2016 H2 Maths Paper 2 Q6 — Worked Solution

Part (i)(a)

Since the sample is stratified by department and gender, the number of males in the sample is found in proportion to the total number of males among all $2345 + 1013 + 237 + 344 + 867 + 679 + 591 + 523 = 6599$ employees.

\text{no. of males} = \frac{2345 + 1013 + 237 + 344}{6599} \times 100 = \frac{3939}{6599} \times 100 = 59.7\ \left( 3\ \text{s.f.} \right)

Since the number sampled must be a whole number, this rounds to $60$ .

$\therefore$ the sample should contain $60$ males.

Part (i)(b)

\text{no. of females from Development} = \frac{679}{6599} \times 100 = 10.3\ \left( 3\ \text{s.f.} \right)

Since the number sampled must be a whole number, this rounds to $10$ .

$\therefore$ the sample should contain $10$ females from the Development department.

Part (ii)

The stratified sample in (i) should not be used, since it is stratified by department and gender, not by age.

Explanation: Stratifying by department and gender gives no assurance about how ages are distributed within each stratum, so the sample in (i) is not designed to be representative of the age distribution of the company’s employees, and is not suitable for testing a belief about the mean age.

Part (iii)

Let $\mu$ be the population mean age, in years, of all employees of the company now, and let $\overline{X}$ be the mean age of a random sample of $80$ employees.

Since the MD believes the mean age now is less than $37$ , we test

H_{0}:\mu = 37\quad\quad\text{against}\quad\quad H_{1}:\mu < 37

at the $5\%$ significance level.

Approach

The population variance is given, so we use a $Z$ -test. Since $n = 80$ is large, by the Central Limit Theorem $\overline{X}$ is approximately normally distributed even though the population distribution of ages is not given, and the test statistic below is approximately standard normal under $H_{0}$ .

Since $n = 80$ is large, by the Central Limit Theorem, under $H_{0}$ ,

Z = \frac{\overline{X} - 37}{\sqrt{140/80}} \sim N(0,1)\ \text{approximately.}

Since $H_{1}$ is one-tailed (lower), we reject $H_{0}$ at the $5\%$ level if $z \leq - 1.6449$ (by GC, $z_{0.05} = 1.6449$ ), i.e. if

\frac{\overline{x} - 37}{\sqrt{140/80}} \leq - 1.6449

\overline{x} \leq 37 - 1.6449\sqrt{\frac{140}{80}}

\leq 34.8\ \left( 3\ \text{s.f.} \right)

We are given that the test result is that the MD’s belief should be accepted, i.e. $H_{0}$ is rejected.

$\therefore$ the set of possible values of $\overline{x}$ is $\{\overline{x} \in \mathbb{R}:\overline{x} \leq 34.8\}$ .

Part (iv)

Approach

Here the sample mean $\overline{x} = 35.2$ is fixed and we are told $H_{0}$ is not rejected at the $\alpha\%$ level; we need the range of $\alpha$ for which this happens, which we find by comparing the (fixed) $p$ -value of this sample to $\alpha/100$ .

With $\overline{x} = 35.2$ ,

z = \frac{35.2 - 37}{\sqrt{140/80}} = - 1.3607\ \left( 5\ \text{s.f.} \right)

By GC, the $p$ -value is

p = P(Z \leq - 1.3607) = 0.0868\ \left( 3\ \text{s.f.} \right)

$H_{0}$ is rejected (MD’s belief accepted) if $p < \frac{\alpha}{100}$ , so the MD’s belief is not accepted if $p \geq \frac{\alpha}{100}$ , i.e.

\alpha \leq 100 \times 0.0868

\alpha \leq 8.68\ \left( 3\ \text{s.f.} \right)

Since $\alpha$ is a significance level expressed as a percentage, $\alpha > 0$ .

$\therefore$ the set of possible values of $\alpha$ is $\{\alpha \in \mathbb{R}:0 < \alpha \leq 8.68\}$ .

2016 H2 Maths Paper 2, Question 6

Part (i)(a)

Part (i)(b)

Part (ii)

Part (iii)

Part (iv)

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