Solution — 2023 Question 11 Sampling| Hypothesis Testing

(a)

$\begin{matrix} Unbiased estimate of population mean \\ = \overline{x} \\ = \frac{\sum (x - 15)}{n} + 15 \\ = \frac{25.5}{50} + 15 \\ = 15.51 \end{matrix}$

$\begin{matrix} Unbiased estimate of population variance \\ = s^{2} \\ = \frac{1}{n - 1} (\sum {(x - 15)}^{2} - \frac{{(\sum (x - 15))}^{2}}{n}) \\ = \frac{1}{50 - 1} (26.4 - \frac{{(25.5)}^{2}}{50}) \\ = 0.273 37 \end{matrix}$

Let $X$ denote the random variable of the distance (in m) thrown by a child in region A
Let $μ$ denote the population mean distance (in m) thrown by a child in region A

$H_{0} : μ = 15.4$
$H_{1} : μ > 15.4$

Under $H_{0},$ at $0.05 %$ level of significance, $\overline{X} \sim N (15.4, \frac{0.273 37}{50})$ $Z = \frac{\overline{X} - 15.4}{\sqrt{\frac{0.273 37}{50}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 50$ is large

From GC,
$p$ -value = $0.068420 > 0.0005 \Rightarrow Do not reject H_{0}$

Hence there is insufficient evidence at the $0.05 %$ level of significance to conclude whether the mean distance thrown has increased after taking Pedro’s course $∎$

(b)

$H_{0} : μ = 15.4$
$H_{1} : μ > 15.4$

Under $H_{0},$ at $0.05 %$ level of significance, $\overline{X} \sim N (15.4, \frac{0.5}{50})$ $Z = \frac{\overline{X} - 15.4}{\sqrt{\frac{0.5}{50}}} \sim N (0,1)$ approximately by the Central Limit Theorem since $n = 50$ is large

Since the null hypothesis is rejected,

\begin{matrix} \frac{\overline{x} - 15.4}{\sqrt{\frac{0.5}{50}}} & \leq - 3.2905 & or \frac{\overline{x} - 15.4}{\sqrt{\frac{0.5}{50}}} & \geq 3.2905 \\ \overline{x} & \leq 15.1 ∎ & \overline{x} & \geq 15.7 ∎ \end{matrix}