DETAILS 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 Sum Degrees Mean p-value of Variation of Squares of Freedom Square Treatments 380 4 95 110.83 0.0000 Error 30 35 0.86 Total 410 39 (a) What hypotheses are implied in this problem? O Ho: H1 = H2 = H3 = H4 = H5 H : Not all the population means are equal. O Ho: H1 = H2 = H3 = HA = H5 Ha:Hi * H2 # 13 # H4 $ 1 5 OHO:H1 # Hz # H3 # H4 # 15 Ha: H1 = H2 = H3 = H4 = H5 O Ho: At least two of the population means are equal. H. : At least two of the population means are different. O Ho: Not all the population means are equal. Ha: H1 = H2 = H3 = H4 = H5 (b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. Because the p-value > a = 0.05, we cannot reject Ho. Because the p-value S a = 0.05, we can reject Ho- O Because the p-value S a = 0.05, we cannot reject Ho. O Because the p-value > a = 0.05, we can reject Ho