In an experiment to compare the tensile strengths of I = 5 different types of copper wire, J = 4 samples of each type were used. The between-samples and within-samples estimates of o² were computed as MSTR = 2676.3 andMSE = 1107.2, respectively. Use the F test at level 0.05 to test Ho: M₁ = ₂ = ... = μ5 versus H₂: at least two μ'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> 0.100O 0.050 < P-value < 0.100O 0.010 < P-value < 0.050O 0.001 < P-value < 0.010O P-value < 0.001State the conclusion in the problem context.O Reject Ho. The data indicates there is not 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 Fail to reject Ho. The data indicates a difference in the mean tensile strengths.O Reject Ho. The data indicates a difference in the mean tensile strengths.

given mean square of treatments(MST)=2676.3mean square of error(MSE)=1107.2level of significance…

Answered: In an experiment to compare the tensile…

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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