| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Standard +0.3 This is a straightforward S4 question on estimator properties requiring standard results about sample means and linear combinations. Parts (a)-(c) involve routine algebraic manipulation of expectations and variances with no novel insight needed. Part (d) requires simple comparison of variances. Slightly above average difficulty only due to the algebraic bookkeeping across multiple parts. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks |
|---|---|
| \(E(\bar{X}) = \mu\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(\bar{X}) = \text{Var}\left(\frac{X_1 + X_2 + X_3 + ... + X_n}{n}\right)\) | ||
| \(= \frac{\sigma^2}{n}\) | B1 | |
| \(E(U) = \frac{1}{n+m}(nE(\bar{X}) + mE(\bar{Y}))\) | M1 | |
| \(= \frac{1}{n+m}(n\mu + m\mu)\) | A1 | |
| \(= \mu \Rightarrow U \text{ is unbiased}\) | A1 | State unbiased |
Total: 2 marks
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(\bar{Y}) = \frac{\sigma^2}{m}\) | B1 | |
| \(\text{Var}(U) = \frac{n^2\text{Var}(\bar{X}) + m^2\text{Var}(\bar{Y})}{(n+m)^2}\) | M1 | |
| \(= \frac{n^2 \cdot \frac{\sigma^2}{n} + m^2 \cdot \frac{\sigma^2}{m}}{(n+m)^2}\) | A1 | |
| \(= \frac{n\sigma^2 + m\sigma^2}{(n+m)^2}\) | ||
| \(= \frac{\sigma^2}{n+m}\) | A1 | cso |
Total: 4 marks
| Answer | Marks |
|---|---|
| \(\frac{n\bar{X} + m\bar{Y}}{n+m}\) is a better estimate since variance is smaller. | B1 B1 |
Total: 2 marks
**Part a:**
$E(\bar{X}) = \mu$ | B1 |
**Part b:**
$\text{Var}(\bar{X}) = \text{Var}\left(\frac{X_1 + X_2 + X_3 + ... + X_n}{n}\right)$ | |
$= \frac{\sigma^2}{n}$ | B1 |
$E(U) = \frac{1}{n+m}(nE(\bar{X}) + mE(\bar{Y}))$ | M1 |
$= \frac{1}{n+m}(n\mu + m\mu)$ | A1 |
$= \mu \Rightarrow U \text{ is unbiased}$ | A1 | State unbiased |
Total: 2 marks
**Part c:**
$\text{Var}(\bar{Y}) = \frac{\sigma^2}{m}$ | B1 |
$\text{Var}(U) = \frac{n^2\text{Var}(\bar{X}) + m^2\text{Var}(\bar{Y})}{(n+m)^2}$ | M1 |
$= \frac{n^2 \cdot \frac{\sigma^2}{n} + m^2 \cdot \frac{\sigma^2}{m}}{(n+m)^2}$ | A1 |
$= \frac{n\sigma^2 + m\sigma^2}{(n+m)^2}$ | |
$= \frac{\sigma^2}{n+m}$ | A1 | cso |
Total: 4 marks
**Part d:**
$\frac{n\bar{X} + m\bar{Y}}{n+m}$ is a better estimate since variance is smaller. | B1 B1 |
Total: 2 marks
---2. The value of orders, in $\pounds$, made to a firm over the internet has distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. A random sample of $n$ orders is taken and $\bar { X }$ denotes the sample mean.
\begin{enumerate}[label=(\alph*)]
\item Write down the mean and variance of $\bar { X }$ in terms of $\mu$ and $\sigma ^ { 2 }$.
A second sample of $m$ orders is taken and $\bar { Y }$ denotes the mean of this sample.\\
An estimator of the population mean is given by
$$U = \frac { n \bar { X } + m \bar { Y } } { n + m }$$
\item Show that $U$ is an unbiased estimator for $\mu$.
\item Show that the variance of $U$ is $\frac { \sigma ^ { 2 } } { n + m }$.
\item State which of $\bar { X }$ or $U$ is a better estimator for $\mu$. Give a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2007 Q2 [11]}}