| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pooled variance estimation |
| Difficulty | Standard +0.3 This is a straightforward application of standard results about unbiased estimators and variance of linear combinations. Part (a) requires showing E(estimator) = μ using linearity of expectation (routine algebra), part (b) uses the standard variance formula for independent sample means, and part (c) is a simple comparison. All steps are mechanical applications of known formulas with no conceptual challenges or novel insights required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(E\left(\frac{1}{3}\bar{X} + \frac{2}{3}\bar{Y}\right) = \frac{1}{3}E\left(\frac{X_1+X_2+X_3}{3}\right) + \frac{2}{3}E\left(\frac{Y_1+Y_2+Y_3+Y_4}{4}\right)\) | M1 | |
| \(= \frac{1}{3}\mu + \frac{2}{3}\mu = \mu\), therefore unbiased estimator | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(E\left(\frac{5\bar{X}+4\bar{Y}}{9}\right) = \frac{1}{9}(5E(\bar{X}) + 4E(\bar{Y}))\) | M1 | |
| \(= \frac{1}{9}(5\mu + 4\mu) = \mu\), therefore unbiased estimator | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\text{Var}(\bar{X}) = \frac{\sigma^2}{3}\), \(\text{Var}(\bar{Y}) = \frac{\sigma^2}{4}\) | M1, A1 | |
| \(\text{Var}(\hat{\mu}_1) = \frac{1}{9}\cdot\frac{\sigma^2}{3} + \frac{4}{9}\cdot\frac{\sigma^2}{4} = \frac{4\sigma^2}{27}\) | M1, A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{4}{27}\sigma^2 < \frac{37}{243}\sigma^2\) so use \(\hat{\mu}_1\) | B1 | (1) |
| TOTAL | 7 |
## Question 3:
### Part (a)(i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $E\left(\frac{1}{3}\bar{X} + \frac{2}{3}\bar{Y}\right) = \frac{1}{3}E\left(\frac{X_1+X_2+X_3}{3}\right) + \frac{2}{3}E\left(\frac{Y_1+Y_2+Y_3+Y_4}{4}\right)$ | M1 | |
| $= \frac{1}{3}\mu + \frac{2}{3}\mu = \mu$, therefore unbiased estimator | A1 | (2) |
### Part (a)(ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $E\left(\frac{5\bar{X}+4\bar{Y}}{9}\right) = \frac{1}{9}(5E(\bar{X}) + 4E(\bar{Y}))$ | M1 | |
| $= \frac{1}{9}(5\mu + 4\mu) = \mu$, therefore unbiased estimator | A1 | (2) |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\text{Var}(\bar{X}) = \frac{\sigma^2}{3}$, $\text{Var}(\bar{Y}) = \frac{\sigma^2}{4}$ | M1, A1 | |
| $\text{Var}(\hat{\mu}_1) = \frac{1}{9}\cdot\frac{\sigma^2}{3} + \frac{4}{9}\cdot\frac{\sigma^2}{4} = \frac{4\sigma^2}{27}$ | M1, A1 | (2) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{4}{27}\sigma^2 < \frac{37}{243}\sigma^2$ so use $\hat{\mu}_1$ | B1 | (1) |
| **TOTAL** | **7** | |
---3. A population has mean $\mu$ and variance $\sigma ^ { 2 }$.
A random sample of size 3 is to be taken from this population and $\bar { X }$ denotes its sample mean. A second random sample of size 4 is to be taken from this population and $\bar { Y }$ denotes its sample mean.
\begin{enumerate}[label=(\alph*)]
\item Show that unbiased estimators for $\mu$ are given by
\begin{enumerate}[label=(\roman*)]
\item $\hat { \mu } _ { 1 } = \frac { 1 } { 3 } \bar { X } + \frac { 2 } { 3 } \bar { Y }$,
\item $\hat { \mu } _ { 2 } = \frac { 5 \bar { X } + 4 \bar { Y } } { 9 }$.
\end{enumerate}\item Calculate Var $\left( \hat { \mu } _ { 1 } \right)$
\item Given that $\operatorname { Var } \left( \hat { \mu } _ { 2 } \right) = \frac { 37 } { 243 } \sigma ^ { 2 }$, state, giving a reason, which of these two estimators should be\\
used. used.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2006 Q3 [7]}}