乙组：概率与统计［35分］ 根据过去的数据，某校学生完成 H2 数学 A-level 考试试卷的平均时间为 170 分钟。在 2024 年 6 月，陈老师认为这届学生还没有准备好迎接考试，需要超过 170 分钟才能完成该试卷。她将同一份试卷发给 50 名学生，假设 these 学生之前没有见过这份试卷。她记录了每个学生所花费的时间并进行了一项检验。

1. Define the population parameter being tested

Let $\mu$ be the population mean time, in minutes, taken by this cohort of students to complete the examination paper.

2. State the null and alternative hypotheses

$H_0: \mu = 170$
$H_1: \mu > 170$

1. Calculate the unbiased estimate of the population variance

Given the sample variance is $53.8$, we calculate the unbiased estimate of the population variance, $s^2$:
$s^2 = \frac{n}{n-1} \times \text{sample variance} = \frac{50}{49} \times 53.8 = \frac{2690}{49} \approx 54.898$

2. State the test statistic and its distribution

From part (a), the hypotheses are $H_0: \mu = 170$ and $H_1: \mu > 170$.
Since the sample size $n = 50 \ge 30$, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
Under $H_0$, the test statistic is:
$Z = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim N(0, 1) \text{ approximately}$

3. Calculate the value of the test statistic and the corresponding $p$-value

Substitute the values into the test statistic:
$z = \frac{172.1 - 170}{\sqrt{54.898 / 50}} \approx 2.0041$
For a right-tailed test ($H_1: \mu > 170$),
$p\text{-value} = \mathrm{P}(Z > 2.0041) \approx 0.02253$

4. Conclude the hypothesis test based on the $p$-value

Since the $p$-value $= 0.0225 < 0.05$, we reject $H_0$.
There is sufficient evidence at the 5% level of significance to conclude that the mean time taken by students to complete the examination paper is more than 170 minutes. Hence, the teacher's belief is supported.