Unbiased Estimates of Population Mean and Variance (9709 S2 §6.4) — Mathematics

Introduction

When we collect a sample from a population, we rarely have access to every data value. Instead, we use the sample to estimate unknown population parameters — chiefly the population mean $\mu$ and population variance $\sigma^2$ . The key requirement in 9709 is that our estimates are unbiased: although each individual sample will give a slightly different estimate, the estimation process produces the correct value on average across all possible samples. This topic appears regularly in PS2 questions, both as a standalone calculation and as a gateway to hypothesis testing and confidence intervals.

Core Concept

What does "unbiased" mean?

An estimator $T$ of a parameter $\theta$ is called unbiased if

E(T) = \theta.

In plain terms: if you were to repeat the sampling process many times and average all the estimates you obtained, the long-run average would equal the true parameter. No single estimate is guaranteed to be exact, but the method is accurate on average. That is the only level of understanding required by the syllabus.

Unbiased estimate of the population mean

The sample mean $\bar{x}$ is always an unbiased estimate of $\mu$ :

\hat{\mu} = \bar{x} = \frac{\sum x}{n}.

This follows directly from the fact that $E(\bar{X}) = \mu$ , which you met in The Distribution of the Sample Mean.

Unbiased estimate of the population variance

The sample variance computed by dividing by $n$ (rather than $n-1$ ) gives a biased estimate — it systematically underestimates $\sigma^2$ . The correct unbiased estimator is:

\hat{\sigma}^2 = s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2.

An equivalent, and more computationally convenient, form is:

s^2 = \frac{1}{n-1}\!\left(\sum x^2 - \frac{(\sum x)^2}{n}\right).

The expression inside the brackets is the sum of squared deviations (often called $S_{xx}$ ). Dividing by $n-1$ rather than $n$ corrects the bias; the quantity $n-1$ is called the degrees of freedom.

Key rule: always divide by $n-1$ for an unbiased variance estimate. Dividing by $n$ gives the biased version and will lose marks in an exam.

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