A two-factor factorial design has 2 degrees of freedom for factor A variation, 5 degrees of freedom for factor B variation, 10 degrees of freedom for interaction variation, 54 degrees of freedom for random variation, and 71 degrees of freedom for total variation. Suppose also that SSA=70, SSB=50, SSAB=100, SSE=270, SST=490, MSA=35, MSB=10, MSAB=10, and MSE=5.Form the ANOVA summary table and fill in all values in the body of the table The summary of a randomized block experiment with four groups and eight blocks is given below in the following ANOVA summary table. Complete parts (a) and (b).Degrees ofFreedomMean Square(Variance)SourceSum ofFSquaresAmong groups c-1=3Among blocks r-1=7SSA = 210MSA = 70FSTAT = 5.4444SSBL = 540 MSBL = 77.1429FSTAT = 6.0(r- 1)(c- 1) = 21rc -1 = 31ErrorSSE = 270MSE = 12.8571TotalSST = 1020a. At the 0.05 level of significance, is there evidence of a difference among the four group means? Determine the null and alternative hypotheses. Let c represent the number of groups and r represent the number of blocks.Ho: H1 = H2 =... = HcH1: Not all u are equal (where j= 1, 2, .., c)Determine the test statistic.FSTAT = 5.4444 (Type an integer or a decimal.)Calculate the p-value.p-value =(Round to three decimal places as needed.)

Answered: the ANOVA summary table

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

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A two-factor factorial design has 2 degrees of freedom for factor A variation, 5 degrees of freedom for factor B variation, 10 degrees of freedom for interaction variation, 54 degrees of freedom for random variation, and 71 degrees of freedom for total variation. Suppose also that SSA=70, SSB=50, SSAB=100, SSE=270, SST=490, MSA=35, MSB=10, MSAB=10, and MSE=5.

Form the ANOVA summary table and fill in all values in the body of the table

The summary of a randomized block experiment with four groups and eight blocks is given below in the following ANOVA summary table. Complete parts (a) and (b).
Degrees of
Freedom
Mean Square
(Variance)
Source
Sum of
F
Squares
Among groups c-1=3
Among blocks r-1=7
SSA = 210
MSA = 70
FSTAT = 5.4444
SSBL = 540 MSBL = 77.1429
FSTAT = 6.0
(r- 1)(c- 1) = 21
rc -1 = 31
Error
SSE = 270
MSE = 12.8571
Total
SST = 1020
a. At the 0.05 level of significance, is there evidence of a difference among the four group means? Determine the null and alternative hypotheses. Let c represent the number of groups and r represent the number of blocks.
Ho: H1 = H2 =... = Hc
H1: Not all u are equal (where j= 1, 2, .., c)
Determine the test statistic.
FSTAT = 5.4444 (Type an integer or a decimal.)
Calculate the p-value.
p-value =
(Round to three decimal places as needed.)

Expert Solution

Step 1

The ANOVA table is GIven, all SS and degrees of freedom are given

we have to check

1) To test :

H0: Whether the four means are significantly different,

that is

versus $H o : μ_{1} = μ_{2} = μ_{3} = . . . . . μ_{c}$

H1:the four means are not significantly different,

that is

H1: $n o t a l l μ' s a r e e q u a l$

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