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

Population 1:               x11,       x12,       …….. x1n                       :           Treatment 1

Population 2:               x21,       x22,       …….. x2n                       :           Treatment 2

———————————————————————————————–

Population k:               xk1,       xk2,       …….. xkn                       :           Treatment k

Here,

xij 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,

xij = µ + αi + eij  where,i = 1, 2, …….. k and j = 1, 2, …….. n

Here,

µ = Grand Mean

αi = Treatment Effects ()

eij = 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,

µ12 =………. = µk and samples will be obtained by using k population with equal mean.

Equivalently,

αi = 0 where, i = 1, 2, …… k (No special effect)

H1 : αi≠ 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,

Ti.         = 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 = Fcal = MS (Tr)/ MSE

Let L be the level of significance

Hence,

Ftab = Fα, α – 1, k (n – 1)

It’s Conclusion:

Reject H0 if Fcal> Ftab

Accept H0 if Fcal< Ftab

Example No. 1

At random, five students are taken out from Xth 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 = α23 = 0

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

1. Let the value of α = 0.01, k = 3 and n = 5

Ftab = F0.01, 2.12 = 6.93

• Computations:

T1. = 77 + 81 + 71 + 76 + 80 = 385

T2. = 72 + 58 + 74 + 66 + 70 = 340

T3. = 76 + 85 + 82 + 80 + 77 = 400

Thus,

T.. = T1.+ T2.+ T3.= 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,

Fcal> Ftab

H0 is rejected and it proves that students are not same of different schools of class Xth 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

Ftab = F0.05, 3.16= 3.24

• Computations:

T1. = 13 + 16 + 12 + 14 + 15 = 70

T2. = 14 + 16 + 11 + 19 + 15 = 75

T3. = 13 + 18 + 16 + 14 + 18 = 79

T4. = 18 + 10 + 14 + 15 + 12 = 69

Thus,

T.. = T1. + T2. + T3.+ T4.=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 ofVariation 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,

Fcal< Ftab

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,

xij         =          +          (i –  )    +          (xij–  )

Grand Mean                Deviation due to         Error

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