Question

1. In an experiment to compare the tensile strengths of 1 - 6 different types of copper wire, ) = 5 samples of each type were used. The between-samples and within samples estimates of a were computed as MSTR = 2649.3 and MSE - 1169.2, respectively. Use the F test at level 0.05 to test Ho: H1 12 - ... - versus Ha: at least two wi's are unequal Calculate the test statistic (Round your answer to two decimal places.) What can be said about the p-value for the test? O p-value > 0.100 O 0.050 < P-value < 0.100 O 0.010 < P-value < 0.050 O 0.001 < P-value < 0.010 O p-value < 0.001 State the conclusion in the problem context. Reject Ho. The data indicates there is not a difference in the mean tensile strengths. Reject Ho. The data indicates a difference in the mean tensile strengths. O Fail to reject Ho. The data indicates a difference in the mean tensile strengths. Fail to reject Ho. The data indicates there is not a difference in the mean tensile strengths. You may need to use the appropriate table in the Arrendix of Tables to answer this question.

Answer 1

MS TO 2649-3 F =>2.26597 MSE 1169.2 L=0.05 k=6, I= 6,J= 5 N=6x5=30 12) K-1 =) =) 6-1=5 Degrees of freedom 2 =) nak =) 30-6=) 24 P-value 0.0802 0.050 option ® <p-value c 0-100 P-vale (0.08) Accept (0.05) Ho ie Fail to Reject to Ophon Fail to reject Ho. The data Judicates difference in the mean There is not a tensile strengthe CS Scanned with CamScanner