| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Calculating bias of estimator |
| Difficulty | Standard +0.3 This is a straightforward S4 question testing standard definitions and routine calculations of bias and variance for estimators. Students need to apply E(X_i) = np and Var(X_i) = np(1-p), then use linearity of expectation—all mechanical applications of known formulas with no novel insight required. Slightly easier than average due to its formulaic nature. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks |
|---|---|
| (a)(i) \(E(\hat{\theta}) = \theta\) | B1 |
| (ii) \(E(\hat{\theta}) = \theta\) or \(E(\hat{\theta}) \to \theta\) and \(\text{Var}(\hat{\theta}) \to 0\) as \(n \to \infty\) where \(n\) is the sample size | B1; B1 |
| (3) | |
| (b) \(E(\hat{p}_1) = p, \therefore \text{Bias} = 0\) | B1 |
| \(E(\hat{p}_2) = \frac{5p}{6}, \therefore \text{Bias} = \frac{1}{6}p\) | B1 B1 |
| \(E(\hat{p}_3) = p, \therefore \text{Bias} = 0\) | B1 |
| (4) | |
| (c) \(\text{Var}(\hat{p}_1) = \frac{1}{9n^2}\{npq + npq + npq\} = \frac{pq}{3n}\) | M1; A1 |
| \(\text{Var}(\hat{p}_2) = \frac{1}{36n^2}\{npq + 9npq + npq\} = \frac{11pq}{36n}\) | A1 |
| \(\text{Var}(\hat{p}_3) = \frac{1}{36n^2}\{4npq + 9npq + npq\} = \frac{7pq}{18n}\) | A1 |
| (4) | |
| (d)(i) \(\hat{p}_1\); unbiased and smallest variance | B1 dep; B1 |
| (ii) \(\hat{p}_2\); biased | B1 dep; B1 |
| (4) | |
| (15 marks) |
| **(a)(i)** $E(\hat{\theta}) = \theta$ | B1 |
| **(ii)** $E(\hat{\theta}) = \theta$ or $E(\hat{\theta}) \to \theta$ and $\text{Var}(\hat{\theta}) \to 0$ as $n \to \infty$ where $n$ is the sample size | B1; B1 |
| | (3) |
| **(b)** $E(\hat{p}_1) = p, \therefore \text{Bias} = 0$ | B1 |
| $E(\hat{p}_2) = \frac{5p}{6}, \therefore \text{Bias} = \frac{1}{6}p$ | B1 B1 |
| $E(\hat{p}_3) = p, \therefore \text{Bias} = 0$ | B1 |
| | (4) |
| **(c)** $\text{Var}(\hat{p}_1) = \frac{1}{9n^2}\{npq + npq + npq\} = \frac{pq}{3n}$ | M1; A1 |
| $\text{Var}(\hat{p}_2) = \frac{1}{36n^2}\{npq + 9npq + npq\} = \frac{11pq}{36n}$ | A1 |
| $\text{Var}(\hat{p}_3) = \frac{1}{36n^2}\{4npq + 9npq + npq\} = \frac{7pq}{18n}$ | A1 |
| | (4) |
| **(d)(i)** $\hat{p}_1$; unbiased and smallest variance | B1 dep; B1 |
| **(ii)** $\hat{p}_2$; biased | B1 dep; B1 |
| | (4) |
| | (15 marks) |5. (a) Explain briefly what you understand by
\begin{enumerate}[label=(\roman*)]
\item an unbiased estimator,
\item a consistent estimator.\\
of an unknown population parameter $\theta$.
From a binomial population, in which the proportion of successes is $p , 3$ samples of size $n$ are taken. The number of successes $X _ { 1 } , X _ { 2 }$, and $X _ { 3 }$ are recorded and used to estimate $p$.\\
(b) Determine the bias, if any, of each of the following estimators of $p$.
$$\begin{aligned}
& \hat { p } _ { 1 } = \frac { X _ { 1 } + X _ { 2 } + X _ { 3 } } { 3 n } \\
& \hat { p } _ { 2 } = \frac { X _ { 1 } + 3 X _ { 2 } + X _ { 3 } } { 6 n } \\
& \hat { p } _ { 3 } = \frac { 2 X _ { 1 } + 3 X _ { 2 } + X _ { 3 } } { 6 n }
\end{aligned}$$
(c) Find the variance of each of these estimators.\\
(d) State, giving a reason, which of the three estimators for $p$ is
\item the best estimator,
\item the worst estimator.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2004 Q5 [15]}}