How Sample Size Controls Interval Width

Why Sample Size Matters for Interval Width

A confidence interval for a population mean is an estimate with a margin of error on either side. When the sample is more informative, that margin is often smaller, and the interval is narrower. Sample size matters because averages from larger samples tend to vary less from sample to sample. In a t interval, it also affects the degrees of freedom and therefore the critical value.

These effects are easiest to see when comparing intervals at the same confidence level and with the same estimated spread. In actual data, the sample standard deviation can differ from one sample to another, so a larger sample does not guarantee a narrower interval in every comparison. We will distinguish the predictable effect of sample size from that additional variation.

Definition: The width of a confidence interval is its upper endpoint minus its lower endpoint. For a one-sample t interval for a population mean, the estimated standard error is \(SE=s/\sqrt{n}\), the margin of error is \(ME=t^*_{df}SE\), and the interval width is \(W=2ME=2t^*_{df}s/\sqrt{n}\). Here \(s\) is the sample standard deviation, \(n\) is the sample size, \(df=n-1\), and \(t^*_{df}\) is the positive t critical value for the chosen confidence level and degrees of freedom.

The formula shows two ways \(n\) enters the width: through the denominator \(\sqrt{n}\) in the standard error and through the degrees of freedom in the critical value. At a fixed confidence level, \(t^*_{df}\) decreases as the degrees of freedom increase. Thus, if \(s\) is held constant, both effects act to narrow the interval as \(n\) increases.

Why the Standard Error Contains the Square Root of Sample Size

A standard error describes the typical sample-to-sample variation of a statistic. For the sample mean, the square root in \(s/\sqrt{n}\) follows from how variances of independent observations combine. The argument below gives the reason for the sample-size factor; it does not require the population standard deviation to be known in practice.

Derivation: Suppose \(X_1,\ldots,X_n\) are independent observations from a population with mean \(\mu\) and variance \(\sigma^2\). The sample mean is \(\bar{x}=(X_1+\cdots+X_n)/n\). Independence means the variance of the sum is the sum of the variances, so

\operatorname{Var}(\bar{x}) =\operatorname{Var}\left(\frac{X_1+\cdots+X_n}{n}\right) =\frac{1}{n^2}\left(n\sigma^2\right) =\frac{\sigma^2}{n}.

Taking the square root gives the standard deviation of the sampling distribution of the mean, \(\sigma/\sqrt{n}\). Since \(\sigma\) is generally unknown, a one-sample t interval estimates this standard error with \(s/\sqrt{n}\). If the sample standard deviation stays the same, multiplying the sample size by four halves the estimated standard error, because \(\sqrt{4n}=2\sqrt{n}\). It does not reduce it to one-quarter.

Comparing Widths at a Fixed Confidence Level

A one-sample t interval for \(\mu\) has the form \(\bar{x}\pm t^*_{df}s/\sqrt{n}\). The center \(\bar{x}\) determines where the interval sits, but does not affect its width. Subtracting the lower endpoint from the upper endpoint cancels the center:

\left(\bar{x}+t^*_{df}\frac{s}{\sqrt{n}}\right) -\left(\bar{x}-t^*_{df}\frac{s}{\sqrt{n}}\right) =2t^*_{df}\frac{s}{\sqrt{n}}.

This equation is a useful comparison tool, but it must be applied with care. If the confidence level and \(s\) are held fixed, increasing \(n\) reduces \(s/\sqrt{n}\), and the larger degrees of freedom also reduce the t critical value. If two actual samples have different standard deviations, however, the factor \(s\) changes too. The observed width depends on all three quantities: \(s\), \(n\), and \(t^*_{df}\).

Worked Example: Quadrupling the Sample Size

Suppose two hypothetical random samples measure the same quantity in hours. Both have sample standard deviation \(s=12\) hours, and both intervals use 95% confidence. The first sample has \(n=25\), and the second has \(n=100\). Assume each sample is independent, the sampling design satisfies the 10% condition if sampling without replacement, and the data are reasonably compatible with using a t interval. Compare the widths.

For \(n=25\), \(df=24\), and the 95% critical value is approximately \(t^*_{24}=2.064\). The estimated standard error, margin of error, and width are

SE=\frac{12}{\sqrt{25}}=2.4\text{ hours}, \qquad ME=2.064\times2.4=4.9536\approx4.954\text{ hours},

W=2\times4.9536=9.9072\approx9.907\text{ hours}.

For \(n=100\), \(df=99\), and \(t^*_{99}\approx1.984\). Then

SE=\frac{12}{\sqrt{100}}=1.2\text{ hours}, \qquad ME=1.984\times1.2=2.3808\approx2.381\text{ hours},

W=2\times2.3808=4.7616\approx4.762\text{ hours}.

With the same estimated spread and confidence level, the interval based on 100 observations is about 4.762 hours wide, compared with about 9.907 hours for 25 observations. The standard error is halved; the t critical value also decreases slightly. These calculations describe interval width, not where either interval is centered.

The Effects of the Standard Error and the Critical Value

The standard error is usually the stronger and more visible sample-size effect. The change in \(t^*_{df}\) also matters, particularly at small sample sizes, but it becomes less pronounced as the degrees of freedom grow. As degrees of freedom increase, the t critical value approaches the corresponding standard normal critical value. It does not keep shrinking without limit: at a fixed confidence level it approaches a positive value.

