1. Analyses of Variance

2. Simple Situation Genotype AGenotype B 13534

3. Simple Situation Genotype AGenotype B 13534 11576 10283 11064 115.564.2

4. Observations and Questions  From the replicated means, genotype A is “better” than genotype B.  What is the probability that this result will be repeated if this test were done say 100 times?  Could this result have occurred by random chance?

5. t-test |x 1 -x 2 |  2[(  1 2 +  2 2 )/(n 1 +n 2 )] t =t =t =t =

6. Replicate Genotype StephensLambart 15578 26691 34997 46482 57085 66877 t-test

7. Replicate Genotype StephensLambart 123456556649647068789197828577  x mean x 3726251085

8. Replicate Genotype StephensLambart 123456556649647068789197828577  x mean x  x 2 (  x) 2 /n SS(x)3726223,40223,0643385108543,65243,350302

9. Replicate Genotype StephensLambart 123456556649647068789197828577  x mean x  x 2 (  x) 2 /n SS(x)df3726223,40223,06433855108543,65243,3503025 222276.660.4

10. t-test Stephens = 62 bushels Lambart = 85 bushels Significant?

11. t-test Stephens = 62 bushels Lambart = 85 bushels |x Stephens -x Lambart |  2[(  St 2 +  La 2 )/(n St +n La )] t =t =t =t = t = |85-62|/  2[(60+77)/12] = 4.82 cw t 10df : exceeds 99% table value

12. More than two treatments Rep. Genotype BrundageLambertCroftStephens 164787555 272919366 368977849 477827164 556856370 695777668

13. Multiple t-tests  Brundage v Lambert; Brundage v Croft; Brundage v Stephens; Lambert v Croft; Lambert v Stephens; Croft v Stephens.  Problems?  If all tests were done at 95% significance level, and one difference was significant, we have done 6 tests and would expect 1/20 to be significant, at random.

14. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295

15. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295  x 2 (  x) 2 /n ss(x) df 31,994 31,104 890 5 43,652 43,350 302 5 35,144 43,656 488 5 23,402 23,064 338 5 134,192 132,174 2,018 20

16. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295  x 2 (  x) 2 /n ss(x) df 31,994 31,104 890 5 43,652 43,350 302 5 35,144 43,656 488 5 23,402 23,064 338 5 134,192 132,174 2,018 20 22 178.060.497.676.6100.9

17. Analysis of Variance  From pooled variance we can estimate a pooled SED between any two means =  (2)(100.9)/6 = + 5.80, and use this in all t-tests.  Alternatively an analysis of variance could be carried out.  This form of analysis was first proposed by Fisher in the 1920’s

18. Analysis of Variance  Is an elegant and quicker way to calculate a pooled error term.  Analysis is simple in simple designs but can be complicated and lengthy in some designs (i.e. rectangular lattices).  In some experimental designs the ANOVA is the only method to estimate a pooled error term.

19. Analysis of Variance  It can provide an F-test to tests specific hypotheses. (i.e. to test general differences between different treatments).  Can be an invaluable initial contribution to interpretation of experiments.

20. Theory of Analysis of Variance  Consider a simple CRB design.  Four treatments (n = 4).  With all treatments replicated 5 times (k = 5).  The total experiment would be n x k = 20 experimental units.

21. Theory of Analysis of Variance Rep. Treatment ABCD 1x 11 x 21 x 31 x 41 2x 12 x 22 x 32 x 42 3x 13 x 23 x 33 x 43 4x 14 x 24 x 34 x 44 5x 15 x 25 x 35 x 45

22. Theory of Analysis of Variance Rep. Treatment Mean ABCD 1x 11 x 21 x 31 x 41 x.1 2x 12 x 22 x 32 x 42 x.2 3x 13 x 23 x 33 x 43 x.3 4x 14 x 24 x 34 x 44 x.4 5x 15 x 25 x 35 x 45 x.5

23. Theory of Analysis of Variance Rep. Treatment Mean ABCD 1x 11 x 21 x 31 x 41 x.1 2x 12 x 22 x 32 x 42 x.2 3x 13 x 23 x 33 x 43 x.3 4x 14 x 24 x 34 x 44 x.4 5x 15 x 25 x 35 x 45 x.5 Meanx 1. x 2. x 3. x 4. x..

24. Theory of Analysis of Variance TSS =  i  j (x ij -x.. ) 2 TMS =  i  j (x ij -x.. ) 2 /(nk-1)  i  j (x ij -x.. ) 2 =  i  j [(x ij -x i. ) + (x i. -x.. )] 2  i  j [(x ij -x i. ) 2 +2(x ij -x i. )(x i. -x.. )+(x i. -x.. ) 2 ]

25. Theory of Analysis of Variance  i  j [(x ij -x i. ) 2 +2(x ij -x i. )(x i. -x.. )+(x i. -x.. ) 2 ] 2  i  j (x ij -x i. )(x i. -x.. )  i [2n (x i. -x.. )  j (x ij -x i. )] But!  j (x ij -x i. ) = 0 2  i  j (x ij -x i. )(x i. -x.. ) = 0

26. Theory of Analysis of Variance  i  j [(x ij -x i. ) 2 +2(x ij -x i. )(x i. -x.. )+(x i. -x.. ) 2 ]  i  j [(x ij -x i. ) 2 + (x i. -x.. ) 2 ]  i  j (x ij -x.. ) 2 =  i  j (x ij -x i. ) 2 +k  i (x i. -x.. ) 2 ] k  i (x i. -x.. ) 2 = Between Treatment SS  i  j (x ij -x i. ) 2 = Within Treatment SS

