(a)     After a chemical spillage at sea, a scientist measures the amount, x units, of the chemical in the water at 15 randomly chosen sites. The results are summarised in the form \(\sum {x = 18} \) and \(\sum {{x^2} = 28.94} \). Before the spillage occurred the mean level of the chemical in the water was 1.1. Test at the 5 % significance level the hypothesis that there has been an increase in the amount of the chemical in the water.

(b)     Six months later the scientist returns and finds that the mean amount of the chemical in the water at the 15 randomly chosen sites is 1.18. Assuming that this sample came from a normal population with variance 0.0256, find a 90 % confidence interval for the mean level of the chemical.

(a)     \(\bar x = \frac{{\sum x }}{n} = 1.2\)     (A1)

\(s_{n - 1}^2 = 0.524 \ldots \)     (A1)

it is a one tailed test

\({{\text{H}}_0}:\mu = 1.1,{\text{ }}{{\text{H}}_1}:\mu > 1.1\)     A1

EITHER

\(t = \frac{{1.2 - 1.1}}{{\sqrt {\frac{{0.524 \ldots }}{{15}}} }} = 0.535\)     (M1) A1

\(v = 14\)     (A1)

\({t_{crit}} = 1.761\)     A1

since \(0.535 < {t_{crit}}\) we accept \({{\text{H}}_0}\) that there is no increase in the amount of the chemical     R1

OR

\(p = 0.301\)     A4

since \(p > 0.05\) we accept \({{\text{H}}_0}\) that there is no increase in the amount of the chemical     R1

[8 marks]

(b)     90 % confidence interval \( = 1.18 \pm 1.645\sqrt {\frac{{0.0256}}{{15}}} \)     (M1)A1A1A1

\( = [1.11,{\text{ }}1.25]\)     A1     N5

[5 marks]

Total [13 marks]

This question also proved accessible to a majority of candidates with many wholly correct or nearly wholly correct answers seen. A few candidates did not recognise that part (a) was a t-distribution and part (b) was a Normal distribution, but most recognised the difference. Many candidates received an accuracy penalty on this question for not giving the final answer to part (b) to 3 significant figures.