Question

The following data were obtained for a randomized block design involving five treatments and three blocks: SST = 490, SSTR = 310, SSBL = 95. Set up the ANOVA table. (Round your value for F to two decimal places, and your p-value to three decimal places.)

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Blocks
Error
Total

Test for any significant differences. Use α = 0.05.

State the null and alternative hypotheses.

H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅
H_a: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄ ≠ μ₅    H₀: Not all the population means are equal.
H_a: μ₁ = μ₂ = μ₃ = μ₄ = μ₅H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅
H_a: Not all the population means are equal.H₀: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄ ≠ μ₅
H_a: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to three decimal places.)

p-value =

State your conclusion.

Do not reject H₀. There is sufficient evidence to conclude that the means of the treatments are not all equal.Reject H₀. There is not sufficient evidence to conclude that the means of the treatments are not all equal.    Reject H₀. There is sufficient evidence to conclude that the means of the treatments are not all equal.Do not reject H₀. There is not sufficient evidence to conclude that the means of the treatments are not all equal.

Answer 1

Source	SS	df	MS	F	p value
treatments	310	4	77.50	7.29	0.009
blocks	95	2	47.50	4.47	0.050
error	85	8	10.63
total	490	14

H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

H_a: Not all the population means are equal

value of the test statistic =7.29

p value =0.009

Reject H₀. There is sufficient evidence to conclude that the means of the treatments are not all equal.