Question

An experiment was carried out to investigate the effect of species (factor A, with I = 4) and grade (factor B, with J = 3) on breaking strength of wood specimens. One observation was made for each species—grade combination—resulting in SSA = 445.0, SSB = 429.6, and SSE = 124.4. Assume that an additive model is appropriate.

(a) Test H0: α1 = α2 = α3 = α4 = 0 (no differences in true average strength due to species) versus Ha: at least one αi ≠ 0 using a level 0.05 test.

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 H0. The data does not suggest any difference in the true average strength due to species.

Fail to reject H0. The data suggests that true average strength of at least one of the species is different from the others.

Fail to reject H0. The data does not suggest any difference in the true average strength due to species.

Reject H0. The data suggests that true average strength of at least one of the species is different from the others.

(b) Test H0: β1 = β2 = β3 = 0 (no differences in true average strength due to grade) versus Ha: at least one βj ≠ 0 using a level 0.05 test. 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 H0. The data does not suggest any difference in the true average strength due to grade.

Reject H0. The data suggests that true average strength of at least one of the grades is different from the others.

Fail to reject H0. The data suggests that true average strength of at least one of the grades is different from the others.

Fail to reject H0. The data does not suggest any difference in the true average strength due to grade.

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

SOLUTION::::::

Solutfon: - DFA = 4-1 = 3, DFB = 3-1 = 2 DET = 4*3-1 = 11 DFE = 11 - 3-2 = 6 SSA = 445.0, SSB = 429.6, SSE -124.4. @ MSA t-MSA – SSA / DFA. MSE SSE | DFE 445.0/3 124,416 = 148.33 20.73 = 7.155 ** p-value = pl f316 > 7.155) = 0.0229 0.010 < P-value <0.050

* Reject to. The data Suggests that lie toue avecrage Strength of at least one of the Species is different from the othens [ option (iu)] MSE o f MSB SSB / DFB SSE / DFE 499.6/2 124.4 16 214.8 20.73 = 10.36179 f = MSB ~ 10.36 5 - MSE ** R Value = p (faro > 10.36179) = 0.0125 0.010 < P- value <0,0501 ** Reĝect to. The data suggests that true average Strength of at least one of the grades as different from the others [option (;)]