One Way Annova

ANALYSIS OF VARIANCE

ANOVA

Example - 1

• To test the significance of variation in the retail price of a commodity in three cities Mumbai, Kolkata and Delhi ,4 shops were treatment were chosen at random in each city and price observed are as follows

• Do the data indicate that the null hypothesis in 3 cities are significantly different

Mumbai 16 8 12 14

Kolkata 14 10 10 6Delhi 4 10 8 8

Example -1 Solution• Let us take the hypothesis that there is no

significant difference in the prices of a commodity in three cities. Calculations for analysis of variance are as under

x1 x1² x2 x2² x3 x3²16 256 14 196 4 168 64 10 100 10 10012 144 10 100 8 6414 196 6 36 8 64Σx1=50 Σx1²=660 Σx2=40 Σx2²=432 Σx3=30 Σx3²=244

Sample1 Sample2 Sample 3Mumbai Kolkata Delhi

Example -1 Solution

• There are r=3 treatments (samples) with n1=4, n2=4, n3=4 and n=12

• T=Sum of all the observations in three samples• = Σx1+Σx2+Σx3 = 50+40+30 = 120• CF=correction factor = T²/n = 120²/12 = 1200• SSy=total sum of squares• =(Σx1²+Σx2²+Σx3²)-CF = (660+432+244) - 1200• = 136

Example -1 Solution

• SSx=sum of squares between samples (Σx1)² + (Σx2)² + (Σx3)² CF n1 n2 n3

(50)² + (40)² + (30)² 1200 4 4 4 (2500) + (1600) +(900) 1200 = 50 4 4 4 SSE = SSy - SSx = 136-50 = 86

=

=

=

Example -1 Solution

• Degrees of freedom df1=r-1=3-1=2; df2=n-r=12-3=9• Thus MSx=SSx/df1=50/2=25• MSE=SSE/df2=86/9=9.55

Source of variation

Sum of squares

Degree of freedom

Mean squares

Test statistic

Between samples

50 2 25 F=25/9.55=2.617

Within samples

86 9 9.55

Total 136 11

Example -1 Solution

• The table value of F for df1=2, df2=9 and α=5% level of significance is 4.26. Since calculated value of F is less than its critical value , the null hypothesis is accepted. Hence we conclude that the prices of commodity in three cities have no significant difference

Example 2

• Study investigated the perception of corporate ethical values among individuals specializing in marketing . Using α =0.05 and the following data (higher scores indicate higher ethical values), test for significance in perception among three groups

Example 2 -SolutionMarketing Manager Marketing research Advertising

6 5 65 5 74 4 65 4 56 5 64 4 6

Example 2-Solution Marketing Manager

Marketing research

Advertising

x1 x1² x2 x2² x3 x3²

Total= Total= Total= Total= Total= Total=

Example 2-Solution Marketing Manager

Marketing research

Advertising

x1 x1² x2 x2² x3 x3²6 36 5 25 6 365 25 5 25 7 494 16 4 16 6 365 25 4 16 5 256 36 5 25 6 364 16 4 16 6 36Total=30 Total=154 Total=27 Total=123 Total=36 Total=218

Example 2-Solution

• There are r=3 treatments(samples)withn1=n2=n3 and n=18

• T= ΣX1 + ΣX2+………. +ΣXn =30+27+36=93• CF=correction factor=T²/n=(93)²/18=480.50• SSy=Total sum of squares =(Σx1²+Σx2²+Σx3²)-CF =(154+123+218)- 480.50 =14.50

Example -2 Solution

• SSx=sum of squares between samples (Σx1)² + (Σx2)² + (Σx3)² CF n1 n2 n3 (30)² + (27)² + (36)² 480.50 6 6 6 (900) + (729) + (1296) 480.50 = 7 6 6 6 SSerror = SSy - SSx = 14.50-7=7.50

=

=

=

Example -2 Solution• Degrees of freedom df1=r-1=3-1=2; df2=n-r=18-3=15• Thus MSx=SSx/df1=7/2=3.5• MSe=SSe/df2=7.50/15=0.5 ANOVA TABLE

Source of variation

Sum of squares

Degree of freedom

Mean squares

Test statistic

Between samples

7 2 3.5 F=3.5/0.5=7

Within samples

7.5 15 0.5

Total 14.5 117

Example -2 Solution

• The table value of F for df1=2, df2=15 and α=5% level of significance is 3.68. Since calculated value of F is more than its critical value , the null hypothesis is rejected. Hence we conclude that there is significant difference in ethical values among individuals specializing in marketing

Example -3

• As head of the department of a consumer’s research organization , you have the responsibility for testing and comparing lifetimes of four brands of electric bulbs. Suppose you test the life time of three bulbs of each of the four brands. The data shown below , each entry representing the lifetime of an electric bulb, measured in hundreds of hours

Example -3

• Can we infer that the mean lifetimes of the four brands of electric bulbs are equal ?

BRAND A BRAND B BRAND C BRAND D20 25 24 2319 23 20 2021 21 22 20

Example -3 Solution• Let us take the null hypothesis that the mean

lifetime of four brands of electric bulbs are equal

SAMPLE A

SAMPLE B

SAMPLE C

SAMPLE D

x1 x1² x2 x2² x3 x3² x4 x4²

20 400 25 625 24 576 23 52919 361 23 529 20 400 20 40021 441 21 441 22 484 20 400Total= 60

Total =1202

Total = 69

Total =1595

Total =66

Total =1460

Total = 63

Total=1329

Example 3-Solution

• There are r=4 treatments(samples)with n1=n2=n3 =n4=3 and n=12

• T= ΣX1 + ΣX2+………. +ΣXn = 60+69+66+63=258• CF= correction factor=T²/n=(258)²/12=5547• SSy=Total sum of squares =(Σx1²+Σx2²+Σx3²+ Σx4²)-CF =(1202+1595+1460+1329)- 5547 = 39

Example -3 Solution

• SSx=sum of squares between samples (Σx1)² + (Σx2)² + (Σx3)² + (Σx4)² CF n1 n2 n3 n4 (60)² + (69)² + (66)²+ (63)² 5547

3 3 3 3 (3600) + (4761) + (4356) + (3969) 5547= 15 3 3 3 3 SSerror = SSy- SSx = 39-15=24

=

=

=

Example -3 Solution• Degrees of freedom df1=r-1=4-1=3; df2,= n-r = 12-4 = 8• Thus MSx=SSx/df1 = 15/3 = 5• MSe = SSe/df2 = 24/8 = 3 ANOVA TABLE

Source of variation

Sum of squares

Degree of freedom

Mean squares

Test statistic

Between samples

15 3 5 F=5/3=1.66

Within samples

24 8 3

Total 39 11

Example -3 Solution

• The table value of F for df1=3, df2=8 and α=5% level of significance is 4.0662. Since calculated value of F is less than its critical value , the null hypothesis is accepted. Hence we conclude that the difference in mean life time of four brands of bulbs is not significant and we infer that the average lifetime of four bulbs is equal

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