Margin of error vs standard error

Q: Can a 99% margin of error be smaller than a 90% margin of error?

Yes, when they come from different samples. With p-hat = 0.50, a sample of 1500 gives a 99% margin of about 0.0333, while a sample of 400 gives a 90% margin of about 0.0411. Only with the standard error held fixed does a higher level always mean a bigger margin.

By Jude Wallis · Published August 11, 2026

The standard error estimates how much a statistic varies from sample to sample. The margin of error is a critical value times that standard error, so it carries a confidence level and the standard error does not. One standard error of 0.03036 gives margins of 0.0499, 0.0595, and 0.0782.

AP Statistics: Unit 3 (topics 3.2 Sampling Distributions for Sample Proportions, 3.3 Constructing a Confidence Interval for a Population Proportion, 4.2 Constructing a Confidence Interval for a Population Mean or Population Mean Difference). In the Fall 2026 AP Statistics course, topic 3.2 covers the sampling distribution of a sample proportion, the spread that a standard error estimates, and topic 3.3 constructs the proportion interval from it. Topic 4.2 constructs the interval for a population mean the same way, so the same critical-value-times-standard-error structure appears in both Unit 3 and Unit 4.

Margin of error vs standard error: the short answer

One multiplication separates them. The standard error measures how much a statistic, such as a sample proportion or a sample mean, moves from one random sample to the next. The margin of error is that standard error multiplied by a critical value:

\text{margin of error} = (\text{critical value}) \times (\text{standard error})

The critical value is where the confidence level enters, and it is the only place it enters. A standard error is a property of the data and the sample size, so asking for 90%, 95%, or 99% confidence does not move it at all. The margin of error changes with every one of those choices. That is the whole difference: the margin of error carries a confidence level, and the standard error does not.

What the standard error measures

The standard error estimates the standard deviation of a statistic's sampling distribution. Draw one random sample and compute $\hat{p}$ ("p-hat", the sample proportion). Draw another sample the same way and you get a slightly different value. The standard error says how large those sample-to-sample differences typically are.

Two formulas cover most of the course. For a sample proportion,

SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}

and for a sample mean, writing $s$ for the sample standard deviation and $n$ for the sample size,

SE_{\bar{x}} = \frac{s}{\sqrt{n}}

Both are built from the sample and nothing else. Neither has anywhere to put 95%. One formula that looks like these but is not one of them is the denominator of a one-proportion significance test, which uses the null value $p_0$ in place of $\hat{p}$ : $\sqrt{p_0(1 - p_0)/n}$ is the standard deviation of $\hat{p}$ under $H_0$ , not an estimate from your sample. The step before this one, separating the standard error from the standard deviation of the raw data, is worked through in standard error vs standard deviation. This page starts where that one ends: it takes the standard error as the input and asks what the critical value does to it.

What the margin of error measures

The margin of error is the reach on each side of an estimate, so it is exactly half the width of a confidence interval. The AP formula sheet writes a confidence interval this way:

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

Everything after the $\pm$ sign is the margin of error. The critical value is how many standard errors you travel to bracket the middle $C\%$ of the sampling distribution, so you read it off a normal or $t$ curve once you have chosen $C$ . For a proportion at 95% confidence it is $z^* = 1.960$ ; at 90% it is $1.645$ and at 99% it is $2.576$ .

So a margin of error reports two things at once: the sampling variability, which comes from the standard error, and how much confidence you demanded, which comes from the critical value. That is why a margin quoted without its confidence level cannot be checked, and why a standard error on its own does not tell a reader how wide an interval was. Building one from scratch is covered in how to calculate a confidence interval.

One standard error, three margins

Take a random sample of 250 students in which 160 eat breakfast, so $\hat{p} = \frac{160}{250} = 0.64$ . The standard error is

SE = \sqrt{\frac{0.64 \times 0.36}{250}} = \sqrt{0.0009216} \approx 0.03036

That single number feeds three different margins.

Confidence level	Critical value $z^*$	Margin of error	Interval
90%	1.645	0.0499	0.5901 to 0.6899
95%	1.960	0.0595	0.5805 to 0.6995
99%	2.576	0.0782	0.5618 to 0.7182

The standard error is 0.03036 in every row. Only the critical value changes, and the margin changes in exact proportion to it: the 99% margin is $\frac{2.576}{1.645} \approx 1.57$ times the 90% margin. Nothing about the sample enters that ratio, so any interval built from these same $z^*$ values stretches by the same factor of 1.57 when you move from 90% to 99%.

For a mean the critical value is $t^*$ rather than $z^*$ , read at $n - 1$ degrees of freedom, and the structure is identical. A sample of 20 with $s = 4.5$ has $SE = \frac{4.5}{\sqrt{20}} \approx 1.0062$ , and at 19 degrees of freedom the 95% critical value is $t^* = 2.093$ , giving a margin of $2.093 \times 1.0062 \approx 2.106$ . That one standard error would feed a 90% margin of $1.729 \times 1.0062 \approx 1.740$ and a 99% margin of $2.861 \times 1.0062 \approx 2.879$ without moving. Swapping in $t^*$ also changes the stretch factor: $\frac{2.861}{1.729} \approx 1.65$ rather than 1.57. The values come off the t-table.

