1. Assumptions of Variance Analysis
12/2/2018
By Rudi HM

2. Assumptions: All population have homogenous variance
Mean and variance are independent
eij ~ independent N(0, 1)
Block and treatment effects are additive, i.e. no interaction between treatments and block
12/2/2018
By Rudi HM

3. Homogenity of variance
t test 2i homogen combined = 2p
Analysis of varian => MSError = 2pooled
How to detect the homogenity of variance > two population? F test?
12/2/2018
By Rudi HM

4. If the variance not equal?
Heterogeneous error variances pose much more serious problem than non normality of the error variances.
If ignore the unequal variances and calculate the same F or t test ?
The usual test are quite good if:
sampel size are all equal
larger sample sizes with larger variances.
Conclusion will be bias
12/2/2018
By Rudi HM

5. Hartley’s Test (homogenity of var.)
Variance of samples: 21, 22, 23,…, 2t
F max =H =
Percentile of H statistic distribution
H table=H((1-);t,r) or
Percentage points of the maximum F ratio
F(; k; dfmax)
k, t=number of treatment
r= replication
12/2/2018
By Rudi HM

6. The Variance are homogenous
Example 1
A research used four teatments, each treatments were repeated 10 time. The variance of each treatment as follow:
21=193, 22=146, 23= 215, and 24=128
Answer:
Ho = 21= 22= 23= 24
H1= Not all 2 homogenous
=0.05 => H(0.95;4;10)=6.31
F max=
Conclusion:
The Variance are homogenous
12/2/2018
By Rudi HM

7. Example
Data about time required for production of component of car in a plant with three shift. The employee in each shift: 20, 17, 21 person.
What is the all variance homogenous?
Shift
2i 1
415 2
698 3
384
12/2/2018
By Rudi HM

8. BARTLETT’S TEST (homogenity of variance)
If 21, 22, 23,…,2t of r normal population
MSE (Aritmatic MSE)=
Geometric MSE =
GMSE MSE
If 21 = 22= 23=…=2t GMSE = MSE
12/2/2018
By Rudi HM

9. Log MSE/GMSE  2 , df =t-1
B= 2 =
12/2/2018

10. Example
Data about time required for production of component of car in a plant with three shift. The employee in each shift: 20, 17, 21 person.
What is the all variance homogenous?
Shift
σ2i 1
415 2
698 3
384
12/2/2018

11. 2(95%;3-1)=5.99 Conclusion: All varian are homogenous
Shift
2i
ri-1
(ri-1) 2i
Log 2i
(ri-1)log 2i 1
415 19
7885
2.618
49.74 2
698 16
11168
2.843
45.50 3
384 20
7680
2.584
51.68  55
26733
146.92
MSE=
Log MSE=2.686
2(95%;3-1)= Conclusion: All varian are homogenous
12/2/2018
By Rudi HM

12. Levene’s test
Proposed doing a one way analysis of variance on the variables eij=|Yij-Yi|
In Two way clasification :
Analysis of variance for eij by design of original data
If the F-statistic is significant
Maap: Ho rejected (Not all variances homogenous)
12/2/2018
By Rudi HM

13. Steps of Box’s test
The sample corresponding to each treatment group is partitioned into subsamples (equal size and random)
The variance of each sample is determined logs of variances
One way analysis variance on log of variance
If F-statistic is significant homogenity of variance is rejected
12/2/2018
By Rudi HM

14. TEST of NORMALITY ij~N(0,1) ? 2 test Lilliefors test 12/2/2018
By Rudi HM

15. Step of 2 test for normality
Make the frequency distribution of data
Determined: the number of data include in each classes
Determined: mean and variance
Upper and lower level of each classes are standardized
Probability of each classes?
Expected frequency of each classes?
2 test: Observed and expected significant
difference?
H0: ij~N
H1: Data distribution not normal
12/2/2018
By Rudi HM

