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}Â Â B_{2}——B_{j}———B_{m}  Means  
Treatment 1
Treatment 2 
y_{11}Â Â y_{12}——y_{1j}———y_{1m}
Y_{21}Â Â y_{22}——y_{2j}———y_{2m} 
_{.}_{1} 
Treatment i  Y_{i1}Â Â 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} = Âµ + Î±_{i} + Î²_{j} + e_{ij}Â where, i = 1, 2, â€¦â€¦.. n and j = 1, 2, â€¦â€¦.. m
Here,
Âµ = Grand Mean
Î±_{i} = ithTreatment Effects ()
Î²_{j} = jthTreatment Effects ()
e_{ij} = Random Errors, these are identically distributed {N (0, Ïƒ^{2})
Hypothesis:
To get the statistic hypothesis, it can be written as
H_{0} :Î±_{1} = Î±_{2} = â€¦â€¦â€¦. = Î±_{n} = 0
Or,
H_{0} :Î²_{1} = Î²_{2} = â€¦â€¦â€¦. = Î²_{n} = 0
H_{1} :At least one value of these effects should not be zero
Constructing the ANOVA Table:
Â
Source of Variation  Degree of Freedom  Sum of
Squares 
Mean
Square 
F 
Treatments  n â€“ 1  SS (Tr)  MS (Tr)
= 
F_{Tr} = 
Blocks  m â€“ 1  SS (Bl)  MS (Bl)
= 
F_{Bl} = 
Error  (n â€“ 1) (m â€“ 1)  SSE  MSE
= 

Total  nm â€“ 1  SST 
Where,
SSEÂ Â Â Â Â Â = Error Sum of Squares
SSTÂ Â Â Â Â Â = SS (Tr) + SSE
It can be calculated as,
SSTÂ Â Â Â Â Â = Total Sum of Square
=
SS(Tr)Â Â =
SS(Bl)Â Â =
cÂ Â Â Â Â Â Â Â Â =
Where,
T_{i.}Â Â Â Â Â Â Â Â = Total of m Observations for ith Treatment
T_{.j}Â Â Â Â Â Â Â Â = Total of n Observations for jthBlock
T_{..}Â Â Â Â Â Â Â Â = Grand Total of all â€˜nmâ€™ Observations
So,
SSEÂ Â Â Â Â Â = SST â€“ SS (Tr) â€“ SS (Bl)
Set up the value of Î±
Itâ€™s Conclusion:
Reject H_{0}:Â Â Â Â Â Â Â Î±_{i} = 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â iÂ Â Â Â Â Â Â Â Â Â if F_{Tr}> F_{Î±.n â€“ 1. (n â€“ 1) (m â€“ 1)}
RejectH_{0}:Â Â Â Â Â Â Â Â Î²_{j}= 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â jÂ Â Â Â Â Â Â Â Â Â if F_{BL}> F_{Î±.m â€“ 1. (n â€“ 1) (m â€“ 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:
 As there are three equivalent tests,
H_{0} :Î±_{1} = Î±_{2} = Î±_{3} = 0
H_{1} :Î±_{i}â‰ 0 for at least one value of i
Since studentsâ€™ IQs are equivalent
H_{0} :Î²_{1} = Î²_{2} = Î²_{3} = Î²_{4} = 1340 = 0
H_{1} :Î²_{j}â‰ 0 for at least one value of j
 Given:Î± = 0.01, n = 3 and m = 4
F_{tab}(Tr) = F_{0.01, 2, 6}= 10.92
F_{tab}(Bl) = F_{0.01, 3, 6 }= 9.78
 Computations:
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 â€“ 56718.75 = 1136.25
SS (Tr) = 226925/ 4 â€“ 56718.75
= 12.5
SS (Bl)Â = 173045/ 3 â€“ 56718.75
= 962.92
Hence,
SSE = SST â€“ SS (Tr) â€“ 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 
Â
 Conclusion:
Here,
F_{cal} (Tr)<F_{tab}(Tr)
Hence,
H_{0} :Î±_{1} = Î±_{2} = Î±_{3} = 0 is accepted for all the three tests as equivalent
And,
F_{cal} (Bl) >F_{tab} (Bl)
Hence,
H_{0} :Î²_{1} = Î²_{ 2} = Î²_{ 3} = Î²_{4} = 0 is rejected for all the three tests as studentsâ€™ 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:
 Three samples are tabulated below. To obtain the hypothesis, consider the population means are equal and level of significance as 5%:
A  B  C 
8
10 7 14 11 
7
5 10 9 9 
12
9 13 12 14 
 From the group of three main brands of certain powder, 120 sales are examined. Allocate these brands of powder considering the table below:
Brands  Groups  
A  B  C  D  
I
II III 
0
5 18 
4
8 19 
8
13 11 
15
6 13 
 Fours batches of electric bulbs with 1000 hrs. of lives are tabulated below:
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.
 After conducting five tests in Computer Science paper, the following table has been tabulated with their test scores.
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.
 Construct a twoway ANOVA table using the following data
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.
 In a factory, four different workers are hired for working on five different machines. After taking proper observations, some data has been made. You have to calculate in two different ways whether all men are similar in production or not along with the efficiency if each machines. The following has been given
 Sum of Squares for Variance between machines = 35.2
 Sum of Square for Variance between workmen = 53.8
 Sum of Squares for Total Variance = 174.2
Answers:
 Reject H_{0} : Âµ_{1} = Âµ_{2} = Âµ_{3}, F_{cal} = 4
 No such difference in the brands
 Reject H_{0} as the means are different
 ANOVA Table:
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 
Â
 Workersâ€™ productivity is similar to mean productivity and it is same for all the f machines.
Here,
F (Workers) = 2.53
F (Machines) = 1.24
Links of Previous Main Topic:
 Introduction to statistics
 Knowledge of central tendency or location
 Definition of dispersion
 Moments
 Bivariate distribution
 Theorem of total probability addition theorem
 Random variable
 Binomial distribution
 What is sampling
 Estimation
 Statistical hypothesis and related terms
 Analysis of variance introduction
 Completely randomized design c r d one way classification
Links of NextÂ Statistics Topics:
 Definition of stochastic process
 Introduction operations research
 Introduction and mathematical formulation in transportation problems
 Introduction and mathematical formulation
 Queuing theory introduction
 Inventory control introduction
 Simulation introduction
 Time calculations in network
 Introduction of game theory