S4 June 2012 Q6
6. 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.
A second sample of size \(m\) is taken and the number of tree seeds, \(Y\), which germinate is recorded.
Given that \(\hat{p}_2 = \dfrac{Y}{m}\) and that \(\hat{p}_3 = a(3\hat{p}_1 + 2\hat{p}_2)\) is an unbiased estimator of \(p\),
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{E}(\hat{p}_1) = \mathrm{E}\left(\dfrac{X}{n}\right)\) \(= \dfrac{1}{n}\mathrm{E}(X)\) \(= \dfrac{1}{n} \times np\) | M1 |
| \(= p\) unbiased | A1cso |
| (ii) \(\mathrm{Var}(\hat{p}_1) = \mathrm{Var}\left(\dfrac{X}{n}\right)\) \(= \dfrac{1}{n^2}\mathrm{Var}(X)\) | M1 |
| \(= \dfrac{1}{n^2} \times np(1-p)\) \(= \dfrac{p(1-p)}{n}\) | A1 |
| (4) |
Notes
(i) M1 either \(\dfrac{1}{n}\mathrm{E}(X)\) or \(\dfrac{1}{n} \times np\)
A1 cso
(ii) M1 either \(\dfrac{1}{n^2}\mathrm{Var}(X)\) or \(\dfrac{1}{n^2} \times np(1-p)\)
A1 cso
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{E}(\hat{p}_3) = 3a\,\mathrm{E}(\hat{p}_1) + 2a\,\mathrm{E}(\hat{p}_2)\) \(= 3ap + 2ap\) \(= 5ap\) | M1 |
| \(5ap = p\) | M1 |
| \(a = \dfrac{1}{5}\) | A1 |
| (ii) \(\mathrm{Var}(\hat{p}_3) = \dfrac{9}{25}\mathrm{Var}(\hat{p}_1) + \dfrac{4}{25}\mathrm{Var}(\hat{p}_2)\) | M1 |
| \(= \dfrac{9p(1-p)}{25n} + \dfrac{4p(1-p)}{25m}\) | M1d |
| \(= \dfrac{p(1-p)}{25}\left(\dfrac{9}{n} + \dfrac{4}{m}\right)\) | A1 |
| (6) |
Notes
(i) M1 For either \(3a\,\mathrm{E}(\hat{p}_1) + 2a\,\mathrm{E}(\hat{p}_2)\) or \(3ap + 2ap\)
M1 Putting their \(\mathrm{E}(\hat{p}_3) = p\)
(ii) M1 for \(\dfrac{9}{25}\mathrm{Var}(\hat{p}_1) + \dfrac{4}{25}\mathrm{Var}(\hat{p}_2)\)
M1d for substituting (aii) for \(\mathrm{Var}(\hat{p}_1)\) and (aii) with \(m\) instead of \(n\) for \(\mathrm{Var}(\hat{p}_2)\)
A1 cso
| Scheme | Marks |
|---|---|
| \(\dfrac{p(1-p)}{25}\left(\dfrac{9}{n} + \dfrac{4}{m}\right) \lt \dfrac{p(1-p)}{n}\) \(9m + 4n \lt 25m\) \(4n \lt 16m\) \(\dfrac{n}{m} \lt 4\) | M1 |
| \(\dfrac{p(1-p)}{25}\left(\dfrac{9}{n} + \dfrac{4}{m}\right) \lt \dfrac{p(1-p)}{m}\) \(9m + 4n \lt 25n.\) \(9m \lt 21n\) \(\dfrac{9}{21} \lt \dfrac{n}{m}\) or \(\dfrac{3}{7} \lt \dfrac{n}{m}\) | M1 |
| \(\dfrac{3}{7} \lt \dfrac{n}{m} \lt 4\) | A1 |
| (3) |
Notes
M1 Putting \(\mathrm{Var}(\hat{p}_3) \lt\) their \(\mathrm{Var}(\hat{p}_1)\) leading to an inequality of the form \(\dfrac{n}{m} \lt a\) or \(\dfrac{n}{m} \gt a\) where a is a constant.
M1 Putting \(\mathrm{Var}(\hat{p}_3) \lt\) their \(\mathrm{Var}(\hat{p}_2)\) leading to an inequality of the form \(\dfrac{n}{m} \gt a\) or \(\dfrac{n}{m} \lt a\) where a is a constant.
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(\hat{p}_1) = 0.05\,p(1-p)\) \(\mathrm{Var}(\hat{p}_2) = 0.0167\,p(1-p)\) \(\mathrm{Var}(\hat{p}_3) = 0.0207\,p(1-p)\) Or since \(\dfrac{1}{3}\) is not in the range \(\dfrac{9}{21} \lt \dfrac{n}{m} \lt 4\) \(\mathrm{Var}(\hat{p}_3)\) is not the smallest variance. \(\mathrm{Var}(\hat{p}_1) = 0.05\,p(1-p)\) \(\mathrm{Var}(\hat{p}_2) = 0.0167\,p(1-p)\) | M1 |
| Therefore \(\hat{p}_2\); is the best estimator as it has the smallest variance | A1ft; A1ft |
| (3) | |
| (16 marks) |
Notes
1/3 is not in their range in part(c)
M1 attempt to find all 3 variances or eliminating \(\mathrm{Var}(\hat{p}_3)\) with reason and finding the other 2 variances.
A1ft correct estimator chosen.
A1ft correct supporting reason from correct working for their var formulae
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 epen.