SOLUTION: 1)In a recent random audit of 300 technical reports, 86 had at least 1 major error. With 99% confidence, what proportion of all reports have no major errors?
2)Suppose in a rando
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Question 541925: 1)In a recent random audit of 300 technical reports, 86 had at least 1 major error. With 99% confidence, what proportion of all reports have no major errors?
2)Suppose in a random sample of 120 sets of tires, we observe an average stopping distnace from a speed of 60mph of 250 ft with a standard dev. of 20 feet. If we use the sample standard dev. as an estimate of the true standard dev. what should we estimate as out average stopping distance for this brand of tire at 98% confidence?
What formula do I use?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1)In a recent random audit of 300 technical reports, 86 had at least 1 major error. With 99% confidence, what proportion of all reports have no major errors?
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P(at least one error) = 1 - P(no errors)
(86/300) = 1 - P(none)
p-hat = P(none) = 0.713
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ME = 2.5758*sqrt[0.713*0.287/300] = 0.067
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99% CI: 0.713-0.067 < p < 0.713+0.067
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2)Suppose in a random sample of 120 sets of tires, we observe an average stopping distnace from a speed of 60mph of 250 ft with a standard dev. of 20 feet. If we use the sample standard dev. as an estimate of the true standard dev. what should we estimate as our average stopping distance for this brand of tire at 98% confidence?
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Same process as with # 1 above
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Cheers,
Stan H.
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