27. Theory of Analysis of Variance df [WTSS] = nk-n : df [BTSS] = n-1 MS = SS/df WTMS ~  2 nk-n df : B TMS ~  2 n-1 df k  i (x i. -x.. ) 2 = Between Treatment SS  i  j (x ij -x i. ) 2 = Within Treatment SS

28. Theory of Analysis of Variance WTMS ~  2 nk-n df : B TMS ~  2 n-1 df Y ij =  + g i + e ij g i = BTMS : e ij = WTMS Assumption is homogeneity of error variance between treatments.

29. Theory of Analysis of Variance Source of variationdfEMS Between treatmentsn-1  e 2 + k  t 2 Within treatmentsnk-n e2e2 Totalnk-1 [  e 2 + k  t 2 ]/  e 2 = 1, if k  t 2 = 0

30. Analysis of Variance of CRB SourcedfSS Between treatments k-1[G 1 2 /n 1 + G 2 2 /n 2 … G k 2 /n k ] - CF Within treatments jk-kBy difference Totaljk-1[x 11 2 + x 12 2 + … + x jk 2 ] - CF CF = [  x ij ] 2 /jk

31. More than two treatments Rep. Genotype BrundageLambertCroftStephens 164787555 272919366 368977849 477827164 556856370 695777668

32. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295 TSS = ∑(64 2 + 72 2 + 68 2 +.... + 68 2 ) - CF CF = ∑(64 + 72 + 68 +.... + 68) 2 /24 BSS = ∑(432 2 /6 + 510 2 /6 + 456 2 /6 + 372 2 /6) - CF WSS = TSS - BSS

33. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295 TSS = ∑(64 2 + 72 2 + 68 2 +.... + 68 2 ) - CF CF = ∑(64 + 72 + 68 +.... + 68) 2 /24 BSS = ∑(432 2 + 510 2 + 456 2 + 372 2 )/6 - CF WSS = TSS - BSS

34. More than two treatments # Genotype Total BrundageLambertCroftStephens xxxx 432 72 510 85 456 76 372 62 1,770 295 TSS = ∑(64 2 + 72 2 + 68 2 +.... + 68 2 ) - CF CF = ∑(64 + 72 + 68 +.... + 68) 2 /24 BSS = ∑(432 2 /6 + 510 2 /6 + 456 2 /6 + 372 2 /6) - CF WSS = TSS - BSS

35. Example of Analysis of Variance SourcedfSSMSF Between genotypes31636.5545.55.41 ** Within genotypes202018.0100.9 Total233654.5 ** = 0.01 > P > 0.001

36. Rep. Treatment ABCD 1x 11 x 21 x 31 x 41 2x 12 x 22 x 32 x 42 3x 13 x 23 x 33 x 43 4-x 24 x 34 x 44 5--x 35 - Analysis of Variance of CRB

37. Rep. Treatment ABCD 1x 11 x 21 x 31 x 41 2x 12 x 22 x 32 x 42 3x 13 x 23 x 33 x 43 4-x 24 x 34 x 44 5--x 35 - TotalG1G1 G2G2 G3G3 G4G4 Analysis of Variance of CRB

38. Rep. Treatment ABCD 1x 11 x 21 x 31 x 41 2x 12 x 22 x 32 x 42 3x 13 x 23 x 33 x 43 4-x 24 x 34 x 44 5--x 35 - TotalG1G1 G2G2 G3G3 G4G4 Analysis of Variance of CRB

39. SourcedfSS Between treatments k-1[G 1 2 /n 1 + G 2 2 /n 2 … G k 2 /n k ] - CF Within treatments jk-kBy difference Totaljk-1[x 11 2 + x 12 2 + … + x jk 2 ] - CF CF = [  x ij ] 2 /jk

40. Assumptions behind the ANOVA  Assumption of data being normally distributed.  Homogeneity of error variance.  Additivity of variance effects.  Data collected from a properly randomized experiment.

41. Dealing with Wrongful Data  It is usually assumed that the data collected is correct!.  Why would data not be correct? Mis-recording, mis-classification, transcription errors, errors in data entry. Mis-recording, mis-classification, transcription errors, errors in data entry. Outliers. Outliers.

42. Dealing with Wrongful Data  What things can help? Keep detailed records, on each experimental unit. Keep detailed records, on each experimental unit. Decide beforehand what values would arouse suspision. Decide beforehand what values would arouse suspision.

43. Dealing with Wrongful Data  What do you do with suspicios data? If correct, and it is discarded, then valuable information is lost. This will bias the results. If correct, and it is discarded, then valuable information is lost. This will bias the results. If wrong and included, will bias results and may have extreme consequences. If wrong and included, will bias results and may have extreme consequences.

44. Checking ANOVA Accurucy  Coefficient of variation: [  e /  ]x100.  CV=(√100.9/73.75)*100=13.6%  R 2 value = {[TSS-ESS]/TSS}x100.  R 2 = (1654/3654)*100 = 44.7%.  Compare the effect of blocking or sub-blocking (discussed later).

45. Next Class ANOVA of RCB Designs ANOVA of Latin Square Designs