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:
H_{0} : α_{1} = α_{2} =α_{3} = 0
H_{1} : α_{i}≠ 0 for at least one value of i
F_{tab} = F_{0.01, 2.12 }= 6.93
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 |
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_{1} : There should be at least two of them should not be equal.
α = 0.05, k = 4 and n = 5
F_{tab} = F_{0.05, 3.16}= 3.24
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 |
Here,
F_{cal}< F_{tab}
H_{0} is accepted and it means that typing mistakes can be attributed to chance.
Note:
x_{ij} = + (i – ) + (x_{ij}– )
Grand Mean Deviation due to Error
Treatment
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