Based on observations over a long period, the mean time taken by male students in Griffles Junior College (GJC) to complete a 2.4 km run is known to be 11.3 minutes with a standard deviation of 2.2 minutes. As part of a review of the current physical training programme, the Physical Education (PE) department in GJC decided to test, at the 2.5% level of significance, whether the mean time taken by male students in GJC to complete a 2.4 km run has changed. The time taken, $x$ minutes, by a random sample of 8 Year 5 male students from GJC to complete a 2.4 km run, are as follows. $\begin{matrix} 11 & 11.5 & 10.8 & 11.2 & 11.4 & 11 & 11.8 & 12.5 \end{matrix}$

1. State the null and alternative hypotheses

$H_0: \mu = 11.3$
$H_1: \mu \neq 11.3$

2. Define the symbol used for the population mean

where $\mu$ is the population mean time taken by male students in GJC to complete a 2.4 km run.

1. State the necessary assumption for the hypothesis test.

Since the sample size is small ($n = 8 < 30$), the Central Limit Theorem cannot be applied to approximate the distribution of the sample mean. Therefore, we must assume that the population follows a normal distribution.

Assumption: The time taken by a male student in GJC to complete a 2.4 km run follows a normal distribution.

2. Find the critical region for the test in terms of the sample mean, $\bar{x}$.

The test is two-tailed (testing whether the mean time has changed) at the $2.5\%$ level of significance, so the probability in each tail is $\frac{0.025}{2} = 0.0125$.

Under $H_0$, the sample mean $\bar{X} \sim \text{N}\left(11.3, \frac{2.2^2}{8}\right)$.

The critical values $c_1$ and $c_2$ are such that $\mathrm{P}(\bar{X} \le c_1) = 0.0125$ and $\mathrm{P}(\bar{X} \ge c_2) = 0.0125$.
Using the inverse normal function on a graphing calculator:
$c_1 \approx 9.5566$
$c_2 \approx 13.0434$

The critical region is $\bar{x} \le 9.56$ or $\bar{x} \ge 13.0$ (to 3 s.f.).

3. Calculate the sample mean from the given data.

From the given sample of 8 students, we compute the sample mean $\bar{x}$:
$\bar{x} = \frac{11 + 11.5 + 10.8 + 11.2 + 11.4 + 11 + 11.8 + 12.5}{8}$
$\bar{x} = \frac{91.2}{8} = 11.4$

4. State the conclusion of the test in the context of the question.

Since $\bar{x} = 11.4$ does not lie in the critical region ($9.5566 < 11.4 < 13.0434$), we do not reject $H_0$.

There is insufficient evidence at the $2.5\%$ level of significance to conclude that the mean time taken by male students in GJC to complete a 2.4 km run has changed.

1. State the null and alternative hypotheses

Let $\mu$ be the population mean time taken by female students to complete a 2.4 km run.
$H_0: \mu = 14.5$
$H_1: \mu < 14.5$

2. Determine the unbiased estimate of population variance and standard error

The standard deviation of the sample is given as $1.5$, meaning the sample variance is $1.5^2$.
The unbiased estimate of the population variance is:
$s^2 = \frac{n}{n-1}(1.5^2)$
Thus, the variance of the sample mean $\bar{X}$ is $\frac{s^2}{n} = \frac{1.5^2}{n-1}$, and the standard error of the mean is $\frac{1.5}{\sqrt{n-1}}$.

3. Set up the inequality for the test statistic using the critical value

Under $H_0$, since $n$ is large, the Central Limit Theorem applies and the test statistic is $Z = \frac{\bar{X} - \mu}{\frac{1.5}{\sqrt{n-1}}} \sim N(0, 1)$.
For a left-tailed test at the 3% level of significance, the critical value is $z = \text{invNorm}(0.03) \approx -1.8808$.
Since $H_0$ is rejected (the department concludes $\mu < 14.5$), the calculated test statistic must fall within the critical region:
$\frac{14.2 - 14.5}{\frac{1.5}{\sqrt{n-1}}} < -1.8808$

4. Solve the inequality for $n$

$\frac{-0.3}{1.5} \sqrt{n-1} < -1.8808$
$-0.2 \sqrt{n-1} < -1.8808$
$\sqrt{n-1} > 9.404$
$n - 1 > 88.435$
$n > 89.435$
Since the sample size $n$ must be an integer, $n \ge 90$.