The differences side by side

Feature	Standard error	Margin of error
What it is	Estimated spread of a statistic across samples	Half the width of a confidence interval
Depends on the confidence level	No	Yes, through the critical value
Formula	$s/\sqrt{n}$ for a mean, $\sqrt{\hat{p}(1-\hat{p})/n}$ for a proportion	(critical value)(standard error)
Where it shows up	Inside every margin of error, and in the denominator of a $t$ test for a mean; a proportion test divides by $\sqrt{p_0(1-p_0)/n}$ instead, built from the null value	After the $\pm$ in a confidence interval
Effect of a larger $n$	Falls in proportion to $\frac{1}{\sqrt{n}}$	Falls with the standard error
How it is usually reported	Regression output and error bars	A poll's plus or minus

The pattern down the table is that the margin of error inherits everything the standard error has, then adds the confidence level on top. Sample size acts on both identically, because it acts through the standard error. The confidence level acts on one of them only. For a proportion you can compute either at the margin of error calculator; for a mean use the confidence interval calculator.

The classic mix-up and how to avoid it

The signature error is treating a standard error as if it were a margin of error and attaching 95% to it. An error bar drawn one standard error out on each side is not a 95% interval. When the sampling distribution is roughly normal, one standard error each way covers about 68% of the possible statistics, and the 95% version reaches out 1.960 standard errors, nearly twice as far. Calling the short bar a 95% interval overstates the precision badly.

Two habits keep them apart. First, look for a confidence level. A spread quoted with a percentage attached is a margin of error; a spread quoted as a bare property of the estimate is a standard error. Second, check the confidence level before comparing two margins. With $\hat{p} = 0.50$ , a 99% margin from a sample of 1500 is about 0.0333, tighter than the 0.0411 that a 90% margin from a sample of 400 gives, because the larger sample shrank the standard error by more than the higher level widened it. Only when the standard error is held fixed does raising the level always widen the result, which is the point the confidence interval coverage interactive makes visible.

The same standard error at three confidence levels

A random sample of 250 students finds that 160 eat breakfast. Find the standard error of the sample proportion, then find the margin of error at 90%, 95%, and 99% confidence, and identify which quantity changed.

Find the sample proportion: $\hat{p} = \frac{160}{250} = 0.64$ , so $1 - \hat{p} = 0.36$ .
Write the standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.64 \times 0.36}{250}}$ .
Compute inside the root: $0.64 \times 0.36 = 0.2304$ , and $\frac{0.2304}{250} = 0.0009216$ , so $SE = \sqrt{0.0009216} \approx 0.03036$ .
At 90% confidence $z^* = 1.645$ , so the margin is $1.645 \times 0.03036 \approx 0.0499$ , and the interval runs from $0.64 - 0.0499 = 0.5901$ to $0.64 + 0.0499 = 0.6899$ .
At 95% confidence $z^* = 1.960$ , so the margin is $1.960 \times 0.03036 \approx 0.0595$ , and the interval runs from 0.5805 to 0.6995.
At 99% confidence $z^* = 2.576$ , so the margin is $2.576 \times 0.03036 \approx 0.0782$ , and the interval runs from 0.5618 to 0.7182.
Check what moved. The standard error stayed at 0.03036 in all three lines, because it is computed from $\hat{p}$ and $n$ alone. Only the critical value changed, and each margin is that critical value times 0.03036.

The standard error is about 0.03036 and does not change with the confidence level. The margin of error is about 0.0499 at 90%, 0.0595 at 95%, and 0.0782 at 99%. One standard error produced three margins, because the confidence level enters only through the critical value that multiplies it.

Reading a reported margin backward to the standard error

A poll reports 47% support with a margin of error of 3 percentage points at 95% confidence. Recover the standard error, then find what the margin would have been at 99% confidence from the same sample.

Write the relationship the other way round: margin of error $= z^* \times SE$ , so $SE = \frac{\text{margin of error}}{z^*}$ .
Put the margin on the proportion scale: 3 percentage points is 0.03.
At 95% confidence $z^* = 1.960$ , so $SE = \frac{0.03}{1.960} \approx 0.01531$ .
The standard error comes from the sample, not from the level, so 0.01531 is also the standard error behind a 99% interval from that same poll.
At 99% confidence $z^* = 2.576$ , so the margin becomes $2.576 \times 0.01531 \approx 0.0394$ , about 3.9 percentage points.
Check the direction. Raising the level from 95% to 99% raises the critical value, so the margin should grow, and 3.9 points is larger than 3 points.

The standard error is about 0.01531, or 1.53 percentage points. At 99% confidence the same poll would report a margin of about 0.0394, or 3.9 percentage points. The reported 3 points and the 3.9 points are two different multiples of one unchanged standard error.

Frequently asked questions

Is the margin of error just another name for the standard error?

No. The margin of error is a critical value times the standard error, so the two would match only if the critical value were 1. Reaching one standard error on each side of an estimate corresponds to about 68% confidence, not the 95% that most reported margins use.

Does the standard error change when I change the confidence level?

No. The standard error is computed from the sample data and the sample size, and the formula has no slot for a percentage. Going from 95% to 99% leaves the standard error alone and raises the critical value from 1.960 to 2.576, which is what widens the interval.

Which one goes into a confidence interval?

Both, as a product. The AP formula sheet writes an interval as statistic $\pm$ (critical value)(standard error), and that product is the margin of error. Compute the standard error from your sample, take the critical value from your confidence level, and multiply.

Can a 99% margin of error be smaller than a 90% margin of error?

Yes, when they come from different samples. With $\hat{p} = 0.50$ , a sample of 1500 gives a 99% margin of about 0.0333, while a sample of 400 gives a 90% margin of about 0.0411. Only with the standard error held fixed does a higher level always mean a bigger margin.