In a completely randomized design, eight experimental units were used for each of the five levels of the factor. Consider the following ANOVA table.

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments	380	4	95	30.23	0.0000
Error	110	35	3.14
Total	490	39

(a)What hypotheses are implied in this problem?H₀: ₁ = ₂ = ₃ = ₄ = ₅ H_a: _{1 2 3 4 5}H₀: _{1 2 3 4 5} H_a: ₁ = ₂ = ₃ = ₄ = ₅ H₀: Not all the population means are equal. H_a: ₁ = ₂ = ₃ = ₄ = ₅H₀: ₁ = ₂ = ₃ = ₄ = ₅ H_a: Not all the population means are equal.H₀: At least two of the population means are equal. H_a: At least two of the population means are different.

(b)At the = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain.

Because the p-value > = 0.05, we cannot reject H₀.

Because the p-value = 0.05, we can reject H₀.

Because the p-value > = 0.05, we can reject H₀.

Because the p-value = 0.05, we cannot reject H₀.