陈先生（学校的行政经理）认为他所在学校的 JC 学生在学校食堂购买并享用午餐平均需要 22 分钟。 在数据收集过程中，陈先生使用学校的电脑程序，根据学生的法定姓名首字母顺序，为他所在学校的每个 JC 学生分配一个号码，并使用随机数生成器抽取 50 名学生。被选中的学生接受了一项关于他们在某一天在学校食堂吃午餐所花时间的调查。 设随机变量 $X$ 表示在学校食堂吃午餐所花的时间（以分钟为单位），得到以下结果：$\sum x = 1117$，$\sum x^2 = 25061$。

1. State the conclusion and justify using the definition of a random sample.

Yes, it is a random sample. By using a random number generator to select the students, every JC student in the school has an equal chance of being selected.

1. Calculate the unbiased estimate for the population mean

$\bar{x} = \frac{\sum x}{n}$
$\bar{x} = \frac{1117}{50} = 22.34$

2. Calculate the unbiased estimate for the population variance

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$
$s^2 = \frac{1}{50-1}\left(25061 - \frac{1117^2}{50}\right)$
$s^2 = \frac{1}{49}\left(25061 - 24953.78\right)$
$s^2 = \frac{107.22}{49} = 2.18816\dots$
$s^2 \approx 2.19 \text{ (to 3 s.f.)}$

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean time taken by JC students to buy and consume their lunch.

$H_0: \mu = 22$
$H_1: \mu \neq 22$

Level of significance, $\alpha = 0.05$

2. Retrieve the unbiased estimates and calculate the test statistic and p-value.

From part (ii), the unbiased estimates are:
$\bar{x} = \frac{1117}{50} = 22.34$
$s^2 = \frac{1}{49}\left(25061 - \frac{1117^2}{50}\right) = \frac{107.22}{49} \approx 2.1882$

Under $H_0$, since the sample size $n = 50$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - \mu}{s / \sqrt{n}} \approx \frac{22.34 - 22}{\sqrt{\frac{2.1882}{50}}}$

Using a graphing calculator (Z-Test):
$z = 1.6253 \approx 1.63$ (to 3 s.f.)
$p\text{-value} = 0.10411 \approx 0.104$ (to 3 s.f.)

3. State the conclusion of the test in the context of the question.

Since $p\text{-value} = 0.104 > 0.05$, we do not reject $H_0$.

There is insufficient evidence at the $5\%$ level of significance to conclude that the average time spent by students for lunch is not 22 minutes. Thus, Mr Tan's belief is valid.