Suppose k independent random samples. Each of them has size n, then the result from k different populations will be-

 

Population 1:               x₁₁,       x₁₂,       …….. x_1n                       :           Treatment 1

Population 2:               x₂₁,       x₂₂,       …….. 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 σ².

 

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, σ²)

 

Hypothesis:

To get the statistic hypothesis, it can be written as-

H₀ : Population Mean will be equal

Hence,

µ₁ =µ₂ =………. = 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)

H₁ : 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 H₀ if F_cal> F_tab

Accept H₀ 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:

1. As there is no difference between the schools,

H₀ : α₁ = α₂ =α₃ = 0

H₁ : α_i≠ 0 for at least one value of i

1. 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

 

1. Conclusion:

Here,

F_cal> F_tab

H₀ 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:

1. H₀ : µ₁ = µ₂ = µ₃ = µ₄

H₁ : There should be at least two of them should not be equal.

1. 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

 

1. Conclusion:

Here,

F_cal< F_tab

H₀ is accepted and it means that typing mistakes can be attributed to chance.

 

Note:

1. 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.
2. Here, the observations can be written as,

 

x_ij         =          +          (i –  )    +          (x_ij–  )

Grand Mean                Deviation due to         Error

Treatment