2021 H2 Maths Paper 2 Q8 — Worked Solution

Part (a)

3 marks

The sales manager wants to test only whether the tyres last longer than $20\, 000$ miles (a specific direction), not merely whether the life span differs from $20\, 000$ miles in either direction. This one-directional claim is why a $1$ -tail test is appropriate.

Let $\mu$ thousand miles be the population mean life span of the front tyres.

H_{0}:\ \mu = 20

H_{1}:\ \mu > 20

where the test is carried out at the $5\%$ significance level.

$\therefore$ a $1$ -tail test is appropriate, with $H_{0}:\mu = 20$ tested against $H_{1}:\mu > 20$ at the $5\%$ level of significance.

Part (b)

2 marks

With $n = 50$ , $\Sigma(x - 20) = 9.4$ and $\Sigma(x - 20)^{2} = 38.76$ , we find the unbiased estimates from the coded data.

\overline{x} = 20 + \frac{\Sigma(x - 20)}{n}

= 20 + \frac{9.4}{50}

= 20.188

s^{2} = \frac{1}{n - 1}\left\lbrack \Sigma(x - 20)^{2} - \frac{\left( \Sigma(x - 20) \right)^{2}}{n} \right\rbrack

= \frac{1}{49}\left\lbrack 38.76 - \frac{{9.4}^{2}}{50} \right\rbrack

= \frac{36.9928}{49}

= 0.755\ \left( 3\ \text{s.f.} \right)

$\therefore$ the unbiased estimate of the population mean is $\overline{x} = 20.188$ thousand miles, and the unbiased estimate of the population variance is $s^{2} = 0.755$ (thousand miles) $^{2}$ (3 s.f.).

Part (c)

3 marks

Since $n = 50$ is large, by the Central Limit Theorem $\overline{X}$ is approximately normal even though the population distribution of tyre life spans is not stated. Under $H_{0}$ ,

\overline{X} \sim N\left( 20,\ \frac{s^{2}}{n} \right)\ \text{approximately}

Using the sample data, the test statistic is

z = \frac{\overline{x} - 20}{\sqrt{s^{2}/n}}

= \frac{0.188}{\sqrt{0.75496/50}}

= 1.53\ (3\ \text{s.f., by GC)}

For a $1$ -tail test at the $5\%$ level, the critical value is $z = 1.6449$ , i.e. the critical region is $z > 1.6449$ .

To check your answer: since

z \approx 1.53 < 1.6449

, the test statistic does not fall in the critical region, consistent with

\overline{x} = 20.188

being only modestly above

20

relative to the spread of the data. ✓

Since $z \approx 1.53 < 1.6449$ , we do not reject $H_{0}$ .

$\therefore$ there is insufficient evidence, at the $5\%$ level of significance, to conclude that the mean life span of front tyres is more than $20\, 000$ miles.

Part (d)

1 mark

The test in (c) would no longer be valid.

Explanation: with only $15$ tyres, the sample is too small for the Central Limit Theorem to guarantee that $\overline{X}$ is approximately normal. Since the population distribution of tyre life spans is not given to be normal, $\overline{X}$ can no longer be assumed approximately normal, so the $z$ -test in (c) breaks down.

$\therefore$ the test would be inappropriate, since the Central Limit Theorem no longer applies for a sample of only $15$ tyres.

2021 H2 Maths Paper 2, Question 8

Part (a)

Part (b)

Part (c)

Part (d)

This is how we mark, and how we teach.