How Sample Size Affects Interval Width

Sample Size Changes the Width

In How Confidence Level Affects Interval Width, we compared intervals from the same sample and saw that changing the confidence level changes the critical value \(z^*\). Now hold the confidence level fixed and consider what happens when the sample size changes. A larger sample usually gives a smaller margin of error and a narrower interval.

For a one-proportion \(z\)-interval, the margin of error depends on the critical value, the sample proportion, and the sample size. If the confidence level and sample proportion stay fixed, the sample size affects the estimated standard error: increasing \(n\) makes that standard error smaller. The interval remains centered at \(\hat{p}\), but its endpoints move closer to the center.

Key relationship: For the same confidence level and the same \(\hat{p}\), multiplying the sample size by \(k\) divides the margin of error and interval width by \(\sqrt{k}\). In particular, multiplying \(n\) by 4 halves both.

ME=z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \qquad \text{and} \qquad \text{Interval width}=2ME

This relationship follows from the one-proportion \(z\)-interval formula covered in Structure of a One-Proportion z-Interval. The standard error contains \(\sqrt{1/n}\), so it decreases according to the square root of the sample size—not in direct proportion to the sample size. For example, quadrupling \(n\) does not divide the margin of error by 4; it divides it by 2.

To isolate that effect, first compare intervals with the same confidence level and the same sample proportion. In practice, separate samples will not necessarily produce identical \(\hat{p}\) values. We will look at that complication after the controlled comparison.

Compare \(n=100\) and \(n=400\)

Suppose two hypothetical random samples are taken from the same town to estimate the proportion of residents who have signed up for emergency text alerts. One sample contains 100 residents, and 60 report signing up. The other contains 400 residents, and 240 report signing up. Both samples have \(\hat{p}=0.60\). For this comparison, construct a 95% one-proportion \(z\)-interval from each sample.

The samples are stated to be random. If the town has 20,000 residents and sampling is without replacement, the 10% condition is met for both samples: \(100\leq2{,}000\) and \(400\leq2{,}000\). For the \(n=100\) sample, there are 60 successes and 40 failures; for \(n=400\), there are 240 successes and 160 failures. Each success and failure count is at least 10, so the Large Counts condition is met for both intervals.

The 95% critical value is \(z^*=1.96\). For \(n=100\), the estimated standard error and margin of error are:

SE_{\hat{p}} =\sqrt{\frac{(0.60)(0.40)}{100}} =\sqrt{0.0024} \approx0.04899

ME=1.96(0.04899)\approx0.09602

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

SE_{\hat{p}} =\sqrt{\frac{(0.60)(0.40)}{400}} =\sqrt{0.0006} \approx0.02449

ME=1.96(0.02449)\approx0.04801

The sample size is four times as large in the second sample, while the confidence level and \(\hat{p}\) are held fixed. The margin of error is approximately half as large: it decreases from 0.09602 to 0.04801. The intervals and their widths are:

Sample size	\(\hat{p}\)	Margin of error	95% interval	Width
100	0.60	0.09602	(0.50398, 0.69602)	0.19204
400	0.60	0.04801	(0.55199, 0.64801)	0.09602

Both intervals are centered at 0.60. With \(n=400\), the margin of error and full interval width are each half their \(n=100\) values. The intervals are estimates for the same population proportion, but the larger sample produces the narrower range because its estimated standard error is smaller.

Worked Example: Four Times the Sample Size

Use the emergency-alert samples to compare 95% confidence intervals for the proportion of town residents who have signed up for alerts. Give the margin of error for each sample and quantify the change.

State.
Let \(p\) be the true proportion of residents in the town who have signed up for emergency text alerts. The \(n=100\) sample has 60 successes, and the \(n=400\) sample has 240 successes. Both sample proportions are \(0.60\).

Plan and check conditions.
Both samples are stated to be random. For sampling without replacement from 20,000 residents, \(100\leq0.10(20{,}000)=2{,}000\) and \(400\leq2{,}000\), so the 10% condition holds. The success and failure counts are 60 and 40 for \(n=100\), and 240 and 160 for \(n=400\); all are at least 10, so the Large Counts condition holds. Use a 95% one-proportion \(z\)-interval for each sample, with \(z^*=1.96\).

