11-26-2005, 06:47 PM
This is the problem and I'm at a loss at what to do. I've tried a few things but feel like it's not correct. Can anyone help? Thanks!
A manufacturer finds that in a random sample of 100 of its CD players, 96 have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to axaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition?
His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed.
11-27-2005, 04:32 PM
Here is an inelegant solution.
Treat the problem as a hypothesis test. Start with an arbitrary hypothesis. For example, assume that the true proportion of non-defective units is P. Then, compute the region of acceptance. If the sample result (p = .96) falls within the region of acceptance, increase P by a small amount. If it falls outside the region of acceptance, reduce P by a small amount. Repeat this procedure until you find the largest value of P for which the sample result lies within the region of acceptance.
One approach would be to use a normal approximation to the binomial. I tried this (it took several iterations), and found that the region of acceptance is 0.96 to 1.00 when the true proportion of non-defective units is assumed to be 0.982. Therefore, the manufacturer could honestly claim that 98% of his units are non-defective, assuming an 0.05 level of significance. (You could do more iterations and carry the answer out to more decimal places.)
You can find an explanation of how to test a hypothesis about a population proportion at http://stattrek.com/Lesson5/Proportion.aspx . (Truth in advertising: I may not be entirely objective about the StatTrek.com web site, since I am its web master. But you can judge for yourself whether it is helpful.)
A better approach might be to use the binomial distribution, rather than the normal approximation to the binomial. The question you would ask is: What would the true population proportion be if the probability of getting 96 or fewer non-defective units was 0.05? You can use a binomial calculator to answer this question. You can find an online binomial calculator at http://stattrek.com/Tables/Binomial.aspx . That calculator shows that the probability of getting 96 or fewer non-defectives is 0.05, assuming that true proportion of non-defective units is 0.9862, the sample size is 100, and the population size is much larger than 100. Using the binomial distribution, the manufacturer could honestly claim that 98.6 percent of his units are non-defective, based on the sample results.
The result with the normal approximation to the binomial is very close to the result with the binomial.
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