| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Compare estimator properties |
| Difficulty | Standard +0.3 This is a standard S4 question on properties of estimators (unbiasedness and variance) with straightforward algebraic manipulation. Parts (a) and (b) involve routine application of expectation and variance formulas for binomial distributions and linear combinations. Part (c) requires solving inequalities but follows directly from the variance expressions. Part (d) is a simple numerical comparison. While it's a multi-part question worth 16 marks, each step is methodical with no novel insights required—slightly easier than the average A-level question due to its predictable structure. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks |
|---|---|
| \(E(\hat{p}_1) = E\left(\frac{X}{n}\right)\) | M1 |
| \(= \frac{1}{n}E(X)\) | M1 |
| \(= \frac{1}{n} \times np\) | M1 |
| \(= p\) unbiased | A1cso |
| (4 marks) |
| Answer | Marks |
|---|---|
| \(\text{Var}(\hat{p}_1) = \text{Var}\left(\frac{X}{n}\right)\) | M1 |
| \(= \frac{1}{n^2}\text{Var}(X)\) | M1 |
| \(= \frac{1}{n^2} \times np(1-p)\) | M1 |
| \(= \frac{p(1-p)}{n}\) | A1 |
| (4 marks) |
| Answer | Marks |
|---|---|
| \(E(\hat{p}_3) = 3a E(\hat{p}_1) + 2a E(\hat{p}_2)\) | M1 |
| \(= 3ap + 2ap\) | M1 |
| \(= 5ap\) | M1 |
| \(5ap = p\) | M1 |
| \(a = \frac{1}{5}\) | A1 |
| (4 marks) |
| Answer | Marks |
|---|---|
| \(\text{Var}(\hat{p}_3) = \frac{9}{25}\text{Var}(\hat{p}_1) + \frac{4}{25}\text{Var}(\hat{p}_2)\) | M1 |
| \(= \frac{9p(1-p)}{25n} + \frac{4p(1-p)}{25m}\) | M1d |
| \(= \frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right)\) | A1 |
| (6 marks) |
| Answer | Marks |
|---|---|
| \(\frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right) < \frac{p(1-p)}{n}\) | M1 |
| \(9m + 4n < 25m\) | M1 |
| \(4n < 16m\) | M1 |
| \(\frac{n}{m} < 4\) | A1 |
| \(\frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right) < \frac{p(1-p)}{m}\) | M1 |
| \(9m + 4n < 25n\) | M1 |
| (3 marks) |
| Answer | Marks |
|---|---|
| \(\text{Var}(\hat{p}_1) = 0.05 p(1-p)\) | M1 |
| \(\text{Var}(\hat{p}_2) = 0.0167 p(1-p)\) | M1 |
| \(\text{Var}(\hat{p}_3) = 0.0207 p(1-p)\) | M1 |
| Or since \(\frac{1}{3}\) is not in the range \(\frac{9}{21} < \frac{n}{m} < 4\); \(\text{Var}(\hat{p}_3)\) is not the smallest variance. \(\text{Var}(\hat{p}_1) = 0.05 p(1-p)\) \(\text{Var}(\hat{p}_2) = 0.0167 p(1-p)\) | A1ft; A1ft |
| Therefore \(\hat{p}_2\) is the best estimator as it has the smallest variance | A1ft; A1ft |
| (3 marks) |
**Part (a)(i)**
| $E(\hat{p}_1) = E\left(\frac{X}{n}\right)$ | M1 |
| $= \frac{1}{n}E(X)$ | M1 |
| $= \frac{1}{n} \times np$ | M1 |
| $= p$ unbiased | A1cso |
| | **(4 marks)** |
**Part (a)(ii)**
| $\text{Var}(\hat{p}_1) = \text{Var}\left(\frac{X}{n}\right)$ | M1 |
| $= \frac{1}{n^2}\text{Var}(X)$ | M1 |
| $= \frac{1}{n^2} \times np(1-p)$ | M1 |
| $= \frac{p(1-p)}{n}$ | A1 |
| | **(4 marks)** |
**Part (b)(i)**
| $E(\hat{p}_3) = 3a E(\hat{p}_1) + 2a E(\hat{p}_2)$ | M1 |
| $= 3ap + 2ap$ | M1 |
| $= 5ap$ | M1 |
| $5ap = p$ | M1 |
| $a = \frac{1}{5}$ | A1 |
| | **(4 marks)** |
**Part (b)(ii)**
| $\text{Var}(\hat{p}_3) = \frac{9}{25}\text{Var}(\hat{p}_1) + \frac{4}{25}\text{Var}(\hat{p}_2)$ | M1 |
| $= \frac{9p(1-p)}{25n} + \frac{4p(1-p)}{25m}$ | M1d |
| $= \frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right)$ | A1 |
| | **(6 marks)** |
