A randomized block design has three levels of factor A and f

A randomized block design has three levels of factor A and five levels of factor B where six replicates for each combination are examined. The results include the following sum of square terms: Construct an ANOVA table. (Leave no cells blank - be certain to enter \"0\" wherever required. Round \"p\" value to 3 decimal places and all other answers except \"df\" and \"SS\" to 2 decimal places.) At the 1% significance level, can you conclude that there is interaction between factor A and factor B? At the 1% significance level, can you conclude that the factor A means differ? At the 1% significance level, can you conclude that the factor B means differ? Entries in this table provide the values of that correspond to a given upper-tail area alpha and a specified number of degrees of freedom in the numerator df_1 and degrees of freedom in the denominator df_2. For example, for alpha = 0.05, df_1 = 8, and df_2 = 6

Solution

ANOVA Sources of variation

Sum of squares

    degrees of freedom

  Mean Sum of Squares

   F

F critical at 1%

Rows

SSA=1042

p-1=2

MSA=SSA/(p-1)

=521

F_A=MSA/MSE

=521/1.3733

=379.378

F_0.01,2,75 =4.89988

Columns

SSB=358

q-1=4

MSB=SSB/(q-1)

=89.5

F_B=MSB/MSE

=89.5/1.3733

=65.171

F_0.01,4,75 =3.58011

Interaction

SSAB=39

(p-1)*(q-1)=8

MSAB=SSAB/

(p-1)*(q-1)               

=4.875

F_AB =MSAB/MSE

=4.875/1.3733

=3.5498

F_0.01,8,75 =2.75804

Error

SSE=103

p*q*(r-1)=75

MSE=SSE/(p*q*(r-1)) =1.3733

Total

SST=1542

p*q*r-1=89

Here p=no of levels of factor A

q=no of levels of factor B

r=no. of replicates

p=3,q=5,r=6

b)

Here we are interested in testing H₀:There is any interaction between the factors A and B against H₁:not H₀

In the light of the given observations at 1% level of significance we reject H0 and conclude that there is no interaction between A and B since observed(FAB)>F_0.01,8,75 .

c)At 1% level of significance we cannot conclude about the means of the factor A

d)At 1% level of significance we cannot conclude about the means of the factor B

since the test for interaction effect is rejected.

ANOVA Sources of variation	Sum of squares	degrees of freedom	Mean Sum of Squares	F	F critical at 1%
Rows	SSA=1042	p-1=2	MSA=SSA/(p-1) =521	FA=MSA/MSE =521/1.3733 =379.378	F0.01,2,75 =4.89988
Columns	SSB=358	q-1=4	MSB=SSB/(q-1) =89.5	FB=MSB/MSE =89.5/1.3733 =65.171	F0.01,4,75 =3.58011
Interaction	SSAB=39	(p-1)*(q-1)=8	MSAB=SSAB/ (p-1)*(q-1)                =4.875	FAB =MSAB/MSE =4.875/1.3733 =3.5498	F0.01,8,75 =2.75804
Error	SSE=103	p*q*(r-1)=75	MSE=SSE/(p*q*(r-1)) =1.3733
Total	SST=1542	p*q*r-1=89