Anova
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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