Question

7) In a completely randomized design, 4 experimental units were used for each of the seven levels of the factor. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Treatment 385.12 Error Total 1563.71

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatment	385.12
Error
Total	1563.71

i. Complete the ANOVA table.

ii. Find the F critical, and use the critical value approach at α = 0.05 to test whether the population means for the four levels of the factors are the same. (1 mark)

Answer 1

Solution:

SSerror = SStotal - SStreatment = 1563.71-385.12

SSerror = 1178.59

No. of levels of factor k = 4

dftreatment = k-1 = 4-1

dftreatment = 3

Total sample size N = 5×4 = 20

dferror = N-k = 20-4

dferror = 16

dftotal = N-1 = 20-1 = 19

MStreatment = SStreatment/dftreatment = 385.12/3

MStreatment = 128.3733

MSerror = SSerror/dferror = 1178.59/16

MSerror = 73.6614

F = MStreatment/MSerror = 128.37/73.66

F = 1.743

The ANOVA table is :

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatment	385.12	3	128.3733	1.743
Error	1178.59	16	73.6614
Total	1563.71	19

Test statistics is F = 1.743

Significance level, = 0.05

Critical value:

At 0.05 significance of the critical value of F with df1 = 3 and df2 = 16 is 3.2389

Decision rule :

If FSTAST > 3.2389, reject null hypothesis, otherwise fail to reject

Since F < 3.2389, we do not reject null hypothesis

Decision : Fail to reject Ho.

Conclusion:

There is insufficient evidence to conclude that population means of four levels of factors are the same at 0.05 significance level.