2024 H2 Mathematics Paper 2 Question 8

Hypothesis Testing

Answers

Lee should carry out a one-tailed test because he is interested to test whether the mean number of birds has reduced.

(b)

$H_{0} : μ = 17.3$
$H_{1} : μ < 17.3.$

(c)

$x = 16 .$
$s^{2} = \frac{510}{31} .$
Minimum integer $α = 4.$

(d)

There is sufficient evidence at the $α %$ level of significance to conclude that the mean number of birds visiting the new bird feeder while he eats his breakfast has reduced.

(e)

It is insufficient as the distribution of the number of birds visiting his feeder is not known to be normal. Hence, for Lee to carry out his hypothesis test, he needs to use Central Limit Theorem to justify that the mean number of birds visiting his feeder is normally distributed approximately. $n = 10 < 30$ so this sample size is not sufficient.

Full Solutions

(a)

Lee should carry out a one-tailed $∎$ test because he is interested to test whether the mean number of birds has reduced

(b)

Let $μ$ denote the population mean number of birds visiting the feeder in 20 minutes each morning

$H_{0} : μ = 17.3$
$H_{1} : μ < 17.3$ $∎$

(c)

\begin{aligned} Unbiased estimate of population mean \\ = x \\ = \frac{\sum x}{n} \\ = \frac{512}{32} \\ = 16 ∎ \end{aligned}

\begin{aligned} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum x^{2} - \frac{{(\sum x)}^{2}}{n}) \\ = \frac{1}{32 - 1} (8702 - \frac{{(512)}^{2}}{32}) \\ = \frac{510}{31} ∎ \end{aligned}

Under $H_{0},$

Z = \frac{X - 17.3}{\sqrt{\frac{{\frac{510}{31}}^{2}}{32}}} \sim N (0,1)

approximately by the Central Limit Theorem since

n = 32

is large

By GC,

p -value = 0.034911

Since $H_{0}$ is rejected,

\begin{aligned} p -value & \leq α % \\ 0.034911 & \leq \frac{α}{100} \\ α & \geq 3.4911 \end{aligned}

Since $α$ is an integer,

Minimum integer α = 4 ∎

(d)

There is sufficient evidence at the $α %$ level of significance to conclude that the mean number of birds visiting the new bird feeder while he eats his breakfast has reduced $∎$

(e)