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 = 444.0, SSB = 424.6, and SSE = 122.4. 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 = 1 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 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 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. (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 = 4 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. Fail to 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.

Answer 1

Total Size, N = 4*3 = 12

Level of Factor A, a = 4

Level of Factor B, b = 3

df(A) = a-1 = 3

df(B) = b-1 = 2

df(error) = (a-1)*(b-1) = 6

df(total) = N-1 = 11

SSA = 444

SSB = 424.6

SSE = 122.4

SS(total) = SSA + SSB + SSE = 991

MSA = SSA/df(A) = 148

MSB = SSB/df(B) = 212.3

MSE = SSE/df(error) = 20.4

F for factor A = MSA/MSE = 7.2549

p-value = F.DIST.RT(7.2549, 3, 6) = 0.0202

F for Factor B = MSB/MSE = 10.40686

p-value = F.DIST.RT(10.4069, 2, 6) = 0.0112

ANOVA
Source of Variation	SS	df	MS	F	P-value
Factor A	444	3	148	7.2549	0.0202
Factor B	424.6	2	212.3	10.4069	0.0112
Error	122.4	6	20.4
Total	991	11

(a) Test H0: α1 = α2 = α3 = α4 = 0 versus Ha: at least one αi ≠ 0 using a level 0.05 test.

Test statistic.

F = 7.25

P-value for the test = 0.0202

0.010 < P-value < 0.050

Conclusion in the problem context:

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 versus Ha: at least one βj ≠ 0 using a level 0.05 test.

Test statistic:

F = 10.41

P-value for the test:

0.010 < P-value < 0.050

Conclusion in the problem context:

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