ASSUMPTION OF ANALYSIS OF VARIANCE

1. ASSUMPTION OF ANALYSIS OF VARIANCE Erlina Ambarwati

2. Assumptions: All population have equal of variance Mean and variance were independent εij ~ independent N ~ (0, 1) Block and treatment effects are additive, i.e. there is no interactionbetween treatments and block Erlina Ambarwati

3. Homogenity of variance • t test σ2i homogen combined =σ2p • Analysis of varian => MSError = σ2p • How to detect the homogenity of variance more than two population? F test? Erlina Ambarwati

4. If the variance not equal? Conclusion will be bias • Hetergeneous 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: • sample size are all equal • larger sample sizes with larger variances. Erlina Ambarwati

5. Hartley’s Test (homogenity of var.)(equal replication number for CRD) • Variance of samples: σ21, σ22, σ23,…, σ2t • F max =H = ; H = Percentile of H statistic distribution H table=H[(1-);t,n] (Hartley’s table), or Percentage points of the maximum F ratio F(; k; dfmax) k, t=number of treatment n = replication

6. Example 1 Answer: Ho = 21= 22= 23= 24 H1= Not all 2 homogenous =0.05 => H(0.95;4;10)=6.31 F max= A research used four teatments, each treatments were repeated 10 time. The variance of each treatment as follow: σ21=193, σ22=146, σ23= 215, and σ24=128 Conclusion: The Variance are homogenous

7. BARTLETT’S TEST (homogenity of variance)(unequal replications for CRD) • If σ21, σ22, σ23,…,σ2t of t normal population • MSE (Aritmatic MSE)= • Geometric MSE = • GMSE MSE • If σ21 =σ22= σ23=…=σ2t GMSE = MSE Erlina Ambarwati

8. Log MSE/GMSE  2 ,df =t-1 B= 2 = n n

9. 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? Erlina Ambarwati

10. MSE= Log MSE=2.686 2(95%;3-1)=5.99 Conclusion: All varian are homogenous Erlina Ambarwati

11. BARTLETT’S TEST (homogenity of variance)(unequal replications for CRD) • If σ21, σ22, σ23,…,σ2t of t normal population •  = • Distribution of and = •   χ2 with df = (t-1) ErlinaAmbarwati

12. Levene’s test • Proposed doing a one way analysis of variance on the variables εij=|Yij-Yi.| • In Two way clasification : • Analysis of variance for εij by design of original data If the F-statistic is significant Ho rejected (Not all variances homogenous) Erlina Ambarwati

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 away analysis variance on log of variance If F-statistic is significant homogenity of variance is rejected Erlina Ambarwati

14. TEST of NORMALITY ij~N(0,1) ? 2test Lilliefors test Erlina Ambarwati

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 Erlina Ambarwati

16. 2test Example: • The telephone number drawn randomly as follow: Ho: normal distribution H1: distribution not normal Erlina Ambarwati

17. Step: 1. Made the frequency distribution Data grouped into classes: 2. Mean? Variance? Erlina Ambarwati

18. 3. Standardized upper and lower level 2tabel= (0,05;4-1-2)=3.841 Data normally distribute Erlina Ambarwati

19. Contohujinormalitas (program) Univariate Procedure Variable=TT Moments Quantiles(Def=5) N 50 Sum Wgts 50 100% Max 97 99% 97 Mean 55.04 Sum 2752 75% Q3 68 95% 89 Std Dev 19.00479 Variance 361.182 50% Med 57.5 90% 79 Skewness 0.158196 Kurtosis -0.60557 25% Q1 40 10% 30 USS 169168 CSS 17697.92 0% Min 23 5% 24 CV 34.52905 Std Mean 2.687683 1% 23 T:Mean=0 20.4786 Pr>|T| 0.0001 Range 74 Num ^= 0 50 Num > 0 50 Q3-Q1 28 M(Sign) 25 Pr>=|M| 0.0001 Mode 58 Sgn Rank 637.5 Pr>=|S| 0.0001 W:Normal 0.964217 Pr<W 0.2309 Extremes Lowest Obs Highest Obs 23( 6) 81( 30) 23( 1) 87( 35) 24( 11) 89( 40) 27( 16) 93( 45) 29( 21) 97( 50) Erlina Ambarwati

20. 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) Number of value  Xi S(x)= Total data Erlina Ambarwati

21. Frequency Distribution |F*(x)-S(x)| |F*(x)-S(x)| > Liliefor’s frequency distribution Erlina Ambarwati

22. Lilliefor’s test L max < L tabel=L(0.05;12)=0,242 Normal distributed

23. NON ADITIVITY TEST SS error= eij2 SS residual=SS error- SSNA Ho:  = 0 H1:   0 Erlina Ambarwati

24. Anova Erlina Ambarwati

26. Erlina Ambarwati

27. F(0.05; 1; 8)=5.319 Erlina Ambarwati • Dependent Variable: PROTEIN • Source DF Sum of Squares Mean Square F Value Pr > F • BLOCK 3 24.00000000 8.00000000 2.12 0.1681 • TREAT 3 104.00000000 34.66666667 9.18 0.0042 • Error 9 34.0000 3.7777777 • NA 1 7.41 7.41 3.32 0.15 • Residual 8 26.59 4.19 • Corrected Total 15 162.000

28. SOAL LATIHAN Ujilah apakah data mengikuti asumsi berdistribusi normal dengan menggunakan uji χ2 Erlina Ambarwati