Question 1184863: Use a 0.05 significance level and conduct a full hypothesis test. List the null and alternative hypotheses, the z-score, the P-value, and whether to reject or not reject the null hypothesis. Round z-scores to the nearest tenth.
In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 36 modified components produce a mean time between failures of 960 hours, with a standard deviation of 52 hours. Test the claim that for the modified components, the mean time between failures is greater than 937 hours.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! assumed mean = 937
sample size = 36
sample mean = 960
sample standard deviation = 52
significance level = .05
test is for greater than.
one sides critical z-score = z-score with .05 area under the normal distribution to the right of it.
that z-score = 1.645.
test z-score = (x - m) / s
x is the raw score being compared to the assumed mean
m is the assumed mean
s is the standard error.
standard error = standard deviation / square root of sample size = 52/sqrt(36) = 8.67.
test z-score = (960 - 937) / 8.67 = 23/8.67 = 2.65.
2.65 is greater than the critical z-score of 1.645, therefore the test is significant and the conclusion is that the mean time between failures can be assumed to be greater than 937.
if you were to test at a higher significance level, like .01, the critical z-score would have been 2.33.
the test would still have been significant and the conclusion would have been the same.
if you're are working off of p-value, the critical p-value is .05 and the test p-value, with a z-score of 2.65 is /.004.
since .004 is less than .05, the test is significant.
the critical z-score and the critical p-value go hand in hand.
if your test z-score is greater than the critical z-score, then your test p-value will also be less than the critical p-value.
you can use either criteria and you will get the same result.
the critical p-value and the test p-value, in this case, was the area to the right of the critical z-score and the test z-score, respectively.
some additional information:
since you are using the sample standard deviation, then the use of the t-score, rather than the z-score, is indicated.
in this case, it didn't matter.
the results were the same anyway.
the t-score would have been done with 35 degrees of freedom (1 less than the sample size).
the critical t-score would have been 1.69.
the test t-score would have still been 2.65.
since the test t-score was higher than the critical t-score, the results were still significant.
here's a reference on when to use the t-score versus when to use the z-score.
https://math.stackexchange.com/questions/1817980/how-to-know-when-to-use-t-value-or-z-value
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