Do.
For \(n=100\), \(SE_{\hat{p}}=\sqrt{(0.60)(0.40)/100}\approx0.04899\), so \(ME=1.96(0.04899)\approx0.09602\). The interval is \(0.60\pm0.09602=(0.50398,0.69602)\), with width \(2(0.09602)=0.19204\). For \(n=400\), \(SE_{\hat{p}}=\sqrt{(0.60)(0.40)/400}\approx0.02449\), so \(ME=1.96(0.02449)\approx0.04801\). The interval is \(0.60\pm0.04801=(0.55199,0.64801)\), with width \(2(0.04801)=0.09602\). The margin of error is reduced by about 0.04801, or 50%; the width is also reduced by 50%.

Conclude.
We are 95% confident that the true proportion of town residents who have signed up for emergency text alerts is between 0.50398 and 0.69602 using the sample of 100, or between 0.55199 and 0.64801 using the sample of 400. In this comparison, the larger sample gives a margin of error and interval width about half as large. The 95% confidence level describes the long-run capture rate of the method, not the probability that either particular interval contains \(p\).

The conclusion about halving the margin of error relies on holding both the confidence level and \(\hat{p}\) fixed. The examples above do that to display the sample-size effect clearly. The next example shows why comparisons between actual, different samples may not match the exact ratio.

The Square-Root Rule

More generally, if the confidence level and sample proportion are held fixed, the margin of error changes by the factor \(1/\sqrt{k}\) when \(n\) is multiplied by \(k\). The width has the same factor because it is twice the margin of error.

\frac{ME_{\text{new}}}{ME_{\text{old}}} = \sqrt{\frac{n_{\text{old}}}{n_{\text{new}}}} = \frac{1}{\sqrt{k}} \quad \text{when } n_{\text{new}}=kn_{\text{old}}

For a fourfold increase in sample size, \(k=4\), so the factor is \(1/\sqrt{4}=1/2\). For a twofold increase, the factor is \(1/\sqrt{2}\approx0.7071\). That means the new margin of error is about 70.71% of the old one—a reduction of about 29.29%, not 50%.

The rule predicts the change in margin of error under a controlled comparison. It does not say that quadrupling a sample size makes an interval four times as narrow, and it does not guarantee that every interval from a larger, different sample will be narrower. Different samples can have different sample proportions, which affect their estimated standard errors.

Worked Example: Doubling the Sample Size

A hypothetical random sample of 100 students finds that 40 use a school’s bike-share program. A second random sample of 200 students finds that 80 use it. Compare the 95% margins of error for estimating the proportion of students who use the program.

Let \(p\) be the true proportion of students at the school who use the bike-share program. The school has 8,000 students. Both samples are random, and the 10% condition holds because \(100\leq800\) and \(200\leq800\). The success and failure counts are 40 and 60 for \(n=100\), and 80 and 120 for \(n=200\); all are at least 10, so the Large Counts condition holds. Both sample proportions are \(0.40\), and the 95% critical value is \(1.96\).

For \(n=100\), the estimated standard error is \(\sqrt{(0.40)(0.60)/100}=\sqrt{0.0024}\approx0.04899\), giving \(ME=1.96(0.04899)\approx0.09602\). For \(n=200\), the estimated standard error is \(\sqrt{(0.40)(0.60)/200}=\sqrt{0.0012}\approx0.03464\), giving \(ME=1.96(0.03464)\approx0.06790\). The new margin is about \(0.7071\) times the old one, consistent with \(1/\sqrt{2}\). It decreases by about \(0.02812\), or 29.29%.

The intervals are \(0.40\pm0.09602=(0.30398,0.49602)\) for \(n=100\) and \(0.40\pm0.06790=(0.33210,0.46790)\) for \(n=200\). Their widths are approximately 0.19204 and 0.13579. Doubling the sample size makes the margin of error smaller, but it does not halve it.

When the Sample Proportions Differ

The exact square-root comparison assumes the confidence level and \(\hat{p}\) stay fixed. If separate samples have different proportions, both the sample size and \(\hat{p}\) affect the estimated standard error. The standard error is largest when \(\hat{p}\) is near 0.50 and smaller when \(\hat{p}\) is closer to 0 or 1.

This means the sample-size effect is a useful prediction, not a guarantee about the observed widths from any two samples. A larger sample often leads to a narrower interval, but the actual comparison depends on both samples’ results. Check the conditions for each interval as well; a large sample does not compensate for a sample that is not random or otherwise representative.

