Question 181119This question is from textbook
: 4. In tests of a computer component, it is found that the mean time between failures is 530 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. Determine the mean and standard deviation for the data set. (5 points)
b. At the 0.01 level of significance, test the claim that for the modified components, the mean time between failures is greater than 530 hours. You may assume the time between failures is normally distributed. (5 points)
This question is from textbook
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 530 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
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a. Determine the mean and standard deviation for the data set. (5 points)
mean: add the numbers and divide by 10. = 537.1
stan. dev.: s = 20.7013...
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b. At the 0.01 level of significance, test the claim that for the modified components, the mean time between failures is greater than 530 hours. You may assume the time between failures is normally distributed. (5 points)
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Ho: u = 530
Ha: u > 530
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With a right-tail test with alpha = 1% the critical value is z = 2.326.
Test statistic: z(537.1) = (537.1-530)/[20.7013/sqrt(10)] = 1.0845
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Conclusion: Since t.s. < 2.326, Fail to reject Ho.
The test does not support reject the belief that u = 530
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Cheers,
Stan H.
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