In statistics, why is n−1 used in the sample variance denominator?

In statistics, why is n−1 used in the sample variance denominator?

• A. It minimizes the variance of the estimator.
• B. It provides an unbiased estimator of the population variance.
• C. It has no impact on the bias of the estimator.
• D. It is used to estimate the population mean.

Answer: (B)

The key idea is that we adjust the denominator to account for using the sample mean as an estimate of the population mean. When you compute deviations from the sample mean, that mean is itself random and tied to the data, which reduces the number of independent pieces of information by one. To obtain an unbiased estimate of the population variance, you divide by n−1 (this is Bessel’s correction). With this choice, the expected value of the sample variance equals the true variance of the population, assuming independent observations with finite variance. If you used n in the denominator, you would systematically underestimate the variance because E[(1/n) sum (X_i − X̄)^2] = (n−1)/n · σ².

The key idea is that we adjust the denominator to account for using the sample mean as an estimate of the population mean. When you compute deviations from the sample mean, that mean is itself random and tied to the data, which reduces the number of independent pieces of information by one. To obtain an unbiased estimate of the population variance, you divide by n−1 (this is Bessel’s correction). With this choice, the expected value of the sample variance equals the true variance of the population, assuming independent observations with finite variance. If you used n in the denominator, you would systematically underestimate the variance because E[(1/n) sum (X_i − X̄)^2] = (n−1)/n · σ².