a completely randomized design, seven experimental units were used for each of the five levels of the factor. Consider the llowing ANOVA table. Source Sum Degrees Mean p-value of Variation of Squares of Freedom Square Treatments 360 4 90 38.57 0.0000 Error 70 30 2.33 Total 430 34 a) What hypotheses are implied in this problem? H,: Not all the population means are equal. H.i Hq = Hz = Hz = H4 = Hs Ho: At least two of the population means are equal. : At least two of the population means are different. Ho: H1 # H2# Hz * H4* Hs Hi Hq = Hz = H3 = H4 = H5 O Ho: H1 = H2 = H3 = H4 = Hs : Not all the population means are equal. Ho: H1 = H2 = Hz = H4 = Hs Hi Hy # Hz# H3 * H4 # Hs b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. Because the p-value s a = 0.05, we can reject H,. Because the p-value s a = 0.05, we cannot reject Ho. Because the p-value > a = 0.05, we cannot reject Ho: Because the p-value > a = 0.05, we can reject Ho:

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In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Consider the
following ANOVA table.
Source
Sum
Degrees
of Squares of Freedom
Mean
F
p-value
of Variation
Square
Treatments
360
4
90
38.57
0.0000
Error
70
30
2.33
Total
430
34
(a) What hypotheses are implied in this problem?
O Ho: Not all the population means are equal.
H3i Hq = Hz = Hz = H4 = Hs
O Ho:
: At least two of the population means are equal.
: At least two of the population means are different.
O Ho: H1 # H2 * Hz# Mq# M5
Hai Hy = H2 = H3 = H4 = H5
O Ho: H1 = H2 = !3 = H4 = Hs
: Not all the population means are equal.
O Ho: H1 = H2 = !3 = H4 = Hs
(b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain.
O Because the p-value s a = 0.05, we can reject H,.
Because the p-value s a = 0.05, we cannot reject Ho:
O Because the p-value > a = 0.05, we cannot reject Ho:
Because the p-value > a = 0.05, we can reject H:
Transcribed Image Text:In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Consider the following ANOVA table. Source Sum Degrees of Squares of Freedom Mean F p-value of Variation Square Treatments 360 4 90 38.57 0.0000 Error 70 30 2.33 Total 430 34 (a) What hypotheses are implied in this problem? O Ho: Not all the population means are equal. H3i Hq = Hz = Hz = H4 = Hs O Ho: : At least two of the population means are equal. : At least two of the population means are different. O Ho: H1 # H2 * Hz# Mq# M5 Hai Hy = H2 = H3 = H4 = H5 O Ho: H1 = H2 = !3 = H4 = Hs : Not all the population means are equal. O Ho: H1 = H2 = !3 = H4 = Hs (b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. O Because the p-value s a = 0.05, we can reject H,. Because the p-value s a = 0.05, we cannot reject Ho: O Because the p-value > a = 0.05, we cannot reject Ho: Because the p-value > a = 0.05, we can reject H:
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