**Part (c)**
| $\frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right) < \frac{p(1-p)}{n}$ | M1 |
| $9m + 4n < 25m$ | M1 |
| $4n < 16m$ | M1 |
| $\frac{n}{m} < 4$ | A1 |
| $\frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right) < \frac{p(1-p)}{m}$ | M1 |
| $9m + 4n < 25n$ | M1 |
| | **(3 marks)** |
**Part (d)**
| $\text{Var}(\hat{p}_1) = 0.05 p(1-p)$ | M1 |
| $\text{Var}(\hat{p}_2) = 0.0167 p(1-p)$ | M1 |
| $\text{Var}(\hat{p}_3) = 0.0207 p(1-p)$ | M1 |
| Or since $\frac{1}{3}$ is not in the range $\frac{9}{21} < \frac{n}{m} < 4$; $\text{Var}(\hat{p}_3)$ is not the smallest variance. $\text{Var}(\hat{p}_1) = 0.05 p(1-p)$ $\text{Var}(\hat{p}_2) = 0.0167 p(1-p)$ | A1ft; A1ft |
| Therefore $\hat{p}_2$ is the best estimator as it has the smallest variance | A1ft; A1ft |
| | **(3 marks)** |
**Guidance notes for 6(a)(i):**
- M1 either $\frac{1}{n}E(X)$ or $\frac{1}{n} \times np$
- A1 cso
**Guidance notes for 6(a)(ii):**
- M1 either $\frac{1}{n^2}\text{Var}(X)$ or $\frac{1}{n^2} \times np(1-p)$
- A1 cso
**Guidance notes for 6(b)(i):**
- M1 For either $3a E(\hat{p}_1) + 2a E(\hat{p}_2)$ or $3ap + 2ap$
- M1 Putting their $E(\hat{p}_3) = p$
- A1 ft correct estimator chosen.
- A1 ft correct supporting reason from correct working for their var formulae
- SC if 1/3 is in their range in part(c) they may get B1 for stating $\hat{p}_3$
- B1dependent on the previous B being awarded- stating smallest variance award first two marks on open.
**Guidance notes for 6(b)(ii):**
- M1 for $\frac{9}{25}\text{Var}(\hat{p}_1) + \frac{4}{25}\text{Var}(\hat{p}_2)$
- M1d for substituting (aii) for Var($\hat{p}_1$) and (aii) with $m$ instead of $n$ for Var($\hat{p}_2$)
- A1 cso
**Guidance notes for 6(c):**
- M1 Putting Var($\hat{p}_3$) $<$ their Var($\hat{p}_1$) leading to an inequality of the form $\frac{n}{m} < a$ or $\frac{n}{m} > a$ where $a$ is a constant.
- M1 Putting Var($\hat{p}_3$) $<$ their Var($\hat{p}_2$) leading to an inequality of the form $\frac{n}{m} > a$ or
**Guidance notes for 6(d):**
- 1/3 is not in their range in part(c)
- M1 attempt to find all 3 variances or eliminating Var($\hat{p}_3$) with reason and finding the other 2 variances.
- A1ft correct estimator chosen.
- A1ft correct supporting reason from correct working for their var formulae
**Total: 16 marks**When a tree seed is planted the probability of it germinating is $p$. A random sample of size $n$ is taken and the number of tree seeds, $X$, which germinate is recorded.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\hat{p}_1 = \frac{X}{n}$ is an unbiased estimator of $p$.
\item Find the variance of $\hat{p}_1$. [4]
\end{enumerate}
A second sample of size $m$ is taken and the number of tree seeds, $Y$, which germinate is recorded.
Given that $\hat{p}_2 = \frac{Y}{m}$ and that $\hat{p}_3 = a(3\hat{p}_1 + 2\hat{p}_2)$ is an unbiased estimator of $p$,
\item show that
\begin{enumerate}[label=(\roman*)]
\item $a = \frac{1}{5}$,
\item $\text{Var}(\hat{p}_3) = \frac{p(1-p)}{25}\left(\frac{9}{n} + \frac{4}{m}\right)$. [6]
\end{enumerate}
\item Find the range of values of $\frac{n}{m}$ for which
$$\text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_1) \text{ and } \text{Var}(\hat{p}_3) < \text{Var}(\hat{p}_2)$$ [3]
\item Given that $n = 20$ and $m = 60$, explain which of $\hat{p}_1$, $\hat{p}_2$ or $\hat{p}_3$ is the best estimator. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2012 Q6 [16]}}