Calculate a $99\%$ confidence interval for $\mu$. Give your answer correct to three decimal places.

Interpret the result found in (a).

Find the confidence level of the interval that corresponds to halving the width of the $99\%$ confidence interval. Give your answer as a percentage to the nearest whole number.

the (unbiased) estimate of $\mu$ is 9.793     (A1)

the $99\%$ CI is $9.793 \pm 2.576\frac{{0.03}}{{\sqrt 6 }}$     (M1)(A1)

$= [9.761,{\text{ }}9.825]$     A1

Note:     Accept $9.762$ and $9.824$.

[4 marks]

if this process is carried out a large number of times     A1

(approximately) $99\%$ of the intervals will contain $\mu$     A1

Note:     Award A1A1 for a consideration of any specific large value of times $(n \ge 100)$.

[2 marks]

METHOD 1

If the interval is halved, $2.576$ becomes $1.288$     M1

normal tail probability corresponding to $1.288 = 0.0988 \ldots$     A1

confidence level $= 80\%$     A1

METHOD 2

half width $= 0.5 \times 0.063$ or $0.062$ or $0.064 = 0.0315$ or $0.031$ or $0.032$     M1

$\frac{{2z \times 0.03}}{{\sqrt 6 }} = 0.0315$ or $0.031$ or $0.032$

giving $z = 1.285 \ldots$ or $1.265 \ldots$ or $1.306 \ldots$     A1

confidence level $= 80\%$ or $79\%$ or $81\%$     A1

Note:     Follow through values from (a).

[3 marks]

Total [9 marks]

The intention in (a) was that candidates should input the data into their calculators and use the software to give the confidence interval. However, as in Question 2, many candidates calculated the mean and variance by hand and used the appropriate formulae to determine the confidence limits. Again valuable time was used up and opportunity for error introduced.

Answers to (b) were extremely disappointing with the vast majority giving an incorrect interpretation of a confidence interval. The most common answer given was along the lines of ‘There is a 99% probability that the interval [9.761, 9.825] contains $\mu$’. This is incorrect since the interval and $\mu$ are both constants; the statement that the interval [9.761, 9.825] contains $\mu$ is either true or false, there is no question of probability being involved. Another common response was ‘I am 99% confident that the interval [9.761, 9.825] contains $\mu$’. This is unsatisfactory partly because 99% confident is really a euphemism for 99% probability and partly because it answers the question ‘What is a 99% confidence interval for $\mu$’ by simply rearranging the words without actually going anywhere. The expected answer was that if the sampling was carried out a large number of times, then approximately 99% of the calculated confidence intervals would contain $\mu$. A more rigorous response would be that a 99% confidence interval for $\mu$ is an observed value of a random interval which contains $\mu$ with probability 0.99 just as the number $\bar x$ is an observed value of the random variable $\bar X$. The concept of a confidence interval is a difficult one at this level but confidence intervals are part of the programme and so therefore is their interpretation. In view of the widespread misunderstanding of confidence intervals, partial credit was given on this occasion for interpretations involving 99% probability or confidence but this will not be the case in future examinations.

Many candidates solved (c) correctly, mostly using Method 2 in the mark scheme.