Suppose k independent random samples. Each of them has size n, then the result from k different populations will be-
Population 1: x_{11}, x_{12}, …….. x_{1n }: Treatment 1
Population 2: x_{21}, x_{22}, …….. x_{2n }: Treatment 2
———————————————————————————————–
Population k: x_{k1}, x_{k2}, …….. x_{kn }: Treatment k
Here,
x_{ij} refers to j-th value of i-th population
Hence, the rest of the random variables will be independent normally distributed. The common variance will be σ^{2}.
So, consider the following model as,
x_{ij} = µ + α_{i} + e_{ij} where,i = 1, 2, …….. k and j = 1, 2, …….. n
Here,
µ = Grand Mean
α_{i} = Treatment 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} : Population Mean will be equal
Hence,
µ_{1} =µ_{2} =………. = µ_{k} and samples will be obtained by using k population with equal mean.
Equivalently,
α_{i} = 0 where, i = 1, 2, …… k (No special effect)
H_{1} : α_{i}≠ 0 for at least one value of i
Constructing the ANOVA Table:
Source of Variation | Degree of Freedom | Sum of
Squares |
Mean
Square |
F
(Test Statistics) |
Treatments | k – 1 | SS (Tr) | MS (Tr) | |
Error | k (n – 1) | SSE | MSE | |
Total | kn – 1 | SST |
Where,
SSE = Error Sum of Squares
SST = SS (Tr) + SSE
It can be calculated as,
SST = Total Sum of Square
=
SS(Tr) = Treatment Sum of Squares
=
Where,
T_{i.} = Total of Values Obtained for i-th Treatment
T_{..} = Grand Total of all ‘nk’ Observations
MS (Tr) = Treament Mean Square
= SS (Tr)/ (k – 1)
MSE = Error Mean Square
= SSE/ k (n – 1)
Now, For F-Distribution,
F = F_{cal} = MS (Tr)/ MSE
Let L be the level of significance
Hence,
F_{tab} = F_{α}_{, }_{α}_{ – 1, k (n – 1)}
It’s Conclusion:
Reject H_{0} if F_{cal}> F_{tab}
Accept H_{0} if F_{cal}< F_{tab}
Example No. 1
At random, five students are taken out from X^{th }standard of 3 different schools. A test was given to each of them and the individual scores are tabulated below.
School I | 77 | 81 | 71 | 76 | 80 |
School II | 72 | 58 | 74 | 66 | 70 |
School III | 76 | 85 | 82 | 80 | 77 |
Do analysis of variance and state the conclusions.
Solution:
- As there is no difference between the schools,
H_{0} : α_{1} = α_{2} =α_{3} = 0
H_{1} : α_{i}≠ 0 for at least one value of i
- Let the value of α = 0.01, k = 3 and n = 5
F_{tab} = F_{0.01, 2.12 }= 6.93
- Computations:
T_{1.} = 77 + 81 + 71 + 76 + 80 = 385
T_{2.} = 72 + 58 + 74 + 66 + 70 = 340
T_{3.} = 76 + 85 + 82 + 80 + 77 = 400
Thus,
T_{..} = T_{1.}+ T_{2.}+ T_{3.}= 1125
Now,
So,
SST = 85041 – 1/ 1.5 (1125)^{2} = 666
SS (Tr) = 1/ 5 {(385)^{2} + (340)^{2} + (400)^{2}} – 1/ 1.5 (1125)^{2}
= 390
Hence,
SSE = SST – SS (Tr)
= 666 – 390
= 276
ANOVA Table:
School I | 77 | 81 | 71 | 76 | 80 |
School II | 72 | 58 | 74 | 66 | 70 |
School III | 76 | 85 | 82 | 80 | 77 |
- Conclusion:
Here,
F_{cal}> F_{tab}
H_{0} is rejected and it proves that students are not same of different schools of class X^{th} standard.
Example No. 2
Four typists are working for a publishing company. A table has been created for the number of mistakes made by them in five successive weeks.
Typist I | 13 | 16 | 12 | 14 | 15 |
Typist II | 14 | 16 | 11 | 19 | 15 |
Typist III | 13 | 18 | 16 | 14 | 18 |
Typist IV | 18 | 10 | 14 | 15 | 12 |
Take level of significance as 0.05 and state whether their difference of mistakes can attribute to chance or not.
Solution:
- H_{0} : µ_{1} = µ_{2} = µ_{3} = µ_{4}
H_{1} : There should be at least two of them should not be equal.
- Given:
α = 0.05, k = 4 and n = 5
F_{tab} = F_{0.05, 3.16}= 3.24
- Computations:
T_{1.} = 13 + 16 + 12 + 14 + 15 = 70
T_{2.} = 14 + 16 + 11 + 19 + 15 = 75
T_{3.} = 13 + 18 + 16 + 14 + 18 = 79
T_{4.} = 18 + 10 + 14 + 15 + 12 = 69
Thus,
T_{..} = T_{1.} + T_{2.} + T_{3.}+ T_{4.}=293
Now,
So,
SST = 4407 – 1/ 20 (293)^{2} = 114.55+
SS (Tr) = 1/ 5 {(70)^{2} + (75)^{2} + (79)^{2}+ (69)^{2}} – 1/ 20 (293)^{2}
= 12.95
Hence,
SSE = SST – SS (Tr)
= 101.6
ANOVA Table:
Sources of
Variation |
Degrees of Freedom | Sum of Squares | Mean Square | F |
Treatment | 3 | 12.95 | 4.32 | 0.68 |
Error | 16 | 101.6 | 6.35 | |
Total | 19 | 114.55 |
- Conclusion:
Here,
F_{cal}< F_{tab}
H_{0} is accepted and it means that typing mistakes can be attributed to chance.
Note:
- As there is no such relation between variance and change of origin (independent of each other), it is possible to shift the origin to arbitrary data. While calculating the variance ratio, it is independent of change of scale. Thus, both the changes could be applied if necessary to perform any simpler arithmetic.
- Here, the observations can be written as,
x_{ij} = + (i – ) + (x_{ij}– )
Grand Mean Deviation due to Error
Treatment