An interval is a claim about p

Every confidence interval in the course has the same shape:

$$\text{point estimate} \pm (\text{critical value}) \times (\text{standard error}).$$

For one proportion that reads

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

The critical value $z^{*}$ comes from the confidence level: 1.645 for 90%, 1.96 for 95%, 2.576 for 99%. It is the number of standard errors that captures the middle of that percentage of a normal curve, so a higher confidence level means a larger $z^{*}$ and a wider interval.

The standard error is the estimated standard deviation of $\hat{p}$. Note which proportion goes into it: the true $p$ is unknown here, so $\hat{p}$ stands in. That substitution is specific to intervals; the test in Topic 3.7 uses the hypothesized $p_0$ instead, because there a value of $p$ is being assumed.

Of 400 randomly selected voters, 248 support a measure, so $\hat{p} = 0.62$ and

$$SE = \sqrt{\frac{0.62(0.38)}{400}} \approx 0.0243, \qquad 0.62 \pm 1.96(0.0243) = 0.62 \pm 0.0476,$$

giving roughly $(0.572, 0.668)$.

The same three conditions as the previous topic, with one adjustment: since $p$ is unknown, the large-counts check uses the observed counts.

1. Random: the data come from a random sample or randomized experiment. Name how the sample was actually drawn.
2. 10%: $n \le 0.10N$. For 400 voters from a district of many thousands, satisfied.
3. Large counts: $n\hat{p} \ge 10$ and $n(1 - \hat{p}) \ge 10$. Here $n\hat{p} = 248$ and $n(1 - \hat{p}) = 152$, which are simply the successes and failures actually observed.

Writing the numbers is the check. "Large counts: 248 and 152, both at least 10" can be verified by a reader; "conditions met" cannot. And each condition still protects its own property: random centers the interval's method on $p$, the 10% condition keeps the standard error from running too large, and the counts deliver the normal shape that $z^{*}$ assumes.

The margin of error $z^{*}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ moves in exactly two ways.

Confidence level. Raising it raises $z^{*}$ and widens the interval. There is no free lunch: a 99% interval catches the parameter more often precisely because it claims less about where the parameter is. Taken to the extreme, the interval from 0 to 1 has 100% confidence and no content.

Sample size. Raising $n$ narrows the interval, through a square root. The 400-voter interval has a margin of error of about 0.048; reaching 0.024 takes $n = 1600$. Quadruple the data for half the width, the same arithmetic as every other spread in this unit.

What does not change the width is the population's size. A sample of 1,000 gives about the same precision for a city and for a country, as long as the 10% condition holds, because $N$ appears nowhere in the formula. That surprises people, and it is the reason national polls run about a thousand respondents.

Two different sentences are required, and mixing them is the most heavily penalized error in the unit.

The interval. "We are 95% confident that the interval from 0.572 to 0.668 captures the true proportion of all voters in the district who support the measure." It names the population, names the parameter, and gives the interval. Three tempting versions are all wrong:

1. "95% of voters support between 0.572 and 0.668" describes the individuals, not the parameter.
2. "There is a 95% probability $p$ lies in this interval" treats $p$ as random. The parameter is a fixed number; it is either in this interval or not.
3. "95% of future sample proportions will land in this interval" describes future statistics, not the parameter.

The level. "If we took many random samples of this size and built an interval from each, about 95% of those intervals would capture the true proportion." The 95% is a property of the method over repeated sampling, not of the one interval in hand, which has already either caught $p$ or missed it.