一名 17 岁学生完成 5 公里长跑所需的时间 $T$（以分钟为单位）是一个随机变量，其均值为 30。针对这些学生引入新的训练计划后，抽取了一个容量为 $n$ 的随机样本。发现该样本的平均时间和标准差分别为 28.9 分钟和 4.0 分钟。

1. Calculate the sample variance from the given sample standard deviation

$\text{Sample variance} = 4.0^2 = 16$

2. Calculate the unbiased estimate of the population variance

$s^2 = \frac{n}{n-1} \times \text{Sample variance} = \frac{n}{n-1}(16) = \frac{16n}{n-1}$

1. State the hypotheses and define the population mean symbol.

Let $\mu$ be the population mean time taken (in minutes) by a 17-year-old student to complete a 5-km run after the new training programme.
$H_0: \mu = 30$
$H_1: \mu \neq 30$

2. Calculate the unbiased estimate of the population variance and state the test statistic.

From part (a), the unbiased estimate of the population variance is $s^2 = \frac{n}{n-1} \sigma_n^2$.
For $n=40$ and sample standard deviation $\sigma_n = 4.0$:
$s^2 = \frac{40}{39}(4.0)^2 = \frac{640}{39} \approx 16.410$
Under $H_0$, since $n=40$ is large ($n \ge 30$), the sample mean $\bar{X}$ is approximately normally distributed by the Central Limit Theorem.
The test statistic is $Z = \frac{\bar{X} - \mu_0}{S/\sqrt{n}} \sim N(0, 1)$ approximately.
$z = \frac{28.9 - 30}{\sqrt{\frac{640/39}{40}}} = \frac{-1.1}{\sqrt{\frac{16}{39}}} \approx -1.71737$

3. Calculate the p-value for the two-tailed test.

$p\text{-value} = 2 \times P(Z \le -1.71737)$
$p\text{-value} \approx 0.085906 \approx 0.0859 \text{ (to 3 s.f.)}$

4. Compare the p-value with the significance level and state the conclusion.

Since the $p$-value $= 0.0859 < 0.10$, we reject $H_0$.
There is sufficient evidence at the 10% level of significance to conclude that the mean time taken has changed.

1. State the contextual meaning of the $p$-value.

The $p$-value is the probability of obtaining a sample mean time of 28.9 minutes or a value more extreme, assuming that the true population mean time taken by the students is 30 minutes.