Estimating a population mean or variance and justifying a normal mean test are separate tasks. The standard unbiased estimators do not require a normally distributed population.
Under an independent random sample with finite variance
\[\overline{X}=\frac{1}{n}\sum_{i=1}^nX_i,\qquad S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2.\] For \(n>1\), \(E(\overline{X})=\mu\) and \(E(S^2)=\sigma^2\). The divisor \(n-1\) belongs to the unbiased variance estimate.
| Claim | Correct interpretation |
|---|---|
| “This estimator is unbiased.” | Across repeated random samples, its expected value equals the population parameter. |
| “This particular estimate must equal the parameter.” | False: individual sample estimates vary. |
| “The unbiased variance estimate is a known population variance.” | False: it is calculated from a sample and remains an estimate. |
Bridge the idea
From nine independent randomly sampled durations, can you calculate an unbiased variance estimate without assuming normality? Yes, under the stated sampling conditions. Does that automatically justify a small-sample H2 normal mean test? No: estimation alone does not supply the distribution or known-variance conditions for that test.
Exam wording: “Population normality is not required for these unbiased estimators.” When using the estimate later, explicitly distinguish it from a supplied population variance.
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