2024 H2 Maths Paper 2 Q8 — Worked Solution

Part (a)

1 mark

Lee should carry out a one-tailed test (lower-tailed).

Explanation: Lee suspects specifically that the mean number of birds has decreased, which is a one-directional claim about $\mu$ . A two-tailed test is used only when the claim is that the mean has simply changed, so here a one-tailed test is appropriate.

Part (b)

2 marks

Let $\mu$ be the population mean number of birds visiting the new feeder in the 20-minute interval while Lee eats his breakfast.

H_{0}:\mu = 17.3

H_{1}:\mu < 17.3

Part (c)

3 marks

We first find unbiased estimates of the population mean and variance from the sample.

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

= \frac{512}{32}

= 16

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

= \frac{1}{31}\left( 8702 - \frac{512^{2}}{32} \right)

= \frac{1}{31}(8702 - 8192)

= \frac{510}{31}

Since $n = 32$ is large, we do not need to assume the population is normal: by the Central Limit Theorem, $\overline{X}$ is approximately normal, and we use the test statistic

z = \frac{\overline{x} - \mu_{0}}{\sqrt{s^{2}/n}}

= \frac{16 - 17.3}{\sqrt{\frac{510/31}{32}}}

= - 1.813\ \left( 4\ \text{s.f., by GC} \right)

For $H_{0}$ to be rejected at the $\alpha\%$ level (one-tailed, lower), the $\alpha\%$ significance level must exceed the $p$ -value:

p\text{-value} = P(Z < - 1.813)

= 1 - \Phi(1.813)

= 0.0349\ \left( 3\ \text{s.f., by GC} \right)

$H_{0}$ is rejected when $\frac{\alpha}{100} > p\text{-value}$ , so

\frac{\alpha}{100} > 0.03491

\Rightarrow \alpha > 3.491

Since $\alpha$ is a positive integer, the least value satisfying this is $\alpha = 4$ .

To check your answer: at

\alpha = 4

,

p = 0.0349 < 0.04

, so

H_{0}

is rejected ✓; at

\alpha = 3

,

p = 0.0349 > 0.03

, so

H_{0}

would not be rejected — consistent with

\alpha_{\min} = 4

. ✓

$\therefore$ $\overline{x} = 16$ , $s^{2} = \frac{510}{31}\ ( \approx 16.5)$ , and the least possible value of $\alpha$ is $4$ .

Part (d)

1 mark

Since $H_{0}$ is rejected, there is sufficient evidence, at the $4\%$ significance level, to conclude that the population mean number of birds visiting the new feeder during the 20-minute interval while Lee eats his breakfast has decreased (is less than $17.3$ ).

Part (e)

1 mark

No.

Explanation: the population distribution of the number of birds visiting the feeder is not stated (or known) to be normal, and $n = 10$ is too small for the Central Limit Theorem to guarantee that $\overline{X}$ is approximately normal. Without either of these, the test in (c) could not be carried out validly on only 10 mornings.

2024 H2 Maths Paper 2, Question 8

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

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