Solution — 2022 Question 10 Sampling| Hypothesis Testing

(a)

$μ$ is the population mean checking time (in seconds) of a metal disc $∎$

$H_{0} : μ = 16.2$
$H_{1} : μ \neq 16.2$

\begin{matrix} Unbiased estimate of population mean \\ = \overline{t} \\ = \frac{\sum t}{n} \\ = \frac{645.6}{40} \\ = 16.14 \end{matrix}

\begin{matrix} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum t^{2} - \frac{{(\sum t)}^{2}}{n}) \\ = \frac{1}{40 - 1} (10 421 - \frac{{(645.6)}^{2}}{40}) \\ = 0.026 051 \end{matrix}

Under $H_{0},$ $\overline{T} \sim N (16.2, \frac{0.026 051^{2}}{40})$ $Z = \frac{\overline{T} - 16.2}{\sqrt{\frac{0.026 051^{2}}{40}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 40$ is large

From GC, $p -value = 0.018 719$

Since there is insufficient evidence to reject $H_{0},$

\begin{matrix} p -value & > α % \\ 0.018 719 & > \frac{α}{100} \\ 1.8719 & > α \\ α & < 1.8719 \\ α & < 1.87 ∎ (to 3 s.f.) \end{matrix}

(b)

Let $X$ denote the random variable of the length (in cm) of a metal rod
Let $μ$ denote the population mean length (in cm) of a metal rod

$H_{0} : μ = 12$
$H_{1} : μ < 12$

From the question, $\overline{x} = 11.8, σ^{2} = 1.1, n = 60$

Under $H_{0},$ $\overline{X} \sim N (12, \frac{{1.1}^{2}}{60})$ $Z = \frac{\overline{X} - 12}{\sqrt{\frac{{1.1}^{2}}{60}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 60$ is large

From GC,
$p$ -value = $0.069825 > 0.05 \Rightarrow Do not reject H_{0}$

Hence there is insufficient evidence at the $5 %$ level of significance to conclude whether the mean length of the rods is less than 12 cm $∎$