One-Proportion z-Interval

One-Proportion z Confidence Interval

By the end of this lesson, you will be able to construct and interpret a one-proportion z confidence interval, check all conditions, and determine the required sample size.

The Setting

Use a one-proportion z interval when you want to estimate a single population proportion p based on sample data.

Statistic: $\hat{p} = X/n$ (sample proportion of successes)

Parameter: p (unknown population proportion)

Conditions (State and Check Every Time)

1. Random: Data came from a random sample (SRS) or randomized experiment.

2. Large Counts: $n\hat{p} \geq 10$ AND $n(1-\hat{p}) \geq 10$

• This ensures the sampling distribution of $\hat{p}$ is approximately normal
• Use $\hat{p}$ (not p) because p is unknown

3. Independent: Either sampling with replacement, OR population size ≥ 10n (the 10% condition)

The Formula

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

Common critical values:

Confidence Level	z*
90%	1.645
95%	1.960
99%	2.576

Standard error: $SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})/n}$

Four-Step AP Solution

State: "I will construct a one-proportion z confidence interval for p, the [describe parameter in context]."

Plan: "Conditions: (1) Random — [verify]. (2) Large Counts — $n\hat{p} = [value] \geq 10$ ✓ and $n(1-\hat{p}) = [value] \geq 10$ ✓. (3) Independent — [verify 10% condition]."

Do: Compute the interval.

Conclude: "We are [C]% confident that the true proportion of [context] is between [L] and [U]."

Worked Example

In a random sample of 400 high school students, 240 reported using social media for more than 3 hours daily. Construct a 95% confidence interval for the true proportion of high school students who use social media this much.