Question

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. Use the P-value method of testing hypotheses.
H0:
H1:
Test Statistic:
Critical Value:
Do you reject H0?
Conclusion:
If you were told that the p-value for the test statistic for this hypothesis test is 0.014, would you reach the same decision that you made for the Rejection of H0 and the conclusion as above?

Answer 1

Answer:

As the calculated value of t is greater than critical value reject H0. The tests supports the claim at ∝= 0.05

If the p-value for the test statistic for this hypothesis test is 0.014, then the critical region is t ( with df=9) for a right tailed test is 2.821

then we would accept H0. The test would not support the claim at ∝= 0.01

Step-by-step explanation:

Mean x`= 518 +548 +561 +523 + 536 + 499+  538 + 557+ 528 +563 /10

x`= 537.1

The Variance is  = 20.70

H0 μ≤ 520

Ha μ > 520

Significance level is set at ∝= 0.05

The critical region is t ( with df=9) for a right tailed test is 1.8331

The test statistic under H0 is

t=x`- x/ s/ √n

Which has t distribution with n-1 degrees of freedom which is equal to 9

t=x`- x/ s/ √n

t = 537.1- 520 / 20.7 / √10

t= 17.1 / 20.7/ 3.16227

t= 17.1/ 6.5459

t= 2.6122

As the calculated value of t is greater than critical value reject H0. The tests supports the claim at ∝= 0.05

If the p-value for the test statistic for this hypothesis test is 0.014, then the critical region is t ( with df=9) for a right tailed test is 2.821

then we would accept H0. The test would not support the claim at ∝= 0.01