Why Is My Confidence Interval So Wide?

Q: Why does a proportion near 0.5 give the widest interval?

Because the standard error contains p-hat(1 - p-hat), which is a downward parabola peaking at p-hat = 0.5 with a value of 0.25. Any proportion further from 0.5 makes that product smaller, so the standard error and the margin of error both shrink.

By Jude Wallis · Published July 31, 2026

Four things control the width: your sample size, your confidence level, how spread out the data are, and for proportions how close the sample proportion is to 0.5. Sample size is the strongest lever, but width shrinks like 1 over the square root of n, so quadrupling n only halves it.

Interval width is part of constructing and justifying confidence intervals in the Fall 2026 AP Statistics course: Unit 3 topics 3.3 and 3.4 for proportions, and Unit 4 topic 4.2 for means.

Where the width comes from

Every confidence interval on the AP course has the same shape:

\text{statistic} \pm (\text{critical value}) \times (\text{standard error})

The part after the plus-or-minus sign is the margin of error, and the full width of the interval is twice it. So anything that changes the width has to work through one of two factors: the critical value, or the standard error.

The critical value ( $z^*$ or $t^*$ ) depends only on your confidence level, and on the degrees of freedom when it is $t^*$ .
The standard error depends on the sample size and on how variable the data are.

That gives four levers in total, and nothing else. If your interval is wider than you wanted, the cause is one of them. Work through them in order below.

A wide interval is not a mistake, by the way. It is the interval telling you the truth: with this much data, this range of parameter values is plausible. Reporting a narrow interval you did not earn would be the error.

Lever 1: sample size, and why quadrupling only halves the width

Sample size sits under a square root in every standard error on the course:

SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \qquad SE_{\bar{x}} = \frac{s}{\sqrt{n}}

Both are proportional to $1/\sqrt{n}$ . That single fact governs how much extra data buys you.

Take $\hat{p} = 0.60$ and a 95% interval with $z^* = 1.960$ , and watch the margin of error as $n$ grows:

$n$	$SE$	Margin of error	Interval	Width
100	0.0490	0.0960	(0.504, 0.696)	0.1920
400	0.0245	0.0480	(0.552, 0.648)	0.0960
1600	0.0122	0.0240	(0.576, 0.624)	0.0480
6400	0.0061	0.0120	(0.588, 0.612)	0.0240

Read the pattern down the last column. Each row multiplies $n$ by 4 and the width drops by exactly a factor of 2, because $\sqrt{4} = 2$ . Going from 100 to 6400 is 64 times the data for 8 times the precision.

The practical consequence: if you want to cut a margin of error in half, you need four times the sample size, not twice. To cut it to a third, nine times. Precision gets expensive fast, and that is why large surveys stall around a two or three point margin instead of pushing lower.

One caution about big samples. A larger $n$ shrinks the interval around whatever your sampling method was aiming at, so it makes a biased estimate more precisely wrong. Does a bigger sample fix bias explains why the answer is no.

Lever 2: confidence level

Higher confidence means a wider net. The critical value grows with the confidence level, and the margin of error is directly proportional to the critical value.

Hold $\hat{p} = 0.60$ and $n = 400$ , so $SE = 0.024495$ for all three rows:

Confidence level	Critical value $z^*$	Margin of error	Width
90%	1.645	0.04029	0.08059
95%	1.960	0.04801	0.09602
99%	2.576	0.06310	0.12620

Moving from 95% to 99% widens the interval by a factor of $2.576/1.960 = 1.314$ , about 31% wider, for four extra percentage points of confidence. Moving down to 90% narrows it by a factor of $1.645/1.960 = 0.839$ .

This is a real tradeoff, not a free choice. A 99% interval is right more often across many samples, and it is less useful for pinning the parameter down. A 100% interval would be all possible values and would tell you nothing. Pick the level before you compute, for the same reason you pick $\alpha$ before you compute. See confidence level vs confidence interval and what 95% confidence means.

For a mean the critical value is $t^*$ , which also depends on degrees of freedom, so small samples get hit twice: once through $\sqrt{n}$ in the denominator and again through a larger $t^*$ . At 95% confidence, $t^* = 2.776$ when $df = 4$ , $2.262$ when $df = 9$ , $2.064$ when $df = 24$ , and $1.984$ when $df = 99$ . It never drops below $z^* = 1.960$ . Why the t-distribution is used for small samples covers the reason.

Lever 3: variability in the data

For a mean, the sample standard deviation $s$ sits in the numerator of the standard error, so width scales with it directly. Double the variability and you double the margin of error.

