1. Analysis of Variance

2. Analysis of Variance : the simultaneous comparizon of several population means
F-distribution : used to test whether two samples are from population having equal variances and it is also applied when we want to compare several population means simultaneously.

3. Comparing two population variance2
Ho : σ = σ
H1 : σ = σ
Test statistic for comparing two variances :
F = ~ F- distribution 2 2 1 2 2 1 2 2 s 1 2 s 2

4. Contoh :
Sean Lammers, president of the Lammers Limos Service Company, membandingkan variasi waktu tempuh (menit) dari Cityhall di Toledo, Ohio ke Metro Airport di Detroit melalui 2 rute ( via US -25 dan Interstate-75), dengan taraf α = 10 %
US-25 route
Interstate-75 route 52 67 56 45 70 54 64 59 60 61 51 63 57 65

5. US-Route-25
Interstate-75
Rata-rata = 28.29
Std deviasi =
Rata-rata = 59.00
Std deviasi =
US-Route25 is more variation then Interstate-75 route,This is somewhat consistent with his knowledge of two route; US-25 contain more stoplights, I-75 is limited-access interstate highway, but I-75 is several miles longer.

6. Nilai F = 4.23,
Nilai tabel F distribution pada Appendix G pada buku Douglas A. Lind,dkk,Statistical Techniques in Business of Economics,International Edition,yaitu : nilai F dengan α/2 = 5 %, dan df numerator 6 dan df denominator 7 adalah sebesar 3.87.
Hasil ini menunjukkan bahwa Ho di tolak
Berikan kesimpulan anda.

7. The Anova Test
Total Variation : the sum of the squared differences between each observation and the overall mean (grand mean); SS-total
Treatment Variation : the sum of the squared differences between each treatment mean and the overall mean (grand mean); SST
Random Variation : the sum of the squared differences between each observation and its treatment mean; SSE

8. SS-total = Σ(X-XG)2
SSE = Σ(X-XC)2
SST = Σ(Xc -XG)2
SST = SS-total - SSE

9. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom
Mean Square F
Treatments
Error
Total
SST
SSE
SS- Tolal
k-1
n-k
n-1
SST/(k-1) = MST
SSE/(n-k) = MSE
MST/ MSE

10. Example :
Prof. James had students in his marketing class rate his performance as Excellent, Good, Fair or Poor. A graduate student collected the rate and from records office, Prof. James was matched with his or her course grade. The sample information is reported below. Is there a difference in the mean score of the students in each of the four rating categories ? Use the 0.01 significance level. (note : The rating is the treatment variable)
Formulasikan hipotesis, buat tabel anova, tentukan nilai F tabel
dan buat keputusan !

11. Course Grades
Excellent
Good
Fair
Poor
Total 94 90 85 80 75 68 77 83 88 70 73 76 78 65 72 74
Column
349
391
510
414
1664 n 4 5 7 6 22
Mean
87.25
78.20
72.86
69.00
75.64

12. Ho : μ1= μ2 = μ3 = μ4
H1 : The mean scores are not all equal

13. Berikan kesimpulan anda !
ANOVA Table
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square F
Treatments
Error
Total
890.68
594.41SSE
SS- Tolal 3 18 21
296.89
33.02
8.99
Nilai F tabel untuk α = 0.01, df numerator = 3 dan df denominator =18, adalah 5.09
Berikan kesimpulan anda !

14. Two-way Anova
We have the second treatment variable, that is Blocking variable
Blocking variable : A second treatment variable that when included in the ANOVA analysis will have the effect of reducing the SSE term

15. SSB = k Σ(Xb-XG)2
k is the number of treatment
b is the number of blocks
Xb is the sample mean of block b
XG is the overall or grand mean
SSE = SS total – SST - SSB

16. ANOVA Table Source of Variation Sum of Squares Degrees of Freedom
Mean Square F
Treatments
Blocks
Error
Total
SST
SSB
SSE
SS- Tolal
k-1
b-1
(k-1)(b-1)
n-1
SST/(k-1) = MST
SSB/(b-1) =MSB
SSE/(k-1)(b-1) = MSE
MST/ MSE
MSB/MSE

17. Example
The Chapin Manufacturing Company operates 24 hours a day, five days a week. The workers rotate shifts each week. Management is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the 0.05 significance level, can we conclude there is a difference in the mean production rate by shift or by employee ?

18. Employee
Units Produced
Day
Afternoon
Night
Skaff 31 25 35
Lum 33 26
Clark 28 24 30
Treece 29
Morgan 27

19. ANOVA Table For Block : For treatment : Ho :μ1 = μ2 = μ3 = μ4 = μ5
H1 : Not all means equal
Reject if F > 4.46
For Block :
Ho :μ1 = μ2 = μ3 = μ4 = μ5
H1 : Not all means equal
Reject if F > 3.84
ANOVA Table
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square F
Treatment (rotate shift)
Blocks (employee)
Error
Total
62.53
33.73
43.47
139.73 2 4 8 14
8.4325
5.4338
5.75
1.55
There is a difference in shifts but not by employees.