2. Part III The General Linear Model Chapter 10 GLM. ANOVA.

3. Chapter 10.1 Single Sample t-test

4. GLM, applied to ANOVA Single Sample t-test. Sleep data example William Sealy Gosset AKA Student

5. 1. Construct Model Verbal – Drug A increases time slept. Graphical model – Define quantity of interest D V G F T=T DrugA T Control –

6. 1. Construct Model D V G F Formal – Response: – Explanatory: – Formal:

7. 2. Execute analysis Data=Model+ResRes 2 T=βoβo +εε2ε2 0.7=0.75+-0.050.002 -1.6=0.75+-2.355.523 -0.2=0.75+-0.950.903 -1.2=0.75+-1.953.803 -0.1=0.75+-0.850.723 3.4=0.75+2.657.023 3.7=0.75+2.958.703 0.8=0.75+0.050.003 0=0.75+-0.750.563 2=0.75+1.251.563 ∑=0∑=28.809 ?

8. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent?

9. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

10. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

11. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

12. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

13. 4. State population and whether sample is representative. a)Differences in time slept relative to control for all possible subjects b)All possible differences in time slept relative to control, given the experimental protocol c)Differences in time slept relative to control, given the experimental protocol

14. 5. Decide on mode of inference. Is hypothesis testing appropriate? Does Drug A affect hours of sleep? – Don’t know if the answer is yes or no 6. State H A / H o pair, with tolerance for Type I error H A : H o : State test statistic: Distribution of test statistic: Tolerance for Type I error:

15. 7. ANOVA- Calculate df, SS, MS, according to model GLM:T=βoβo +ε df: 1019 SS: 34.435.6228.81

16. 7. ANOVA- Calculate df, SS, MS, according to model GLM:T=βoβo +ε df:1019 SS:34.435.6228.81 ANOVA Table dfSSMSFP-value DrugA0.218 Res Total

17. 8. Recompute p-value if necessary. Assumptions met so no need to recompute 9. Declare and report decision about model terms (compare p to α). p = 0.218 < α = 0.05, so reject H A : β o ≠ 0 Report decision: – There is no significant difference in extra time slept, for drug A (F 1,9 = 1.76, p = 0.218) – But might Type II error be a problem here?

18. There may be a difference, but it is hidden in the variance Power analysis: – Compute the minimum detectable difference T = 0.75 hours F = 1.76 p = 0.218 T = 1.00 hours F = 3.12 p = 0.111 T = 1.25 hours F = 4.88 p = 0.054 T = 1.28 hours F = 5.12 p = 0.050 9. Declare and report decision about model terms (compare p to α).

19. Another experiment, with more subjects, should be considered before concluding there is no evidence of an effect Power analysis: – Compute sample size needed to detect a difference n = 10 F = 1.76 p = 0.218 n = 20 F = 3.71 p = 0.0692 n = 24 F = 4.49 p = 0.0451 9. Declare and report decision about model terms (compare p to α).

20. Parameters are not of interest there because there appears to be no difference BUT, with this sampling effort and variability, the study needs to be repeated to be conclusive. 10. Report and interpret parameters of biological interest.

21. Chapter 10.2 Two Sample t-test

22. GLM, applied to ANOVA Two Sample t-test. Sleep data example William Sealy Gosset AKA Student

23. 1. Construct Model D V G F

24. 2. Execute analysis lm1 <- lm(diff~drug, data=drugs) diffdrugfitsres 0.7DrugA0.75-0.05 -1.6DrugA0.75-2.35 -0.2DrugA0.75-0.95 -1.2DrugA0.75-1.95 -0.1DrugA0.75-0.85 3.4DrugA0.752.65 3.7DrugA0.752.95 0.8DrugA0.750.05 0DrugA0.75-0.75 2DrugA0.751.25 1.9DrugB2.33-0.43 0.8DrugB2.33-1.53 1.1DrugB2.33-1.23 0.1DrugB2.33-2.23 -0.1DrugB2.33-2.43 4.4DrugB2.332.07 5.5DrugB2.333.17 1.6DrugB2.33-0.73 4.6DrugB2.332.27 3.4DrugB2.331.07

25. 2. Execute analysis lm1 <- lm(diff~drug, data=drugs)

26. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent?

27. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

28. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

29. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

30. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA X

31. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA X

32. 4. State population and whether sample is representative. All possible differences in time slept relative between the two groups 5. Decide on mode of inference. Is hypothesis testing appropriate? Yes. The question is whether one drug is better than the other. It is not clear whether the greater hours of sleep due to the one drug is more than just chance.

33. 6. State H A / H o pair, with tolerance for Type I error H A : H o : State test statistic: Distribution of test statistic: Tolerance for Type I error:

35. 7. ANOVA- Calculate df, SS, MS, according to model GLM:T=βoβo +β Drug Drug+ε Source:TotalDrugRes df: SS:77.3764.89

36. 8. Recompute p-value if necessary. When assumptions not met, recompute if: – n small (n = 19, so _____) – p near α (p =0.079, so _____) Colquhoun (1971) carried out a randomization test – p = 0.0813 (976/12000)

37. 9. Declare and report decision about model terms (compare p to α). p = 0.0813< α = 0.05, so reject H A Report decision: – There is no significant difference in extra time slept for the two drugs (F 1,18 = 3.46, p = 0.081) – Again, Type II error may be a problem Run a Power Analysis to guide future study

38. Parameters are not of interest there because there appears to be no difference BUT, with this sampling effort and variability, the study needs to be repeated to be conclusive. – The inclusion of 10 more samples may allow the detection of a significant difference 10. Report and interpret parameters of biological interest.

39. Chapter 10.3 One way ANOVA, Fixed Effects

40. GLM, applied to ANOVA One way ANOVA, Fixed Effects Pea section growth data, from Box 9.4 in Sokal and Rohlf (1995). Does growth depend on treatment (control versus 4 different sugars with auxin present)? = + ε

41. 1. Construct Model

42. 2. Execute analysis lm1 <- lm(len~trt, data=peas) lentrtfitsres 75Control70.14.9 67Control70.1-3.1 70Control70.1-0.1 75Control70.14.9 65Control70.1-5.1 71Control70.10.9 67Control70.1-3.1 67Control70.1-3.1 76Control70.15.9 68Control70.1-2.1 57Glucose59.3-2.3 58Glucose59.3-1.3 60Glucose59.30.7 59Glucose59.3-0.3 62Glucose59.32.7 60Glucose59.30.7 60Glucose59.30.7 57Glucose59.3-2.3 59Glucose59.3-0.3 61Glucose59.31.7 58Fructose58.2-0.2 61Fructose58.22.8 56Fructose58.2-2.2

43. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent?

44. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

45. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

46. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

47. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

48. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

49. 3. Evaluate Model □ Straight line model ok? □ Errors homogeneous? □ Errors normal? □ Errors independent? NA

50. 4. State population and whether sample is representative. Population is all possible measurements, given the method of applying treatments and the protocol for taking measurements. It is taken to be representative (i.e. not biased) 5. Decide on mode of inference. Is hypothesis testing appropriate? Yes. We want to know if there are any differences between treatments ANOVA tells us if there are ANY differences in variance among groups

51. 6. State H A / H o pair Research Hypothesis (H A ) Treatment effects differ μ C ≠ μ G ≠ μ F ≠ μ G+F ≠ μ S OR var( β Trt Trt) > 0 Null Hypothesis (H o ) Treatment effects do not differ μ C = μ G = μ F = μ G+F = μ S OR var( β Trt Trt) = 0

52. 6. H A / H o pairs for planned comparisons 5·5 = 25 possible comparisons We usually have some expectations about the direction of the contrasts among groups – Based on these, we can undertake planned or a priori comparisons – Examples: H A : β C ≠ (1/4)(β G + β F + β G+F + β S ) [Control vs. Treatment] H A : β G+F ≠ (1/3)(β G + β F + β S )[Mixed vs. Pure] H A : β S ≠ (1/2)(β G + β F )[Poly vs Mono]

53. 6. H A / H o pair, test statistic, distribution, tolerance of Type I error State test statistic :F-ratio Distribution of test statistic: F-distribution Tolerance for Type I error: 5% BUT we need to adjust α for planned comparisons – 5% = 1 in 20; hence, we would likely reject one true H A if we did 20 comparisons – Adjust α level for three comparisons (use Dunn- Sidak method) α expwise = 1 – (1 – α) k = 1 – (1 – 0.05) 3 = 0.017

54. 7. ANOVA GLM:Len – β o =β Trt Trt+ε Source:TotalTrtError df:n – 1 49 5-1 4 49-4 45 SS:1322.8245.51077.3 n = 50

55. 8. Recompute p-value if necessary. Assumptions met, skip 9. Declare decision about model terms. p < 0.001 p < 0.05 so accept H A That μ C ≠ μ G ≠ μ F ≠ μ G+F ≠ μ S Pea section length differs significantly among the 5 groups (control and 4 treatments). – F 4,45 = 49.37, p < 0.0001

56. Where are the differences, among the 5 groups? – Two approaches: A priori A posteriori First planned comparison: Growth in treated media differs from that in untreated 10. Report and interpret parameters of biological interest.

