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Analysis Of Variance

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Assumptions

Each sample is drawn from a normalpopulation and the sample statistics tendto reflect the characteristics of thepopulation

The populations from which the samplesare drawn have identical means andvariances1=2=3=4 1= 2= 3= 4

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Computation One way classification Two way classification

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One way classification

Single variable of interest Ho: 1=2=3== k where k is the number

of samples drawn

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1) Calculate the variance betweenthe samples(SSB)

Calculate mean of each sample Calculate grand mean

Difference between grand mean and eachsample mean =deviation Square deviations

Divide total by d.f ; v=k-1, where k is thenumber of samples

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2) Variance within samples(SSW)

Calculate mean of each sample Take deviation of each observation from

respective mean Square deviations; find total Divide by degrees of freedom; v=n-k; n is

the total no. of observations ;k is thenumber of samples

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3) Calculate the F ratio

F = Variance between the samplesVariance within the samples

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4) Compare to the table value

If F cal > F tab, Reject Ho

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To assess the significance of possible variation in the testscores of four candidates of a company,a battery of 5 tests

was administered on them.The results are given below.Make an Analysis of Variance of data

Marks A B C D8 12 18 1310 11 12 912 9 16 128 14 6 167 4 8 15

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Ho : 1=2=3=4. MarksX1 X2 X3 X4

8 12 18 13

10 11 12 912 9 16 128 14 6 16

7 4 8 15Total 45 50 60 65Mean 9 10 12 13Grand Mean = 11

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Variance between samplesSample 1 Sample 2 Sample 3 Sample 4(X1 - 11) (X2 - 11) (X3 - 11) (X4 - 11)4 1 1 44 1 1 44 1 1 4

4 1 1 44 1 1 420 5 5 20

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Sum of squares between samples(SSB)= 50

Mean sum of squares betweensamples(MSB) = 50/4-1 = 50/3 = 16.7

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Variance within the samplesSample 1 Sample 2 Sample 3 Sample 4X1 (X1-X1 ) X2 (X2-X2 ) X3 (X3-X3 ) X4 (X4-X4 )

8 1 12 4 18 36 13 010 1 11 1 12 0 9 1612 9 9 1 16 16 12 1

8 1 14 16 6 36 16 97 4 4 36 8 16 15 416 58 104 30

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Total sum of squares withinsamples = 208

Mean sum of squares within the samples =208 / 20 4= 13

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ANOVA TableSource ofVariation

Sum ofSquares

Degrees OfFreedom

MeanSquare

Betweensamples

50 3 16.7

Withinsamples

208 16 13

Total 258 19

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F = 16.7 = 1.2813

Tabulated value for F with v 1 = 3 andv2=16; at 5 % L.O.S = 3.24

As F cal < F tab , Accept Ho

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q

To test the significance of variation in theretail prices of a commodity in threeprincipal cities, Mumbai, Kolkata andDelhi, four shops were chosen at randomin each city and the prices observed inrupees were as follows :

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Mumbai 16 8 12 14Kolkata 14 10 10 6Delhi 4 10 8 8

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Ho : there is no significant difference in the pricesof a commodity in the three cities

Sample 1 Sample 2 Sample 3Mumbai Kolkata Delhix1 x12 x2 x22 x3 X3 2

16 256 14 196 4 168 64 10 100 10 10012 144 10 100 8 6414 196 6 36 8 6450 660 40 432 30 244

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T = sum of all observations= 50 + 40 + 30 = 120

Correction factor = T 2N

= 120 / 12 = 1200

SST = Total Sum of Squares= x1 + x2 + x3 - CF= (660 + 432 + 244 ) 1200= 136

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Sum of squares between the samples

=x

1 + x

2 + x

3 - CF

n1

n2

n3

= 1250 1200

= 50

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Anova TableSource ofVariation

Sum ofSquares

Degrees OfFreedom

MeanSquares

Test Statistic

Betweensamples

50 r-1=2 25 F = 25 / 9.55

= 2.617Withinsamples

86 n-r=9

Total 136 11

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Tabulated value of F for d.f 2,9 = 4.26 at 5%l.o.s

Accept Ho

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Core marketing Marketing research Advertising

6 5 65 5 74 4 6

5 4 56 5 64 4 6

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Ho: No significant difference in ethicalvalues among the three groups.

