Section B: Probability and Statistics [35 marks] From past data, it is known that the mean time taken by students at a school to complete a H2 Mathematics A-level examination paper was 170 minutes. In June 2024, Ms Tan believes that this cohort of students is not ready for the examination and will take more than 170 minutes to complete the paper. She gives the same examination paper to 50 students, assuming that these students have not seen the paper before. She notes the time taken by each student and carries out a test.

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.