2023 H2 Maths Paper 2 Q8 — Worked Solution

Part (a)(i)

2 marks

Let $\mu$ be the population mean mass, in kg, of granulated sugar in bags filled by the new machine.

H_{0}:\mu = 1

H_{1}:\mu \neq 1

This is a two-tailed test, since the manager wants to check whether the machine is correctly adjusted, i.e. whether the mean has shifted away from $1$ kg in either direction.

Part (a)(ii)

2 marks

Reason 1: the sample size is only $10$ , too small for the Central Limit Theorem to guarantee that $\overline{X}$ is approximately normal, unless the mass of sugar per bag is already known to be normally distributed.

Reason 2: the sample is not random — always taking the first bag produced each morning and afternoon may not be representative of typical production, e.g. if the machine’s fill amount drifts systematically over the course of a run.

Part (b)(i)

5 marks

Approach

With $n = 40$ large, the Central Limit Theorem lets us treat $\overline{X}$ as approximately normal, using the sample variance $s^{2}$ in place of the (unknown) population variance. We find the $z$ boundary for the lower $2.5\%$ tail, then convert it into a boundary on $\overline{x}$ .

Let $\mu$ be the population mean mass, in kg, of sugar in the larger bags.

H_{0}:\mu = 2

H_{1}:\mu < 2

This is a one-tailed test, since the concern is specifically that the mean may be less than $2$ kg.

Since $n = 40$ , we first find the unbiased estimate of the population variance:

\bar{x} = \frac{\sum x}{n}

= \frac{78.88}{40}

= 1.972

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

= \frac{1}{39}\left( 155.6746 - \frac{{78.88}^{2}}{40} \right)

= 0.003159\ \left( 4\text{ s.f.} \right)

To check your answer:

\bar{x} = 1.972

kg is indeed slightly below the target of

2

kg, consistent with the customers’ suspicion motivating the test ✓

Since $n = 40$ is large, by the Central Limit Theorem $\overline{X} \sim N\left( 2,\frac{s^{2}}{n} \right)$ approximately under $H_{0}$ . The standard error is

\frac{s}{\sqrt{n}} = \sqrt{\frac{0.003159}{40}}

\approx 0.008888

By GC, at the $2.5\%$ significance level (one-tailed, lower tail), the critical value of $Z$ is

z_{crit} = - 1.9600\ \left( 4\text{ d.p.} \right)

The critical region is $\bar{x} < 2 + z_{crit} \times \frac{s}{\sqrt{n}}$ :

\bar{x} < 2 - 1.9600(0.008888)

\approx 1.9826

$\therefore$ the critical region is $\bar{x} < 1.9826\ \left( 4\text{ d.p.} \right)$ .

Part (b)(ii)

2 marks

The observed sample mean is $\bar{x} = 1.972$ , which lies in the critical region since $1.972 < 1.9826$ . Equivalently, the test statistic

z = \frac{1.972 - 2}{0.008888}

\approx - 3.15

is far below $z_{crit} = - 1.96$ .

To check your answer: the test statistic and the boundary comparison agree — both place

\bar{x}

well inside the rejection region, not just barely across it ✓

Since $\bar{x}$ falls in the critical region, there is sufficient evidence, at the $2.5\%$ level of significance, to reject $H_{0}$ .

$\therefore$ there is sufficient evidence, at the $2.5\%$ level, that the mean mass of sugar in the (larger) bags is less than $2$ kg, supporting the customers’ suspicion.

2023 H2 Maths Paper 2, Question 8

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

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