Question

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 36) 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 At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours.

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Answer 1

Sample size = n = 10

Sample mean = = 537.1

Standard deviation = s = 20.7013

Claim: The modified components, the mean time between failures is greater than 520 hours.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

Degrees of freedom = n - 1 = 10 - 1 = 9

Critical value = 1.833 ( Using t table)

Test statistic > critical vaue we reject null hypothesis.

Conclusion: the modified components, the mean time between failures is greater than 520 hours.