Hypothesis Testing: Sample size; conclusion — NJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
The production manager of a food container manufacturing company wishes to take a random sample of a certain type of lunch box from the thousands produced one day at his factory, for quality control purposes. He wishes to check that the mean height of the lunch boxes is 12.8 cm.
- State what it means for a sample to be random in this context.
- Explain why the manager should take a sample of at least 30 lunch boxes to carry out a hypothesis test.
The heights, \(x\) cm, of a random sample of 48 lunch boxes are summarised as follows: \[ \textstyle\sum(x-13) = -3.6 \qquad \text{and} \qquad \sum(x-13)^2 = 12.02. \]
- Calculate the unbiased estimates of the population mean and variance of the height of the lunch boxes.
The production manager claims that the mean height of the lunch boxes is 12.8 cm.
- Test his claim at the 10% level of significance, stating the hypotheses clearly.
In the hope of keeping the mean height of the lunch boxes not to exceed 12.8 cm, the production manager decides to replace the production line such that the population variance is reduced to 0.04 cm\(^2\). He then carries out a hypothesis test at 2% level of significance using another random sample of \(n\) lunch boxes.
- Assuming that \(n\) is sufficiently large, find the critical region for this test in terms of \(n\).
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(i)
(ii)
(iii)
(iv)
Let the population mean height of the lunch boxes be \(\mu\) cm.
Test \(H_0: \mu = 12.8\) against \(H_1: \mu \neq 12.8\) at \(10\%\) level.
Under \(H_0\), since \(n=48>30\) is large, by CLT: \[ \bar{X} \sim \mathrm{N}\!\left(12.8,\; \frac{0.25}{48}\right)\text{ approximately}. \]
By GC, \(p\)-value \(= 0.0833 < 0.10\). (Or \(|z_{\text{stat}}| = 1.732 > 1.645 = z_{\text{crit}}\).)
Reject \(H_0\): there is sufficient evidence at the \(10\%\) level that the population mean height is not 12.8 cm.
(v)
Test \(H_0: \mu = 12.8\) against \(H_1: \mu > 12.8\) at \(2\%\) level.
Under \(H_0\): \(\bar{X}\sim\mathrm{N}\!\left(12.8,\,\dfrac{0.04}{n}\right)\) approximately by CLT.
Let the critical region be \(\bar{x} \geq a\): \[\begin{aligned} \mathrm{P}\!\left(Z \geq \frac{a-12.8}{0.2/\sqrt{n}}\right) &= 0.02\\[4pt] \frac{(a-12.8)\sqrt{n}}{0.2} &= 2.0537489\\[4pt] a &= 12.8 + \frac{0.411}{\sqrt{n}} \end{aligned}\]