| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Variance of estimators |
| Difficulty | Challenging +1.2 This is a Further Maths S4 question on estimator properties requiring standard variance calculations for binomial distributions, algebraic manipulation to compare variances, and application of a given inequality hint. While it involves multiple steps and some algebraic work, the techniques are routine for this level and the inequality hint significantly guides part (ii). |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_1\) and \(T_2\) both unbiased. \(\text{Var}(T_1) = \frac{\theta^2(1-\theta)}{n_1} + \frac{\theta^2(1-\theta)}{n_2}\), \(\text{Var}(T_2) = \frac{\theta(1-\theta)}{n_1 + n_2}\) | M1 M1 M1 M1 A1 A1 | For \(T_1 = \frac{1}{2}\left(\frac{X_1}{n_1} + \frac{X_2}{n_2}\right)\): \(E(T_1) = \frac{1}{2}(\theta + \theta) = \theta\). For \(T_2 = \frac{X_1 + X_2}{n_1 + n_2}\): \(E(T_2) = \frac{n_1\theta + n_2\theta}{n_1 + n_2} = \theta\). Both unbiased. \(\text{Var}(T_1) = \frac{1}{4}\left[\frac{\theta(1-\theta)}{n_1} + \frac{\theta(1-\theta)}{n_2}\right] = \frac{\theta(1-\theta)}{4}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\). \(\text{Var}(T_2) = \frac{\theta(1-\theta)}{n_1 + n_2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\text{Var}(T_1)}{\text{Var}(T_2)} = \frac{(n_1 + n_2)}{4(n_1 n_2)} \cdot \frac{1}{(n_1 + n_2)} \cdots\) leading to result. \(T_2\) is more efficient when \(n_1 \neq n_2\) | M1 M1 M1 M1 A1 | \(\frac{\text{Var}(T_1)}{\text{Var}(T_2)} = \frac{\frac{\theta(1-\theta)}{4}\left(\frac{n_1 + n_2}{n_1n_2}\right)}{\frac{\theta(1-\theta)}{n_1+n_2}} = \frac{(n_1+n_2)^2}{4n_1n_2}\). Since \((n_1 - n_2)^2 > 0\), we have \(n_1^2 + n_2^2 + 2n_1n_2 > 4n_1n_2\), so \((n_1+n_2)^2 > 4n_1n_2\), meaning \(\frac{\text{Var}(T_1)}{\text{Var}(T_2)} > 1\). Therefore \(T_2\) is the more efficient estimator. |
**(i) Show that $T_1$ and $T_2$ are unbiased, and calculate their variances**
| $T_1$ and $T_2$ both unbiased. $\text{Var}(T_1) = \frac{\theta^2(1-\theta)}{n_1} + \frac{\theta^2(1-\theta)}{n_2}$, $\text{Var}(T_2) = \frac{\theta(1-\theta)}{n_1 + n_2}$ | M1 M1 M1 M1 A1 A1 | For $T_1 = \frac{1}{2}\left(\frac{X_1}{n_1} + \frac{X_2}{n_2}\right)$: $E(T_1) = \frac{1}{2}(\theta + \theta) = \theta$. For $T_2 = \frac{X_1 + X_2}{n_1 + n_2}$: $E(T_2) = \frac{n_1\theta + n_2\theta}{n_1 + n_2} = \theta$. Both unbiased. $\text{Var}(T_1) = \frac{1}{4}\left[\frac{\theta(1-\theta)}{n_1} + \frac{\theta(1-\theta)}{n_2}\right] = \frac{\theta(1-\theta)}{4}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)$. $\text{Var}(T_2) = \frac{\theta(1-\theta)}{n_1 + n_2}$ |
**(ii) Find $\frac{\text{Var}(T_1)}{\text{Var}(T_2)}$ and determine more efficient estimator**
| $\frac{\text{Var}(T_1)}{\text{Var}(T_2)} = \frac{(n_1 + n_2)}{4(n_1 n_2)} \cdot \frac{1}{(n_1 + n_2)} \cdots$ leading to result. $T_2$ is more efficient when $n_1 \neq n_2$ | M1 M1 M1 M1 A1 | $\frac{\text{Var}(T_1)}{\text{Var}(T_2)} = \frac{\frac{\theta(1-\theta)}{4}\left(\frac{n_1 + n_2}{n_1n_2}\right)}{\frac{\theta(1-\theta)}{n_1+n_2}} = \frac{(n_1+n_2)^2}{4n_1n_2}$. Since $(n_1 - n_2)^2 > 0$, we have $n_1^2 + n_2^2 + 2n_1n_2 > 4n_1n_2$, so $(n_1+n_2)^2 > 4n_1n_2$, meaning $\frac{\text{Var}(T_1)}{\text{Var}(T_2)} > 1$. Therefore $T_2$ is the more efficient estimator. |8 The independent random variables $X _ { 1 }$ and $X _ { 2 }$ have the distributions $\mathrm { B } \left( n _ { 1 } , \theta \right)$ and $\mathrm { B } \left( n _ { 2 } , \theta \right)$ respectively. Two possible estimators for $\theta$ are
$$T _ { 1 } = \frac { 1 } { 2 } \left( \frac { X _ { 1 } } { n _ { 1 } } + \frac { X _ { 2 } } { n _ { 2 } } \right) \text { and } T _ { 2 } = \frac { X _ { 1 } + X _ { 2 } } { n _ { 1 } + n _ { 2 } } .$$
(i) Show that $T _ { 1 }$ and $T _ { 2 }$ are both unbiased estimators, and calculate their variances.\\
(ii) Find $\frac { \operatorname { Var } \left( T _ { 1 } \right) } { \operatorname { Var } \left( T _ { 2 } \right) }$. Given that $n _ { 1 } \neq n _ { 2 }$, use the inequality $\left( n _ { 1 } - n _ { 2 } \right) ^ { 2 } > 0$ to find which of $T _ { 1 }$ and $T _ { 2 }$ is the more efficient estimator.
\hfill \mbox{\textit{OCR S4 2015 Q8 [12]}}