Question

Consider the following summary data on the modulus of elasticity (x 106 psi) for lumber of three different grades. Grade j Xi. Si 1 11 1.61 0.22 2 11 1.55 0.23 3 11 1.42 0.21 Use this data and a significance level of 0.01 to test the null hypothesis of no difference in mean modulus of elasticity for the three grades. 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 0.010 < P-value < 0.050 O 0.001 < P-value < 0.010 O p-value < 0.001 What can you conclude? Reject Ho. At least two of the three grades appear to differ significantly. Reject Ho. The three grades do not appear to differ significantly. Fail to reject Ho. The three grades do not appear to differ significantly. Fail to reject Ho. At least two of the three grades appear to differ significantly. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Consider the following summary data on the modulus of elasticity (✕ 10⁶ psi) for lumber of three different grades.

Grade	J	xi.	si
1	11	1.61	0.22
2	11	1.55	0.23
3	11	1.42	0.21

Use this data and a significance level of 0.01 to test the null hypothesis of no difference in mean modulus of elasticity for the three grades. 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 > 0.1000.050 < P-value < 0.100    0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001

What can you conclude?

Reject H₀. At least two of the three grades appear to differ significantly.Reject H₀. The three grades do not appear to differ significantly.    Fail to reject H₀. The three grades do not appear to differ significantly.Fail to reject H₀. At least two of the three grades appear to differ significantly.

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

Answer 1

We perform one way ANOVA

We test Ho : μ₁ = μ₂ = μ₃ , that is all three means are equal

vs H1: atleast one mean is different from the others

The test statistic is ratio of Between groups mean square error to within groups mean square error

That is F = MSE (between groups) / MSE (within groups)

F follows F distribution with (2,30) degrees of freedom

We perform the ANOVA in R software and the ANOVA table is given below:

Source Of Variation	Sum of Squares	d.f	MSE	F	p
Between Groups	0.2075	2	0.1038	2.141	0.1352
Within Groups	1.454	30	0.0485
Total	1.6615	32

The test statistic is f = 2.14

p value is 0.1352

So, p value > 0.100

The significance level of the test is 0.01

So, the conclusion is

Fail to reject H₀. The three grades do not appear to differ significantly