In tests of a computer component, the mean time between failure has been 519 hou

Question

In tests of a computer component, the mean time between failure has been 519 hours. A modification is made which is supposed to increased the time between failures. Tests on a randow sample of 10 modifies components resulted in the following times (in hours) between failures:

518      548      561     523      536     499      538     557      528        563

Based on these data, the sample mena is calculated to be 537.1 hours and the sample standard deviation is 20.7 hours. At the 0.05 significance level, test the claim that for the modified components, the mena time between failures is greater than 519 hours. Use the critical value method of testing hypotheses.

a) H0:

b) H1:

c) Test statistic:

d) Critical value:

e) Do you reject H0 ?

f) If you were told that the p-value for the test statistics for this hypothesis test is 0.014, would you reach the same decision that you made in (e) and (f)?        Yes                   No

Explanation / Answer

H0: u <= 519

H1: u > 519

Test statistics

t = (537.1 - 519)/(20.7/sqrt(10))
t = 2.7651

Here Degrees of freedom = 9
critical value = 1.833 (using t-table for 0.05 significance level)

As t-value is greater than critical value, we reject the null hypothesis.

If p value is 0.014, which is less than the significance level of 0.05, we reject the null hypothesis. This means decision made will be same.

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