Confidence Interval Simulator

Set up the population and the procedure:

Then build intervals with Draw 1, Draw 50 or Draw 200, and clear them with Reset. Changing p, n or the confidence level also clears everything, so every interval on screen comes from the same procedure.

Each draw takes a random sample, computes its $\hat{p}$, and builds the interval $\hat{p} \pm z^*\sqrt{\hat{p}(1-\hat{p})/n}$, just as you would with real data. On the plot each interval is a horizontal bar on a 0 to 1 axis, with a small circle at $\hat{p}$ when the rows are tall enough. A dashed vertical line marks the true p. Teal bars capture it; coral bars miss. The newest interval is on top and up to 120 are shown, but the counters include every interval.

The three counters show Intervals, the number captured and the number missed, each with its percentage to one decimal.

With p = 0.50 and n = 100 at 95%, several rounds of Draw 200 give capture rates that hover near 95%. Switch to 99% and the bars get wider and miss less often. Now try p = 0.05 and n = 20, where the expected number of successes is only 1. Even at 95%, the capture rate settles far below 95% (around 64% in the long run), because the Large Counts condition fails and the formula breaks down.

A one-sample z-interval for a population proportion:

Conditions: a random sample; $n \le 10\%$ of the population when sampling without replacement; and Large Counts, at least 10 successes and 10 failures in the sample ($n\hat{p} \ge 10$ and $n(1-\hat{p}) \ge 10$).

Width moves in two ways. A higher confidence level means a larger $z^*$ and a wider interval. A larger $n$ shrinks the standard error by $\sqrt{n}$, so four times the sample size halves the margin of error.

Problem: In a random sample of 400 students at a large university, 248 support a later start to classes. Construct and interpret a 95% confidence interval for the proportion of all students at the university who support it.

Step 1: Check conditions. The sample is random. 400 is less than 10% of the students at a large university. There are 248 successes and 152 failures, both at least 10.

Step 2: Point estimate and SE. $\hat{p} = 248/400 = 0.62$. $\text{SE} = \sqrt{\dfrac{0.62(0.38)}{400}} \approx 0.0243$.

Step 3: Margin of error. $1.96 \times 0.0243 \approx 0.0476$.

Step 4: Interval. $0.62 \pm 0.0476$ gives $(0.572, 0.668)$.

Step 5: Interpret. We are 95% confident that the interval from 0.572 to 0.668 captures the true proportion of all students at this university who support a later start. With the same data, a 90% interval is $(0.580, 0.660)$ and a 99% interval is $(0.557, 0.683)$.

Check it in the simulator: set p to 0.62 and n to 400 at 95%. Each interval is about 0.095 wide, like yours. Draw 200 a few times: roughly 95 in every 100 intervals are teal. Your one real interval is like one of those bars; you cannot tell whether it is teal.

Confidence intervals for a proportion are topic 3.3, and the general form, statistic ± (critical value)(standard error of statistic), is on the AP Statistics formula sheet. Free-response questions often ask for the full process (conditions, calculation and interpretation in context) or for an interpretation of the confidence level itself.