16. 2 test
Example:
The telephone number drawn randomly as follow: 23 24 27 29 31 32 33 35 36 37 40 42 43 44 45 48 54 56 57 58 59 61 62 63 64 65 66 68 70 73 74 75 77 81 87 89 93 97
Ho: normal distribution
H1: distribution not normal
12/2/2018
By Rudi HM

17. Step: 1. Made the frequency distribution Data grouped into classes:
2. Mean? Variance?
Class
Frequency
20-40 12
40-60 18
60-80 15
80-100 5  50
12/2/2018
By Rudi HM

18. 3. Standardized upper and lower level
Class
Xp=(Xi-X)/
F(Xp) P*
20-40
-1.88 (~)
-0.813
0.21
40-60
0.256
0.60
0.39
60-80
1.33
0.91
0.31
80-100
2.4 1
0.09
Class
Expected (Ei)
Observed (Oi) P*
(Ei-Oi)2/Ei
<40
0.21x50=10.5 12
0.21
0,214
40-60
0.39x50=19.5 18
0.39
0,115
60-80
0.31x50=15.5 15
0.31
0,016
>100
0.09x50=4.5 5
0.09
0,056 
0,401
2tabel= (0,05;4-1-2)= Data normally distribute
12/2/2018
By Rudi HM

19. Lilliefor’s Test (Harus diperbaiki)
Draw a graph of the standard normal distribution function (F*(x)) and also empirical distribution function of normalized sample (Zi)
Find the maximum vertical distance between two graph F*(x) and S(x)
Srandardized data=
F(Zi)=P(Z<Zi)
S(x)=Number of value  Xi
Total data
12/2/2018
By Rudi HM

20. Frequency Distribution
|F*(x)-S(x)|
|F*(x)-S(x)| > Liliefor’s frequency distribution
12/2/2018
By Rudi HM

21. Lilliefor’s test Normal distribution Xi Zi F(Zi) S(Zi)
Lo=|F(Zi)-S(Zi)| 23
-1.65
0.0495
0.088
0.0338 27
-1.41
0.0793
0.1667
0.0874 33
-1.05
0.1469
0.2500
0.1031 40
-0.62
0.2676
0.3333
0.0657 48
-0.14
0.4443
0.500
0.0557 57
0.40
0.6554
0.5833
0.0721 59
0.53
0.7019
0.6667
0.0352 62
0.71
0.7612
0.7500
0.0112 68
1.07
0.8577
0.8333
0.0244 69
1.13
0.8708
0.9167
0.0459 70
1.19
0.8830 1
0.1170
X=50.3 , =16.55
12/2/2018
By Rudi HM
L max < L tabel=L(0.05;12)
Normal distribution

22. NON ADITIVITY TEST SS residual=SS error- SSNA SS error=  eij2
Ho:  = 0
H1:   0
12/2/2018
By Rudi HM

23. Anova Source of Variance df SS MS F calc F table Block (r-1) SSB MSB
MSB/MSE
Treatment
(t-1)
SST
MST
MST/MSE
Error
(r-1)(t-1)
SSE
MSE NA 1
SSNA
MSNA
MSNA/MSPE
Pure error
[(r-1)(t-1)]-1
SSPE
MSPE
Total
rt-1
12/2/2018

24. i j Xij m di bJ eij di2 bJ2 dibjYij 7,41 1 18 10 4 2 16 144 13 -1 -52
-52 3 11 -2
-44 14 8 -3 9
-48 24 5 15 7
-20 12
S  ( ….)
160
104 34
S( …..)2
1762
1600
SSNA =
7,41
TSS CF
SST
 SSB
SSE
SS residu=
26,59
12/2/2018
By Rudi HM

25. 12/2/2018
By Rudi HM

26. Dependent Variable: PROTEIN
Source DF Sum of Squares Mean Square F Value Pr > F
BLOCK
TREAT
Error NA
Residual
Corrected Total
F(0.05; 1; 8)=5.319
12/2/2018
By Rudi HM