| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Calculating bias of estimator |
| Difficulty | Standard +0.3 This is a straightforward application of standard estimator theory formulas. Parts (a)-(b) test definitions, while (c)-(d) require routine calculation of E(T) and Var(T) using linearity of expectation and independence. Part (e) involves comparing mean squared error. All steps are mechanical with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| The distribution of all possible values of the statistic \(T\) (computed from samples of the same size) | B1 | Must mention all possible values / probability distribution of \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(T) \neq \theta\), i.e. the expected value of the estimator does not equal the parameter | B1 | Must state \(E(T)\neq\theta\) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(\hat{\mu}_1)=\frac{E(X_3)+E(X_5)+E(X_7)}{3}=\frac{3\mu}{3}=\mu\), bias \(= 0\) | B1 | Unbiased |
| \(E(\hat{\mu}_2)=\frac{5\mu+2\mu+\mu}{6}=\frac{8\mu}{6}=\frac{4\mu}{3}\), bias \(= \frac{\mu}{3}\) | M1A1 | Correct expectation and bias |
| \(E(\hat{\mu}_3)=\frac{3\mu-\mu}{3}=\frac{2\mu}{3}\), bias \(= -\frac{\mu}{3}\) | M1A1 | Correct expectation and bias |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(\hat{\mu}_1)=\frac{\sigma^2+\sigma^2+\sigma^2}{9}=\frac{3\sigma^2}{9}=\frac{\sigma^2}{3}\) | B1M1 | Correct use of variance of sum |
| \(\text{Var}(\hat{\mu}_2)=\frac{25\sigma^2+4\sigma^2+\sigma^2}{36}=\frac{30\sigma^2}{36}=\frac{5\sigma^2}{6}\) | M1A1 | |
| \(\text{Var}(\hat{\mu}_3)=\frac{9\sigma^2+\sigma^2}{9}=\frac{10\sigma^2}{9}\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\hat{\mu}_1\) is the best estimator: it is unbiased and has the smallest variance \(\frac{\sigma^2}{3}\) | B1B1 | Both conditions needed |
| (ii) \(\hat{\mu}_3\) is the worst estimator: it is biased and has the largest variance \(\frac{10\sigma^2}{9}\) | B1 |
# Question 6:
## Part (a)
| The distribution of all possible values of the statistic $T$ (computed from samples of the same size) | B1 | Must mention all possible values / probability distribution of $T$ |
**1 mark**
## Part (b)
| $E(T) \neq \theta$, i.e. the expected value of the estimator does not equal the parameter | B1 | Must state $E(T)\neq\theta$ or equivalent |
**1 mark**
## Part (c)
| $E(\hat{\mu}_1)=\frac{E(X_3)+E(X_5)+E(X_7)}{3}=\frac{3\mu}{3}=\mu$, bias $= 0$ | B1 | Unbiased |
| $E(\hat{\mu}_2)=\frac{5\mu+2\mu+\mu}{6}=\frac{8\mu}{6}=\frac{4\mu}{3}$, bias $= \frac{\mu}{3}$ | M1A1 | Correct expectation and bias |
| $E(\hat{\mu}_3)=\frac{3\mu-\mu}{3}=\frac{2\mu}{3}$, bias $= -\frac{\mu}{3}$ | M1A1 | Correct expectation and bias |
**4 marks total**
## Part (d)
| $\text{Var}(\hat{\mu}_1)=\frac{\sigma^2+\sigma^2+\sigma^2}{9}=\frac{3\sigma^2}{9}=\frac{\sigma^2}{3}$ | B1M1 | Correct use of variance of sum |
| $\text{Var}(\hat{\mu}_2)=\frac{25\sigma^2+4\sigma^2+\sigma^2}{36}=\frac{30\sigma^2}{36}=\frac{5\sigma^2}{6}$ | M1A1 | |
| $\text{Var}(\hat{\mu}_3)=\frac{9\sigma^2+\sigma^2}{9}=\frac{10\sigma^2}{9}$ | M1A1 | |
**6 marks total**
## Part (e)
| (i) $\hat{\mu}_1$ is the best estimator: it is unbiased **and** has the smallest variance $\frac{\sigma^2}{3}$ | B1B1 | Both conditions needed |
| (ii) $\hat{\mu}_3$ is the worst estimator: it is biased **and** has the largest variance $\frac{10\sigma^2}{9}$ | B1 | |
**3 marks total**\begin{enumerate}
\item (a) Explain what is meant by the sampling distribution of an estimator $T$ of the population parameter $\theta$.\\
(b) Explain what you understand by the statement that $T$ is a biased estimator of $\theta$.
\end{enumerate}
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 $\mu$.
$$\begin{aligned}
& \hat { \mu } _ { 1 } = \frac { X _ { 3 } + X _ { 5 } + X _ { 7 } } { 3 } \\
& \hat { \mu } _ { 2 } = \frac { 5 X _ { 1 } + 2 X _ { 2 } + X _ { 9 } } { 6 } \\
& \hat { \mu } _ { 3 } = \frac { 3 X _ { 10 } - X _ { 1 } } { 3 }
\end{aligned}$$
(d) Find the variance of each of these three estimators.\\
(e) State, giving a reason, which of these three estimators for $\mu$ is\\
(i) the best estimator,\\
(ii) the worst estimator.\\
\hfill \mbox{\textit{Edexcel S4 2014 Q6 [15]}}