S4 January 2006 Q3
3. A population has mean \(\mu\) and variance \(\sigma^2\).
A random sample of size 3 is to be taken from this population and \(\overline{X}\) denotes its sample mean.
A second random sample of size 4 is to be taken from this population and \(\overline{Y}\) denotes its sample mean.
(a) Show that unbiased estimators for \(\mu\) are given by
(i) \(\hat{\mu}_1 = \dfrac{1}{3}\overline{X} + \dfrac{2}{3}\overline{Y}\), (2)
(ii) \(\hat{\mu}_2 = \dfrac{5\overline{X} + 4\overline{Y}}{9}\). (2)
(b) Calculate \(\mathrm{Var}(\hat{\mu}_1)\) (2)
(c) Given that \(\mathrm{Var}(\hat{\mu}_2) = \dfrac{37}{243}\sigma^2\), state, giving a reason, which of these two estimators should be used. (1)
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{E}\left(\frac{1}{3}\overline{X} + \frac{2}{3}\overline{Y}\right) = \dfrac{1}{3}\mathrm{E}\left(\dfrac{X_1 + X_2 + X_3}{3}\right) + \dfrac{2}{3}\mathrm{E}\left(\dfrac{Y_1 + Y_2 + Y_3 + Y_4}{4}\right)\) | M1 |
| \(= \dfrac{1}{3}\mu + \dfrac{2}{3}\mu\) \(= \mu\) therefore unbiased estimator | A1 |
| (2) | |
| (ii) \(\mathrm{E}\left(\dfrac{5\overline{X} + 4\overline{Y}}{9}\right) = \dfrac{1}{9}\left(5\mathrm{E}(\overline{X}) + 4\mathrm{E}(\overline{Y})\right)\) | M1 |
| \(= \dfrac{1}{9}(5\mu + 4\mu)\) \(= \mu\) therefore unbiased estimator | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(\overline{X}) = \dfrac{\sigma^2}{3},\quad \mathrm{Var}(\overline{Y}) = \dfrac{\sigma^2}{4}\) \(\mathrm{Var}(\hat{\mu}_1) = \dfrac{1}{9} \cdot \dfrac{\sigma^2}{3} + \dfrac{4}{9} \cdot \dfrac{\sigma^2}{4} = \dfrac{4\sigma^2}{27}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{4}{27}\sigma^2 \lt \dfrac{37}{243}\sigma^2\) so use \(\hat{\mu}_1\). | B1 |
| (1) | |
| (7 marks) |