Hypothesis Testing: Multi-stage hypothesis test — TJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
At a school canteen, the average queuing time (in minutes) for a student to be served at a stall is known to be 4.5 minutes. The school operation manager claims that with a recent change in students' timetable, there is an improvement in the average queuing time.
A random sample of 10 students' queuing time is observed after the change in students' timetable and the data is shown below. \[4.2,\quad 5.2,\quad 3.1,\quad 3.5,\quad 4.6,\quad 4.8,\quad 3.8,\quad 3.7,\quad 4.5,\quad 5.0\]
Given that the population variance of queuing time is known to be \(0.48\ \text{min}^2\), a one-tailed test is carried out to determine whether the manager's claim is justified.
- State an assumption needed to carry out the hypothesis test.
- Test whether the manager's claim is justified at 5% level of significance.
A student leader Jason suspected that with the change in students' timetable, the average queuing time for a student to be served at a stall is in fact more than \(\lambda\) minutes and also that the population variance of queuing time is no longer \(0.48\ \text{min}^2\). To investigate this, Jason recorded the queuing time of the first 50 students who visited the stall from 12 pm to 1 pm.
- Give a reason why the sample collected is not a random sample.
Jason decided to collect another set of data to do his investigation. He recorded the queuing time, \(x\), in minutes, of a random sample of 50 students. The summarised data is as shown below. \[ \bar{x} = 4.6 \qquad \text{and} \qquad \sum x^2 = 1068 \]
- Find an unbiased estimate of the population variance.
- Explain why the assumption in (a) is no longer necessary.
- By finding the critical region in terms of \(z\)-values, find the range of values of \(\lambda\) such that Jason's suspicion is justified.
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(a) The queuing times follow a normal distribution (population).
(b) Let \(\mu\) be the population mean queuing time and \(X\) the queuing time of a randomly chosen student.
\(\mathrm{H_0}: \mu = 4.5\) \(\mathrm{H_1}: \mu < 4.5\) (manager's claim: improvement \(=\) decrease)
Using GC: \(\bar{x} = 4.24\) (exact).
At 5% significance level. Under \(\mathrm{H_0}\), \(\bar{X}\sim\mathrm{N}\!\left(4.5,\;\dfrac{0.48}{10}\right)\).
Test statistic: \(Z = \dfrac{\bar{X}-4.5}{\sqrt{0.48/10}}\sim\mathrm{N}(0,1)\).
From GC: \(p\)-value \(= 0.118 > 0.05\).
Therefore we do not reject \(\mathrm{H_0}\). There is insufficient evidence at the 5% level of significance to conclude that the manager's claim is justified.
(c) The sample is not random because the first 50 students who visited the stall from 12 pm to 1 pm were chosen. Not all students from the school were given an equal chance of being selected (e.g. only students who happened to visit during that specific hour were included).
(d) Unbiased estimate of population variance: \[\begin{aligned} s^2 &= \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) \\[6pt] &= \frac{1}{49}\left(1068 - \frac{(4.6 \times 50)^2}{50}\right) \\[6pt] &= 0.20408 \approx 0.204 \end{aligned}\]
(e) The assumption in (a) (that the queuing times follow a normal distribution) is no longer necessary because the sample size \(n=50\) is large. By the Central Limit Theorem, the sample mean queuing time \(\bar{X}\) follows a normal distribution approximately, regardless of the underlying population distribution.
(f) \(\mathrm{H_0}: \mu = \lambda\) \(\mathrm{H_1}: \mu > \lambda\) (Jason's suspicion)
At 2% significance level. Under \(\mathrm{H_0}\), \(\bar{X}\sim\mathrm{N}\!\left(\lambda,\;\dfrac{0.20408}{50}\right)\) approximately (CLT).
Test statistic: \(Z = \dfrac{\bar{X}-\lambda}{\sqrt{0.20408/50}}\sim\mathrm{N}(0,1)\) approximately.
Critical region: \(\{z\in\mathbb{R}: z\geq 2.05375\}\) (upper tail test at 2%).
For Jason's suspicion to be justified, \(\mathrm{H_0}\) must be rejected, so the observed test statistic must fall in the critical region: \[ \frac{4.6-\lambda}{\sqrt{0.20408/50}} \geq 2.05375 \] \[ 4.6-\lambda \geq 2.05375\times\sqrt{\frac{0.20408}{50}} = 2.05375\times 0.06389 \approx 0.13127 \] \[ \lambda \leq 4.6 - 0.13127 = 4.4687 \approx 4.47 \]
Since \(\lambda\) is a positive mean queuing time: