- A bag contains a large number of marbles of which an unknown proportion, \(p\), is yellow.
Three random samples of size \(n\) are taken, and the number of yellow marbles in each sample, \(Y _ { 1 } , Y _ { 2 }\) and \(Y _ { 3 }\), is recorded.
Two estimators \(\hat { \mathrm { p } } _ { 1 }\) and \(\hat { \mathrm { p } } _ { 2 }\) are proposed to estimate the value of \(p\)
$$\begin{aligned}
& \hat { p } _ { 1 } = \frac { Y _ { 1 } + 3 Y _ { 2 } - 2 Y _ { 3 } } { 2 n }
& \hat { p } _ { 2 } = \frac { 2 Y _ { 1 } + 3 Y _ { 2 } + Y _ { 3 } } { 6 n }
\end{aligned}$$
- Show that \(\hat { \mathrm { p } } _ { 1 }\) and \(\hat { \mathrm { p } } _ { 2 }\) are both unbiased estimators of \(p\)
- Find the variance of \(\hat { p } _ { 1 }\)
The variance of \(\hat { \mathrm { p } } _ { 2 }\) is \(\frac { 7 p ( 1 - p ) } { 18 n }\)
- State, giving a reason, which is the better estimator.
The estimator \(\hat { p } _ { 3 } = \frac { Y _ { 1 } + a Y _ { 2 } + 3 Y _ { 3 } } { b n }\) where \(a\) and \(b\) are positive integers.
- Find the pair of values of \(a\) and \(b\) such that \(\hat { \mathrm { p } } _ { 3 }\) is a better unbiased estimator of \(p\) than both \(\hat { \mathrm { p } } _ { 1 }\) and \(\hat { \mathrm { p } } _ { 2 }\)
You must show all stages of your working.