A poll finds that 60% of 150 adults read a book last month. The true proportion of all adults is unknown, and a different sample would give a different $\hat{p}$. The sampling distribution of $\hat{p}$ tells us how far $\hat{p}$ typically lands from $p$, and the inverse normal tells us how many standard deviations cover the middle 95%. Together they give an interval of plausible values.
Key idea
A confidence interval for a proportion is
$$\hat{p} \pm z^*\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}},$$
estimate ± margin of error. The critical value $z^*$ sets the confidence level: $1.645$ for 90%, $1.960$ for 95%, $2.576$ for 99%. Use it when the sample is random and there are at least 10 successes and 10 failures.
Why it works
About 95% of all samples give a $\hat{p}$ within $1.96\sigma_{\hat{p}}$ of the true $p$. Distance is symmetric: if $\hat{p}$ is within $1.96\sigma_{\hat{p}}$ of $p$, then $p$ is within $1.96\sigma_{\hat{p}}$ of $\hat{p}$. So the interval $\hat{p} \pm 1.96\sigma_{\hat{p}}$ captures $p$ for about 95% of samples. We do not know $p$, so we estimate $\sigma_{\hat{p}}$ with the standard error $\sqrt{\hat{p}(1 - \hat{p})/n}$. The $1.96$ is the z with 97.5% of the area to its left, leaving 2.5% in each tail.
Worked example: Building the interval
In a random sample of 150 adults, 90 read a book last month. Find a 95% confidence interval for the proportion of all adults who did.
- $\hat{p} = \dfrac{90}{150} = 0.6$. Conditions: random sample; 90 successes and 60 failures, both at least 10.
- $$SE = \sqrt{\dfrac{0.6 \times 0.4}{150}} = \sqrt{0.0016} = 0.04$$
- Margin of error: $1.96 \times 0.04 = 0.0784$.
- Interval: $0.6 \pm 0.0784$, from $0.5216$ to $0.6784$.
- We are 95% confident that between about 52% and 68% of all adults read a book last month.
Answer: $(0.522, 0.678)$
Worked example: Higher confidence
Find a 99% interval from the same sample.
- $z^* = 2.576$; margin $2.576 \times 0.04 \approx 0.103$.
- Interval: $0.6 \pm 0.103$, from about $0.497$ to $0.703$.
- More confidence costs a wider interval.
Answer: $(0.497, 0.703)$
Worked example: What "95% confident" means
A student says: "There is a 95% probability that the true proportion is between 0.522 and 0.678." What is wrong?
- The true proportion is a fixed number. This interval either contains it or it does not; there is nothing random left.
- The 95% describes the method: if we took many samples and built an interval from each, about 95% of those intervals would capture $p$.
Answer: The 95% is a property of the method, not of one interval.
Common mistake
Saying there is a 95% chance that the true value is in this particular interval. Check your sentence: put the confidence on the method ("95% of intervals built this way capture $p$") or use the phrase "we are 95% confident that…" for this interval.
The interval only accounts for random sampling error. A biased sample can miss the truth by far more than any margin of error. The econometrics course has a machine that builds 100 intervals at a time.