57. Run t-test on Control vs. Treatment 10. Report and interpret parameters of biological interest. lentrtcomp1 75Control 67Control 70Control 75Control 65Control 71Control 67Control 67Control 76Control 68Control 57GlucoseTreatment 58GlucoseTreatment 60GlucoseTreatment 59GlucoseTreatment 62GlucoseTreatment 60GlucoseTreatment 60GlucoseTreatment 57GlucoseTreatment 59GlucoseTreatment 61GlucoseTreatment 58FructoseTreatment Analysis of Variance Table Response: len Df Sum Sq Mean Sq F value Pr(>F) comp1 1 832.32 832.32 81.45 6.516e-12 *** Residuals 48 490.50 10.22 lm2 <- lm(len~comp1, data=peas) anova(lm2)

58. First work out UCL ≤ mean ≤ LCL Control: 67.6 ≤ 70.1 ≤ 72.6 units (n=40) Treatment: 58.9 ≤ 59.9 ≤ 60.9 units (n=40) Then work out degree of difference (μ Con – μ Trt )/ μ Con (70.1 - 59.9)/70.1 = 15% Sugar supressed growth by 15% 10. Report and interpret parameters of biological interest.

59. Conclusions from the 3 planned comparisons: – A 2% sugar solution reduces growth F 1,45 = 152.564, α = 0.017 > p < 0.0001 – Mixed glucose + fructose reduces growth relative to pure sugars F 1,45 = 8.82, α = 0.017 > p = 0.00476 – The monosaccharides (fructose, glucose) suppress growth more than the polysaccharide (sucrose) F = 34.98, α = 0.017 > p = 0.00000417 Conclusions reaffirmed using CL 10. Report and interpret parameters of biological interest.

60. Chapter 10.4 One way ANOVA, Random Effects

61. GLM, applied to ANOVA One way ANOVA, Random Effects Example. Box 9.1 of Sokal and Rohlf 1995, p. 210. Does tick size, as measured by scutum width, differ among hosts (rabbits)? Random effects example – contrast with fixed effects

62. 1. Construct Model

63. β H = [ +12.55, -5.33, -4.4, +1.6] β o = 359.7 SS Total

64. β H = [ +12.55, -5.33, -4.4, +1.6] β o = 359.7 SS H

65. β H = [ +12.55, -5.33, -4.4, +1.6] β o = 359.7 SS Res

66. 2.Execute analysis. 3.Evaluate model. 4.State the population and whether the sample is representative. – Fixed vs. Random Fixed: All possible measurement of scutum widths from ticks on these four rabbits only – A fixed factor has levels that are the only ones of interest (e.g. different sugar treatments). Random: All possible measurement of scutum widths from ticks found on all possible rabbits – A random factor has levels that are considered a sample from some larger population of levels (such as rabbits)

67. More Fixed vs. Random Because we want to infer to a larger population level Because we’re usually interested in specific contrasts Depends on context

68. 5.Decide on mode of inference. Is hypothesis testing appropriate? 6.State H A / H o pair, test statistic, distribution, tolerance for Type I error. – Q: Is additional variation in size due to their host? – H A : Var(β H ·H) > 0 – H o : Var(β H ·H) = 0 7.ANOVA - Compute and partition the df in the response variable according to the model 0.004 Recommendation: Cross-check your workings with the computer generated ANOVA table. This ensures the computer did what you wanted it to do!

69. 8.Recompute p-value by randomization if necessary. 9.Declare and report statistical decision, with evidence – The variance among hosts exceeds variance within hosts (F 3,33 = 5.26, p = 0.004) 10.Report and interpret parameters of biological interest. – In this example the interest was in whether there was variance among the hosts. – There was no stated interest in which hosts differed, or by how much.