Question 1066130: Six percent of the computer chips are produced by Cheap Chips are defective. Each month a random sample of 200 chips manufactured is taken. Let X = the number of defective chips in the sample. What are the mean and the standard deviation of X? Find the probability that 8 or fewer chips in a sample of 200 are defective. In an effort to reduce the percentage of defective chips, the company contracts with a new supplier for some important components. In the next month, the sample of 200 chips has only 8 defective ones. Can the company conclude that the new components have reduced the rate of defective chips produced? Explain.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Six percent of the computer chips are produced by Cheap Chips are defective. Each month a random sample of 200 chips manufactured is taken. Let X = the number of defective chips in the sample.
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What are the mean and the standard deviation of X?
mean = n*p = 200*0.06 = 12
std = sqrt(npq) = sqrt(12*0.4) = sqrt(4.8) = 2.19
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Find the probability that 8 or fewer chips in a sample of 200 are defective.
P(x <= 8) = binomcdf(200,0.06,8) = 0.1470
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In an effort to reduce the percentage of defective chips, the company contracts with a new supplier for some important components. In the next month, the sample of 200 chips has only 8 defective ones. Can the company conclude that the new components have reduced the rate of defective chips produced? Explain.
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Ho: p1-p2 = 0 (no change)
Ha: p1-p2 > 0 (rate reduced)
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Sample diff:: 0.06-8/200 = 0.02
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z(0.02) = 0.02/sqrt[[(0.06*0.04/200)+(0.04*0.06/200)] = 0.02/0.0049 = 4.08
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If you check the p-value you will find it is extremely small
Since it is so small you can reject Ho
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Conclusion:: Since the test statistic is in the reject interval,
reject Ho. The new coponents have reduced the rate of defective chips>
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Cheers,
Stan H.
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