Question 269147: Please help answer the following question. Thank you.
In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures.
518 548 561 523 536
499 538 557 528 563
a. Find x-bar and s for this data set.
b. At the 5% significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours? You may assume the times between failures are normally distributed. Use the traditional method to make your decision.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures.
518 548 561 523 536
499 538 557 528 563
a. Find x-bar = 537.1
s = 20.70 for this data set.
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b. At the 5% significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours? You may assume the times between failures are normally distributed. Use the traditional method to make your decision.
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Ho: u <= 520
H1: u > 520 (claim)
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Cv for right-tail test with alpha = 5% and df = 9 is invT(0.05,9) = 1.8331
Ts = t(537.1) = (537.1-520)/[20.70/sqrt(10)] = 2.6123
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Conclusion: Since ts is greater than cv, reject Ho.
The test supports the claim at the 5% significance level.
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cheers,
Stan H.
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