Find how many responses a survey needs for your confidence level and margin of error, or the margin for a sample you have.
For a survey of a yes or no question, the sample size is n = z² × p × (1 − p) ÷ e², rounded up, where z is 1.645, 1.96 or 2.576 for 90%, 95% or 99% confidence, p is the share you expect to answer yes, and e is the margin of error as a decimal. At 95% confidence, a 5% margin and p = 50%, that is 1.96² × 0.5 × 0.5 ÷ 0.05² = 3.8416 × 0.25 ÷ 0.0025 = 384.16, which rounds up to 385 responses. The page opens on this example and shows the same steps for your numbers. Changing the margin has the biggest effect: at 95% confidence, 3% needs 1,068 responses and 1% needs 9,604.
The product p × (1 − p) is largest at p = 50%, so 50% gives the biggest sample and is the safe choice when you have no earlier figure. If you expect only 20% to answer yes, the same 95% and 5% settings need 246 responses instead of 385. Only use a lower figure when an earlier survey or a pilot gives you a real estimate. The page accepts proportions strictly between 0 and 100, since at exactly 0 or 100 the formula gives a sample size of zero.
The formula above assumes a very large population. When you survey a known group, such as 1,000 employees, the sample can be smaller: n = n₀ ÷ (1 + (n₀ − 1) ÷ N), where n₀ is the first result and N is the population size. With n₀ = 384.16 and N = 1,000 the result is 384.16 ÷ 1.38316 = 277.74, which rounds up to 278. The page rounds up once, at the end, from the unrounded value. Rounding n₀ to 385 first would give 279 for the same inputs. The correction barely matters when the population is much larger than the sample: for N = 1,000,000,000,000 the answer stays 385.
Choose Margin of error to go the other way: e = z × √(p × (1 − p) ÷ n), multiplied by √((N − n) ÷ (N − 1)) when you give a population. For 385 responses at 95% and p = 50% that is 1.96 × 0.0255 = 4.99%. For 1,000 responses it is about 3.1%. The result also shows the range around your expected proportion. What the page does not cover: it assumes a simple random sample of completed responses, so it does not allow for a low response rate, clustering, weighting or a biased list. Invite more people than the sample size shown, and treat the figure as the least you need.
At 95% confidence and p = 50%, 385 completed responses. At 90% it is 271 and at 99% it is 664. If your group has 1,000 people, 95% confidence needs only 278 responses.
n = z² × p × (1 − p) ÷ e², with z = 1.96 for 95% confidence. For a 5% margin and p = 50% this is 1.96² × 0.25 ÷ 0.05² = 384.16, rounded up to 385. For a known population N, divide by 1 + (n − 1) ÷ N.
Rounding down would give a margin of error slightly larger than the one you asked for. 384.16 rounded to 384 would give 5.001%, so the page rounds up to 385. Fractions of a person cannot be surveyed either.
Use 50%. It gives the largest sample, 385 at 95% confidence and a 5% margin, so the real margin will be no worse than 5% whatever the true share turns out to be.
Hardly. A population of 100,000 gives 383 responses and 1,000,000 gives 385, the same as leaving the box empty. The correction only matters when the group is small, such as 1,000 people.
No. The numbers are processed in your browser and are not uploaded. They are not stored on your device either, and reloading the page puts the example back. The page only counts that a result was shown, without any of your numbers.