Why does the sample variance use n−1 instead of n?

What Is Sample Variance?

The formula for sample variance is:

\[ s^2 = \frac{1}{n – 1} \sum_{i=1}^n (x_i – \bar{x})^2 \]

This is different from the population variance:

\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i – \mu)^2 \]

The key difference is that the population mean \( \mu \) is known, while the sample mean \( \bar{x} \) is estimated from the data.

Why Does Sample Variance Use n − 1?

1. The sample mean uses up one degree of freedom

When you compute the sample mean, the data lose one degree of freedom. Once \( n – 1 \) deviations from the mean are known, the last one is forced.

2. Using n underestimates the true variance

If you divide by n, the estimator systematically pulls downward because the deviations are measured from the sample mean, which is closer to the data than the true mean.

3. Dividing by n − 1 makes the estimator unbiased

The expected value of the sample variance equals the true population variance:

\[ E[s^2] = \sigma^2 \]

4. The correction is called Bessel’s correction

Bessel’s correction adjusts for the fact that the sample mean is estimated from the data.

5. It ensures correct inference

Confidence intervals, t‑tests, and ANOVA all rely on the unbiased sample variance.

How the n − 1 Correction Works (Step by Step)

Step 1: Write deviations from the sample mean

The deviations must sum to zero:

\[ \sum_{i=1}^n (x_i – \bar{x}) = 0 \]

This constraint removes one degree of freedom.

Step 2: Recognize that the sample mean is random

Because \( \bar{x} \) is computed from the sample, the deviations are smaller on average than deviations from the true mean.

Step 3: Compute the expected value of the naive estimator

If you divide by n:

\[ E\left[\frac{1}{n}\sum (x_i – \bar{x})^2\right] = \frac{n – 1}{n}\sigma^2 \]

This shows the estimator is biased downward.

Step 4: Solve for the unbiased version

To correct the bias, multiply by \( \frac{n}{n – 1} \), which leads to:

\[ s^2 = \frac{1}{n – 1}\sum (x_i – \bar{x})^2 \]

Step 5: Interpret the correction

Dividing by n − 1 inflates the variance just enough to offset the downward pull caused by estimating the mean.

Step 6: Connect to degrees of freedom

The sample has n observations but only n − 1 independent deviations. The denominator reflects the number of independent pieces of information.

Numerical Example

Suppose the sample is:

\[ \{4,\ 7,\ 9\} \]

Step 1: Compute the sample mean

\[ \bar{x} = \frac{4 + 7 + 9}{3} = 6.67 \]

Step 2: Compute squared deviations

• (4 − 6.67)² = 7.11
• (7 − 6.67)² = 0.11
• (9 − 6.67)² = 5.44

Sum = 12.66

Step 3: Compute sample variance

\[ s^2 = \frac{12.66}{3 – 1} = 6.33 \]

Step 4: Compare to dividing by n

\[ \frac{12.66}{3} = 4.22 \]

Dividing by n underestimates the true variability.