How to Interpret a Confidence Interval for a Proportion

The two sentences that earn credit

Almost every exam question about a confidence interval for a proportion wants one of two sentences, and they are not interchangeable. One interprets the interval you computed. The other interprets the confidence level, which is a property of the method that produced the interval.

Here is the running example. A transit agency draws a random sample of 500 weekday riders and finds 210 who use the mobile ticketing app, so $\hat{p} = \frac{210}{500} = 0.42$ (read "p-hat", the sample proportion). The 95% confidence interval works out to 0.377 to 0.463, and the arithmetic is below.

Throughout, $p$ with no hat is the parameter: the true proportion of all weekday riders on this system who use the app. That number is fixed and unknown. $\hat{p}$ is the statistic you computed from one sample, and it shifts from sample to sample.

Sentence 1: interpret the interval in context

We are 95% confident that the interval from 0.377 to 0.463 captures the true proportion of all weekday riders on this transit system who use the mobile ticketing app.

Four pieces have to be present:

• The confidence level, here 95%.
• The two endpoints, or the phrase "the interval from a to b".
• The parameter named as a proportion, not a count and not a mean.
• The population and the response variable, in the words of the problem.

Either "captures" or "contains" works. The AP course describes this reading as being C% confident that the interval contains the true value of the parameter for the population, with details about the population it represents.

The most common near-miss is a sentence with no context: "we are 95% confident the true proportion is between 0.377 and 0.463". Nothing in it says which population or which response variable, so a reader cannot tell what was measured. Adding the phrase "of all weekday riders on this system who use the mobile ticketing app" is the whole fix.

Sentence 2: interpret the confidence level

If this sampling procedure were repeated many times, taking random samples of 500 weekday riders and building an interval from each one, about 95% of those intervals would capture the true proportion of riders who use the app.

This sentence describes the method, not your one interval. Your interval either contains $p$ or it does not, and you cannot know which. What you can say is that the recipe succeeds 95% of the time in the long run.

The AP course phrases it as repeated random sampling with the same sample size, in which approximately C% of the intervals calculated will capture the population proportion. The phrase "same sample size" matters, because a different $n$ produces intervals of a different typical width. What 95% confidence actually means draws the many-intervals picture behind this sentence.

Six wrong readings, each corrected

"95% of the data fall between 0.377 and 0.463." Wrong. The interval estimates one proportion, not the spread of individual riders. Each rider either uses the app or does not, so there is no distribution of data for the interval to cover.

**"There is a 95% probability that $p$ is between 0.377 and 0.463."** Wrong. Once the interval is computed, both endpoints and $p$ are fixed numbers, so that probability is 0 or 1 and you cannot tell which. The randomness lives in the sampling, which is why the 95% attaches to the procedure instead.

"95% of all sample proportions fall between 0.377 and 0.463." Wrong. That describes the sampling distribution of $\hat{p}$, which is centered at $p$, not at your $\hat{p}$. It is a different interval answering a different question.

**"We are 95% confident that $\hat{p}$ is between 0.377 and 0.463."** Wrong. You know $\hat{p}$ exactly: it is 0.42, and it sits inside the interval by construction. Confidence statements are only ever about the unknown parameter $p$.

"The true proportion is 0.42, give or take 0.043." Wrong as written. 0.42 is the estimate, not the truth. Say the interval captures the true proportion rather than asserting that your estimate is it.

"We are 95% confident the true proportion is between 0.377 and 0.463." Incomplete rather than false. It names no population and no response variable, so it falls short of full credit. Attach the context and it becomes sentence 1.

Using an interval to test a claim

A confidence interval doubles as a two-sided test. Any value inside the interval is plausible for $p$; any value outside is not supported by your data at that confidence level.

The vendor claims half of weekday riders use the app, so the claim is $p = 0.50$. The interval runs from 0.377 to 0.463, and 0.50 sits outside it. The sample therefore gives convincing evidence that the true proportion is not 0.50, and the interval also shows the direction: every plausible value is below 0.50.

Compare that with 0.40, which lies inside the interval. Your data are consistent with $p = 0.40$, so you would not rule it out. "Consistent with" is as strong as the claim can get, because an interval never proves a value correct.

A C% interval corresponds to a two-sided test at $\alpha = 1 - C$, so a 95% interval matches $\alpha = 0.05$ and a 99% interval matches $\alpha = 0.01$. The match is close but not exact for proportions: the interval uses $\hat{p}$ in the standard error while the test uses the null value $p_0$. Borderline cases can therefore disagree, and what does a p-value mean covers the test side.

Where the endpoints come from

The one-sample z-interval for a population proportion is

where $z^*$ is the critical value for your confidence level. Check three conditions first: the sample is random, the sample is under 10% of the population (here the system needs at least 5,000 weekday riders), and the observed counts are large enough. The observed counts are 210 successes and $500 - 210 = 290$ failures, both at least 10.

Note that the interval checks the observed counts $n\hat{p}$ and $n(1 - \hat{p})$, while a hypothesis test checks the expected counts $np_0$ and $n(1 - p_0)$. An interval has no null value to work from, so it estimates everything from the sample. Why 10 successes and 10 failures explains the threshold, and how to calculate a confidence interval covers the mean version.

Now the numbers. With $\hat{p} = 0.42$ and $n = 500$, the standard error is $\sqrt{\frac{0.42(0.58)}{500}} = \sqrt{0.0004872} \approx 0.02207$. At 95% confidence $z^* = 1.96$, so the margin of error is $1.96(0.02207) \approx 0.0433$. The interval is $0.42 \pm 0.0433$, which runs from 0.377 to 0.463, and the confidence interval calculator will confirm it.

What changes at 90% and 99% confidence

Only the critical value changes, and it changes the width. Use $z^* = 1.645$ for 90% confidence, $1.96$ for 95%, and $2.576$ for 99%.

With the same sample, the margins of error become $1.645(0.02207) \approx 0.0363$ and $2.576(0.02207) \approx 0.0569$. Those give a 90% interval of 0.384 to 0.456 and a 99% interval of 0.363 to 0.477. The margin of error calculator handles the middle step.

Both sentences change only in the number: "90% confident" or "99% confident", and "about 90% of intervals" or "about 99% of intervals". Higher confidence buys a wider interval, which rules out fewer values but captures $p$ more often. Even the 99% interval stops short of 0.50, so the evidence against the vendor's claim survives at that level. The full AP treatment sits at constructing a confidence interval for a population proportion.