Ugam Sheth Posted April 23, 2014 Report Share Posted April 23, 2014 (edited) This is the question:A factory makes wine glasses. The manager claims that on average 2% of the glasses are imperfect. A random sample of 200 glasses is taken and 8 of these are found to be imperfect. Test the manager’s claim at a 1% level of significance using a one-tailed test.I did this question and I got 0.029160179 as the p-value. However, the mark-scheme says the answer is 0.0493 (using binomial distribution) or 0.0511 (using Poisson distribution). Is my answer wrong? or is the mark scheme wrong?Could you perhaps show the method you used to work out the answer?Thanks in advance! Edited April 23, 2014 by Ugam Sheth Reply Link to post Share on other sites More sharing options...
ctrls Posted April 23, 2014 Report Share Posted April 23, 2014 I'm not 100% certain about my method, but my answer agrees with the mark scheme. Here's my working,I set my null and alternate hypotheses as H0: p=0.02, H1: p>0.02, where p is the probability of a glass being imperfect. Note that my alternative hypothesis is that the probability is larger as opposed to smaller, because 4% of the glasses from the sample were imperfect (I would presume the question mentioned this to make it clear, but a one tailed test could also test whether the failure rate is smaller).Assuming H0, the number of imperfect glasses from the sample can be modelled using a binomial distribution, X~B(200,0.02). Since it is a one tailed test, I'm looking at the probability of eight or more imperfect glasses, which is equal to P(X>=8)=0.0493, as required. I'm guessing here that you only calculated P(X=8) here, since that gives the value you found.On a side note, I think the mark scheme approximates the binomial using a possion for the second answer of 0.0511, but I'm not sure why they included that since it's not part of the syllabus. It's probably best to ignore the second method. Reply Link to post Share on other sites More sharing options...
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