Margin of Error and Interval Precision

What the Margin of Error Measures

A confidence interval reports a range of plausible values for a population proportion \(p\), based on a sample. Its width tells us how precise that estimate is: a narrower interval is more precise, and a wider interval is less precise. The margin of error gives a simple way to describe that width for a symmetric interval. It is the distance from the sample estimate at the interval’s center to either endpoint.

Definition: For a symmetric confidence interval centered at \(\hat{p}\), the margin of error is the interval’s half-width. If the interval is written as \(\hat{p} \pm E\), then \(E\) is the margin of error, the lower endpoint is \(\hat{p}-E\), and the upper endpoint is \(\hat{p}+E\).

For the one-proportion Wald interval, the margin of error is the critical value multiplied by the estimated standard error:

E=z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Here \(z^*\) is the positive critical value for the chosen confidence level, \(\hat{p}\) is the sample proportion, and \(n\) is the sample size. The square root is the estimated standard error \(SE\) of \(\hat{p}\). Thus \(E=z^*SE\). Since proportions are unitless, the margin of error is expressed in proportion units as well. For example, a margin of error of \(0.04\) is four percentage points.

The Model Behind the Formula

The formula applies to the Wald interval for a binary outcome, where each sampled individual either has the characteristic of interest (a success) or does not (a failure). The usual model treats the outcomes as independent Bernoulli observations with the same success probability \(p\). Under this model, \(\hat{p}\) estimates \(p\), and its standard error is estimated by \(\sqrt{\hat{p}(1-\hat{p})/n}\). The normal approximation used for the Wald interval requires adequate numbers of sampled successes and failures; the success-failure condition is a standard check for that approximation. Random sampling and independence must also be justified for the study design. The examples below assume a random sample from a population large enough that sampling a small fraction of it is reasonable, and check the success and failure counts.

The margin of error describes the interval produced by the method and the sample at hand. It is not the known distance between \(\hat{p}\) and the unknown \(p\). That actual estimation error cannot generally be calculated, because \(p\) is unknown. Nor does the margin of error mean that the estimate must be within \(E\) of \(p\) in every sample.

Worked Examples

Worked Example: Finding a 95% Margin of Error

Suppose a hypothetical simple random sample of 400 people is drawn from a large population. Of these, 240 report using a particular service. Find the 95% Wald interval’s margin of error and interval.

The sample proportion is \(\hat{p}=240/400=0.600\). There are 240 successes and 160 failures, so both counts meet the usual minimum of 10 for the success-failure check. The random-sampling design is given, and the population is large relative to the sample, making independence a reasonable approximation.

For a 95% confidence level, \(z^*=1.96\). First calculate the estimated standard error, then multiply it by the critical value:

SE=\sqrt{\frac{0.600(1-0.600)}{400}} =\sqrt{\frac{0.240}{400}} =\sqrt{0.000600} \approx 0.02449

E=1.96(0.02449)\approx 0.0480

The margin of error is \(0.0480\), or about 4.80 percentage points. The interval is centered at \(0.600\):

0.600\pm0.0480=(0.552,\ 0.648)

The endpoints are about 55.2% and 64.8%. The margin of error is the distance from the estimate, 60.0%, to either endpoint, not the full interval width. The full width is \(0.648-0.552=0.096\), twice the margin of error.

Worked Example: How Sample Size Affects Precision

Compare two hypothetical random samples with the same estimated proportion, \(\hat{p}=0.600\), and the same 95% confidence level. One has \(n=100\), with 60 successes and 40 failures; the other has \(n=400\), with 240 successes and 160 failures. Assume both are drawn from large populations under designs that support independence. The success-failure counts in both samples are adequate.

For \(n=100\), the estimated standard error and margin of error are:

SE_{100}=\sqrt{\frac{0.600(0.400)}{100}} =\sqrt{0.0024} \approx 0.04899, \qquad E_{100}=1.96(0.04899)\approx 0.0960

For \(n=400\), the calculation is:

SE_{400}=\sqrt{\frac{0.600(0.400)}{400}} =\sqrt{0.000600} \approx 0.02449, \qquad E_{400}=1.96(0.02449)\approx 0.0480

The sample size has been multiplied by four, while the estimated proportion and confidence level stay fixed. The margin of error has been cut in half. This follows from the \(1/\sqrt{n}\) factor in the standard error: multiplying \(n\) by four divides the standard error, and thus the margin of error, by \(\sqrt{4}=2\). The larger sample gives a narrower interval, assuming the sampling method and other conditions remain suitable.

Worked Example: How the Estimated Proportion Affects Precision

Consider two hypothetical random samples of the same size, \(n=400\), using a 95% confidence level. One has \(\hat{p}=0.500\), or 200 successes and 200 failures; the other has \(\hat{p}=0.100\), or 40 successes and 360 failures. Assume random sampling, a sufficiently large population for independence to be reasonable, and adequate success and failure counts in both samples.

