The weights, \(X\) grams, of tomatoes may be assumed to be normally distributed with mean \(\mu \) grams and standard deviation \(\sigma \) grams. Barry weighs \(21\) tomatoes selected at random and calculates the following statistics.\[\sum {x = 1071} \) ; \(\sum {{x^2} = 54705} \]

  (i)     Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\) .

  (ii)     Determine a \(95\%\) confidence interval for \(\mu \) .

The random variable \(Y\) has variance \({\sigma ^2}\) , where \({\sigma ^2} > 0\) . A random sample of \(n\) observations of \(Y\) is taken and \(S_{n - 1}^2\) denotes the unbiased estimator for \({\sigma ^2}\) .

By considering the expression

\({\rm{Var}}({S_{n - 1}}) = {\rm{E}}(S_{n - 1}^2) - {\left\{ {E\left. {({S_{n - 1}})} \right\}} \right.^2}\) ,

show that \(S_{n - 1}^{}\) is not an unbiased estimator for \(\sigma \) .

(i)     \(\overline x  = \frac{{1071}}{{21}} = 51\)     A1

\(S_{n - 1}^2 = \frac{{54705}}{{20}} - \frac{{{{1071}^2}}}{{20 \times 21}} = 4.2\)     M1A1

(ii)     degrees of freedom \( = 20\) ; \(t\)-value \( = 2.086\)     (A1)(A1)

\(95\%\) confidence limits are

\(51 \pm 2.086\sqrt {\frac{{4.2}}{{21}}} \)     (M1)(A1)

leading to \(\left[ {50.1,51.9} \right]\)     A1

[8 marks]

\({\rm{Var}}({S_{n - 1}}) > 0\)     A1

\(E(S_{n - 1}^2) = {\sigma ^2}\)     (A1)

substituting in the given equation,

\({\sigma ^2} - E(S_{n - 1}^{}) > 0\)     M1

it follows that

\(E(S_{n - 1}^{}) < \sigma \)     A1

this shows that \({S_{n - 1}}\) is not an unbiased estimator for \(\sigma \) since that would require = instead of \( < \)     R1

[5 marks]

Most candidates attempted (a) although some used the normal distribution instead of the \(t\)-distribution.

Many candidates were unable even to start (b) and many of those who did filled several pages of algebra with factors such as \(n\) / \((n - 1)\) prominent. Few candidates realised that the solution required only a few lines.