A2 June 2024 Q7
7. Two organisations are each asked to carry out a survey to find out the proportion, \(p\), of the population that would vote for a particular political party.
The first organisation finds that out of \(m\) people, \(X\) would vote for this particular political party.
The second organisation finds that out of \(n\) people, \(Y\) would vote for this particular political party.
An unbiased estimator, \(Q\), of \(p\) is proposed where
\[Q = k\left(\frac{X}{m} + \frac{Y}{n}\right)\]A second unbiased estimator, \(R\), of \(p\) is proposed where
\[R = \frac{aX}{m} + \frac{bY}{n}\]Given that \(m = 100\) and \(n = 200\) and that \(R\) is a better estimator of \(p\) than \(Q\)
Show your working clearly. (7)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(Q) = k\left(\dfrac{\mathrm{E}(X)}{m} + \dfrac{\mathrm{E}(Y)}{n}\right)\) | ||
| \(\mathrm{E}(Q) = k\left(\dfrac{mp}{m} + \dfrac{np}{n}\right)\) | M1 | 3.3 |
| \(2kp = p \qquad\) therefore \(k = \dfrac{1}{2}\)* | A1* | 1.1b |
| (2) |
Notes
M1: For selecting the correct models for \(X\) and \(Y\) and subst into \(\mathrm{E}(Q) = k\left(\dfrac{\mathrm{E}(X)}{m} + \dfrac{\mathrm{E}(Y)}{n}\right)\)
Both \(mp\) and \(np\) must be seen or used as the expected values for \(X\) and \(Y\).
Allow to be implied by \(k\left(\mathrm{E}\left(\dfrac{X}{m}\right) + \mathrm{E}\left(\dfrac{Y}{n}\right)\right) \Rightarrow k(p + p)\) for this mark.
Cannot be implied by \(k(p + p)\)
A1*: Cao sets their expression in \(k\) and \(p\) equal to \(p\) before achieving the given answer with no errors.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(R) = \dfrac{amp}{m} + \dfrac{bnp}{n}\) | M1 | 3.4 |
| \(\dfrac{amp}{m} + \dfrac{bnp}{n} = p \quad \therefore a + b = 1\)* | A1* | 1.1b |
| (2) |
Notes
M1: Using the model to find \(\mathrm{E}(R)\) in terms of \(a\) and \(b\).
Both \(mp\) and \(np\) must be seen or used as the expected values for \(X\) and \(Y\).
Must see \(\dfrac{amp}{m} + \dfrac{bnp}{n}\) for this mark.
Cannot be implied by \(ap + bp\)
A1*: Cao sets their expression in \(a\), \(b\) and \(p\) equal to \(p\) before achieving the given answer with no errors.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{Var}(Q) = \dfrac{mp(1-p)}{\text{‘}4\text{’}m^2} + \dfrac{np(1-p)}{\text{‘}4\text{’}n^2}\) \(\left[= \text{‘}\frac{1}{4}\text{’}p(1-p)\left(\dfrac{1}{m} + \dfrac{1}{n}\right)\right]\) | M1 | 2.1 |
| \(\mathrm{Var}(R) = \dfrac{a^2mp(1-p)}{m^2} + \dfrac{b^2np(1-p)}{n^2}\) \(\left[= p(1-p)\left(\dfrac{a^2}{m} + \dfrac{b^2}{n}\right)\right]\) | M1 | 2.1 |
| \(\left(\dfrac{a^2}{m} + \dfrac{b^2}{n}\right) \lt \text{‘}\frac{1}{4}\text{’}\left(\dfrac{1}{m} + \dfrac{1}{n}\right)\) | M1 | 1.1b |
| \(\left(\dfrac{a^2}{100} + \dfrac{(\text{“}1-a\text{”})^2}{200}\right) \lt \text{‘}\frac{1}{4}\text{’}\left(\dfrac{1}{100} + \dfrac{1}{200}\right)\) | M1 | 1.1b |
| \(12a^2 - 8a + 1 \lt 0 \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = \dfrac{1}{6}\) or \(\dfrac{1}{2}\) | A1 | 1.1b |
| \(\dfrac{1}{6} \lt a \lt \dfrac{1}{2}\) | A1ft | 2.2a |
| (7) | ||
| (11 marks) |
Notes
M1: For a correct attempt at \(\mathrm{Var}(Q)\) with at least two of \(m\), \(n\) and 2 being squared on the denominator. May be implied if they cancel by \(m\) or \(n\)
M1: For a correct attempt at \(\mathrm{Var}(R)\) in terms of \(a\) and \(b\) with at least one of \(a\) and \(b\) being squared and at least one of \(m\) and \(n\) being squared. May be implied if they cancel by \(m\) or \(n\)
Expression may be in \(a\) only
e.g. \(\mathrm{Var}\left(\dfrac{aX}{m} + \dfrac{bY}{n}\right) = \mathrm{Var}\left(\dfrac{aX}{m} + \dfrac{(1-a)Y}{n}\right) = \dfrac{a^2mp(1-p)}{m^2} + \dfrac{(1-a)^2np(1-p)}{n^2}\)
Note attempting \(\mathrm{Var}\left(\dfrac{aX}{m} + \dfrac{bY}{n}\right) = \mathrm{Var}\left(\dfrac{aX}{m} + \dfrac{(1-a)Y}{n}\right) = \mathrm{Var}\left(\dfrac{aX}{m} + \dfrac{Y}{n} - \dfrac{aY}{n}\right)\) is M0
M1: using their \(\mathrm{Var}(R) \lt\) their \(\mathrm{Var}(Q)\) condone = instead of < (there must be a term in \(a^2\) in their equation or inequality)
M1: substituting their \(b = 1 - a\) may be scored earlier
M1: forming and solving correctly a 3 term quadratic in \(a\). condone = instead of <
A1: correct values
A1ft: Dep. on all previous M marks and for selecting the right range using their values of \(a\) which must be between 0 and 1