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 an MSE = 1107.2, respectively. Use the F test at level 0.05 to test Ho: #₁ = #₂ == #5 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 > 0.100 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 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.

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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 and
MSE = 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> 0.100
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 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.
Transcribed Image Text: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 and MSE = 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> 0.100 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 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.
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