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

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

Given the hypotheses

\[{{\text{H}}_0}:\mu  = 15;{\text{ }}{{\text{H}}_1}:\mu  \ne 15,\]

find the p-value of the above results and state your conclusion at the 1 % significance level.

\(\bar x = 14\)     A1

\(s_{n - 1}^2 = \frac{{3977.57}}{{19}} - \frac{{{{280}^2}}}{{380}}\)     (M1)

\( = 3.03\)     A1

[3 marks]

Note: Accept any notation for these estimates including \(\mu \) and \({\sigma ^2}\).

Note: Award M0A0 for division by 20.

the 95% confidence limits are

\(\bar x \pm t\sqrt {\frac{{s_{n - 1}^2}}{n}} \)     (M1)

Note: Award M0 for use of z.

ie, \(14 \pm 2.093\sqrt {\frac{{3.03}}{{20}}} \)     (A1)

Note:FT their mean and variance from (a).

giving [13.2, 14.8]     A1

Note: Accept any answers which round to 13.2 and 14.8.

[3 marks]

Use of t-statistic \(\left( { = \frac{{14 - 15}}{{\sqrt {\frac{{3.03}}{{20}}} }}} \right)\)     (M1)

Note:FT their mean and variance from (a).

Note: Award M0 for use of z.

Note: Accept \(\frac{{15 - 14}}{{\sqrt {\frac{{3.03}}{{20}}} }}\).

\( = - 2.569 \ldots \)     (A1)

Note: Accept \(2.569 \ldots \)

\(p{\text{ - value}} = 0.009392 \ldots  \times 2 = 0.0188\)     A1

Note: Accept any answer that rounds to 0.019.

Note: Award (M1)(A1)A0 for any answer that rounds to 0.0094.

insufficient evidence to reject \({{\text{H}}_0}\) (or equivalent, eg accept \({{\text{H}}_0}\) or reject \({{\text{H}}_1}\))     R1

Note:FT on their p-value.

[4 marks]

In (a), most candidates estimated the mean correctly although many candidates failed to obtain a correct unbiased estimate for the variance. The most common error was to divide \(\sum {{x^2}} \) by \(20\) instead of \(19\). For some candidates, this was not a costly error since we followed through their variance into (b) and (c).

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.