Question 1112694: A manufacturing company manufacturing computer chips finds that 8% of all chips
manufactured are defective. Management is concerned that high employee turnover is
partially responsible for the high defect rate. In an effort to decrease the percentage of
defective chips, management decides to provide additional training to those employees
hired within the last year. After training was implemented, a sample of 450 chips
revealed only 27 defects. In order to determine whether the additional training was
effective in lowering the defect rate:
(a) State the null and the alternative hypotheses for this test. (3 marks)
(b) Calculate the value of the test statistic for this test. (2 marks)
(c) Using α = 0.01, identify the critical region for this test. (2 marks)
(d) State the conclusion of this test. Give a reason for your answer. (3 marks)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: proportion of defects after=proportion of defects before.
Ha: proportion of defects after is different from proportion before (2 sided test here is more conservative). One could make it one sided, if one were certain the training might not make things worse.
alpha=0.01 P{reject Ho|Ho true}
Test statistic is a z= (p after-known p)/sqrt (p*(1-p)/n)
Critical value is |z|>2.576
=(0.06-0.08) divided by sqrt (.0 8*.92/450); sqrt term is 0.0128
=-0.02/0.0128
=-1.56, which does not fall in the rejection zone. p-value is 0.11.
Given a defect rate of 8%, 6% or fewer could be found in a sample of 450 11% of the time. That is not 1%, so the result could have been by chance.
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