How Sample Size Affects Coefficient Interval Width

Why Sample Size Matters for a Coefficient Interval

A regression coefficient is estimated from a sample, so it varies from sample to sample. A confidence interval expresses that sampling uncertainty: its width reflects both the estimated standard error and the critical value used to form the interval. When other relevant features of the data are kept comparable, a larger sample usually makes a slope estimate more precise and its interval narrower.

The phrase “when other features are kept comparable” matters. In simple linear regression, sample size affects slope precision through the amount of information in the predictor values, summarized by \(S_{xx}\), as well as through the degrees of freedom used to choose the critical value. To isolate the sample-size effect, this tutorial compares samples with the same predictor sample standard deviation and the same residual standard deviation. It does not claim that simply collecting more observations guarantees a narrower interval if those other quantities change substantially.

The Slope Standard Error and Interval Width

Consider the simple linear regression model \(Y_i=\beta_0+\beta_1x_i+\varepsilon_i\), for \(i=1,\ldots,n\). Here \(\beta_0\) and \(\beta_1\) are the population intercept and slope, and \(\varepsilon_i\) is the error for observation \(i\). For the conventional slope confidence interval, assume the errors are independent, have mean zero and a common variance, and are normally distributed conditional on the observed predictor values. The predictor values must have nonzero spread. These conditions support the usual \(t\)-based interval; the residual standard deviation \(s\) estimates the common error standard deviation.

Write \(b_1\) for the fitted slope, \(s_x\) for the sample standard deviation of the predictor, and \(S_{xx}=\sum_{i=1}^n(x_i-\bar{x})^2\) for its sum of squared deviations. Since \(S_{xx}=(n-1)s_x^2\), the estimated standard error of the simple-regression slope is \(SE(b_1)=s/\sqrt{S_{xx}}\). A two-sided confidence interval with confidence level \(1-\alpha\) has degrees of freedom \(n-2\): two parameters, the intercept and slope, are estimated.

This expression separates three influences. A larger residual standard deviation \(s\) increases the standard error. A larger predictor spread \(s_x\) decreases it. Holding both fixed, increasing \(n\) increases \(S_{xx}\) and decreases the standard error. The critical value also depends on \(n\), because the degrees of freedom change. It generally decreases toward its large-sample limit as \(n\) increases, making the interval narrower still.

Why the Sample-Size Relationship Is Approximately a Square Root

If \(s\) and \(s_x\) are fixed across samples, the standard error has an exact dependence on sample size through \(1/\sqrt{n-1}\). The confidence-interval width has that factor too, but it also contains \(t^*\), which depends on \(n-2\). Thus the width is not exactly proportional to \(1/\sqrt{n}\) for every finite sample size.

The standard-error comparison can be derived directly. Suppose two samples have sizes \(n_1\) and \(n_2\), with the same \(s\) and \(s_x\). Dividing their standard errors cancels those shared quantities. The ratio of their interval widths has one additional factor: the ratio of their critical values. This derivation shows both the familiar square-root pattern and why a precise interval comparison must include the critical values.

For example, doubling the sample size does not generally halve the standard error. If \(n_2\) is about twice \(n_1\), the standard-error ratio is approximately \(1/\sqrt{2}\), provided the predictor spread and residual standard deviation stay fixed. The interval-width ratio is usually a little smaller because the critical value also decreases. For small samples, however, the critical-value change can be noticeable, so it should not be ignored in a numerical comparison.

Comparing Samples While Holding Other Quantities Fixed

A fair comparison of the sample-size effect uses the same predictor spread and residual variability for each sample. For the next example, this is a controlled, hypothetical comparison—not a claim that two independently collected data sets will automatically have identical \(s_x\) and \(s\). The \(t\) critical values differ because the degrees of freedom differ.

What Happens When the Sample Size Doubles?

The square-root relationship is useful for setting expectations, but it is not an exact rule for the full interval width. The next example doubles the sample size and calculates both widths rather than assuming that one must be a fixed fraction of the other.

Interpreting the Comparison in Practice

A narrower interval means greater precision for the coefficient under the model. It does not establish that the estimated association is important, that a causal effect exists, or that the model describes the data well. The interval remains tied to the response and predictor units: a slope interval for score in points per hour has width in points per hour. Widths from coefficients with different units should not be compared without care.

The fixed-spread comparison also has a practical qualification. In a real data set, adding observations can change the predictor spread and the estimated residual standard deviation. The formula uses both: \(SE(b_1)=s/[s_x\sqrt{n-1}]\). If new predictor values are concentrated near the existing mean, the overall \(s_x\) might differ from the comparison assumed here. If the added observations come from a population with more residual variability, \(s\) might increase. Those changes can offset some of the precision gain associated with a larger \(n\). The central lesson is therefore conditional: with comparable predictor spread and residual variability, larger samples tend to produce narrower slope intervals.

Sample size also does not repair violations of the regression assumptions. A large sample cannot make dependent errors independent or turn an inappropriate linear mean relationship into an appropriate one. The conventional interval should be interpreted in light of the model and data-collection process, not just the number of observations.

Common Mistake

A common mistake is to reason, “The sample size quadrupled, so the confidence interval must be exactly one-half as wide.” The standard error would be approximately halved under the fixed-spread, fixed-residual-variability comparison because its sample-size factor is \(1/\sqrt{n-1}\). But the interval width also multiplies the standard error by \(2t^*\), and \(t^*\) depends on \(df=n-2\). The exact comparison must use both critical values and both standard errors, as in the worked examples.

A related mistake is to attribute every narrower interval in a real analysis to sample size alone. The formula shows why that conclusion may fail: differences in \(s\) or \(s_x\) also change the standard error. State which quantities are being held fixed before attributing a width difference to \(n\).