blank   no correction: 385 completed responsesno response rate        invitations not worked outz rounded to 1.96       n0 = 384.16, 385 completed responses, the same as above

Completed responses for an unbounded population and expected 50%
confidence	3 pts	4 pts	5 pts	6 pts
90%	752	423	271	188
95%	1,068	601	385, your settings	267
99%	1,844	1,037	664	461

Confidence level (%)

Margin (± points)

Expected proportion (%)

Population (optional)

Response rate (%, optional)

The margin is in percentage points either side of the estimate: 5 means plus or minus 5 points. A blank population is treated as unbounded; a blank response rate leaves the invitations out.

Completed responses

385

Unbounded population

385

384.1459

Invitations

no rate entered

z = 1.959964, p = 0.5, e = 0.05population left blank, so unbounded
n0 = 384.1459, rounded up to 385
z rounded to 1.96three decimals, as printed tables give it
n0 = 384.16, rounded up to 385

Accuracy. The textbook sample size for estimating a proportion, worked in full: z comes from your confidence level through the inverse normal, n0 = z² × p(1 - p) / e², a population you enter divides that by 1 + (n0 - 1) / N, and only the final counts are rounded up. It assumes a simple random sample of independent responses and the normal approximation, and applies no design effect. It cannot make a sample representative, remove nonresponse or other bias, or stand in for a clinical-trial power calculation, and a blank population is treated as unbounded rather than guessed.

Common questions

How do you calculate survey sample size with Cochran's formula?: Cochran's formula for estimating one proportion is n0 = z² × p(1 - p) / e², where z comes from the confidence level, p is the proportion you expect and e is the margin as a proportion. At 95% confidence z is 1.959964, and with p = 0.5 and e = 0.05 that is 3.841459 × 0.25 / 0.0025 = 384.1459, rounded up to 385 completed responses. For a population of N the page divides n0 by 1 + (n0 - 1) / N and rounds up again: 278 for a population of 1,000 and 370 for 10,000.
How can I calculate survey sample size in Excel?: Use A1 for confidence as a percentage such as 95%, A2 for the expected proportion such as 50%, and A3 for the margin as a proportion such as 5%. In B1 enter =NORM.S.INV((1+A1)/2)^2*A2*(1-A2)/A3^2 for the unrounded n0. =ROUNDUP(B1,0) gives the unbounded completed count. With a population in A4, =ROUNDUP(B1/(1+(B1-1)/A4),0) gives the corrected completed count. Put that count in B2 and a response rate such as 50% in A5; =ROUNDUP(B2/A5,0) gives invitations. The confidence, margin, proportion, population and rate have the same meanings as the fields here.
Why is the answer 385 and not 384?: Because a sample size is rounded up, never to the nearest. n0 at 95% confidence, a 5-point margin and an expected 50% is 384.1459, and 384 responses would give a margin a little wider than 5 points. The page rounds only the final counts: the population correction works on the unrounded 384.1459, which is why a population of 1,000 gives 278 rather than the 279 that correcting 385 would give.
What expected proportion do I enter if I have no estimate?: 50% is the conventional entry, because p(1 - p) is largest there, at 0.25, so 50% gives the largest sample of any value. At 95% confidence and a 5-point margin it gives 385, against 246 for an expected 20% and 139 for 10%. The page uses whatever you enter and says how it read it: a bare 0.5 is read as 0.5%, and the note under the result says so.
Does the population size matter?: Much less than people expect once the population is large. At 95% confidence and a 5-point margin, a population of 1,000 gives 278 completed responses, 10,000 gives 370, 1,000,000 gives 384 and an unbounded one 385. Leave the population field blank and the page treats it as unbounded rather than guessing a size.
Why do other calculators give a slightly different number?: Two things move the result: how z was rounded and whether the population correction was applied. Printed tables give 1.96 for 95%; this page computes 1.959964 and also shows the 1.96 version, where n0 is 384.16 instead of 384.1459. Often the rounded-up count is the same, but not always: for a population of 1,000,000 it is 384 with 1.959964 and 385 with 1.96.
How many invitations does a response rate mean?: Enter a response rate and the page divides the completed responses by it and rounds up: 278 completed responses at 50% is 278 / 0.5 = 556 invitations. A rate of 0 is refused, because no number of invitations reaches the sample, and when the invitations come to more than the population you entered, the page says so. The rate only scales the count: it cannot tell whether the people who answer differ from the people who do not.
How do I choose sample size for a confidence interval?: For one survey proportion, the margin is half the planned interval width: a margin of 5 percentage points means a total width of 10 points. The confidence level determines z, and the calculator finds the completed count for that width and expected proportion. This is sample-size planning; an interval calculated from observed responses requires the observed count or proportion as well.
Is this the minimum sample size for a representative survey?: The result is the smallest whole count reaching the chosen margin under this proportion formula and its sampling assumptions. It does not establish representativeness. A large convenience sample can still miss parts of the population; increasing the count does not fix selection or nonresponse bias.
Can I use this for power analysis, A/B tests or a clinical trial?: This page sizes a sample to estimate one proportion using a simple random sample of independent responses and the normal approximation. It applies no design effect for clustered or weighted samples. Power analysis for an A/B test or a comparison of means needs a target difference, a significance threshold and a chosen power; those are different inputs and a different calculation. Clinical-trial sample sizes also depend on the study design and outcome, which this page does not model.

The textbook sample size for estimating a proportion, worked in full: z comes from your confidence level through the inverse normal, n0 = z² × p(1 - p) / e², a population you enter divides that by 1 + (n0 - 1) / N, and only the final counts are rounded up. It assumes a simple random sample of independent responses and the normal approximation, and applies no design effect. It cannot make a sample representative, remove nonresponse or other bias, or stand in for a clinical-trial power calculation, and a blank population is treated as unbounded rather than guessed.