In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Consider the following ANOVA table. Sum Degrees of Freedom Source Mean p-value of Variation of Squares Square Treatments 300 4 75 17.31 0.0000 Error 130 30 4.33 Total 430 34 (a) What hypotheses are implied in this problem? O Ho: Not all the population means are equal. Hai H1 = H2 = H3 = H4 = H5 O Ho: H1 = H2 = H3 = H4 = H5 H: Not all the population means are equal. O Ho: H1 + H2 * H3 # H4 # H5 Hai H1 = H2 = H3 = H4 = H5 O Ho: At least two of the population means are equal. H3: At least two of the population means are different. O Ho: H1 = H2 = H3 = H4 = Hs 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 s a = 0.05, we can reject Ho: O Because the p-value > 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 cannot reject Ho:

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1.
DETAILS
MY NOTES
ASK YOUR TEA
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
Mean
Degrees
of Freedom
F
p-value
of Variation
of Squares
Square
Treatments
300
4
75
17.31
0.0000
Error
130
30
4.33
Total
430
34
(a) What hypotheses are implied in this problem?
O Ho: Not all the population means are equal.
Ha: H1 = H2 = H3 = H4= H5
Hoi H1 = H2 = H3 = H4= H5
H: Not all the population means are equal.
%3D
Ho: H1 + H2 # H3
* H5
Ha: H1 = H2 = H3 = H4= H5
%3D
Ho: At least two of the population means are equal.
H: At least two of the population means are different.
Ho: H1 = H2 = H3 = H4 = H5
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 can reject Ho:
Because the p-value > a = 0.05, we can reject Ho.
Because the p-value < a = 0.05, we cannot reject Ho:
Because the p-value > a = 0.05, we cannot reject H,.
O O O O
Transcribed Image Text:1. DETAILS MY NOTES ASK YOUR TEA 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 Mean Degrees of Freedom F p-value of Variation of Squares Square Treatments 300 4 75 17.31 0.0000 Error 130 30 4.33 Total 430 34 (a) What hypotheses are implied in this problem? O Ho: Not all the population means are equal. Ha: H1 = H2 = H3 = H4= H5 Hoi H1 = H2 = H3 = H4= H5 H: Not all the population means are equal. %3D Ho: H1 + H2 # H3 * H5 Ha: H1 = H2 = H3 = H4= H5 %3D Ho: At least two of the population means are equal. H: At least two of the population means are different. Ho: H1 = H2 = H3 = H4 = H5 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 can reject Ho: Because the p-value > a = 0.05, we can reject Ho. Because the p-value < a = 0.05, we cannot reject Ho: Because the p-value > a = 0.05, we cannot reject H,. O O O O
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