The next comparison separates the two roles. The estimated spread and confidence level remain fixed, so the changes in the calculation come from \(n\) and its associated degrees of freedom.

Worked Example: Separating the Two Sample-Size Effects

Consider hypothetical measurements in millilitres, with \(s=8\) millilitres in each sample and 90% confidence. One sample has \(n=16\); another has \(n=64\). Assume random, independent observations, the 10% condition where relevant, and no severe skewness or outliers that would make the t interval inappropriate.

For \(n=16\), \(df=15\), and \(t^*_{15}\approx1.753\). Thus

SE=\frac{8}{\sqrt{16}}=2\text{ mL}, \qquad ME=1.753\times2=3.506\text{ mL}, \qquad W=2\times3.506=7.012\text{ mL}.

For \(n=64\), \(df=63\), and \(t^*_{63}\approx1.669\). The calculation is

SE=\frac{8}{\sqrt{64}}=1\text{ mL}, \qquad ME=1.669\times1=1.669\text{ mL}, \qquad W=2\times1.669=3.338\text{ mL}.

The sample size is four times as large, so the standard error is half as large. The critical value falls from about 1.753 to 1.669 as well. The resulting width falls from 7.012 mL to 3.338 mL. The interval is narrower for both reasons, though most of the change comes from the standard error.

Why “Generally Narrower” Is More Accurate Than “Always Narrower”

The formula includes the sample standard deviation \(s\), which is calculated from the observed data. Two samples from the same population need not have the same \(s\). A larger sample can, by chance, contain more varied observations and produce a larger standard deviation. That increase can offset some or all of the narrowing associated with \(n\).

The following example compares two hypothetical samples with different observed spreads. It illustrates that a larger sample can still produce a narrower interval even when its sample standard deviation is higher; the calculation, rather than sample size alone, settles the comparison.

Worked Example: A Larger Sample With More Observed Variability

Suppose measurements are recorded in minutes. A hypothetical sample of \(n=9\) has \(\bar{x}=52.0\) minutes and \(s=5.0\) minutes. A second hypothetical sample has \(n=36\), \(\bar{x}=51.4\) minutes, and \(s=7.0\) minutes. Use 95% confidence. Assume each sample is random and independent, satisfies the 10% condition if sampled without replacement, and has a distribution suitable for its t interval. For the small sample, this requires no strong skewness or influential outliers.

For the first sample, \(df=8\) and \(t^*_{8}\approx2.306\). Its standard error and margin of error are

SE=\frac{5.0}{\sqrt{9}}=1.6667\text{ min}, \qquad ME=2.306\times\frac{5.0}{3} \approx3.8433\text{ min}.

The interval is approximately \(52.0\pm3.8433\), or \((48.1567,\ 55.8433)\) minutes. Its width, calculated before rounding the margin, is

W=2\times2.306\times\frac{5.0}{3} \approx7.6867\text{ min}.

For the second sample, \(df=35\) and \(t^*_{35}\approx2.030\). Although its sample standard deviation is larger, its standard error is

SE=\frac{7.0}{\sqrt{36}}=1.1667\text{ min}, \qquad ME=2.030\times\frac{7.0}{6} \approx2.3683\text{ min}.

The interval is approximately \(51.4\pm2.3683\), or \((49.0317,\ 53.7683)\) minutes, with width

W=2\times2.030\times\frac{7.0}{6} \approx4.7367\text{ min}.

Here the larger sample produces the narrower interval despite having greater observed variability. But this particular outcome does not establish that every larger sample must yield a narrower interval: another sample might have a sufficiently large \(s\) to produce a wider one.

Why Width Matters and a Common Mistake

Width describes the range of parameter values covered by the interval. In a fixed setting, a narrower interval gives a more precise estimate in the sense that its endpoints are closer together. It does not, by itself, show that the interval is centered closer to the true population mean. The sample mean determines the center, and different samples can produce different centers.

A common mistake is to say, “The sample size doubled, so the interval width must be cut in half.” That reasoning treats the standard error as the entire width and ignores the square root, the t critical value, and possible changes in \(s\). Even when \(s\) is held fixed, doubling \(n\) multiplies the standard error by \(1/\sqrt{2}\), not by \(1/2\). The t critical value changes too, so the total width must be calculated using \(2t^*_{df}s/\sqrt{n}\).

Key takeaway: At a fixed confidence level, if the estimated spread is held constant, a larger sample generally yields a narrower one-sample t interval: \(s/\sqrt{n}\) decreases and \(t^*_{n-1}\) also decreases. When comparing actual samples, check the observed \(s\) as well as \(n\); sample size alone does not determine which interval is narrower.

Check Your Understanding

Use the interval-width formula and explain your reasoning in context.

If the sample size increases from 25 to 100 while \(s\) stays fixed, by what factor does \(s/\sqrt{n}\) change?
For a fixed confidence level and fixed \(s\), name the two parts of a t interval calculation that change as \(n\) increases.
A larger sample has a substantially larger sample standard deviation than a smaller sample. What quantities should you compare before deciding which interval is narrower?
Why does the sample mean affect the location of a t interval but not its width?
Explain why quadrupling \(n\) does not, in general, quarter the t interval’s width.