2024 JPJC P2 Q10

2024 JPJC P2 Q10

12 marks
Prelims

In a drinks factory, a machine is programmed to dispense \(500\) ml of green tea into empty bottles.

  1. After a routine check of the machine, the production manager suspects that the machine is dispensing more green tea than expected. A random sample of \(50\) bottles is taken for a hypothesis test at \(\alpha \%\) level of significance and the data is as follows:
    Taking \(\alpha =2\), test the production manager’s suspicion and state the meaning of the \(p\)-value in the context of the question.[5]

    Volume of green tea (correct to nearest ml)

    \(498\)

    \(499\)

    \(500\)

    \(501\)

    \(502\)

    \(503\)

    \(504\)

    Number of bottles

    \(5\)

    \(11\)

    \(10\)

    \(11\)

    \(10\)

    \(1\)

    \(2\)

    1. Determine the set of values of \(\alpha \) for which the production manager’s suspicion is valid.[1]
    2. Explain why the production manager is able to carry out the hypothesis test without knowing anything about the distribution of the volume of green tea dispensed by the machine.[1]
  2. The machine is being recalibrated to dispense \(500\) ml of green tea. Another random sample of \(50\) bottles of green tea is taken and the mean and standard deviation of this sample are \(502\) ml and \(k\) ml respectively. A hypothesis test, at the \(5\%\) significance level, concluded that the recalibration is done accurately. Find the set of values that \(k\) can take.[5]

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Answer:(a) From GC, \({p\text{-value}}\) is \({0.0259}\). (3 s.f.) Since \({p\text{-value} = 0.0259 > 0.02}\), we do not reject \({H_0}\) at \({2\%}\) level of significance and conclude that there is insufficient evidence that the production manager's suspicion is valid. The \({p\text{-value}}\) is \({0.0259}\) and it means there is a probability of \({0.0259}\) of observing a test statistic, \({z \geq 1.95}\), given that the population mean green tea dispensed in a bottle is \({500\text{ ml}}\). (i) \({2.59 < \alpha < 100}\) (3 s.f.) (ii) Since the sample size \({= 50}\) is large enough, Central Limit Theorem can be applied for sample means to follow a normal distribution approximately. (b) \({k > 7.14}\) (3 s.f.)

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