The following table gives yields (in 100 Kgs) per hectares of three varieties of wheat, each grown in four plots. Test the claim that there is no difference among the mean yields of the wheat varieties at 5% significance level. Also Make a ANOVA table for the data given.
Step 1:H
_{o} : μ
_{1} = μ
_{2} = μ
_{3} (Claim the average yields are equal)
H
_{1} : At least one mean is different from others.
Step 2: Find the critical value
N = 12 and k =3
d.f.N = k 1 = 2
d.f.D = N  K = 12  3 =9
Critical value
F_{2,9} at α = 0.05 from F table =
4.26Step 3: Computing the test value
To calculate the mean and variance of each group a table is made as follows:
 X_{1}

X_{2}

X_{3}

X_{12} 
X_{22}

X_{32} 
 7 
6 
6 
49 
36 
36 
 8 
6 
5 
64 
36 
25 
 4 
4 
4 
16 
16 
16 
 8 
7 
5 
64 
49 
25 
Total  27 
23 
20 
193 
137 
102 
Means for each category
X_{}_{1} = `27/4` = 6.75
X_{}_{2}= `23/4` = 5.75
X_{3} = `20/4` = 5
Computed variances for each category
$S_{1}^{2}$ = 3.58 $S_{2}^{2}$ = 1.58 $S_{3}^{2}$ = 0.67
Grand Mean X
_{GM} = `70/12` = 5.83
Sum of Squares Between Group SS
_{B} = ∑n
_{i}(
X_{i} 
X_{GM})
^{2} = 4(6.75  5.83)
^{2} + 4(5.75 5.83)
^{2} + 4(5  5.83)
^{2} = 1.5417
Sum of Squares Within Groups SS
_{W} = ∑(n
_{i}  1)S
_{i2} = 3(3.58) + 3(1.58) + 3(0.67) = 17.49
Variance Between Groups $S_{B}^{2}$ = SS
_{B} /k1 = `1.5417/2` = 0.77085
Variance Within Groups $S_{W}^{2}$ = SS
_{W} /N  k = `17.49/9` = 1.943
F test value = $\frac{S_{B}^{2}}{S_{W}^{2}}$ = `0.77085/1.943` =
0.397Step 4:The computed test value 0.397 < The critical value 4.26
There is no sufficient evidence to reject the claim the mean yields are equal for the three varieties of wheat A,B and C.
The ANOVA Summary Table for the test is as follows:
Source 
Sum of Squares 
Degrees of Freedom d.f 
Mean Squares

F

Between 
1.54

2

0.771_{} 
0.397_{} 
Within 
17.49

9

1.943_{} 
Total  19.03  11
 