Hypothesis Testing: Mean; conclusion change — HCI 2025 H2 Math Prelim Paper 2
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Question
A food producer claims that the mean mass of a can of beans it produces is \(425\) g. Following customer feedback, the production manager wishes to test if the mean mass of a can of beans is indeed \(425\) g.
The production manager took a random sample of size \(50\) and the mass of each can, in \(x\) g, is recorded and the results are shown below: \[ \sum x=21209,\quad \sum (x-424.18)^{2}=522 \]
(a) State what it means for a sample to be random in this context.
(b) Find the unbiased estimates for the population mean and variance.
(c) State the hypotheses for the manager's test, defining any parameters you use. Carry out the test at the \(5\%\) level of significance, giving your conclusion in the context of the question.
The production manager wishes to test whether the mean mass of a can of beans has increased using the alternative manufacturing process. He finds that the mean mass of \(55\) randomly chosen cans is \(426.5\) g. He carries out a hypothesis test at \(10\%\) level of significance.
(d) Explain, with justification, how the population standard deviation of the mass of a can produced under the alternative process will affect the conclusion made by the production manager.
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(a) Every can in the population has an equal chance of being selected and is selected independently of the others.
(b) Unbiased estimate for the population mean: \[ \bar x=\frac{\sum x}{50}=\frac{21209}{50}=424.18. \]
Unbiased estimate for the population variance: \[\begin{aligned} s^{2} &= \frac{50}{50-1}\cdot\frac{\sum(x-\bar x)^{2}}{50} = \frac{\sum(x-424.18)^{2}}{49} = \frac{522}{49}\ \text{ or }\ 10.65306122. \end{aligned}\]
(c) Let \(X\) be the mass (in grams) of a can of beans.
Let \(\mu\) and \(\sigma^{2}\) be the population mean and variance of \(X\).
Hypotheses: \[ H_{0}:\mu=425,\qquad H_{1}:\mu\neq 425. \]
Here \(n=50,\ \bar x=424.18,\ s^{2}=\dfrac{522}{49}.\)
Under \(H_{0}\), since \(n=50\) is large, by the Central Limit Theorem, \[ \bar X\sim\mathrm{N}\!\left(\mu,\frac{\sigma^{2}}{n}\right)\text{ approximately}. \] Test statistic \[ Z=\frac{\bar X-\mu}{\sqrt{S^{2}/n}}\sim\mathrm{N}(0,1)\text{ approximately}. \]
Level of significance: \(5\%\). Reject \(H_{0}\) if \(p\text{-value}\le 0.05\).
Assuming \(H_{0}\) is true, by G.C., \(p\text{-value}=0.0757\) (3 s.f.).
Since \(p\text{-value}=0.0757>0.05\), we do not reject \(H_{0}\) at the \(5\%\) level of significance and conclude that there is insufficient evidence to say that the population mean mass of a can of beans is not \(425\) g.
(d) Let \(Y\) be the mass (in grams) of a can of beans produced using the alternative process.
Let \(\mu\) and \(\sigma^{2}\) be the population mean and variance of \(Y\).
Hypotheses: \(H_{0}:\mu=425,\ H_{1}:\mu>425.\)
\(n=55,\ \bar y=426.5\).
Under \(H_{0}\), since \(n=55\) is large, by the Central Limit Theorem, \(\bar Y\sim\mathrm{N}\!\left(\mu,\dfrac{\sigma^{2}}{n}\right)\) approximately.
Test statistic \(Z=\dfrac{\bar Y-\mu}{\sqrt{\sigma^{2}/n}}\sim\mathrm{N}(0,1)\) approximately.
At the \(10\%\) level of significance, reject \(H_{0}\) if \(z\ge 1.281551567\). So \(H_{0}\) is rejected when \[\begin{aligned} \frac{426.5-425}{\sqrt{\sigma^{2}/55}}&\ge 1.281551567 \\ \text{Since }\sigma\ge 0,\ \sqrt{\sigma^{2}}&\le \frac{1.5\sqrt{55}}{1.281551567} \\ &\Leftrightarrow\quad\sigma \le 8.680335632 \\ &\Leftrightarrow\quad\sigma \le 8.68\ (3\,\text{s.f.}) \\ &\Leftrightarrow\quad 0<\sigma \le 8.68\ (3\,\text{s.f.}). \end{aligned}\]
At the \(10\%\) level of significance, if \(0<\sigma\le 8.68\), then \(H_{0}\) is rejected and the production manager may conclude that mean mass of a can of beans produced using the alternative process has increased.