Mr Tan, who is the school Admin Manager, believes that JC students in his school take 22 minutes on average to buy and consume their lunch in the school canteen. In the data collection process, Mr Tan uses the school's computer programme to assign every JC students in his school a number based on the alphabetical order of their statutory names and select 50 students using a random number generator. The selected students do a survey on the duration they spent for lunch on a particular day in the school canteen. The random variable $X$, which denotes the time spent for lunch in the school canteen, in minutes gives the following results: $\sum x = 1117$, $\sum x^2 = 25061$.

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.