Interval for a Proportion — AP® Statistics Lesson

Part 1 · Hook

Why this matters

A dietitian surveys 400 randomly chosen students and finds that 62% regularly skip breakfast. She could report "62%" and stop, but another sample would give 59% or 65%. A confidence interval reports a range instead, built by a method that captures the true campus proportion a known share of the time.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $p = 0.5$ and $n = 400$ , what is $\sigma_{\hat{p}}$ ?

0.025
0.000625
0.5
0.25

Show the answer

$\sqrt{\frac{0.5(0.5)}{400}}$ $= \sqrt{0.000625}$ $= 0.025$ .

Correct: 0.025:
0.000625:
0.5:
0.25:

2. Under the standard normal curve, what value has an area of 0.975 below it?

1.96
1.645
2.576
0.975

Show the answer

$P(Z \le 1.96) = 0.975$ , so 95% of the area lies between $-1.96$ and $1.96$ .

Correct: 1.96:
1.645:
2.576:
0.975:

Part 4 · See it

See it first

Top: the normal sampling distribution of the sample proportion, centered at the unknown true proportion p, with the middle 95% shaded between p minus 1.96 standard deviations and p plus 1.96 standard deviations. Below it, three sample proportions are drawn with their intervals, each the sample proportion plus or minus 1.96 standard deviations. The two sample proportions that land inside the shaded middle 95% give intervals that reach across p; the one that lands in the tail gives an interval that misses p. — In 95% of samples, the sample proportion lands within 1.96 standard deviations of p, so the interval built around it reaches across p. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$\hat{p}$ is approximately normal, centered at $p$ in 95% of random samples, $\hat{p}$ lands within 1.96 standard deviations of $p$
Turn that aroundin 95% of samples, $p$ is within 1.96 standard deviations of $\hat{p}$
We do not know $p$ , so the standard deviation cannot be computedwe use the standard error $\sqrt{\hat{p}(1 - \hat{p})/n}$ in its place
Go $z^*$ standard errors out on each side of $\hat{p}$ $\hat{p} \pm z^*\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ captures $p$ for the stated share of samples

Part 6 · Key ideas

Key ideas

Interval = statistic ± (critical value)(standard error). For one proportion: $\hat{p} \pm z^*\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ .
$z^*$ : 1.645 for 90%, 1.96 for 95%, 2.576 for 99%. Area below $z^*$ is $C + \frac{1 - C}{2}$ .
Conditions in words: random sample; $n$ at most 10% of the population; at least 10 successes and 10 failures in the sample.
Margin of error $= z^* \cdot \text{SE}$ . It shrinks with a larger $n$ (4 times $n$ halves it) and grows with the confidence level.
Name the procedure in full: one-sample z interval for a proportion.

Part 7 · Misconception

A common mistake

The wrong idea: For a 90% interval, find the $z$ with 0.90 of the area below it, so $z^* = 1.28$ .

What actually happens: A 90% interval keeps 90% in the middle and leaves 5% in each tail, so $z^*$ has 0.95 below it: $z^* = 1.645$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Skipping breakfast at a large university

A dietitian at a university with 28,000 students selects a random sample of 400 students. Of them, 248 say they regularly skip breakfast. She wants to estimate the proportion of all students at the university who regularly skip breakfast.

Survey results (invented data)
Response	Number of students
Regularly skip breakfast	248
Do not	152
Total	400

1. Which statement correctly checks the large counts condition for a confidence interval?

248 skip and 152 do not, and both are at least 10
$n = 400$ is at least 30
400 is less than 10% of 28,000
$400(0.5) = 200$ is at least 10

Show the answer

For an interval, $p$ is unknown, so check the observed successes and failures: $n\hat{p} = 248$ and $n(1 - \hat{p}) = 152$ , both at least 10.

Correct: 248 skip and 152 do not, and both are at least 10: Right: for an interval, check the sample's successes and failures.
$n = 400$ is at least 30: The rule of 30 is about means. For a proportion, check successes and failures.
400 is less than 10% of 28,000: That is the 10% condition, which is about independence, not the normal shape.
$400(0.5) = 200$ is at least 10: No value of 0.5 is claimed here. For an interval, use the sample's counts, 248 and 152.

2. Calculate the standard error of $\hat{p}$ . Give four decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$\hat{p} = \frac{248}{400} = 0.62$ and $\text{SE}$ $= \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ $= \sqrt{\frac{0.62(0.38)}{400}}$ $= 0.0243$ . It uses $\hat{p}$ because $p$ is unknown.

Answer: 0.0243

3. Calculate the margin of error for a 95% confidence interval for the proportion of all students who regularly skip breakfast. Give four decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$\text{ME}$ $= z^* \cdot \text{SE}$ $= 1.96\sqrt{\frac{0.62(0.38)}{400}}$ $= 1.96(0.0243)$ $= 0.0476$ .

Answer: 0.0476

4. Construct a 95% confidence interval for the proportion of all students at the university who regularly skip breakfast.

Type the interval as "from a to b".

Show the answer

$0.62$ $\pm 1.96\sqrt{\frac{0.62(0.38)}{400}}$ $= 0.62$ $\pm 0.048$ , which gives the interval from 0.572 to 0.668. Technology (a one-proportion z interval) gives the same.

Answer: from 0.572 to 0.668

5. Type the 95% confidence interval as an expression with the values substituted: the point estimate, then +/-, then the margin of error.

Type the expression as on the exam, with the values substituted.

Show the answer

Statistic $\pm$ (critical value)(standard error): $0.62 \pm 1.96\sqrt{\frac{0.62(0.38)}{400}}$ .

Answer: $0.62 \pm 1.96\,\sqrt{\frac{0.62 \cdot 0.38}{400}}$

6. Find the critical value $z^*$ for a 90% confidence interval. Give three decimal places.

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

A 90% interval leaves 5% in each tail, so $z^*$ has 95% of the standard normal area below it: $z^* = 1.645$ .

Answer: 1.645

7. The dietitian's 95% confidence interval for the proportion of all students at her university who regularly skip breakfast runs from 0.572 to 0.668. Build the interpretation of this interval.

Write the conclusion in context.

Show the answer

The interpretation names the confidence level, the interval, and the parameter in context: the proportion of all students at the university, not the 400 in the sample.

A full-credit answer: We are 95% confident that the interval from 0.572 to 0.668 captures the true proportion of all students at the university who regularly skip breakfast.

Part 9 · Summary

Summary

A confidence interval for

p

is

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

: the point estimate plus or minus a margin of error. Check random, 10% and large counts (sample successes and failures at least 10) in words first. Larger samples shrink the margin of error; higher confidence levels stretch it.

Part 10 · Up next

What comes next

Part 11 · Connections