In a completely randomized design, eight experimental units were used for each of the five levels of the factor. Consider the following ANOVA table. Degrees of Freedom Source Sum Mean of Variation of Squares Square F p-value Treatments 340 4 85 22.88 0.0000 Error 130 35 3.71 Total 470 39 (a) What hypotheses are implied in this problem? Hoi H1# Hz # H3 * H4 * H5 Ha: H1 = H2 = H3= H4 = H5 O Ho: H1 = H2 = H3 = H4= H5 H: Not all the population means are equal. O Ho: Not all the population means are equal. 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: H1 = H2 = H3 = H4 = H5 Hai H1 # Hz # Hz#H4# H5 (b) At the a = 0.05 level of significance, can we reject the null hypothesis in part (a)? Explain. O Because the p-value > a = 0.05, we cannot reject Ho: O Because the p-value < a = 0.05, we cannot reject Ho: O Because the p-value > a = 0.05, we can reject Ho: O Because the p-value < a = 0.05, we can reject Ho:

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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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
of Freedom
Mean
p-value
of Variation
of Squares
Square
Treatments
340
4
85
22.88
0.0000
Error
130
35
3.71
Total
470
39
(a) What hypotheses are implied in this problem?
O Ho: H1 # H2 + Hz# H4# H5
Ha: H1 = H2 = H3 = H4= H5
%3D
O Ho: H1 = H2 = H3
H4 = H5
H: Not all the population means are equal.
%3D
: Not all the population means are equal.
Ho
Ha: H1 = H2 = Hz= H4= H5
Ho: At least two of the population means are equal.
H: At least two of the population means are different.
= 4
Hoi H1 = H2 = H3
45
+ 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 < a = 0.05, we cannot reject Ho:
Because the p-value > a = 0.05, we can reject Ho:
Because the p-value < a = 0.05, we can reject Ho:
Transcribed Image Text: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 of Freedom Mean p-value of Variation of Squares Square Treatments 340 4 85 22.88 0.0000 Error 130 35 3.71 Total 470 39 (a) What hypotheses are implied in this problem? O Ho: H1 # H2 + Hz# H4# H5 Ha: H1 = H2 = H3 = H4= H5 %3D O Ho: H1 = H2 = H3 H4 = H5 H: Not all the population means are equal. %3D : Not all the population means are equal. Ho Ha: H1 = H2 = Hz= H4= H5 Ho: At least two of the population means are equal. H: At least two of the population means are different. = 4 Hoi H1 = H2 = H3 45 + 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 < a = 0.05, we cannot reject Ho: Because the p-value > a = 0.05, we can reject Ho: Because the p-value < a = 0.05, we can reject Ho:
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