| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Challenging +1.2 This is a standard S4 question on estimator properties requiring routine application of E(aX+bY) and Var(aX+bY) formulas, plus Lagrange multipliers or substitution for constrained optimization. While it involves multiple parts and some algebraic manipulation, the techniques are well-practiced in Further Maths Statistics and follow predictable patterns with no novel insight required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(S) = \frac{1}{2}(E(\bar{U}_4) + E(\bar{U}_6))\) | M1 | |
| \(= \frac{1}{2}(\mu + \mu) = \mu\), so \(S\) is unbiased | A1 | With conclusion |
| \(\text{Var}(S) = \frac{1}{4}(\sigma^2/4 + \sigma^2/6)\) | M1 | |
| \(= \frac{5\sigma^2}{48}\) | A1 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(T) = (a+b)\mu = \mu\), \(a+b=1\) | M1 | |
| \(\text{Var}(T) = a^2\sigma^2/4 + b^2\sigma^2/6\) | B1 | |
| Minimise \(y = a^2/4 + b^2/6 = a^2/4 + (1-a)^2/6\) | M1 | |
| EITHER by differentiation OR completing square, OR from sketch graph | M1 | |
| Giving \(a = \frac{2}{5}\), \(b = \frac{3}{5}\) | A1 | |
| Justify minimum value | B1 | Allow from completion of square |
| Variance \(= \sigma^2/10\) | A1 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\) is better since (both are unbiased and) \(\text{Var}(T) < \text{Var}(S)\) | B1 1 | From calculated variances |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sample mean of 10 observations (is also unbiased) with \(\sigma^2/10\) | M1 | Or show that \(T =\) mean of 10 observations |
| They have the same efficiency | A1 2 | |
| [14] |
## Question 7:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(S) = \frac{1}{2}(E(\bar{U}_4) + E(\bar{U}_6))$ | M1 | |
| $= \frac{1}{2}(\mu + \mu) = \mu$, so $S$ is unbiased | A1 | With conclusion |
| $\text{Var}(S) = \frac{1}{4}(\sigma^2/4 + \sigma^2/6)$ | M1 | |
| $= \frac{5\sigma^2}{48}$ | A1 **4** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = (a+b)\mu = \mu$, $a+b=1$ | M1 | |
| $\text{Var}(T) = a^2\sigma^2/4 + b^2\sigma^2/6$ | B1 | |
| Minimise $y = a^2/4 + b^2/6 = a^2/4 + (1-a)^2/6$ | M1 | |
| EITHER by differentiation OR completing square, OR from sketch graph | M1 | |
| Giving $a = \frac{2}{5}$, $b = \frac{3}{5}$ | A1 | |
| Justify minimum value | B1 | Allow from completion of square |
| Variance $= \sigma^2/10$ | A1 **7** | |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T$ is better since (both are unbiased and) $\text{Var}(T) < \text{Var}(S)$ | B1 **1** | From calculated variances |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sample mean of 10 observations (is also unbiased) with $\sigma^2/10$ | M1 | Or show that $T =$ mean of 10 observations |
| They have the same efficiency | A1 **2** | |
| | **[14]** | |7 The continuous random variable $U$ has unknown mean $\mu$ and known variance $\sigma ^ { 2 }$. In order to estimate $\mu$, two random samples, one of 4 observations of $U$ and the other of 6 observations of $U$, are taken. The sample means are denoted by $\bar { U } _ { 4 }$ and $\bar { U } _ { 6 }$ respectively. One estimator $S$, given by $S = \frac { 1 } { 2 } \left( \bar { U } _ { 4 } + \bar { U } _ { 6 } \right)$, is proposed.\\
(i) Show that $S$ is unbiased and find $\operatorname { Var } ( S )$ in terms of $\sigma ^ { 2 }$.
A second estimator $T$ of the form $a \bar { U } _ { 4 } + b \bar { U } _ { 6 }$ is proposed, where $a$ and $b$ are chosen such that $T$ is an unbiased estimator for $\mu$ with the smallest possible variance.\\
(ii) Find the values of $a$ and $b$ and the corresponding variance of $T$.\\
(iii) State, giving a reason, which of $S$ and $T$ is the better estimator.\\
(iv) Compare the efficiencies of this preferred estimator and the mean of all 10 observations.
\hfill \mbox{\textit{OCR S4 2011 Q7 [14]}}