Question

In an experiment to compare the tensile strengths of I = 6 different types of copper wire, J = 5 samples of each type were used. The between-samples and within-samples estimates of σ2 were computed as MSTr = 2678.3 and MSE = 1188.2, respectively. Use the F test at level 0.05 to test H0: μ1 = μ2 = . . . = μ6 versus Ha: at least two μi's are unequal.

1. Calculate the test statistic. (Round your answer to two decimal places.) f =

2. What can be said about the P-value for the test? Select an answer.

A) P-value > 0.100

B) 0.050 < P-value < 0.100

C) 0.010 < P-value < 0.050

D) 0.001 < P-value < 0.010

E) P-value < 0.001

State the conclusion in the problem context. Select an answer

A) Reject H0. The data indicates there is not a difference in the mean tensile strengths.

B)Reject H0. The data indicates a difference in the mean tensile strengths.

C)Fail to reject H0. The data indicates a difference in the mean tensile strengths.

D) Fail to reject H0. The data indicates there is not a difference in the mean tensile strengths.

Answer 1

Source	df	SS	MS	F
between	5	2678.3	535.66	10.82
within	24	1188.2	49.51
total	29	3866.5

1)

test statistic =10.82

2)

E) P-value < 0.001

3)

B)Reject H0. The data indicates a difference in the mean tensile strengths.