Unit 3 · Topic 3.3 · about 25 minutes

Constructing a Confidence Interval for a Population Proportion

Turn one sample proportion into an interval that says, with a stated level of confidence, where the population proportion is.

Predict first

In a random sample of 500 teens, 62% say they get less sleep than they should. Which range is the best guess for the share of ALL teens who would say that?

An estimate plus a margin

A sample proportion $\hat{p}$ is a point estimate: one number, almost certainly a little off. A confidence interval adds a margin of error on each side so the estimate comes with an honest sense of how far off it might be.

Every confidence interval in this course has the same shape: point estimate plus or minus margin of error. For a population proportion the procedure is the one-sample z-interval for a population proportion:

\hat{p} \pm z^{*} \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}

The square root is the standard error of $\hat{p}$ , which estimates how much sample proportions typically vary from the true proportion. The critical value $z^{*}$ sets the confidence level. Their product is the margin of error, which is half the width of the interval.

Critical values $z^{*}$ for common confidence levels
Confidence level	Middle area of the standard normal curve	$z^{*}$
90%	0.90	1.645
95%	0.95	1.960
99%	0.99	2.576

Three conditions before you calculate

The formula only means something when the data meet three conditions. Check them every time, in context, before computing.

Randomization: the data come from a random sample.
10% condition: when sampling without replacement, the population is at least 10 times the sample size, $n \leq 0.10 N$ .
Normality: the sample has at least 10 successes and at least 10 failures, so $n \hat{p} \geq 10$ and $n (1 - \hat{p}) \geq 10$ . That is what makes the sampling distribution of $\hat{p}$ close enough to normal for $z^{*}$ to be right.

If a condition fails, say so. An interval built on a convenience sample can be calculated, but it estimates nothing.

Worked exampleA 95% interval from a national poll

A polling firm randomly selects 1,024 adults in the United States. Of them, 553 say they would rather work from home at least part of the week. Construct and interpret a 95% confidence interval for the proportion of all U.S. adults who would rather work from home at least part of the week.

Name it. One-sample z-interval for $p$ , the proportion of all U.S. adults who would rather work from home at least part of the week.
Check conditions. Random: the firm randomly selected adults. 10%: 1,024 is far less than 10% of all U.S. adults. Normality: there are 553 successes and 471 failures, both at least 10.
Calculate. $\hat{p} = \frac{553}{1024} \approx 0.54004$ and $SE = \sqrt{\frac{0.54004 (0.45996)}{1024}} \approx 0.015575$ . The margin of error is $1.960 (0.015575) \approx 0.03053$ , so the interval is $0.54004 \pm 0.03053$ , which is about $(0.5095, 0.5706)$ .
Interpret in context. We are 95% confident that the interval from 0.5095 to 0.5706 contains the true proportion of all U.S. adults who would rather work from home at least part of the week.

Answer.

About 0.510 to 0.571, interpreted as a statement about all U.S. adults, not about the 1,024 who answered.

Lab

Confidence Interval Capture

Before you trust the phrase "95% confident", see it happen. Build a batch of intervals from a population whose true proportion you can see, and count how many miss.

Open the full Confidence Interval Capture lab

How big a sample do you need?

Rearranging the margin of error formula gives the smallest sample size that guarantees a margin of error $ME$ :

n \geq {(\frac{z^{*}}{ME})}^{2} \hat{p} (1 - \hat{p})

Before collecting data you do not know $\hat{p}$ . Use a reasonable estimate from earlier work if you have one. Otherwise use $\hat{p} = 0.5$ , which makes $\hat{p} (1 - \hat{p})$ as large as it can be, so the answer is big enough no matter what the true proportion turns out to be.

For a margin of error of 0.03 at 95% confidence with no prior estimate: $n \geq {(\frac{1.960}{0.03})}^{2} (0.5) (0.5) \approx 1067.1$ . Always round UP, to 1,068. Rounding down would give a margin of error slightly larger than you asked for.

Check your understanding

In a random sample of 600 voters in a large county, 312 support a new park bond. Find the standard error of the sample proportion. Round to four decimal places.

Your answer

A student wants a 95% confidence interval for the proportion of all students at her school of 1,800 who walk to school. She asks the first 40 students who arrive in the parking lot, and 14 say they walk. Which condition is the most serious problem?

A researcher randomly samples 750 adults in Ohio and asks whether they have a library card. Which is the best description of the parameter being estimated?

A city wants a 99% confidence interval for the proportion of residents who support a new curfew, with a margin of error of no more than 0.04. With no earlier estimate to use, what is the smallest sample size that guarantees this? Use $z^{*} = 2.576$ .

Your answer

A pollster changes a confidence level from 95% to 90% and changes nothing else. What happens to the interval?

Practice

Practice until it is automatic

New numbers every time. Each one is checked the moment you answer, with the full working shown.

Confidence intervals for a proportion practice page

Course alignment, for teachers

AP Statistics topic 3.3, Unit 3: Inference for Categorical Data: Proportions.

Skill 2.C: Identify appropriate statistical inference methods.
Skill 3.E: Calculate appropriate statistical inference method results.
Skill 4.E: Justify the use of a chosen statistical inference method by verifying conditions.