S4 June 2014 Q6
6.
(a) Explain what is meant by the sampling distribution of an estimator \(T\) of the population parameter \(\theta\). (1)
(b) Explain what you understand by the statement that \(T\) is a biased estimator of \(\theta\). (1)
A population has mean \(\mu\) and variance \(\sigma^2\)
A random sample \(X_1, X_2, \ldots, X_{10}\) is taken from this population.
(c) Calculate the bias of each of the following estimators of \(\mu\). \[\hat{\mu}_1 = \frac{X_3 + X_5 + X_7}{3}\] \[\hat{\mu}_2 = \frac{5X_1 + 2X_2 + X_9}{6}\] \[\hat{\mu}_3 = \frac{3X_{10} - X_1}{3}\] (4)
(d) Find the variance of each of these three estimators. (6)
(e) State, giving a reason, which of these three estimators for \(\mu\) is
(i) the best estimator,
(ii) the worst estimator. (3)
| Scheme | Marks |
|---|---|
| It is the probability distribution of \(T\). | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| An estimator is biased if \(\mathrm{E}(T) \ne \theta\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(\hat{\mu}_1) = \frac{\mathrm{E}(X_3) + \mathrm{E}(X_5) + \mathrm{E}(X_7)}{3} = \frac{\mu + \mu + \mu}{3} = \mu \ \therefore\ \text{Bias} = 0\) | M1A1 |
| \(\mathrm{E}(\hat{\mu}_2) = \frac{5E(X_1) + 2E(X_2) + E(X_9)}{6} = \frac{5\mu + 2\mu + \mu}{6} = \frac{4\mu}{3} \ \therefore\ \text{Bias} = \frac{\mu}{3}\) | A1 |
| \(\mathrm{E}(\hat{\mu}_3) = \frac{3\mathrm{E}(X_{10}) - \mathrm{E}(X_1)}{3} = \frac{3\mu - \mu}{3} = \frac{2\mu}{3} \ \therefore\ \text{Bias} = -\frac{\mu}{3}\) | A1 |
| (4) |
Notes
M1 finding \(\mathrm{E}(\hat{\mu})\) A1 bias 0 A1 \(\pm\dfrac{\mu}{3}\) A1 \(\pm\dfrac{\mu}{3}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(\hat{\mu}_1) = \frac{1}{9}\left(\mathrm{Var}(X_3) + \mathrm{Var}(X_5) + \mathrm{Var}(X_7)\right)\) | M1 |
| \(= \frac{1}{9}\left(\sigma^2 + \sigma^2 + \sigma^2\right)\) | |
| \(= \frac{\sigma^2}{3}\) | A1 |
| \(\mathrm{Var}(\hat{\mu}_2) = \frac{1}{36}\left(25\mathrm{Var}(X_1) + 4\mathrm{Var}(X_2) + \mathrm{Var}(X_9)\right)\) | M1 |
| \(= \frac{1}{36}\left(25\sigma^2 + 4\sigma^2 + \sigma^2\right)\) | |
| \(= \frac{5}{6}\sigma^2\) | A1 |
| \(\mathrm{Var}(\hat{\mu}_3) = \frac{1}{9}\left(9\mathrm{Var}(X_{10}) + \mathrm{Var}(X_1)\right)\) | M1 |
| \(= \frac{1}{9}\left(9\sigma^2 + \sigma^2\right)\) | |
| \(= \frac{10\sigma^2}{9}\) | A1 |
| (6) |
Notes
For method marks allow an incorrect variance, M1 squaring 9, M1 Squaring 5 and 2, M1 adding variances. Do not penalise same mistake twice.
| Scheme | Marks |
|---|---|
| (i) \(\hat{\mu}_1\) is the best estimator. It has no bias | B1 |
| (ii) It has same magnitude of bias as \(\hat{\mu}_2\) but it has the largest variance \(\hat{\mu}_3\) is the worst estimator. | B1ft B1dcao |
| (3) | |
| (15 marks) |
Notes
(ii) Must have idea that its bias is the same as another (\(\hat{\mu}_2\)) and state it has largest variance for first B1. ft their values of Var. Second B1 dependent on first B1cao
SC \(\hat{\mu}_3\) because largest variance B1 B0