For \(\hat{p}=0.500\):

SE=\sqrt{\frac{0.500(0.500)}{400}} =\sqrt{0.000625} =0.0250, \qquad E=1.96(0.0250)=0.0490

For \(\hat{p}=0.100\):

SE=\sqrt{\frac{0.100(0.900)}{400}} =\sqrt{0.000225} =0.0150, \qquad E=1.96(0.0150)=0.0294

With the same sample size and confidence level, the estimate of 0.500 gives the larger margin of error. The reason is that \(\hat{p}(1-\hat{p})\) is largest at \(\hat{p}=0.500\) and gets smaller as \(\hat{p}\) moves toward 0 or 1. For these data, the estimated margins of error are 4.90 and 2.94 percentage points, respectively. This comparison concerns the formula’s estimated precision; it does not make the two sample designs or populations identical.

Worked Example: Confidence Level and Interval Width

A hypothetical random sample of 500 people from a large population has 200 successes, so \(\hat{p}=0.400\). The sample has 200 successes and 300 failures. Assume the random-sampling and independence conditions are reasonable. Compare the margins of error for 90%, 95%, and 99% confidence levels.

The estimated standard error is the same for each confidence level:

SE=\sqrt{\frac{0.400(0.600)}{500}} =\sqrt{0.000480} \approx 0.02191

Using the corresponding critical values, the margins of error are:

E_{90}=1.645(0.02191)\approx0.0360,\qquad E_{95}=1.960(0.02191)\approx0.0429,\qquad E_{99}=2.576(0.02191)\approx0.0564

The higher confidence levels use larger critical values, so they produce wider intervals when the sample and estimated standard error are held fixed. The respective half-widths are about 3.60, 4.29, and 5.64 percentage points. The confidence level and precision are connected: greater confidence comes with a larger margin of error under this method. The statistical meaning of a confidence level is distinct from the arithmetic meaning of the margin of error.

Why the Relationships Hold

The formula for the margin of error directly explains how its three inputs affect width. A short derivation makes the half-width interpretation explicit. Under the normal approximation, the centered, standardized sample proportion is approximately within \(z^*\) standard errors of \(p\) at the chosen confidence level. Replacing the unknown standard error with its estimate gives the Wald interval:

\frac{\hat{p}-p}{SE}\ \text{is approximately within}\ -z^*\ \text{and}\ z^* \quad\Longrightarrow\quad \hat{p}-z^*SE\leq p\leq\hat{p}+z^*SE

The endpoints are \(\hat{p}-z^*SE\) and \(\hat{p}+z^*SE\). Subtracting the center \(\hat{p}\) from the upper endpoint gives \(z^*SE\); subtracting the lower endpoint from the center gives the same quantity. Therefore the interval’s half-width is \(E=z^*SE\), as claimed. This approximation relies on the model and conditions described above.

Sample size: holding \(z^*\) and \(\hat{p}\) fixed, the margin of error is proportional to \(1/\sqrt{n}\). More observations reduce the estimated standard error and narrow the interval, but the reduction is not proportional to \(1/n\).
Confidence level: holding the data fixed, a higher confidence level requires a larger \(z^*\), which increases the margin of error and widens the interval.
Estimated proportion: holding \(n\) and the confidence level fixed, the margin of error depends on \(\hat{p}(1-\hat{p})\). This product is greatest at 0.5 and decreases toward either boundary.

These comparisons describe what happens when one input changes and the others are held fixed. In practice, changing a sampling plan can change the observations as well as the sample size, so the resulting margin of error need not follow a simple comparison exactly. Precision also refers to the interval’s width, not to whether the estimate is close to the true population proportion.

Common Mistakes

Calling the margin of error the full width. If an interval is \(0.552\) to \(0.648\), its width is \(0.096\), and its margin of error is half of that: \(0.048\).
Treating the margin as the actual estimation error. The margin of error is calculated from the interval method and sample. The actual distance between \(\hat{p}\) and \(p\) is unknown when \(p\) is unknown.
Saying that a larger sample always halves the margin of error when the sample size doubles. The margin varies with \(1/\sqrt{n}\). Doubling \(n\) multiplies it by \(1/\sqrt{2}\), not by \(1/2\); quadrupling \(n\) halves it, if the other inputs stay fixed.
Assuming the estimated proportion has no effect. Even with the same \(n\) and confidence level, different values of \(\hat{p}\) can produce different estimated standard errors and margins of error.
Expecting higher confidence to make an interval narrower. With the data fixed, a higher confidence level uses a larger critical value and makes the interval wider.

Key takeaway: For a symmetric Wald interval for a proportion, the margin of error is \(z^*\sqrt{\hat{p}(1-\hat{p})/n}\), the distance from the center to either endpoint. It grows with the critical value, shrinks as sample size increases, and is largest when the estimated proportion is near 0.5.

Check Your Understanding

Use the definitions and relationships in this tutorial to answer the following questions.

An interval is reported as \(0.31\) to \(0.43\). What are its center, margin of error, and full width?
If the estimated proportion and confidence level stay fixed, what happens to the margin of error when the sample size is multiplied by four? Explain using the standard error formula.
For the same sample size and confidence level, which estimated proportion gives the larger Wald margin of error: \(0.20\) or \(0.50\)? Explain using \(\hat{p}(1-\hat{p})\).
For fixed data, why does increasing the confidence level widen the interval?
Explain why a margin of error of 0.03 does not establish that the sample proportion is within 0.03 of the true population proportion.