Solution — 2020 Question 12 Sampling| Hypothesis Testing

(a)

Let $X$ denote the random variable of the height (in cm) of a randomly chosen plant
Let $μ$ denote the population mean height (in cm) of a randomly chosen plant

$H_{0} : μ = 28.5$
$H_{1} : μ \neq 28.5$

(b)

From the sample, $\overline{x} = 27.7, σ^{2} = 6.6, n = 40$

Under $H_{0},$ at $0.05 %$ level of significance, $\overline{X} \sim N (28.5, \frac{6.6}{40})$ $Z = \frac{\overline{X} - 28.5}{\sqrt{\frac{6.6}{40}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 40$ is large

From GC,
$p$ -value = $0.048900 > 0.0005 \Rightarrow Do not reject H_{0}$

Hence there is insufficient evidence at the $0.05 %$ level of significance to conclude whether the mean height of plants in region A differs from 28.5 cm $∎$

(c)

$\begin{matrix} Unbiased estimate of population mean \\ = \overline{x} \\ = \frac{\sum x}{n} \\ = \frac{1650}{60} \\ = \frac{55}{2} \\ = 27.5 ∎ \end{matrix}$

$\begin{matrix} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum x^{2} - \frac{{(\sum x)}^{2}}{n}) \\ = \frac{1}{60 - 1} (46170.1 - \frac{{(1650)}^{2}}{60}) \\ = 13.476 \\ = 13.5 ∎ (to 3 s.f.) \end{matrix}$

(d)

$H_{0} : μ = 28.5$
$H_{1} : μ \neq 28.5$

Under $H_{0},$ at $α %$ level of significance, $ ParseError: Can't use function '$' in math mode at position 1: $̲ \bar{X} \sim \mathrm{N} \left( 28.5, \frac{13.476}{60} \right) $$ Z = \frac{\bar{X}-28.5}{\sqrt{\frac{13.476}{60}}} \sim \mathrm{N}(0,1) approximately by the Central Limit Theorem since $ is large

From GC, $p -value = 0.034 855$

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

\begin{align*} p\text{-value} &> \alpha \% \\ 0.034\,855 &> \frac{\alpha }{100} \\ 3.4855 &> \alpha \\ \alpha &< 3.4855 \\ \alpha &< 3.49\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.