ANALYSIS OF VARIANCE (ANOVA)

PROPOSED BY R.A. FISHER

FIGURE 1:

A C

B

D

E

F

SCOPE OF ANOVA

• TESTING THE EQUALITY OF MEANS OF SEVERAL POPULATIONS ALL WITH EQUAL σ

• TECHNIQUE DEVELOPED FOR DATA THAT FOLLOW THE NORMAL DISTRIBUTION

• VALID FOR LARGE SAMPLE SIZES OTHERWISE

ANOVA:EXAMPLE ITHE HOURLY OUTPUT OF THREE MACHINES DURING RANDOM PERIODS OF OBSERVATIONS ARE AS FOLLOWS:

Obs. No MACHINE IMACHINE

IIMACHINE

III1 15 22 182 18 27 243 19 18 194 22 21 165 11 17 226 - - 15

TOTAL 85 105 114SAMPLE

SIZE n1 = 5 n2 =5 n3 = 6

MEAN 17 21 1919OVERALL MEAN

SOLUTION

CONSIDER THE VARIANCE OF THE SAMPLE MEANS:

σ2^ = VAR (Xˉ)=[∑nj*(xj¯- X¯)2 / (k-1)]where Xˉ is the random variable representing the

mean of a random sample, xj¯ is the mean of the jth sample; X¯ is the grand mean and k is the number of samples.

summation is from j = 1 to k.THIS IS CALLED BETWEEN SAMPLE

VARIANCE.

SOLUTION (CONTINUED)

σ2^ = [5*(17-19)2 + 5*(21-19)2 +6*(19-19)]/2= 20 CONSIDER THE jth SAMPLE. ITS VARIANCE IS sj

2 = ∑(xij-xjˉ)2 / (nj-1) (SUMMATION FROM 1 TO nj-1)

EACH IS AN ESTIMATE OF σ2

A POOLED ESTIMATE IS σ2^ = ∑[(nj-1) *sj

2]/(nT-k) (SUMMATION FROM 1 TO k) where nT-k)= n1+ n2+ n3= Total Sample Size

THIS CALLED WITHIN SAMPLE VARIANCE

WITHIN SAMPLE VARIANCE: CALCULATION

Obs. No MACHINE IMACHINE

IIMACHINE

III I II III1 15 22 18 4 1 1

2 18 27 24 1 36 25

3 19 18 19 4 9 0

4 22 21 16 25 0 9

5 11 17 22 36 16 9

6 - - 15 0 0 16

TOTAL 85 105 114 70 62 60SAMPLE

SIZE n1 = 5 n2 =5 n3 = 6

MEAN 17 21 19

s12 s2

2 s32

70/4 62/4 60/5

ESTIMATE (70+62+60)/13 14.7692

HYPOTHESIS TESTING

LET μk=DENOTE THE MEAN OUTPUT PER HOUR OF MACHINE k (k =1 TO 3)

NULL HYPOTHESIS: H0: THE MEAN OUTPUT OF THE THREE MACHINES DIFFER INSIGNIFICANTLY. HENCE μ1=μ2=μ3

ALTERNATIVE HYPOTHESIS: H1: NOT ALL ARE EQUAL

LEVEL OF SIGNIFICANCE:α: 0.05 AND 0.01

TEST STATISTIC= F

= BETWEEN SAMPLE VARIANCE/ WITHIN SAMPLE VARIANCE

DECISION RULE: REJECT H0WHEN F >c WHERE c IS CHOSEN SO THAT

Pr {F>c WHEN H0IS TRUE} = α

HYPOTHESIS TESTING (CONTINUED)

DISTRIBUTION OF F UNDER H0:F FOLLOWS AN F-DISTRIBUTION WITH DEGREES OF k-1 IN THE NUMERATOR AND nT-1 IN THE DENOMINATOR

VALUE OF TEST STATISTIC= 20/14.7692= 1.3542

TABULAR VALUES FOR F2,13

5% LEVEL = 3.81

1% LEVEL = 6.7

INFERENCE: DO NOT REJECT H0.

EXAMPLE 2: SC 11.5

PROMOTION STRATEGY

I II III IV V MEAN

FREE SAMPLE78 87 81 89 85

84

ONE PACK GIFT94 91 87 90 88

90

CENTS OFF73 78 69 83 76

75.8

REFUND BY MAIL79 83 78 69 81

78

GRAND MEAN           81.95

SOLUTION (CONTINUED) MEAN

84 4.202590 64.8025

75.8 37.822578 15.6025

81.95 40.81

BETWEEN SAMPLE VAR = 5*40.81 = 204.05

SAMPLE VARIANCEI 20II 7.5III 27.7IV 29

OVERALL VARIANCE 21.05

WITHIN SAMPLE VAR= 21.05F-RATIO 9.69

INFERENCE