Confidence Interval for One Proportion

Unit 6 · Inference for Categorical Data: Proportions

What AP Stats asks here

A one-proportion CI estimates the population proportion $p$ from a sample proportion $\hat{p}$ . The two recurring AP problems: build the interval (right formula, right $z^*$ , right conditions), and interpret it. The standard interpretation traps are (1) treating the parameter as random and (2) interpreting the interval as something about individuals rather than about the population proportion.

Confidence interval

Estimate ± ME

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

z^*

values

z^*_{90} = 1.645,\;\; z^*_{95} = 1.960,\;\; z^*_{99} = 2.576

Conditions

Random sample

\text{required}

Large counts

n\hat{p} \ge 10 \text{ AND } n(1-\hat{p}) \ge 10

10%

n < 0.10 \cdot N

AP Tip: Quadrupling n halves the margin of error. Diminishing returns set in fast — choose n based on the precision you need, not the precision you wish you had.

Caution: Confidence is about the method (long-run capture rate), not about this particular interval. 'There is a 95% probability the parameter lies in this interval' is the textbook wrong interpretation.

Type 1

Construct the CI

Verify the conditions, plug in $\hat{p}, n$ , choose $z^*$ , and report the interval.

Example 1

An SRS of 400 voters from a large city includes 240 supporters of a transit measure. Which is the 95% CI for the true proportion of supporters? (A)

(0.560, 0.640)

(B)

(0.552, 0.648)

(C)

(0.537, 0.663)

(D)

(0.600, 0.648)

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Type 2

Interpret CI vs confidence level

The interval-interpretation is about the parameter. The confidence-level interpretation is about the method's long-run capture rate.

Example 2

A 95% CI is

(0.552, 0.648)

. Which interpretation is correct? (A) There is a 95% probability that the true proportion is between 0.552 and 0.648. (B) 95% of voters fall between 0.552 and 0.648. (C) We are 95% confident that the true proportion of supporters in this city is between 0.552 and 0.648. (D) The sample proportion is 95% likely to be between 0.552 and 0.648.

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Type 3

Sample size and ME trade-offs

Solve $n \ge (z^*)^2 \hat{p}(1-\hat{p}) / \text{ME}^2$ and always round up.

Example 3

A pollster wants a 95% CI with ME ≤ 0.03 for the proportion of adults who exercise daily. A pilot study gives

\hat{p} \approx 0.25

. What sample size is needed? (A) 800 (B) 801 (C) 1067 (D) 1068

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