| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2023 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Show unbiased estimator |
| Difficulty | Standard +0.3 This is a straightforward application of standard properties of expectation and variance for linear combinations of sample means. Parts (a)-(d) involve routine algebraic manipulation of E(T) and Var(T) formulas, while part (e) requires basic calculus (differentiation) to minimize variance. All techniques are standard textbook exercises for Further Maths statistics with no novel insight required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(E(T_1) = \frac{3E(\bar{X}) + 7E(\bar{Y})}{10}\) | M1 |
| \(E(T_1) = \frac{3\mu + 7\mu}{10}\) | ||
| \(E(T_1) = \mu\), therefore \(T_1\) is an unbiased estimator for \(\mu\). | A1 | Convincing |
| (b) | \(E(T_2) = \frac{E(\bar{X}) + a^2E(\bar{Y})}{1 + a}\) | |
| To be an unbiased estimator for \(\mu\): \(\frac{\mu + a^2\mu}{1 + a} = \mu\) | M1 | Forming an equation in \(\mu\). si |
| \(1 + a^2 = 1 + a\) | A1 | oe |
| \(a = 0\) or \(a = 1\). \(a\) is positive \(\therefore a = 1\) (so \(T_2 = \frac{\bar{x}+\bar{y}}{2}\)) | A1 | Must reject \(a = 0\). If M0, then SC1 for verification only |
| (c) | \(\text{Var}(T_1) = \frac{3^2 \times \text{Var}(\bar{X}) + 7^2 \times \text{Var}(\bar{Y})}{10^2}\) | M1 |
| \(\text{Var}(T_1) = \frac{9 \times \frac{\sigma^2}{20} + 49 \times \frac{k\sigma^2}{25}}{100}\) | M1 | Use of \(\text{Var}(\bar{W}) = \text{Var}(W)/n\) |
| \(\text{Var}(T_1) = \frac{45\sigma^2 + 196k\sigma^2}{10000} = \frac{\sigma^2}{10000}(45 + 196k)\) | A1 | oe, cao. \(\text{Var}(T_1) = \frac{9\sigma^2}{2000} + \frac{49k\sigma^2}{2500}\) |
| \(\text{Var}(T_2) = \frac{1}{4}(\text{Var}(\bar{X}) + \text{Var}(\bar{Y}))\) | M1 | |
| \(\text{Var}(T_2) = \frac{1}{4}\left(\frac{\sigma^2}{20} + \frac{k\sigma^2}{25}\right)\) | M1 | |
| \(\text{Var}(T_2) = \frac{\sigma^2}{400}(5 + 4k)\) | A1 | oe \(\text{Var}(T_2) = \frac{\sigma^2}{80} + \frac{k\sigma^2}{100}\). If left in terms of \(a\): \(\text{Var}(T_2) = \frac{\sigma^2(5 + 4a^2k)}{100(1 + a)^2}\) |
| (d) | \(\frac{\sigma^2}{400}(5 + 4k) = \frac{45\sigma^2 + 196k\sigma^2}{10000}\) | M1 |
| \(\frac{10000}{400}(5 + 4k) = 45 + 196k\) or \(25(5 + 4k) = 45 + 196k\) | m1 | Forming an equation in \(k\) |
| \(125 + 100k = 45 + 196k\) | ||
| \(k = \frac{5}{6}\) | A1 | Convincing. *ag |
| (e) | \(V = \text{Var}(T_3) = (1 - \lambda)^2 \times \text{Var}(\bar{X}) + \lambda^2 \times \text{Var}(\bar{Y})\) | B1 |
| \(V = \text{Var}(T_3) = (1 - \lambda)^2 \times \frac{\sigma^2}{20} + \lambda^2 \times \frac{k\sigma^2}{25}\) | ||
| \(\frac{dV}{d\lambda} = \frac{-2(1-\lambda)\sigma^2}{20} + \frac{2\lambda k\sigma^2}{25}\) | M1 | M1 for expression for \(\frac{dV}{d\lambda}\). At least 1 term correct |
| Smallest variance is when \(\frac{dV}{d\lambda} = 0\) | M1 | M1 for setting \(\frac{dV}{d\lambda} = 0\) and attempt to solve. |
| \(\frac{2k\sigma^2}{25} = \frac{2(1-\lambda)\sigma^2}{20}\) | ||
| \(\lambda k = \frac{5}{4}(1 - \lambda)\) | ||
| \(\lambda k + \frac{5\lambda}{4} = \frac{5}{4}\) | ||
| \(\lambda\left(\frac{4k + 5}{4}\right) = \frac{5}{4}\) | ||
| \(\lambda = \frac{5}{4k + 5}\) | A1 | cao |
| \(\frac{d^2V}{d\lambda^2} = \frac{\sigma^2}{10} + \frac{2k\sigma^2}{25} > 0\) | E1 | E1 for verifying minimum, oe method |
| Therefore, it is a minimum. | ||
| Total [19] |
(a) | $E(T_1) = \frac{3E(\bar{X}) + 7E(\bar{Y})}{10}$ | M1 | |
| $E(T_1) = \frac{3\mu + 7\mu}{10}$ | | |
| $E(T_1) = \mu$, therefore $T_1$ is an unbiased estimator for $\mu$. | A1 | Convincing |
(b) | $E(T_2) = \frac{E(\bar{X}) + a^2E(\bar{Y})}{1 + a}$ | | |
