How to Calculate a Confidence Interval for a Proportion

Estimates the range that likely contains the population proportion as p̂ ± z × √(p̂(1 − p̂) ÷ n). This is what produces the margin of error quoted with opinion polls.

Explanation

This turns a proportion measured in a survey into a range that plausibly contains the true proportion in the whole population. The "margin of error $\pm$ 3%" quoted with opinion polls comes from exactly this calculation.

\hat{p} \pm z \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}

$\hat{p}$ — the sample proportion
$n$ — the sample size
$z$ — set by the confidence level, 1.96 for 95%

Example

Of 1000 people surveyed, 40% approve, so $\hat{p} = 0.4$ and $n = 1000$ , at a 95% confidence level.

\sqrt{\dfrac{0.4 \times 0.6}{1000}} = \sqrt{0.00024} = 0.01549

0.4 \pm 1.96 \times 0.01549 = 0.4 \pm 0.0304

The 95% interval runs from 36.96% to 43.04% — an approval rating of 40% with a margin of error of $\pm$ 3.0%.

The surprising part

The sample size you need barely depends on the size of the population. Polling 1000 people gives a margin of about $\pm$ 3% whether the country has 100 million people or the town has 100,000. That is the answer to the natural objection that 1000 people cannot possibly speak for a nation.

The margin is widest at $\hat{p} = 0.5$ , which lets pollsters state conservatively that with $n = 1000$ the error is at most $\pm$ 3.1%, whatever the result turns out to be.

Watch out

The normal approximation needs both $n\hat{p}$ and $n(1-\hat{p})$ to be at least 5. For very rare opinions, at 1% say, this formula is not trustworthy
The margin covers random sampling error only. It says nothing about biased questions or about the people who refused to answer, and in real surveys those are often the bigger problem