Plus-Four Interval vs Confidence Interval

The plus-four interval is the ordinary one-proportion z-interval computed as though the sample had held two more successes and two more failures than it did. Add 2 to the success count $x$ and 4 to the sample size, giving $\tilde{p} = \frac{x+2}{n+4}$ (p-tilde), then form $\tilde{p} \pm z^*\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n+4}}$. Both the center and the standard error use the padded counts. It is a college-level adjustment, not part of the AP Statistics course effective Fall 2026, so use the plain z-interval on AP work.

What it fixes is undercoverage. The plain interval puts $\hat{p}$ (p-hat) where the unknown $p$ belongs inside the standard error, and that swap costs it some of its stated confidence level. Sum the binomial probabilities of every count whose interval captures $p$ and you get the exact long-run coverage: at $n = 50$ and $p = 0.60$ a nominal 95% interval covers 94.1% of the time, and at $n = 20$ and $p = 0.90$ it covers 87.6%. The plus-four version covers 95.7% in that second case. The site's coverage simulator computes those same numbers.

With 3 successes in 10 trials, $\tilde{p} = 5/14 = 0.3571$, the standard error is $\sqrt{0.3571(0.6429)/14} = 0.1281$, and the 95% interval is $0.3571 \pm 1.96(0.1281)$, or 0.106 to 0.608. The plain interval on the same data runs 0.016 to 0.584.

"It is just a wider interval, padded to be safe" is the wrong picture. On those counts the plus-four interval is narrower, 0.502 against 0.568. The adjustment pulls the center toward 0.5 and divides by $n+4$, so it widens the interval at extreme counts, where 1 success in 10 gives 0.430 against 0.372, and narrows it in the middle. Coverage improves because the center moved, not because the interval was inflated.

The case it rescues outright is $x = 0$ or $x = n$. There $\hat{p}(1-\hat{p})$ is zero, so the plain interval has zero width and cannot contain anything at all. Once the sample carries at least 10 successes and 10 failures the two intervals nearly agree, which is why the AP course checks that condition instead.

A confidence interval turns a single estimate into a range of plausible parameter values by attaching an explicit allowance for sampling error. Every interval in AP Statistics has the same shape:

The product of the last two pieces is the margin of error, and the standard error estimates how much the statistic varies from one sample to the next.

Say 520 of 1000 randomly sampled voters back a measure. The estimate is $\hat{p} = 0.52$ (p hat), the standard error is $\sqrt{\frac{0.52(0.48)}{1000}} = 0.0158$, and a 95 percent interval uses the critical value $z^* = 1.960$ (z star). The margin of error is $1.960 \times 0.0158 = 0.031$, so the interval runs from 0.489 to 0.551, or 48.9 percent to 55.1 percent.

The interpretation that loses marks is what the 95 percent attaches to. There is not a 95 percent chance that the true proportion lies between 0.489 and 0.551. The true proportion is a fixed number: it is either inside those endpoints or it is not, and no probability is left over once the sample is drawn. What varies from sample to sample is the interval itself, since a fresh sample of 1000 voters produces different endpoints. The 95 percent describes the method, which captures the parameter in about 95 percent of all possible samples. The interval also says nothing about where 95 percent of individual voters sit.

Width is the other half of the idea. The margin of error shrinks like $1/\sqrt{n}$, so quadrupling the sample from 1000 to 4000 only halves it, from 0.0310 to 0.0155. Raising confidence widens the interval instead ($z^* = 1.645$ at 90 percent, $2.576$ at 99 percent), so precision and confidence trade against each other at any fixed sample size.

One reading comes free. This interval contains 0.50, so the data do not rule out an even split, which is the same verdict a two-sided test of $H_0: p = 0.5$ at $\alpha = 0.05$ returns on these counts.