| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Statistics vs non-statistics identification |
| Difficulty | Standard +0.3 This is a straightforward S3 question testing definitions and routine calculations. Parts (a) and (b) require recall of basic statistical concepts (statistic vs parameter, unbiased estimator). Parts (c) and (d) involve standard expectation and variance calculations using linearity properties—mechanical application of E(aX+bY) and Var(aX+bY) formulas with no problem-solving insight required. Slightly easier than average A-level due to being purely definitional and computational. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| It is not a statistic as it involves unknown [population parameters] | B1 | Must include "unknown" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| An estimator for \(\mu\) is unbiased if its expected value is equal to \(\mu\) | B1 | Allow \(\mu - E(X) = 0\); "bias = 0" is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(U_1) = 3E(X_1) - 2E(X_2)\) or \(E(U_2) = \dfrac{1}{4}(E(X_1) + 3E(X_2))\) | M1 | For use of \(aE(X_1) + bE(X_2)\); may be implied by \(3\mu - 2\mu\) or \(\frac{1}{4}(\mu+3\mu)\) |
| \(E(U_1) = 3\mu - 2\mu = \mu\) (therefore unbiased) | A1cso | Correct solution for \(E(U_1)\) with no incorrect working; condone missing notation/subscripts |
| \(E(U_2) = \dfrac{1}{4}(\mu + 3\mu) = \mu\) (therefore unbiased) | A1cso | Correct solution for \(E(U_2)\) with no incorrect working; condone missing notation/subscripts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(U_1) = 9\text{Var}(X_1) + 4\text{Var}(X_2)\) or \(\text{Var}(U_2) = \dfrac{1}{16}\text{Var}(X_1) + \dfrac{9}{16}\text{Var}(X_2)\) | M1 | For use of \(a^2\text{Var}(X_1) + b^2\text{Var}(X_2)\); allow \(9\sigma^2 + 4\sigma^2\) |
| \(\text{Var}(U_1) = 13\sigma^2\) | A1 | Allow \(9\sigma^2 + 4\sigma^2\) |
| \(\text{Var}(U_2) = \dfrac{5}{8}\sigma^2\) | A1 | Allow \(\dfrac{1}{16}\sigma^2 + \dfrac{9}{16}\sigma^2\) or \(\dfrac{5}{8}\sigma^2\) |
| As \(\text{Var}(U_1) > \text{Var}(U_2)\), \(U_2\) is the most efficient estimator for \(\mu\) | A1 | For \(U_2\) with a correct reason; NB possible to score M1 A0 A0 A1 if both variances correct |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| It is not a statistic as it involves unknown [population parameters] | B1 | Must include "unknown" |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| An estimator for $\mu$ is unbiased if its expected value is equal to $\mu$ | B1 | Allow $\mu - E(X) = 0$; "bias = 0" is B0 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(U_1) = 3E(X_1) - 2E(X_2)$ or $E(U_2) = \dfrac{1}{4}(E(X_1) + 3E(X_2))$ | M1 | For use of $aE(X_1) + bE(X_2)$; may be implied by $3\mu - 2\mu$ or $\frac{1}{4}(\mu+3\mu)$ |
| $E(U_1) = 3\mu - 2\mu = \mu$ (therefore unbiased) | A1cso | Correct solution for $E(U_1)$ with no incorrect working; condone missing notation/subscripts |
| $E(U_2) = \dfrac{1}{4}(\mu + 3\mu) = \mu$ (therefore unbiased) | A1cso | Correct solution for $E(U_2)$ with no incorrect working; condone missing notation/subscripts |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(U_1) = 9\text{Var}(X_1) + 4\text{Var}(X_2)$ or $\text{Var}(U_2) = \dfrac{1}{16}\text{Var}(X_1) + \dfrac{9}{16}\text{Var}(X_2)$ | M1 | For use of $a^2\text{Var}(X_1) + b^2\text{Var}(X_2)$; allow $9\sigma^2 + 4\sigma^2$ |
| $\text{Var}(U_1) = 13\sigma^2$ | A1 | Allow $9\sigma^2 + 4\sigma^2$ |
| $\text{Var}(U_2) = \dfrac{5}{8}\sigma^2$ | A1 | Allow $\dfrac{1}{16}\sigma^2 + \dfrac{9}{16}\sigma^2$ or $\dfrac{5}{8}\sigma^2$ |
| As $\text{Var}(U_1) > \text{Var}(U_2)$, $U_2$ is the most efficient estimator for $\mu$ | A1 | For $U_2$ with a correct reason; NB possible to score M1 A0 A0 A1 if both variances correct |\begin{enumerate}
\item A random sample of two observations $X _ { 1 }$ and $X _ { 2 }$ is taken from a population with unknown mean $\mu$ and unknown variance $\sigma ^ { 2 }$\\
(a) Explain why $\frac { X _ { 1 } - \mu } { \sigma }$ is not a statistic.\\
(b) Explain what you understand by an unbiased estimator for $\mu$
\end{enumerate}
Two estimators for $\mu$ are $U _ { 1 }$ and $U _ { 2 }$ where
$$U _ { 1 } = 3 X _ { 1 } - 2 X _ { 2 } \quad \text { and } \quad U _ { 2 } = \frac { X _ { 1 } + 3 X _ { 2 } } { 4 }$$
(c) Show that both $U _ { 1 }$ and $U _ { 2 }$ are unbiased estimators for $\mu$
The most efficient estimator among a group of unbiased estimators is the one with the smallest variance.\\
(d) By finding the variance of $U _ { 1 }$ and the variance of $U _ { 2 }$ state, giving a reason, the most efficient estimator for $\mu$ from these two estimators.
\hfill \mbox{\textit{Edexcel S3 2022 Q5 [9]}}