Take $n = 25$ at 95% confidence, so $df = 24$ and $t^* = 2.064$ :

$s$	$SE = s/\sqrt{25}$	Margin of error
4	0.8	1.651
8	1.6	3.302
12	2.4	4.954

Unlike the first two levers, this one is mostly not yours to set. The spread of the population is what it is. What you can sometimes control is the design.

Pair or block. A paired design compares each subject to itself, which takes person-to-person variability out of the standard error. When subjects differ a lot from each other, that is why a paired design usually gives a narrower interval than two independent samples of the same total size. It is not automatic: pairing also costs degrees of freedom, so it only pays when the pairing genuinely removes variability. See when to use a paired t-test.
Reduce measurement error. A noisier instrument inflates $s$ for no informational gain.
Do not trim outliers to narrow the interval. Removing inconvenient data shrinks $s$ and produces an interval that no longer covers the parameter at the stated rate.

One thing that does not belong on this list: the population size. Provided the sample is under 10% of the population, the standard error does not depend on how big the population is. A sample of 1000 gives about the same margin of error for a city and for a country.

Lever 4: for proportions, how close p-hat is to 0.5

The standard error of a proportion contains the product $\hat{p}(1 - \hat{p})$ , and that product is largest exactly at $\hat{p} = 0.5$ .

The reason is simple algebra: $\hat{p}(1 - \hat{p})$ is a downward parabola through 0 at both ends with its peak at the midpoint, where it equals $0.5 \times 0.5 = 0.25$ . Any move away from 0.5 in either direction makes it smaller.

At $n = 400$ with $z^* = 1.960$ :

$\hat{p}$	$\hat{p}(1 - \hat{p})$	$SE$	Margin of error
0.50	0.2500	0.025000	0.04900
0.60	0.2400	0.024495	0.04801
0.75	0.1875	0.021651	0.04244
0.90	0.0900	0.015000	0.02940
0.95	0.0475	0.010897	0.02136

So a near-even split is the hardest case, and a lopsided proportion is estimated more precisely from the same sample size. The curve is flat near the top, which is why 0.50 and 0.60 differ by only about 2% in margin of error, and steep further out.

This has one very useful consequence. When you are planning a study and have no idea what $\hat{p}$ will be, use $p^* = 0.5$ . It maximizes the standard error, so the sample size it gives you is guaranteed to be enough whatever the real proportion turns out to be. That is the conservative choice, and it is what the sample size calculator uses by default.

Note that this lever does not exist for means. There is no equivalent of " $\bar{x}$ near some special value," because $s$ is a free parameter rather than a function of $\bar{x}$ .

Planning a width before you collect data

Run the margin-of-error formula backwards. For a proportion, set the margin of error you want and solve for $n$ :

n = \left(\frac{z^*}{ME}\right)^2 p^*(1 - p^*)

Always round the result up, since rounding down leaves you short of the target. With $p^* = 0.5$ and 95% confidence:

$ME = 0.03$ gives $n = (1.960/0.03)^2 \times 0.25 = 1067.1$ , so $n = 1068$ .
$ME = 0.02$ gives $n = (1.960/0.02)^2 \times 0.25 = 2401$ , so $n = 2401$ .

That jump, from about 1068 to 2401 for one extra percentage point of precision, is the square root law again in the direction that hurts.

One last check before you blame the sample size. If your interval for a difference contains 0, that is not a defect in the interval, it is a finding: 0 is among the plausible values, so the data are consistent with no difference. See two-proportion confidence interval contains zero. And if the interval and a hypothesis test seem to disagree, there is a specific reason and it is usually the standard error.

The same survey at three sample sizes

In a random sample of registered voters, 60% support a ballot measure. Build the 95% confidence interval for the population proportion when the sample size is 100, then 400, then 1600, and describe how the width changes. Assume each sample is under 10% of the population.

