document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #815646 by Theo(13342)\"\" \"About 
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assumed mean = 937
\n" ); document.write( "sample size = 36
\n" ); document.write( "sample mean = 960
\n" ); document.write( "sample standard deviation = 52
\n" ); document.write( "significance level = .05
\n" ); document.write( "test is for greater than.
\n" ); document.write( "one sides critical z-score = z-score with .05 area under the normal distribution to the right of it.
\n" ); document.write( "that z-score = 1.645.
\n" ); document.write( "test z-score = (x - m) / s
\n" ); document.write( "x is the raw score being compared to the assumed mean
\n" ); document.write( "m is the assumed mean
\n" ); document.write( "s is the standard error.
\n" ); document.write( "standard error = standard deviation / square root of sample size = 52/sqrt(36) = 8.67.
\n" ); document.write( "test z-score = (960 - 937) / 8.67 = 23/8.67 = 2.65.
\n" ); document.write( "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.
\n" ); document.write( "if you were to test at a higher significance level, like .01, the critical z-score would have been 2.33.
\n" ); document.write( "the test would still have been significant and the conclusion would have been the same.
\n" ); document.write( "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.
\n" ); document.write( "since .004 is less than .05, the test is significant.
\n" ); document.write( "the critical z-score and the critical p-value go hand in hand.
\n" ); document.write( "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.
\n" ); document.write( "you can use either criteria and you will get the same result.
\n" ); document.write( "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.
\n" ); document.write( "some additional information:
\n" ); document.write( "since you are using the sample standard deviation, then the use of the t-score, rather than the z-score, is indicated.
\n" ); document.write( "in this case, it didn't matter.
\n" ); document.write( "the results were the same anyway.
\n" ); document.write( "the t-score would have been done with 35 degrees of freedom (1 less than the sample size).
\n" ); document.write( "the critical t-score would have been 1.69.
\n" ); document.write( "the test t-score would have still been 2.65.
\n" ); document.write( "since the test t-score was higher than the critical t-score, the results were still significant.\r
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\n" ); document.write( "\n" ); document.write( "here's a reference on when to use the t-score versus when to use the z-score.
\n" ); document.write( "https://math.stackexchange.com/questions/1817980/how-to-know-when-to-use-t-value-or-z-value\r
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