| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Show unbiased estimator |
| Difficulty | Standard +0.3 This is a straightforward Further Maths Statistics question testing standard properties of estimators. Part (i) requires computing E(Z) by integration and showing E(5Z/4)=k. Parts (ii-iv) involve routine variance calculations and efficiency comparison using variance ratios. All steps follow textbook procedures with no novel insight required, though the multi-part structure and integration make it slightly above average difficulty. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Z) = \int_0^k z \cdot \dfrac{4z^3}{k^4}\,dz = \left[\dfrac{4z^5}{5k^4}\right]_0^k = \dfrac{4k}{5}\) | M1A1 | |
| \(E\!\left(\dfrac{5Z}{4}\right) = \dfrac{5}{4}\left(\dfrac{4k}{5}\right) = k\) | M1A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Z^2) = \int_0^k z^2 \cdot \dfrac{4z^3}{k^4}\,dz = \left[\dfrac{4z^6}{6k^4}\right]_0^k = \dfrac{2k^2}{3}\) | M1 | |
| \(\text{Var}(Z) = \dfrac{2k^2}{3} - \left(\dfrac{4k}{5}\right)^2 = \dfrac{2k^2}{75}\) | M1A1 | |
| \(\text{Var}\!\left(\dfrac{5Z}{4}\right) = \dfrac{25}{16} \times \dfrac{2k^2}{75} = \dfrac{k^2}{24}\) | M1A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(cY) = k\), \(c(\tfrac{1}{2}k + \tfrac{1}{2}k) = k\) | M1 | |
| \(c = \frac{2}{3}\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}\,Y = \dfrac{3k^2}{12}\) | B1 | |
| \(\text{Var}\!\left(\dfrac{2Y}{3}\right) = \dfrac{k^2}{9}\) | B1 | |
| \(\dfrac{k^2}{24} < \dfrac{k^2}{9}\), \(\dfrac{5Z}{4}\) more efficient | M1ft, A1 | |
| [4] |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Z) = \int_0^k z \cdot \dfrac{4z^3}{k^4}\,dz = \left[\dfrac{4z^5}{5k^4}\right]_0^k = \dfrac{4k}{5}$ | M1A1 | |
| $E\!\left(\dfrac{5Z}{4}\right) = \dfrac{5}{4}\left(\dfrac{4k}{5}\right) = k$ | M1A1 | |
| | **[4]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Z^2) = \int_0^k z^2 \cdot \dfrac{4z^3}{k^4}\,dz = \left[\dfrac{4z^6}{6k^4}\right]_0^k = \dfrac{2k^2}{3}$ | M1 | |
| $\text{Var}(Z) = \dfrac{2k^2}{3} - \left(\dfrac{4k}{5}\right)^2 = \dfrac{2k^2}{75}$ | M1A1 | |
| $\text{Var}\!\left(\dfrac{5Z}{4}\right) = \dfrac{25}{16} \times \dfrac{2k^2}{75} = \dfrac{k^2}{24}$ | M1A1 | |
| | **[5]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(cY) = k$, $c(\tfrac{1}{2}k + \tfrac{1}{2}k) = k$ | M1 | |
| $c = \frac{2}{3}$ | A1 | |
| | **[2]** | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}\,Y = \dfrac{3k^2}{12}$ | B1 | |
| $\text{Var}\!\left(\dfrac{2Y}{3}\right) = \dfrac{k^2}{9}$ | B1 | |
| $\dfrac{k^2}{24} < \dfrac{k^2}{9}$, $\dfrac{5Z}{4}$ more efficient | M1ft, A1 | |
| | **[4]** | |6 The continuous random variable $Z$ has probability density function
$$f ( z ) = \left\{ \begin{array} { c c }
\frac { 4 z ^ { 3 } } { k ^ { 4 } } & 0 \leqslant z \leqslant k \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ is a parameter whose value is to be estimated.\\
(i) Show that $\frac { 5 Z } { 4 }$ is an unbiased estimator of $k$.\\
(ii) Find the variance of $\frac { 5 Z } { 4 }$.
The parameter $k$ can also be estimated by making observations of a random variable $X$ which has mean $\frac { 1 } { 2 } k$ and variance $\frac { 1 } { 12 } k ^ { 2 }$. Let $Y = X _ { 1 } + X _ { 2 } + X _ { 3 }$ where $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$ are independent observations of $X$.\\
(iii) $c Y$ is also an unbiased estimator of $k$. Find the value of $c$.\\
(iv) For the value of $c$ found in part (iii), determine which of $\frac { 5 Z } { 4 }$ and $c Y$ is the more efficient estimator of $k$.
\hfill \mbox{\textit{OCR S4 2017 Q6 [15]}}