S4 June 2007 Q2
2. The value of orders, in £, made to a firm over the internet has distribution \(\mathrm{N}(\mu, \sigma^2)\). A random sample of \(n\) orders is taken and \(\overline{X}\) denotes the sample mean.
(a) Write down the mean and variance of \(\overline{X}\) in terms of \(\mu\) and \(\sigma^2\). (2)
A second sample of \(m\) orders is taken and \(\overline{Y}\) denotes the mean of this sample.
An estimator of the population mean is given by
\[U = \frac{n\overline{X} + m\overline{Y}}{n + m}.\](b) Show that \(U\) is an unbiased estimator for \(\mu\). (3)
(c) Show that the variance of \(U\) is \(\dfrac{\sigma^2}{n + m}\). (4)
(d) State which of \(\overline{X}\) or \(U\) is a better estimator for \(\mu\). Give a reason for your answer. (2)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(\overline{X}) = \mu\) | B1 |
| \(\mathrm{Var}(\overline{X}) = \mathrm{Var}\left(\dfrac{X_1 + X_2 + X_3 + \ldots + X_n}{n}\right)\) \(= \dfrac{\sigma^2}{n}\) | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(U) = \dfrac{1}{n + m}\left(n\mathrm{E}(\overline{X}) + m\mathrm{E}(\overline{Y})\right)\) | M1 |
| \(= \dfrac{1}{n + m}(n\mu + m\mu)\) | A1 |
| \(= \mu \quad \underline{\Rightarrow U \text{ is unbiased}}\) state unbiased | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(\overline{Y}) = \dfrac{\sigma^2}{m}\) | B1 |
| \(\mathrm{Var}(U) = \dfrac{n^2\mathrm{Var}(\overline{X}) + m^2\mathrm{Var}(\overline{Y})}{(n + m)^2}\) | M1 |
| \(= \dfrac{n^2\dfrac{\sigma^2}{n} + m^2\dfrac{\sigma^2}{m}}{(n + m)^2}\) | A1 |
| \(= \dfrac{n\sigma^2 + m\sigma^2}{(n + m)^2}\) \(= \dfrac{\sigma^2}{n + m}\) * cso | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{n\overline{X} + m\overline{Y}}{n + m}\) is a better estimate since variance is smaller. | B1 B1 |
| (2) |