S4 June 2005 Q7
7. A bag contains marbles of which an unknown proportion \(p\) is red. A random sample of \(n\) marbles is drawn, with replacement, from the bag. The number \(X\) of red marbles drawn is noted.
A second random sample of \(m\) marbles is drawn, with replacement. The number \(Y\) of red marbles drawn is noted.
Given that \(p_1 = \dfrac{aX}{n} + \dfrac{bY}{m}\) is an unbiased estimator of \(p\),
(a) show that \(a + b = 1\). (4)
Given that \(p_2 = \dfrac{(X + Y)}{n + m}\),
(b) show that \(p_2\) is an unbiased estimator for \(p\). (3)
(c) Show that the variance of \(p_1\) is \(p(1 - p)\left(\dfrac{a^2}{n} + \dfrac{b^2}{m}\right)\). (3)
(d) Find the variance of \(p_2\). (3)
(e) Given that \(a = 0.4\), \(m = 10\) and \(n = 20\), explain which estimator \(p_1\) or \(p_2\) you should use. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = np\,;\quad \mathrm{E}(Y) = mp\) both; can be implied | B1 |
| \(\mathrm{E}(p_1) = \dfrac{a\mathrm{E}(X)}{n} + \dfrac{b\mathrm{E}(Y)}{m} = p\,;\ \Rightarrow \dfrac{anp}{n} + \dfrac{bmp}{m} = p\) | M1 A1 |
| \(\Rightarrow (a + b) = 1\) * | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(p_2) = \dfrac{1}{n + m}\left\{\mathrm{E}(X) + \mathrm{E}(Y)\right\}\) | M1 |
| \(= \dfrac{1}{n + m}\left\{np + mp\right\}\) | A1 |
| \(= \dfrac{1}{(n + m)} \cdot p(n + m) = p \Rightarrow p_2\) is unbiased | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(X) = np(1 - p)\,;\quad \mathrm{Var}(Y) = mp(1 - p)\) both; can be implied | B1 |
| \(\mathrm{Var}(p_1) = \dfrac{a^2\mathrm{Var}(X)}{n^2} + \dfrac{b^2\mathrm{Var}(Y)}{m^2}\) Use of \(\mathrm{Var}(aX) = a^2\mathrm{Var}(X)\) | M1 |
| \(= \dfrac{a^2np(1 - p)}{n^2} + \dfrac{b^2mp(1 - p)}{m^2}\) \(= p(1 - p)\left\{\dfrac{a^2}{n} + \dfrac{b^2}{m}\right\}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(p_2) = \dfrac{1}{(n + m)^2}\left\{np(1 - p) + mp(1 - p)\right\}\) | M1 A1 |
| \(= \dfrac{p(1 - p)}{n + m}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(p_1) = 0.044\,p(1 - p)\,;\quad \mathrm{Var}(p_2) = 0.03\dot{3}\,p(1 - p)\) | B1; B1 |
| Use \(p_2\,;\quad \mathrm{Var}(p_2) \lt \mathrm{Var}(p_1)\) | B1; B1dep |
| (4) |
Notes
The scheme links the final two B1 marks: the last B1 is dependent on the previous B1.