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.