Example 1: In an experimental study of the mineral metabolism of pullets for white pullets of the same strain and hatching were used during the period of investigation two pullets were given the ration C which had high Cao content where as the other two pullets were given the ration NC which has comparable ration with C. Apart from the fact its Cao content is low. In the other aspects the pullets are treated alike and an attempt was made to regulate the daily food consumption of the pullets. The following table keeps the rates of Cao in grams found in the whole eggs laid out by the pullets.
C1

C2

NC3 
NC4

2.435

2.155 
2.156 
2.274 
2.395 
1.877 
2.376 
2.352 
2.235 
2.163 
2.147 
1.690 
2.545 
2.088 
1.821 
1.685 
2.842 
2.136 
1.805 
1.254 
2.749 
2.071 
1.858 
0.833 
2.723 
1.895 
1.736 

2.706 
1.870 
1.758 

2.586 
1.724 


2.479 
1.353 


2.673 



2.721 



2.255 



2.318 



state an appropriate linear model to analyze the above data.
Test for the significant difference in the 4 different diets given to pullets.
Solution: The linear model is y$_{ij}$ = $\mu$ + $\alpha_{i}$ + e$_{ij}$
n $_{1}$ = 14, n$_{2}$ = 10, n$_{3}$ = 8, n $_{4}$ = 6
n = n$_{1}$ + n$_{2}$+ n$_{3}$+ n$_{4}$ = 38
Now to test the difference among 4 diets, we need to set up the null hypothesis.
H$_{0}$: There is no significant difference among 4 different diets
H$_{1}$: Atleast one diet is different from others.
y$_{1.}$ = $\sum_{j = 1}^{n_{i}}$ 35.662
y$_{2.}$ = 19.332
y$_{3.}$ = 15.657
y$_{4.}$ = 10.088
$\bar{y_{1.}}$ = 2.5472
$\bar{y_{1.}}$ = 1.9332
$\bar{y_{1.}}$ = 1.9571
$\bar{y_{1.}}$ =1.6813
$\sum_{i = 1}^{n}y_{i}^{2}$ = y$_{1.}^{2}+y_{2.}^{2}+y_{3.}^{2}+y_{4.}^{2}$
=(35.662)$^{2}$+ (19.332)$^{2}+(15.657)^{2}$+(10.088)$^{2}$
=1992.4138
y$_{..}$ = y$_{1.}+ y_{2.}+y_{3.}+y_{4.}$
= 35.662 + 19.332 + 15.657 + 10.088
= 80.739
(y$_{..})^{2}$ = 6518.7861
C.F =
$\frac{y_{..}^{2}}{n}$
=
$\frac{6518.7861}{38}$= 171.5470
TSS =
$\frac{\sum_{i = 1}^{n}}{n_{i}}y_{i.}^{2}$  C.F
=
$\frac{y_{1.}^{2}}{n_{1}}$+
$\frac{y_{2.}^{2}}{n_{1}}$+
$\frac{y_{3.}^{2}}{n_{1}}$+
$\frac{y_{4.}^{2}}{n_{1}}$  C.F
=4.2709
TSS = $\sum_{i = 1}^{t}\sum_{j = 1}^{n_{i}}y_{ij}^{2}$  C.F
= 178.9946  171.5470
= 7.4476
ESS = TSS  Trss
= 7.4476  4.2709
= 3.1767
We present all the above various sum of squares in the following ANOVA table.
Source 
D.f

Sum of Squares 
Mean sum of squares 
fcal

Due to treatment 
4  1 = 3 
4.2708 
1.4236 
15.2419

Due to error 
37  3 = 34 
3.1767 
0.0934 

Total 
38  1 = 37 
7.4476 


F$_{tab}$ = 2.92
As F$_{cal}$ > F$_{tab}$ we reject H$_{0}$ and conclude that atleast one diet is different from others.