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Generated on: July 20, 2026 at 09:33 PM

Confidence Interval for a Proportion Calculator

Inputs

Number of Successes320
Sample Size500
Confidence Level95%

Generated with OneCalc · https://onecalc.app/en/math/confidence-interval-proportion

Mathematics

Confidence Interval for a Proportion Calculator

Find the confidence interval and margin of error for a sample proportion — enter successes and sample size, pick a 90%, 95%, or 99% confidence level, and see both the Wald and Wilson intervals.

Inputs

Sample Data

≥ 1

Results

Enter a value to see results.

%
%

Point Estimate & Margin

%
%

Wilson Score Interval

%
%

Step by Step

\hat p = \dfrac{x}{n} = \dfrac{320}{500} = ?\%
\begin{aligned} E &= z^{*} \sqrt{\dfrac{\hat p\,(1 - \hat p)}{n}} \\ &= ? \sqrt{\dfrac{\frac{320}{500}\left(1 - \frac{320}{500}\right)}{500}} \\ &= ?\% \end{aligned}
\text{Lower} = \hat p - E = ?\% - ?\% = ?\%
\text{Upper} = \hat p + E = ?\% + ?\% = ?\%

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Confidence interval for a proportion

When a survey reports that 64% of respondents favor a policy, that single percentage is only a point estimate from one sample. A confidence interval turns it into a range of plausible values for the true population proportion, together with a stated confidence level. It pairs the sample proportion with a margin of error that reflects how much the estimate could move from one sample to the next.

Formulas

With xx successes out of a sample of size nn, the sample proportion is:

p^=xn\hat{p} = \frac{x}{n}

Standard error:

SE=p^ (1−p^)nSE = \sqrt{\frac{\hat{p}\,(1 - \hat{p})}{n}}

Margin of error:

E=z∗×SE=z∗p^ (1−p^)nE = z^{*} \times SE = z^{*}\sqrt{\frac{\hat{p}\,(1 - \hat{p})}{n}}

Wald confidence interval:

p^±E=[p^−E, p^+E]\hat{p} \pm E = \left[\hat{p} - E,\ \hat{p} + E\right]

Critical z-values (standard normal distribution):

Confidence levelz∗z^{*}
90%1.6449
95%1.9600
99%2.5758

Wald vs. Wilson

The Wald interval above is the formula found in most introductory textbooks: symmetric around p^\hat{p} and easy to compute. Its weakness is that it relies on the normal approximation to the binomial, which breaks down when the sample is small or the proportion is near 0 or 1. In those cases the Wald interval is too narrow, its coverage falls short of the stated level, and it can even produce bounds below 0% or above 100%.

The Wilson score interval fixes this by inverting the score test rather than assuming symmetry:

p^+z∗22n±z∗p^(1−p^)n+z∗24n21+z∗2n\frac{\hat{p} + \dfrac{z^{*2}}{2n} \pm z^{*}\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n} + \dfrac{z^{*2}}{4n^{2}}}}{1 + \dfrac{z^{*2}}{n}}

Wilson bounds always stay inside 0–100%, the interval is asymmetric, and its coverage stays close to the target confidence level even for small nn. Many statisticians now treat Wilson as the default; this calculator reports both so you can compare them side by side.

When is the normal approximation valid?

A common rule of thumb is that the Wald interval is trustworthy only when both

np^≥5andn(1−p^)≥5n\hat{p} \geq 5 \quad \text{and} \quad n(1 - \hat{p}) \geq 5

(some texts require 10). With very few successes, a proportion near 0 or 1, or a small sample, switch to the Wilson score interval or an exact Clopper–Pearson interval.

Interpretation

A 95% confidence interval does not mean there is a 95% probability that the true proportion lies inside a particular range. The true population proportion is a fixed, unknown constant — the interval is the random object, because it is built from a random sample.

The precise meaning is a statement about the procedure: if the same sampling and interval construction were repeated many times, about 95% of the resulting intervals would contain the true proportion. Any single interval either contains it or does not.

Polling and the margin of error

News coverage of a poll typically reports a single "margin of error, ±3 percentage points." That number is exactly E=z∗p^(1−p^)/nE = z^{*}\sqrt{\hat{p}(1-\hat{p})/n} at 95% confidence. Because p^(1−p^)\hat{p}(1-\hat{p}) is largest at p^=0.5\hat{p} = 0.5, pollsters usually quote the margin for p^=0.5\hat{p} = 0.5 so one figure conservatively covers every question on the survey.

The margin of error shrinks with n\sqrt{n}: halving it requires four times the sample size.

Sample sizeMargin of error (95%, p^=0.5\hat{p} = 0.5)
n = 100±9.8
n = 400±4.9
n = 1000±3.1
n = 1600±2.5

Worked example: a support poll

A survey of 500 voters finds 320 in favor of a proposal.

Sample proportion: p^=320/500=0.64\hat{p} = 320 / 500 = 0.64

Standard error: SE=0.64×0.36/500≈0.0215SE = \sqrt{0.64 \times 0.36 / 500} \approx 0.0215

95% margin of error: E=1.96×0.0215≈0.0421E = 1.96 \times 0.0215 \approx 0.0421

Wald interval:

0.64±0.0421=[0.5979, 0.6821]=[59.79%, 68.21%]0.64 \pm 0.0421 = [0.5979,\ 0.6821] = [59.79\%,\ 68.21\%]

The Wilson score interval for the same data is [59.70%, 68.09%][59.70\%,\ 68.09\%] — almost identical here because n=500n = 500 is large and p^=0.64\hat{p} = 0.64 is far from the extremes. The two intervals diverge most when the sample is small or the proportion is near 0 or 1.

Interpretation: "Based on this sample of 500, we estimate that between about 60% and 68% of all voters favor the proposal, with 95% confidence."

Frequently Asked Questions (FAQ)

What is the difference between the Wald and Wilson intervals?

The Wald interval (p̂ ± z*·√(p̂(1−p̂)/n)) is the textbook formula: simple, symmetric, and taught first. It works well when the sample is large and p̂ is near 0.5, but it performs poorly for small samples or proportions close to 0 or 1 — it can even produce bounds below 0% or above 100%. The Wilson score interval solves the same problem by inverting the score test; it stays inside 0–100%, is asymmetric, and has coverage much closer to the stated confidence level for small n. Many statisticians now recommend Wilson as the default. This calculator shows both so you can compare them.

How does this relate to a poll’s "margin of error"?

When a poll reports "52% support, margin of error ±3 percentage points", that ±3 is exactly the margin of error E = z*·√(p̂(1−p̂)/n) at 95% confidence. News organizations usually quote the margin for p̂ = 0.5 (which gives the widest interval for a given n) so a single number covers every question on the survey. Enter the reported percentage as successes over sample size to reproduce the interval.

When is the normal approximation valid?

The Wald interval relies on the normal approximation to the binomial, which is reasonable when both n·p̂ and n·(1 − p̂) are at least 5 (some texts use 10). With few successes or a proportion near 0 or 1 the approximation breaks down and the Wald interval becomes unreliable — that is exactly the situation where the Wilson score interval (or an exact Clopper–Pearson interval) should be used instead.

How should I interpret the resulting interval?

A 95% confidence interval of 59.79% to 68.21% does not mean there is a 95% probability the true proportion lies in that range — the true proportion is fixed. It means the procedure that produced this interval captures the true population proportion 95% of the time across repeated samples. In practice: you can be reasonably confident the true share falls within the stated bounds, and a wider interval signals a smaller sample or a more demanding confidence level.

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