State hypotheses to enable the quality control manager to test the mean weight using a two-tailed test.

(i)     Calculate unbiased estimates of the mean and the variance of the weights of the bricks being produced.

(ii)     Assuming that the weights of the bricks are normally distributed, determine the $p$-value of the above results and state the conclusion in context using a 5% significance level.

The owner is more familiar with using confidence intervals. Determine a 95% confidence interval for the mean weight of bricks produced on that particular day.

${H_0}:{\text{ }}\mu  = 2.2;{\text{ }}{H_1}:{\text{ }}\mu  \ne 2.2$    A1A1

[2 marks]

(i)     UE of mean $= \frac{{42.0}}{{20}}{\text{ = }}2.1$     A1

UE of variance $= \frac{{89.2}}{{19}} - \frac{{20 \times {{2.1}^2}}}{{19}} = 0.0526{\text{ }}\left( {\frac{1}{{19}}} \right)$     (M1)A1

Note:     Award (M0) for division by 20 where there is no subsequent use of $\frac{{20}}{{19}}$.

(ii)

$t =  - 1.95$    (A1)

${\text{DF}} = 19$    (A1)

$p - value = 0.0662$    A1

Note:     Allow follow through from (b)(i). In particular, 0.05 for the variance gives $t =  - 2$ and $p$-value 0.0600.

accept ${H_0}$, or equivalent statement involving ${H_0}$ or ${H_1}$, indicating that the mean weight is 2.2kg     R1

Note:     Follow through the candidate’s $p$-value.

[7 marks]

$[1.99,{\text{ }}2.21]$    A1A1

Note:     Allow follow through from (b)(i). In particular, 0.05 for the variance gives $[2.00,{\text{ }}2.20]$.

[2 marks]

Most candidates stated the correct hypotheses in (a).

In (b)(i), the mean was invariably found correctly, although to find the variance estimate, quite a few candidates divided by 20 instead of 19. Incorrect variances were followed through in the next part of (b)(i). The $t$-test was generally well applied and the correct conclusion drawn. It was, however, surprising to note that many candidate used the appropriate formula to find the value of $t$ and hence the $p$-value as opposed to using their GDC software.

Part (c) was generally well answered.