The time taken, $T$ (in minutes), for a 17-year-old student to complete a 5-km run is a random variable with mean 30. After a new training programme is introduced for these students, a random sample of $n$ students is taken. The mean time and standard deviation for the sample are found to be 28.9 minutes and 4.0 minutes respectively.

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.