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 = 443.0, SSB = 424.6, and SSE = 125.4. 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

H_a: at least one α_i ≠ 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 > 0.1000.050 < P-value < 0.100    0.010 < P-value < 0.0500.001 < P-value < 0.010P-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.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 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.

(b)

Test

H₀: β₁ = β₂ = β₃ = 0

(no differences in true average strength due to grade) versus

H_a: at least one β_j ≠ 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 > 0.1000.050 < P-value < 0.100    0.010 < P-value < 0.0500.001 < P-value < 0.010P-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

We have species, factor A, with I = 4 and grade, factor B, with J = 3

So the degrees of freedom for Factor A = 4-1 = 3 and the degrees of freedom for Factor B = 3-1 = 2 and degrees of freedom for Error = 3*2 = 6

Also given,

SSA = 443.0, SSB = 424.6, and SSE = 125.4

We can complete the ANOVA as below:

Source of variation	SS	df	MS	F	p-value
A	443	3	147.667	7.07	0.0215
B	424.6	2	212.3	10.16	0.0119
Error	125.4	6	20.9
Total	993	11

Answer(a):

We have to test

H₀: α₁ = α₂ = α₃ = α₄ = 0 (no differences in true average strength due to species)

Ha: at least one αi ≠ 0

The test statistic for above test is

F = 7.07

And the p-value = 0.0215 which is greater than 0.01 and less than 0.05.

Hence the correct option is 0.010 < P-value < 0.050

Conclusion:

The p-value or above test is less than 0.05 which indicates that we have enough evidence against null hypothesis to reject it, so we reject the null hypothesis.

Hence the correct option is

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

Answer(b):

We have to test

_H0: β₁ = β₂ = β₃ = 0(no differences in true average strength due to grade)

Ha: at least one βj ≠ 0   

The test statistic for above test is

F = 10.16

And the p-value = 0.0119 which is greater than 0.01 and less than 0.05.

Hence the correct option is 0.010 < P-value < 0.050

Conclusion:

The p-value or above test is less than 0.05 which indicates that we have enough evidence against null hypothesis to reject it, so we reject the null hypothesis.

Hence the correct option is

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