Find the value of \(k\).

(i)     Calculate an unbiased estimate for \(\theta \), using the random sample,

8.3, 4.2, 6.5, 10.3, 2.7, 1.2, 3.3, 4.3.

(ii)     Explain briefly why this is not a good estimate for \(\theta \).

(i)     Show that \({\text{Var}}(U) = \frac{{{\theta ^2}}}{{6n}}\).

(ii)     Show that \({U^2}\) is not an unbiased estimator for \({\theta ^2}\).

(iii)     Find an unbiased estimator for \({\theta ^2}\) in terms of \(U\) and \(n\).

\({\text{E}}(U) = k{\text{E}}(\overline X ) = k{\text{E}}(X)\)    (M1)

\( = \frac{{k\theta }}{2}\)    (A1)

unbiased when \(k = 2\)     A1

[3 marks]

(i)     for the data, \(\Sigma x = 40.8\)     (A1)

\( \Rightarrow \bar x = 5.1\)    (A1)

so that unbiased estimate for \(\theta  = 10.2\)     A1

(ii)     this is impossible because of the sample value 10.3     R1

[4 marks]

(i)     \({\text{Var}}(U) = 4 \times {\text{Var}}(\bar X)\)     (M1)

\( = 4 \times \frac{{{\theta ^2}}}{{24n}}\)    A1

\( = \frac{{{\theta ^2}}}{{6n}}\)    AG

(ii)     \({\text{E}}({U^2}) = {\text{Var}}(U) + {\left( {{\text{E}}(U)} \right)^2}\)     M1

\( = \frac{{{\theta ^2}}}{{6n}} + {\theta ^2}\)    A1

\({\text{E}}({U^2}) \ne {\theta ^2}\)    R1

so not unbiased     AG

(iii)     \({\text{E}}({U^2}) = \frac{{{\theta ^2}}}{{6n}}(1 + 6n)\)     (A1)

\({\text{E}}\left( {\left( {\frac{{6n}}{{1 + 6n}}} \right){U^2}} \right) = {\theta ^2}\)    (A1)

therefore \(\left( {\left( {\frac{{6n}}{{1 + 6n}}} \right){U^2}} \right)\) is an unbiased estimator for \({\theta ^2}\)     A1

[8 marks]

Solutions to (a) were often disappointing with some candidates seeming to be confused by the notation used.

In (b)(i), many candidates evaluated the sample mean as 5.1 but some failed to convert this to the estimate 10.2 even if they had correctly found the value of \(k\).

In (b)(ii), very few candidates realised that \(\theta  = 10.2\) was not a feasible estimate when one of the sample values was 10.3.

Solutions to (c) were generally poor.

In (c)(i), many good answers were seen although some candidates failed to take account of the difference between \({\text{Var}}(X)\) and \({\text{Var}}(\bar X)\).

In (c)(ii), many candidates thought that \({\text{E}}({\bar X^2}) = {\left[ {{\text{E}}(\bar X)} \right]^2}\) although this had the unfortunate consequence of showing that \({U^2}\) is an unbiased estimator for \({\theta ^2}\). Few candidates realized that an expression for \({\text{E}}({U^2})\) could be found by considering the standard result that \({\text{Var}}(U) = {\text{E}}({U^2}) - {\left[ {{\text{E}}(U)} \right]^2}\) or the equivalent expression for \({\text{Var}}(\bar X)\). Part (c)(iii) was inaccessible to candidates who were unable to solve (ii).