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

$X \sim B (10, 0.75)$

\begin{matrix} P (X = 4 or 5) \\ = P (X = 4) + P (X = 5) \\ = 0.016 222 + 0.058 399 \\ = 0.0746 ∎ (to 3 s.f.) \end{matrix}

(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}) \\ = \frac{1}{60 - 1} (8408.5 - \frac{{(708.6)}^{2}}{60}) \\ = 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 $H_{0},$ $\overline{X} \sim N (12, \frac{0.676 85^{2}}{60})$ $Z = \frac{\overline{X} - 12}{\sqrt{\frac{0.676 85^{2}}{60}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 60$ is large

From GC,
$p$ -value = $0.036816 \leq 0.05 \Rightarrow Reject H_{0}$

Hence there is sufficient evidence at the $5 %$ level of significance to conclude that 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 $∎$