Hypothesis Testing: Multiple samples; range — RI 2025 H2 Math Prelim Paper 2
What this question tests
Question
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 physical education programme, the PE department is determining whether the mean time taken by male students in GJC to complete a 2.4 km run has changed. The time takes, \(x\) minutes, a random sample of 8 Year 5 male students in GJC to complete a 2.4 km run are as follows.
\[ 11 \quad 11.5 \quad 10.8 \quad 11.2 \quad 11.4 \quad 11.8 \quad 11.9 \quad 12.5 \]Given that the PE department concludes that the mean time for female students to complete a 2.4 km run is less than 14.5 minutes, find the range of values of \(n\) that \(n\) can take. Give two reasons why you would expect this estimate to be reliable.
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Population variance \(\sigma^2 = 2.2^2 = 4.84\) is known.
Necessary assumption: Since \(n=8\) is small, assume the time \(X\) follows a normal distribution.
Under \(H_0\): \(\bar{X}\sim N\!\left(11.3,\,\dfrac{4.84}{8}\right)\).
For a two-tailed test at 2.5% significance, from GC: \[ \text{Critical region: } \bar{x} < 9.5566 \quad\text{or}\quad \bar{x} > 13.043 \] i.e. \((0,\, 9.56]\cup[13.0,\,\infty)\).
Since \(\bar{x}=11.4\) does not lie in the critical region, we do not reject \(H_0\). There is insufficient evidence at the 2.5% level to conclude that the mean time has changed.
Let \(Y\) be the time taken in minutes by a female student in GJC to complete a 2.4 km run, and let \(\mu_y\) and \(\bar{Y}\) be the population mean time and sample mean time respectively.
From the sample, \(\bar{y} = 14.2\), \(s_y^2 = \dfrac{n}{n-1}\!\left(1.5^2\right)\)
To test \(\mathrm{H}_0: \mu_y = 14.5\) vs \(\mathrm{H}_1: \mu_y < 14.5\)
Under \(\mathrm{H}_0\), since \(n\) is large, \[ \bar{Y} \sim \mathrm{N}\!\left(14.5,\; \frac{\dfrac{n}{n-1}\!\left(1.5^2\right)}{n}\right) \quad\text{i.e.}\quad \bar{Y} \sim \mathrm{N}\!\left(14.5,\;\frac{1.5^2}{n-1}\right) \] approximately by Central Limit Theorem.
For \(\mathrm{H}_0\) to be rejected,
\[ p\text{-value} = \mathrm{P}\!\left(\bar{Y} \le 14.2\right) \le 0.03 \] \[\begin{aligned} \mathrm{P}\!\left(Z \le \frac{14.2 - 14.5}{\sqrt{\dfrac{1.5^2}{n-1}}}\right) &\le 0.03 \\[8pt] \frac{-0.3\sqrt{n-1}}{1.5} &\le -1.8808 \\[6pt] \sqrt{n-1} &\ge 9.404 \\[4pt] n &\ge 89.435 \\[4pt] \therefore n &\ge 90, \quad \text{where } n \in \mathbb{Z}^+ \end{aligned}\]