Solution — 2021 Question 11 Binomial Distribution| Sampling| Hypothesis Testing

(a)

(i)

Let $X$ be the random variable of the number of orders that are delivered within 24 hours out of $10$ orders

$ ParseError: Can't use function '$' in math mode at position 1: $̲ X \sim \mathrm{B} \left( 10, 0.75 \right)

\begin{align*} & \mathrm{P}\left(X = 4 \text{ or } 5\right) \\ &= \mathrm{P}\left(X = 4\right) + \mathrm{P}\left(X = 5\right) \\ &= 0.016\,222 + 0.058\,399 \\ &= 0.0746\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.

(ii)

$\begin{matrix} P (X > 6) & = 1 - P (X \leq 6) \\ = 1 - 0.224 12 \\ = 0.775 88 \\ = 0.7759 ∎ (to 4 s.f.) \end{matrix}$

[probability of X > 6]

(b)

$\begin{matrix} Unbiased estimate of population mean \\ = \overline{x} \\ = \frac{\sum x}{n} \\ = \frac{708.6}{60} \\ = 11.81 \end{matrix}$

$\begin{matrix} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum x^{2} - \frac{{(\sum x)}^{2}}{n}) \\ = 0.676 85 \end{matrix}$

Let $X$ denote the random variable of the delivery time (in hours) for an order
Let $μ$ denote the population mean delivery time (in hours) for an order

$H_{0} : μ = 12$
$H_{1} : μ < 12$

Under $H_{0},$ at $0.05 %$ level of significance, $\overline{X} \sim N (12, \frac{0.676 85}{60})$ $Z = \frac{\overline{X} - 12}{\sqrt{\frac{0.676 85}{60}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 60$ is large

From GC,
$p$ -value = $0.036816 > 0.0005 \Rightarrow Do not reject H_{0}$

Hence there is insufficient evidence at the $0.05 %$ level of significance to conclude whether the mean delivery time is less than 12 hours $∎$

(c)

Since $n = 60$ is large, by the Central Limit Theorem, $\overline{X}$ is approximately normally distributed. Hence it is not necessary to assume that the delivery times are normally distributed $∎$