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 o2 were computed as MSTr = 2654.3 and MSE = 1154.2, respectively. Use the F test at level 0.05 to test Hoi H = H, = ... = Hg versus H: at least two u,'s are unequal. You can use the Distribution Calculators page in SALT to find critical values and/or p-values to answer parts of this question. Calculate the test statistic. (Round your answer to two decimal places.) f = 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. O Reject Ha. 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. O Fail to reject Ho. The data indicates there is not a difference in the mean tensile strengths. O Reject H. The data indicates there is not a difference in the mean tensile strengths.
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 o2 were computed as MSTr = 2654.3 and MSE = 1154.2, respectively. Use the F test at level 0.05 to test Hoi H = H, = ... = Hg versus H: at least two u,'s are unequal. You can use the Distribution Calculators page in SALT to find critical values and/or p-values to answer parts of this question. Calculate the test statistic. (Round your answer to two decimal places.) f = 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. O Reject Ha. 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. O Fail to reject Ho. The data indicates there is not a difference in the mean tensile strengths. O Reject H. The data indicates there is not a difference in the mean tensile strengths.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text: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 o? were computed as MSTr = 2654.3 and
MSE = 1154.2, respectively. Use the F test at level 0.05 to test Ho: 4, = µ, = ... = lg versus H: at least two u's are unequal.
You can use the Distribution Calculators page in SALT to find critical values and/or p-values to answer parts of this question.
Calculate the test statistic. (Round your answer to two decimal places.)
f =
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.
O Reject H. 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.
O Fail to reject Ho: The data indicates there is not a difference in the mean tensile strengths.
O Reject Ho. The data indicates there is not a difference in the mean tensile strengths.
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