Let measurements of n treatments are distributed to m blocks where,
y_{ij} be the observation of ith treatment for jth block
For this, the following layout shall be used,
Blocks:
B_{1} B_{2}——B_{j}———B_{m}  Means  
Treatment 1 Treatment 2  y_{11} y_{12}——y_{1j}———y_{1m} Y_{21} y_{22}——y_{2j}———y_{2m}  _{.}_{1} 
Treatment i  Y_{i1} y_{i2}——y_{ij}———y_{im}  _{.}_{2} 
Treatment n  Y_{n1}y_{n2}——y_{nj}———y_{nm}  _{.}_{j} 
Means  _{.}_{1}_{.}_{2}——_{.}_{j}———_{.}_{m}  _{.}_{m} 
Consider this model as,
y_{ij} = µ + α_{i} + β_{j} + e_{ij} where, i = 1, 2, …….. n and j = 1, 2, …….. m
Here,
µ = Grand Mean
α_{i} = ithTreatment Effects ()
β_{j} = jthTreatment Effects ()
e_{ij} = Random Errors, these are identically distributed {N (0, σ^{2})
Hypothesis:
To get the statistic hypothesis, it can be written as
H_{0} :α_{1} = α_{2} = ………. = α_{n} = 0
Or,
H_{0} :β_{1} = β_{2} = ………. = β_{n} = 0
H_{1} :At least one value of these effects should not be zero
Constructing the ANOVA Table:
Source of Variation  Degree of Freedom  Sum of Squares  Mean Square  F 
Treatments  n – 1  SS (Tr)  MS (Tr) =  F_{Tr} = 
Blocks  m – 1  SS (Bl)  MS (Bl) =  F_{Bl} = 
Error  (n – 1) (m – 1)  SSE  MSE =  
Total  nm – 1  SST 
Where,
SSE = Error Sum of Squares
SST = SS (Tr) + SSE
It can be calculated as,
SST = Total Sum of Square
=
SS(Tr) =
SS(Bl) =
c =
Where,
T_{i.} = Total of m Observations for ith Treatment
T_{.j} = Total of n Observations for jthBlock
T_{..} = Grand Total of all ‘nm’ Observations
So,
SSE = SST – SS (Tr) – SS (Bl)
Set up the value of α
It’s Conclusion:
Reject H_{0}: α_{i} = 0 i if F_{Tr}> F_{α.n – 1. (n – 1) (m – 1)}
RejectH_{0}: β_{j}= 0 j if F_{BL}> F_{α.m – 1. (n – 1) (m – 1)}
Example No. 3
When four students appeared for three different IQ test, the following table has been obtained. The scores are tabulated as
Students  
1  2  3  4  
TestI TestII TestIII  75 83 86  73 72 61  59 56 53  68 69 70 
By using the method of twoway analysis of variance, state the reasons whether these tests are equivalent or not. Assume level of significance = 0.01
Solution:
H_{0} :α_{1} = α_{2} = α_{3} = 0
H_{1} :α_{i}≠ 0 for at least one value of i
Since students’ IQs are equivalent
H_{0} :β_{1} = β_{2} = β_{3} = β_{4} = 1340 = 0
H_{1} :β_{j}≠ 0 for at least one value of j
F_{tab}(Tr) = F_{0.01, 2, 6}= 10.92
F_{tab}(Bl) = F_{0.01, 3, 6 }= 9.78
T_{1.} = 75 + 73 + 59 + 68= 275
T_{2.} = 83 + 72 + 56 + 69 = 280
T_{3.} = 86 + 61 + 53 + 70 = 270
T_{.}_{1} = 75 + 83 +86= 244
T_{.}_{2} = 73 + 72 + 61 = 206
T_{.}_{3} = 59 + 56 + 53 = 168
T_{.}_{4} = 68 + 69 + 70 = 207
Thus,
T_{..} = 825
Now,
c = (825)^{2}/ 12 = 56718.75
So,
SST = 57855 – 56718.75 = 1136.25
SS (Tr) = 226925/ 4 – 56718.75
= 12.5
SS (Bl) = 173045/ 3 – 56718.75
= 962.92
Hence,
SSE = SST – SS (Tr) – SS (Bl)
= 160.83
ANOVA Table:
Sources of Variation  Degrees of Freedom  Sum of Squares  Mean Square  F 
Treatment  2  12.5  12.5/ 2 = 6.25  6.25/ 26.81 = 0.23 
Blocks  3  962.92  962.92/ 3 = 320.97  320.97/ 26.81 = 11.97 
Error  6  160.83  160.83/ 6 = 26.81  
Total  11  1136.25 
Here,
F_{cal} (Tr)<F_{tab}(Tr)
Hence,
H_{0} :α_{1} = α_{2} = α_{3} = 0 is accepted for all the three tests as equivalent
And,
F_{cal} (Bl) >F_{tab} (Bl)
Hence,
H_{0} :β_{1} = β_{ 2} = β_{ 3} = β_{4} = 0 is rejected for all the three tests as students’ IQs are different.
Note:
The R.B.D is greatly accepted above the C.R.D. because of its greater flexibility in solving different types of experimental works. This is because there are no restrictions on numbers of treatments along with numbers of replicates. But the fact is that for the large number of treatments or for the wide variable blocks, the R.B.D. is not suitable.
Problems:
A  B  C 
8 10 7 14 11  7 5 10 9 9  12 9 13 12 14 
Brands  Groups  
A  B  C  D  
I II III  0 5 18  4 8 19  8 13 11  15 6 13 
Batches  
A B C D  1.6 1.58 1.46 1.51  1.61 1.64 1.55 1.52  1.65 1.64 1.6 1.53  1.68 1.7 1.62 1.57  1.7 1.75 1.64 1.6  1.72
1.66 1.68  1.8
1.74

1.82

Show that the significance test is accepted by using the analysis of variance method.
BatchI  BatchII  BatchIII 
86 77 81 84 69  79 78 82 83 68  88 83 86 89 82 
The test was conducted in three batches. Take the level of significance as 5% for the analysis of variance.
Prices of Field  Treatment  
A  B  C  D  
P Q R  45 43 39  40 41 39  38 45 41  37 38 41 
If required, you can shift the origin to 40.
Answers:
Sources of Variation  Degrees of Freedom  Sum of Squares  Mean Square  F 
Between Treatment  3  22.917  7.639  1.0 
Between Fields  2  8.167  4.083  0.534 
Error  6  45.833  7.639  
Total  11  76.917 
Here,
F (Workers) = 2.53
F (Machines) = 1.24
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