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 VariationDegree of FreedomSum of

Squares

Mean

Square

F

(Test Statistics)

Treatmentsk – 1SS (Tr)MS (Tr)
Errork (n – 1)SSEMSE
Totalkn – 1SST

 

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 I7781717680
School II7258746670
School III7685828077

 

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 I7781717680
School II7258746670
School III7685828077

 

  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 I1316121415
Typist II1416111915
Typist III1318161418
Typist IV1810141512

 

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 of

Variation

Degrees of FreedomSum of SquaresMean SquareF
Treatment312.954.320.68
Error16101.66.35
Total19114.55

 

  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

Treatment

Submit Your Assignment