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Specify the population and design
The fixed labelled values are 2, 4, 6 and 8, whose population mean is 5. Select two distinct units by simple random sampling without replacement. The six unordered subsets each have probability 1/6.
Calculate each sample mean
| Unit labels | Observed values | Sample mean |
|---|---|---|
| 1 and 2 | 2 and 4 | 3 |
| 1 and 3 | 2 and 6 | 4 |
| 1 and 4 | 2 and 8 | 5 |
| 2 and 3 | 4 and 6 | 5 |
| 2 and 4 | 4 and 8 | 6 |
| 3 and 4 | 6 and 8 | 7 |
Combine identical estimator outcomes
Two distinct samples produce mean 5, so that outcome has probability 2/6. Means 3, 4, 6 and 7 each have probability 1/6. A sampling distribution concerns the calculated statistic, while a sample table concerns selected units.
Every size-two subset is equally likely; the chart combines subsets that produce the same mean. This is an illustrative example.
View the chart values
| Series | Sample mean | Probability |
|---|---|---|
| Probability | 3 | 0.16666667 |
| Probability | 4 | 0.16666667 |
| Probability | 5 | 0.33333333 |
| Probability | 6 | 0.16666667 |
| Probability | 7 | 0.16666667 |
Reconcile the exact moments
The expected sample mean is 5. The variance is [4 + 1 + 0 + 0 + 1 + 4]/6 = 5/3. The finite-population S² is 20/3, so (1 − 2/4) × (20/3)/2 also gives 5/3. With two independent replacement draws, the value-distribution variance 5 instead gives mean variance 5/2.