| To be an unbiased estimator for $\mu$: $\frac{\mu + a^2\mu}{1 + a} = \mu$ | M1 | Forming an equation in $\mu$. si |
| $1 + a^2 = 1 + a$ | A1 | oe |
| $a = 0$ or $a = 1$. $a$ is positive $\therefore a = 1$ (so $T_2 = \frac{\bar{x}+\bar{y}}{2}$) | A1 | Must reject $a = 0$. If M0, then SC1 for verification only |
(c) | $\text{Var}(T_1) = \frac{3^2 \times \text{Var}(\bar{X}) + 7^2 \times \text{Var}(\bar{Y})}{10^2}$ | M1 | Use of $\text{Var}(cW) = c^2\text{Var}(W)$ |
| $\text{Var}(T_1) = \frac{9 \times \frac{\sigma^2}{20} + 49 \times \frac{k\sigma^2}{25}}{100}$ | M1 | Use of $\text{Var}(\bar{W}) = \text{Var}(W)/n$ |
| $\text{Var}(T_1) = \frac{45\sigma^2 + 196k\sigma^2}{10000} = \frac{\sigma^2}{10000}(45 + 196k)$ | A1 | oe, cao. $\text{Var}(T_1) = \frac{9\sigma^2}{2000} + \frac{49k\sigma^2}{2500}$ |
| $\text{Var}(T_2) = \frac{1}{4}(\text{Var}(\bar{X}) + \text{Var}(\bar{Y}))$ | M1 | |
| $\text{Var}(T_2) = \frac{1}{4}\left(\frac{\sigma^2}{20} + \frac{k\sigma^2}{25}\right)$ | M1 | |
| $\text{Var}(T_2) = \frac{\sigma^2}{400}(5 + 4k)$ | A1 | oe $\text{Var}(T_2) = \frac{\sigma^2}{80} + \frac{k\sigma^2}{100}$. If left in terms of $a$: $\text{Var}(T_2) = \frac{\sigma^2(5 + 4a^2k)}{100(1 + a)^2}$ |
(d) | $\frac{\sigma^2}{400}(5 + 4k) = \frac{45\sigma^2 + 196k\sigma^2}{10000}$ | M1 | M1 for setting their $\text{Var}(T_1) = \text{Var}(T_2)$ |
| $\frac{10000}{400}(5 + 4k) = 45 + 196k$ or $25(5 + 4k) = 45 + 196k$ | m1 | Forming an equation in $k$ |
| $125 + 100k = 45 + 196k$ | | |
| $k = \frac{5}{6}$ | A1 | Convincing. *ag |
(e) | $V = \text{Var}(T_3) = (1 - \lambda)^2 \times \text{Var}(\bar{X}) + \lambda^2 \times \text{Var}(\bar{Y})$ | B1 | cao |
| $V = \text{Var}(T_3) = (1 - \lambda)^2 \times \frac{\sigma^2}{20} + \lambda^2 \times \frac{k\sigma^2}{25}$ | | |
| $\frac{dV}{d\lambda} = \frac{-2(1-\lambda)\sigma^2}{20} + \frac{2\lambda k\sigma^2}{25}$ | M1 | M1 for expression for $\frac{dV}{d\lambda}$. At least 1 term correct |
| Smallest variance is when $\frac{dV}{d\lambda} = 0$ | M1 | M1 for setting $\frac{dV}{d\lambda} = 0$ and attempt to solve. |
| $\frac{2k\sigma^2}{25} = \frac{2(1-\lambda)\sigma^2}{20}$ | | |
| $\lambda k = \frac{5}{4}(1 - \lambda)$ | | |
| $\lambda k + \frac{5\lambda}{4} = \frac{5}{4}$ | | |
| $\lambda\left(\frac{4k + 5}{4}\right) = \frac{5}{4}$ | | |
| $\lambda = \frac{5}{4k + 5}$ | A1 | cao |
| $\frac{d^2V}{d\lambda^2} = \frac{\sigma^2}{10} + \frac{2k\sigma^2}{25} > 0$ | E1 | E1 for verifying minimum, oe method |
| Therefore, it is a minimum. | | |
| **Total [19]** | | |
---2. The random variables $X$ and $Y$ are independent, with $X$ having mean $\mu$ and variance $\sigma ^ { 2 }$, and $Y$ having mean $\mu$ and variance $k \sigma ^ { 2 }$, where $k$ is a positive constant.
Let $\bar { X }$ denote the mean of a random sample of 20 observations of $X$, and let $\bar { Y }$ denote the mean of a random sample of 25 observations of $Y$.
\begin{enumerate}[label=(\alph*)]
\item Given that $T _ { 1 } = \frac { 3 \bar { X } + 7 \bar { Y } } { 10 }$, show that $T _ { 1 }$ is an unbiased estimator for $\mu$.
\item Given that $T _ { 2 } = \frac { \bar { X } + a ^ { 2 } \bar { Y } } { 1 + a } , a > 0$, and $T _ { 2 }$ is an unbiased estimator for $\mu$, prove that $a = 1$.
\item Find and simplify expressions for the variances of $T _ { 1 }$ and $T _ { 2 }$.
\item Show that the value of $k$ for which $T _ { 1 }$ and $T _ { 2 }$ are equally good estimators is $\frac { 5 } { 6 }$.
\item Given that $T _ { 3 } = ( 1 - \lambda ) \bar { X } + \lambda \bar { Y }$, find an expression for $\lambda$, in terms of $k$, for which $T _ { 3 }$ has the smallest possible variance.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2023 Q2 [19]}}