Question

In test of a computer component, it is found that the mean between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 8 modified components resulted in the following times (in hours) between failures. Use a 0.05 significance level to test the claim that the mean time between failures is greater than 520 hours. Assume that the population is normally distributed.

518, 548, 561, 523, 536, 499, 538, 557

a. Compute the mean and standard deviation of the data.

b. State null and alternative hypotheses. Define the parameter.

c. Show the graph and label the p-value in graph.

d. Compare the p-value vs. ?.

e. State the decision and conclusion.

Answer 1

Solution:

x	x2
518	268324
548	300304
561	314721
523	273529
536	287296
499	249001
538	289444
557	310249
∑x=4280	∑x2=2292868

a ) Mean ˉx=∑xn

=518+548+561+523+536+499+538+557/8

=4280/8

=535


Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√2292868-(4280)28/7

=√2292868-2289800/7

=√30687

=√438.2857

=20.9353

b ) This is the right tailed test .

The null and alternative hypothesis is ,

H0 :    =   520

Ha : >  520

Test statistic = t

= ( - ) / S / n

= (535- 520) / 20.93 / 8

= 2.027

Test statistic = t =2.027  

P-value =0.0411

d ) = 0.05  

P-value <

0.0411 < 0.05

e ) Reject the null hypothesis .

There is sufficient evidence to suggest that