2019 H2 Maths Paper 2 Q9 — Worked Solution

Part (i)

2 marks

Since the manager wants to find out whether the mean resistance of the resistors differs from $750$ ohms, without any prior indication of the direction of any difference, a 2-tail test should be used.

Let $\mu$ ohms be the population mean resistance of the $750$ ohm resistors.

H_{0}:\mu = 750

H_{1}:\mu \neq 750

Part (ii)

5 marks

We first find the sample mean.

\overline{x}

= \frac{742 + 771 + 768 + 738 + 769 + 752 + 742 + 766}{8}

= \frac{6048}{8}

= 756

$\therefore$ the mean of the sample of $8$ resistors is $756$ ohms.

Approach

The population variance is given as $100\ {ohms}^{2}$ , so we do not need to estimate it from the sample. Since the individual resistances are normally distributed, $\overline{X}$ is exactly normal for any sample size, and a $Z$ -test is used directly (no Central Limit Theorem needed here).

Under $H_{0}$ ,

\overline{X} \sim N\left( 750,\ \frac{100}{8} \right).

Standardising, the test statistic is

Z = \frac{\overline{X} - 750}{\sqrt{100/8}} \sim N(0,1)\ \text{under}\ H_{0}.

Using $\overline{x} = 756$ ,

z = \frac{756 - 750}{\sqrt{100/8}}

= 1.70\ (3\ s.f.)

By GC, carrying out a 1-sample $Z$ -test with $\overline{x} = 756,\ n = 8,\ \sigma^{2} = 100$ under $H_{0}:\mu = 750$ , the p-value is

p = 0.0897\ (3\ s.f.)

Since $p = 0.0897 > 0.05$ , we do not reject $H_{0}$ at the $5\%$ significance level.

To check your answer:

z = 1.70 < 1.960

, the 2-tail critical value at the 5% level, which agrees with

p > 0.05

. ✓

$\therefore$ there is insufficient evidence, at the $5\%$ significance level, to conclude that the mean resistance of the $750$ ohm resistors differs from $750$ ohms.

Part (iii)

2 marks

Since nothing is known about the distribution of the resistances of the $1250$ ohm resistors, the manager cannot assume they are normally distributed, so the test would need to differ in two ways.

A large sample (e.g. $n \geq 30$ ) would need to be taken, so that, by the Central Limit Theorem, the sample mean $\overline{X}$ is approximately normally distributed regardless of the underlying distribution of the individual resistances.

Since the population variance of the $1250$ ohm resistors is also unknown, the unbiased estimate of the population variance,

s^{2} = \frac{1}{n - 1}\sum\left( x - \overline{x} \right)^{2},

computed from the sample, would be used in place of the population variance in the test statistic.

2019 H2 Maths Paper 2, Question 9

Part (i)

Part (ii)

Part (iii)

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