2025 ACJC P2 Q11

2025 ACJC P2 Q11

Junior College 2
11 marks

In a laboratory, the time in minutes for an inspector to test an electrical circuit board is a continuous random variable \(X\). The standard deviation of \(X\) is \(0.63\) and under ordinary conditions, the expected value of \(X\) is \(5.82\). As a result of the introduction of an incentive scheme, the laboratory supervisor claims that the inspector carries out the test more quickly. It is found that for a random sample of \(150\) tests, the mean time taken to test an electrical circuit board is \(5.74\).

  1. State appropriate hypotheses for the test of the claim, defining any symbols that you use.[2]
  2. Test whether the supervisor's claim is justified at \(5\)% level of significance.[3]
  3. Explain whether there is a need for the laboratory supervisor to know anything about the population distribution of \(X\).[1]

Two years later, a quality control officer took a random sample of \(20\) tests and found that the mean time taken to test an electrical circuit board is \(5.88\). He makes a claim that the mean time taken to test an electrical circuit board differs from \(\mu_0\) mins.

  1. State two assumptions the quality control officer has to make in order to carry out a hypothesis test for this claim.[2]
  2. Given that the officer concludes that there is no reason to reject the null hypothesis at \(1\)% level of significance, find the range of values of \(\mu_0\).[3]
Answer:(a)\(\mu\) represents the population mean time taken to test an electric circuit board. \(H_0 : \mu=5.82, H_1 : \mu<5.82 \) at 5% significance level. (b) Do not reject \(H_0\). (c) No need, since n is large, by Central Limit Theorem \(\bar{x}\) will be approximately normally distributed. (d)The time taken to test an electric circuit board follows a normal distribution. The population standard deviation of the time taken to test an electric circuit board is still 0.63. (e)5.52< \(\mu_0\) <6.24

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