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Marketing manager Marketing Research Advertising

x1 x1 x2 x2 x3 x36 36 5 25 6 365 25 5 25 7 494 16 4 16 6 365 25 4 16 5 256 36 5 25 6 36

4 16 4 16 6 3630 154 27 123 36 218

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T = 93Correction factor = 480.50

SST = total sum of squares= x1

+ x2

+ x3

- CF

= 14.50

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SSTR = sum of squares between thesamples

= (x1

) + (x

2

) + (x

3

) - CF

n1 n2 n3= 7

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Source ofVariation

Sum OfSquares

Degrees offreedom

Meansquares

Test Statistic

Betweensamples

7 2 (no. ofsamples 1)

3.5 3.5/ 0.5

= 7WithinSamples

7.5 15 0.5

Total 14.5 17 (cr-1)

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Tabulated F 2,15 = 3.68 at 5 %Reject Ho

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TWO WAY CLASSIFICATIONSource ofVariation

Sum ofSquares

Degrees offreedom

Meansum ofSquares

ratio

Betweensamples SSColumns C-1 Ssc/c-1 MSC/MSEBetweenrows

SS Rows R-1 SSR/ r-1 MSR/MSE

Residual SSE (c-1)(r-1) SSE/d.f

Total SST N-1 or cr-1

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Q

The following table gives the number ofsubscribers added by four major telecomplayers in India, in the months of August,September, October and November. Thedata is given in lakhs. Find out at 5 % levelof significance

- If the four companies differ in theirperformance

- Is there a significant difference between themonths ?

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CompanyMonths Bharti BSNL Tata

IndicomReliance

August 6 6 2 5September 7 6 2 3October 7 6 6 4November 7 8 7 4

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H1 : there is no sig diff in the performanceof the 4 companies

H2 : there is no sig diff between the months

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Company Month

Months Bharti BSNL TataIndicom

Reliance Total Average

August 6 6 2 5 19 4.75

September 7 6 2 3 18 4.5

October 7 6 6 4 23 5.75

November 7 8 7 426 6.50

Total 27 26 17 16 86 Grandmean =5.375

Average 6.75 6.5 4.25 4

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Correction factor = 86/ 16 = 462.25

Total sum of squares = TSS= 6 + 6 +2 +. 4 - CF

= 51.75Sum of Squares between months= 19 + 18 + 23 + 26 - 462.25

4 4 4 4= 10.25

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Sum of squares between companies= 27 + 26 + 17 + 16 - 462.25

4= 487.5 462.25= 25.25

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Source ofVariation

Sum ofSquares

Degreesoffreedom

MeanSum OfSquares

F cal Ftabulated

Betweenmonths

10.25 3 3.42 3.42/1.81=1.89

3.86

Betweencompanies

25.25 3 8.42 4.65 3.86

Error 16.25 9 1.81

Total 51.75 15

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Reject H1 Accept H2

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Ho : No significant difference in treatments

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Ho : No significant difference in treatmentsHo: No significant difference in plots of landUsing coding method, subtract 40Plot ofLand

Treatment Total A B C D

I -2 0 +1 -1 -2

II +5 +2 +9 -4 12

III 0 -2 +2 +2 +2

Total +3 0 +12 -3 +12

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Correction factor = T = 12= 12N 12

Sum of squares between treatments:= +3 + 0 +12 + (-3) - T = 42

3 3 3 3 Nv = 4-1 =3

S f b l f l d

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. Sum of squares between plots of land:= (-2) + 12 +2 - T= 26

4 4 4 Nv = 3-1 = 2

Total sum of squares

= (-2) + 5+0+0+2+(-2)+1+9 +2+(-1)+(-4)+2 - T = 132

N

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Source ofVariation

Sum ofSquares

Degrees offreedom

Meansum ofSquares

ratio

Betweencolumns 42 C-1 = 3 42/3 = 14 14/10.6= 1.32Betweenrows

26 R-1=2 26/2=13 13/10.67=1.218

Residual 64 6 10.67

Total 132 N-1 or cr-1

=11

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. For F 3,6,0.05 = 4.76 As Fcal < Ftab Accept Ho

For F (2,6)0.05 = 5.14 As Fcal < Ftab Accept Ho

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Q

The following are the number of kilometres /litre which a test driver with three diff typesof cars has obtained randomly on 3 diffoccasions. Examine the hypothesis thatthe diff in the average mileage is due tochance.

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Car 1 15 14.5 14.8 14.9

Car 2 13 12.5 13.6 13.8 14Car 3 12.8 13.2 12.7 12.6 12.9 13