The sample proportion is $\hat{p} = 0.60$ in all three cases, and $z^* = 1.960$ for 95% confidence.
Check the large counts condition at the smallest sample: $n\hat{p} = 100(0.60) = 60 \geq 10$ and $n(1 - \hat{p}) = 100(0.40) = 40 \geq 10$ . It holds, and holds more easily at larger $n$ .
At $n = 100$ : $SE = \sqrt{\frac{0.6(0.4)}{100}} = \sqrt{0.0024} = 0.048990$ . Margin of error $= 1.960 \times 0.048990 = 0.09602$ . Interval $= 0.60 \pm 0.09602 = (0.5040,\ 0.6960)$ , width 0.19204.
At $n = 400$ : $SE = \sqrt{\frac{0.24}{400}} = \sqrt{0.0006} = 0.024495$ . Margin of error $= 1.960 \times 0.024495 = 0.04801$ . Interval $= (0.5520,\ 0.6480)$ , width 0.09602.
At $n = 1600$ : $SE = \sqrt{\frac{0.24}{1600}} = \sqrt{0.00015} = 0.012247$ . Margin of error $= 1.960 \times 0.012247 = 0.024005$ . Interval $= (0.5760,\ 0.6240)$ , width 0.04801.
Compare the widths: 0.19204, then 0.09602, then 0.04801. Each is exactly half the one above it.
Check that against the formula. Multiplying $n$ by 4 divides $\sqrt{n}$ by 2, and the margin of error is proportional to $1/\sqrt{n}$ , so the width halves. Note this is exact, not approximate, because $\hat{p}$ and $z^*$ were held fixed.

The 95% intervals are (0.504, 0.696) at $n = 100$ , (0.552, 0.648) at $n = 400$ , and (0.576, 0.624) at $n = 1600$ . Widths of 0.192, 0.096, and 0.048: each quadrupling of the sample size halves the width, which is the $1/\sqrt{n}$ rule showing up as numbers.

Choosing a sample size for a target margin of error

You are planning a poll and want a 95% confidence interval for the proportion of adults who support a policy, with a margin of error no larger than 3 percentage points. You have no prior estimate of the proportion. How many people do you need? What if you tighten the target to 2 percentage points?

With no prior estimate, use the conservative value $p^* = 0.5$ , which maximizes $p^*(1 - p^*)$ at 0.25 and therefore maximizes the required sample size.
Use $z^* = 1.960$ for 95% confidence and the target $ME = 0.03$ .
Solve $n = \left(\frac{z^*}{ME}\right)^2 p^*(1 - p^*) = \left(\frac{1.960}{0.03}\right)^2 (0.25)$ .
Compute the ratio: $1.960/0.03 = 65.3333$ . Square it: $65.3333^2 = 4268.444$ .
Multiply by 0.25: $4268.444 \times 0.25 = 1067.111$ .
Round up to the next whole person: $n = 1068$ .
Repeat for $ME = 0.02$ : $1.960/0.02 = 98$ , and $98^2 = 9604$ , and $9604 \times 0.25 = 2401$ , so $n = 2401$ .
Compare the two. Cutting the margin of error from 0.03 to 0.02 is a factor of $2/3$ , and $(3/2)^2 = 2.25$ , so the sample size must grow by a factor of 2.25: $1067.111 \times 2.25 = 2401$ , which matches.

You need 1068 people for a 3 percentage point margin of error, and 2401 for 2 percentage points. The second target costs 2.25 times the sample for a one point improvement, which is the square root law read in reverse: precision scales with $\sqrt{n}$ , so cost scales with the square of the precision you want.

Frequently asked questions

How much does the sample size have to grow to halve the margin of error?

It has to quadruple. The margin of error is proportional to $1/\sqrt{n}$ , so dividing it by 2 means multiplying $\sqrt{n}$ by 2, which means multiplying $n$ by 4. To divide the margin of error by 3, multiply $n$ by 9.

Does a higher confidence level make the interval wider or narrower?

Wider. The critical value grows with the confidence level, and the margin of error is proportional to that critical value. At the same sample size, moving from 95% to 99% multiplies the width by $2.576/1.960 = 1.314$ , so the interval gets about 31% wider.

Why does a proportion near 0.5 give the widest interval?

Because the standard error contains $\hat{p}(1 - \hat{p})$ , which is a downward parabola peaking at $\hat{p} = 0.5$ with a value of 0.25. Any proportion further from 0.5 makes that product smaller, so the standard error and the margin of error both shrink.

Does the population size affect the width of a confidence interval?

Not in any AP procedure. As long as the sample is less than 10% of the population, the standard error depends on the sample size and the variability, not on the population total. A sample of 1000 gives roughly the same margin of error for a city of 100,000 and a country of 100 million.

Is a wide confidence interval wrong?

No. It is an honest report that the data leave a wide range of parameter values plausible. The fix is more data, less variable data, or a lower confidence level, chosen before you look. Narrowing an interval after seeing it, by dropping outliers or lowering the confidence level, breaks the coverage the interval claims.