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2020 TJC P2 Q11

2020 TJC P2 Q11

12 marks
Prelims

It is found that certain kinds of meat lose weight as a result of being cooked. A restaurant chef is prepared to accept up to \(10\%\) loss but suspects that the recent consignments have a higher percentage weight loss. She decides to carry out a hypothesis test on a random sample of steaks.

  1. Explain the meaning of ‘a random sample’ in the context of the question and why the chef should sample a large number of steaks.[2]
  2. State suitable hypotheses for the test, defining any symbols that you use.[2]

The chef takes a random sample of \(40\) steaks, and calculate the percentage weight loss of each steak, \(x\) (in percent). The mean and variance of the percentage weight loss of the \(40\) steaks are \(10.48\%\) with variance \(3.37\%\) respectively.

  1. Test, at the \(5\%\) significance level, to determine whether the sample supports the chef’s suspicion.[3]

The chef carries out another test using \(100\) readings obtained from chefs of other restaurants. The percentage weight loss of each steak, \(y\) (in percent), is summarised by:

\(\sum{y=\mathrm{1055}\mathrm{.2}}\), \(\sum{{{y}^{2}}=k}\).

  1. Find the range of values of \(k\) such that the chef’s suspicion is not valid at the \(5\%\) level of significance, giving your answer correct to \(2\) decimal places.[5]

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Answer:(i) A random sample means that each of the steaks is selected independently from each other and each steak has an equal chance of being selected. Since the distribution of the percentage weight loss of the steak after cooking (population) is not known, the chef should take a sample of at least \({30}\) steaks so that the distribution of the sample mean of percentage weight loss of steak after cooking can be approximated by a normal distribution using the Central Limit Theorem. (ii) Let \({X}\) (in percent) be the percentage weight loss as a result of cooking and \({\mu \%}\) be the population mean percentage weight loss of steaks as a result of being cooked, \({H_0 : \mu = 10 \text{;}\hspace{0.5em} H_1 : \mu > 10}\) (iii) From GC, \({p\text{-value} = 0.051235 > 0.05}\). Since \({p\text{-value} >\text{ significance value}}\), we do not reject \({H_0}\). There is insufficient evidence at \({5\%}\) level of significance to conclude that the sample confirms the chief's suspicion. (iv) \({\mathrm{k} > 12249.43}\)

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