2018 H2 Maths Paper 2 Q5 — Worked Solution

Part (i)

2 marks

Since the distribution of the time to failure of an individual fan is not stated to be normal, we cannot assume that $\overline{X}$ is normally distributed for a small sample. Taking a sample of at least $30$ fans lets the manager invoke the Central Limit Theorem, so that the sample mean $\overline{X}$ is approximately normally distributed whatever the distribution of an individual fan’s time to failure, which is what the hypothesis test requires.

The fans should be chosen randomly (e.g. by simple random sampling) from the manufacturer’s production, so that the sample is representative of all such fans and the test is not biased by, say, testing only fans from one batch or one production run.

Part (ii)

2 marks

Let $\mu$ be the population mean time to failure, in hours, of this type of fan (under the accelerated testing procedure). The quality control manager suspects the MTTF is less than the claimed value, so the test is

H_{0}:\ \mu = 65000

H_{1}:\ \mu < 65000

carried out at the $5\%$ significance level.

Part (iii)

3 marks

Approach

We are not given the variance used in the test — instead we are told the conclusion of the test (do not reject $H_{0}$ ), and asked to work backwards for the range of variance values consistent with that conclusion. The population variance is unknown, so let $\sigma^{2}$ denote the (unbiased) estimate of it used in the test statistic. Since $n = 43$ is large, by the Central Limit Theorem, under $H_{0}$ ,

Z = \frac{\overline{X} - 65000}{\sigma/\sqrt{43}} \sim N(0,\ 1)\ \text{approximately.}

This is a lower-tailed test at the $5\%$ level, so $H_{0}$ is rejected if the test statistic $z \leq - 1.6449$ (by GC, the critical value is $- invNorm(0.95) = - 1.6449$ , i.e. $z_{0.05} = 1.6449$ ).

Since the manager finds no reason to reject $H_{0}$ , the test statistic must satisfy $z > - 1.6449$ . Using $\overline{x} = 64230$ ,

z = \frac{64230 - 65000}{\sigma/\sqrt{43}} = \frac{- 770\sqrt{43}}{\sigma}.

Substituting into the non-rejection condition:

\frac{- 770\sqrt{43}}{\sigma} > - 1.6449

\Rightarrow \frac{770\sqrt{43}}{\sigma} < 1.6449\ \text{(multiplying by } - 1\text{, so the inequality reverses)}

\Rightarrow \sigma > \frac{770\sqrt{43}}{1.6449}

\Rightarrow \sigma > 3069.7\ \text{(by GC).}

Since $\sigma > 0$ , squaring preserves the inequality:

\sigma^{2} > {3069.7}^{2}

\Rightarrow \sigma^{2} > 9.42 \times 10^{6}\ \text{(3 s.f.).}

To check your answer: at

\sigma^{2} = 9.42 \times 10^{6}

exactly,

z = \frac{- 770}{\sqrt{9.42 \times 10^{6}/43}} = - 1.645

, right at the critical value; a larger variance gives a

z

closer to

0

, which is further from the critical region, consistent with not rejecting

H_{0}

. ✓

Note: there is no upper bound on $\sigma^{2}$ — as the variance used increases, the test statistic moves towards $0$ and stays out of the (lower-tailed) critical region, so every larger variance is also consistent with not rejecting $H_{0}$ .

$\therefore$ the variance used in calculating the test statistic satisfies $\sigma^{2} > 9.42 \times 10^{6}\ \text{hours}^{2}$ (3 s.f.).

2018 H2 Maths Paper 2, Question 5

Part (i)

Part (ii)

Part (iii)

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