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 = 2623.3 and MSE = 1193.2, respectively. Use the F test at level 0.05 to test Ho: μ₁ = M2 μ6 versus Ha: 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? P-value>0.100 0.050 P-value < 0.100 0.010 P-value < 0.050 0.001 P-value < 0.010 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. Fail to reject Ho. The data indicates a difference in the mean tensile strengths. 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.
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 = 2623.3 and MSE = 1193.2, respectively. Use the F test at level 0.05 to test Ho: μ₁ = M2 μ6 versus Ha: 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? P-value>0.100 0.050 P-value < 0.100 0.010 P-value < 0.050 0.001 P-value < 0.010 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. Fail to reject Ho. The data indicates a difference in the mean tensile strengths. 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.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.1: Measures Of Center
Problem 9PPS
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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 σ² were computed as MSTr = 2623.3
and MSE = 1193.2, respectively. Use the F test at level 0.05 to test Ho: μ₁ = M2 μ6 versus Ha: 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?
P-value>0.100
0.050 P-value < 0.100
0.010 P-value < 0.050
0.001 P-value < 0.010
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.
Fail to reject Ho. The data indicates a difference in the mean tensile strengths.
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.
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