| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2005 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Finding unbiased estimator constraints |
| Difficulty | Standard +0.3 This is a structured S4 question on unbiased estimators that guides students through standard expectation and variance calculations. While it requires understanding of linearity of expectation and variance of independent random variables, each part follows directly from applying well-rehearsed formulas with minimal problem-solving insight needed. The algebraic manipulation is straightforward, making this slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(E(X) = n\mu\); \(E(Y) = m\mu\) | B1 | Both; can be implied |
| \(E(\hat{p}) = \frac{aE(X) + bE(Y)}{b} \Rightarrow \frac{an\mu + bm\mu}{b}\) | M1 A1 | |
| \(\Rightarrow (a+b) = 1\) | A1 (4) | |
| (b) \(E(\hat{p}) = \frac{1}{n+m}\{E(X) + E(Y)\}\) | M1 | |
| \(= \frac{1}{n+m}(n\mu + m\mu)\) | A1 | |
| \(= \frac{1}{n+m} \cdot \mu(n+m) = \mu \Rightarrow \hat{p}\) is unbiased | A1 (3) | |
| (c) \(\text{Var}(X) = n\mu(1-\mu)\); \(\text{Var}(Y) = m\mu(1-\mu)\) | B1 | Both; can be implied |
| \(\text{Var}(\hat{p}) = \frac{a^2\text{Var}(X) + b^2\text{Var}(Y)}{b^2}\) | M1 | Use of \(\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)\) |
| \(= \frac{a^2n\mu(1-\mu) + b^2m\mu(1-\mu)}{b^2}\) | ||
| \(= \mu(1-\mu)\{\frac{a^2}{n} + \frac{b^2}{m}\}\) | A1 (3) | |
| (d) \(\text{Var}(\hat{p}) = \frac{1}{(n+m)^2}\{n\mu(1-\mu) + m\mu(1-\mu)\}\) | M1 A1 | |
| \(= \frac{\mu(1-\mu)}{n+m}\) | A1 (3) | |
| (e) \(\text{Var}(\hat{p}) = 0.0444\mu(1-\mu)\); \(\text{Var}(\hat{p}) = 0.0333\mu(1-\mu)\) | B1; B1 (4) | Use \(p_2\); \(\text{Var}(\hat{p}) \not\leq \text{Var}(\hat{p}_1)\) |
| Use \(p_2\); \(\text{Var}(\hat{p}) \not\leq \text{Var}(\hat{p}_1)\) | B1; B1 (4) | Step |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $E(X) = n\mu$; $E(Y) = m\mu$ | B1 | Both; can be implied |
| $E(\hat{p}) = \frac{aE(X) + bE(Y)}{b} \Rightarrow \frac{an\mu + bm\mu}{b}$ | M1 A1 | |
| $\Rightarrow (a+b) = 1$ | A1 (4) | |
| **(b)** $E(\hat{p}) = \frac{1}{n+m}\{E(X) + E(Y)\}$ | M1 | |
| $= \frac{1}{n+m}(n\mu + m\mu)$ | A1 | |
| $= \frac{1}{n+m} \cdot \mu(n+m) = \mu \Rightarrow \hat{p}$ is unbiased | A1 (3) | |
| **(c)** $\text{Var}(X) = n\mu(1-\mu)$; $\text{Var}(Y) = m\mu(1-\mu)$ | B1 | Both; can be implied |
| $\text{Var}(\hat{p}) = \frac{a^2\text{Var}(X) + b^2\text{Var}(Y)}{b^2}$ | M1 | Use of $\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)$ |
| $= \frac{a^2n\mu(1-\mu) + b^2m\mu(1-\mu)}{b^2}$ | | |
| $= \mu(1-\mu)\{\frac{a^2}{n} + \frac{b^2}{m}\}$ | A1 (3) | |
| **(d)** $\text{Var}(\hat{p}) = \frac{1}{(n+m)^2}\{n\mu(1-\mu) + m\mu(1-\mu)\}$ | M1 A1 | |
| $= \frac{\mu(1-\mu)}{n+m}$ | A1 (3) | |
| **(e)** $\text{Var}(\hat{p}) = 0.0444\mu(1-\mu)$; $\text{Var}(\hat{p}) = 0.0333\mu(1-\mu)$ | B1; B1 (4) | Use $p_2$; $\text{Var}(\hat{p}) \not\leq \text{Var}(\hat{p}_1)$ |
| Use $p_2$; $\text{Var}(\hat{p}) \not\leq \text{Var}(\hat{p}_1)$ | B1; B1 (4) | Step |7. A bag contains marbles of which an unknown proportion $p$ is red. A random sample of $n$ marbles is drawn, with replacement, from the bag. The number $X$ of red marbles drawn is noted.
A second random sample of $m$ marbles is drawn, with replacement. The number $Y$ of red marbles drawn is noted.
Given that $p _ { 1 } = \frac { a X } { n } + \frac { b Y } { m }$ is an unbiased estimator of $p$,
\begin{enumerate}[label=(\alph*)]
\item show that $a + b = 1$.
Given that $p _ { 2 } = \frac { ( X + Y ) } { n + m }$,
\item show that $p _ { 2 }$ is an unbiased estimator for $p$.
\item Show that the variance of $p _ { 1 }$ is $p ( 1 - p ) \left( \frac { a ^ { 2 } } { n } + \frac { b ^ { 2 } } { m } \right)$.
\item Find the variance of $p _ { 2 }$.
\item Given that $a = 0.4 , m = 10$ and $n = 20$, explain which estimator $p _ { 1 }$ or $p _ { 2 }$ you should use.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2005 Q7 [17]}}