Worked Example: Different Sample Proportions

Two hypothetical random samples from a county estimate the proportion of residents who use a public library’s digital lending service. The first sample has 100 residents, 20 of whom use the service. The second has 400 residents, 200 of whom use it. Compare their 95% margins of error and explain why this is not a controlled sample-size comparison.

Let \(p\) be the true proportion of county residents who use the digital lending service. Suppose the county has 25,000 residents and both samples are taken without replacement. Both samples are random. The 10% condition holds because \(100\leq2{,}500\) and \(400\leq2{,}500\). For the first sample, there are 20 successes and 80 failures; for the second, there are 200 successes and 200 failures. All counts are at least 10, so the Large Counts condition holds for both intervals.

The sample proportions are \(\hat{p}_1=20/100=0.20\) and \(\hat{p}_2=200/400=0.50\). Using \(z^*=1.96\), the first margin of error is \(1.96\sqrt{(0.20)(0.80)/100}=1.96(0.04)=0.07840\). The second is \(1.96\sqrt{(0.50)(0.50)/400}=1.96(0.025)=0.04900\). The corresponding intervals are \(0.20\pm0.07840=(0.12160,0.27840)\) and \(0.50\pm0.04900=(0.45100,0.54900)\), with widths 0.15680 and 0.09800.

The larger-sample interval is narrower here, but the exact fourfold-sample-size rule does not apply: the sample proportions differ. The first sample’s proportion is farther from 0.50, making its estimated standard error smaller than it would be if \(\hat{p}_1\) were 0.50. The second sample’s proportion of 0.50 gives a larger estimated standard error for its sample size. To attribute a precise width ratio to sample size alone, hold the confidence level and sample proportion fixed, as in the emergency-alert comparison.

Common Mistakes and AP Exam Tips

Dividing the margin of error by the sample-size multiplier. If \(n\) becomes four times as large, the margin of error is divided by \(\sqrt{4}=2\), not by 4.
Claiming every larger-sample interval must be narrower. The sample proportions may differ. Compare the actual estimated standard errors and margins of error, or state the fixed-\(\hat{p}\) assumption when using the square-root rule.
Confusing margin of error with full width. The margin is the distance from \(\hat{p}\) to one endpoint. The full width is \(2ME\). If the margin is halved, the full width is also halved.
Ignoring the confidence level. A width comparison based on sample size alone should hold the confidence level fixed. Otherwise, \(z^*\) also changes the margin of error.
Assuming a narrower interval removes all uncertainty. A larger sample reduces sampling variability reflected in the margin of error. It does not fix bias from a poor sampling method or other data-collection problems.
Leaving conditions unstated. For each one-proportion \(z\)-interval, identify evidence for randomness, check the 10% condition when sampling without replacement, and verify that both observed counts meet the Large Counts condition.

AP Exam Tip: For a comparison with the same confidence level and sample proportion, show the margin-of-error formula and use the square-root rule. State that increasing \(n\) from 100 to 400 multiplies it by \(\sqrt{100/400}=1/2\), so the margin of error and width are halved. If the sample proportions differ, calculate each margin separately rather than claiming the exact ratio follows from sample size alone.

Key Takeaway

For a one-proportion confidence interval, a larger sample reduces the estimated standard error and usually narrows the interval. When the confidence level and sample proportion are held fixed, increasing \(n\) from 100 to 400 halves the margin of error and interval width. The square-root rule describes that controlled comparison; separate samples with different \(\hat{p}\) values may not follow the exact ratio.

Key takeaway: With confidence level and \(\hat{p}\) fixed, multiplying \(n\) by \(k\) multiplies the margin of error and interval width by \(1/\sqrt{k}\). Check conditions and compare actual margins when sample proportions differ.

Check Your Understanding

Use the relationship between sample size, estimated standard error, and margin of error to answer each question.

If the confidence level and \(\hat{p}\) stay the same, what happens to the margin of error when the sample size increases from 100 to 400?
A 95% interval has a margin of error of 0.08. If the sample size is quadrupled while the confidence level and sample proportion stay fixed, what is the new margin of error?
If the sample size doubles with the confidence level and \(\hat{p}\) fixed, what factor multiplies the margin of error? About what percent does it decrease?
Why might an interval from a sample of 400 have a different margin of error than the one predicted by the fourfold square-root rule when compared with a sample of 100?
For a one-proportion \(z\)-interval based on sampling without replacement, name the conditions to check before comparing its width with another interval.