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 \(p_1 = \frac{aX}{n} + \frac{bY}{m}\) is an unbiased estimator of \(p_1\),
- show that \(a + b = 1\). [4]
Given that \(p_2 = \frac{(X + Y)}{n + m}\)
- [(b)] show that \(p_2\) is an unbiased estimator for \(p\). [3]
- Show that the variance of \(p_1\) is p(1 - \(p\))\(\left(\frac{a^2}{n} + \frac{b^2}{m}\right)\). [3]
- Find the variance of \(p_2\). [3]
- Given that \(a = 0.4\), \(m = 10\) and \(n = 20\), explain which estimator \(p_1\) or \(p_2\) you should use. [4]
(Total 17 marks)