1. Part IV The General Linear Model. Multiple Explanatory Variables Chapter 13.3 Fixed *Random Effects Paired t-test

2. Overview of GLM GLM RegressionANOVAANCOVAOne-Way ANOVATwo-Way ANOVA  Simple regression  Multiple regression  Two categories (t-test)  Multiple categories - Fixed (e.g., treatment, age) - Random (e.g., subjects, litters)  2 fixed factors  1 fixed & 1 random (e.g., Paired t-test) Multi-Way ANOVA

3. GLM: Paired t-test  Two factors (2 explanatory variables on a nominal scale)  One fixed (2 categories)  The other random (many categories) + Fixed factor Random factor Remove var. among units → sensitive test

4. Sleep data example, used by W. Gosset (1908) in the paper that introduced the t-test Are the effects of 2 sleep inducing drugs: Hyoscyamine (Drug A) and L Hyoscine (Drug B), controlled for among subject variation, different? GLM | Paired t-test SubjectDrugADrug B 10.71.9 2-1.60.8 3-0.21.1 4-1.20.1 5-0.1 63.44.4 73.75.5 80.81.6 90.04.6 102.03.4 Data are means

5. 1. Construct Model Response variable: T=hours of extra sleep – ratio scale Explanatory variables: 1. Drug. X D = Drug A, Drug B. Nominal scale Fixed effect 2. Subject. X S = [1,2,…,10]. Nominal scale Random effect Mean value for each subject varies randomly and is not under the control of the investigator

6. 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical:

7. 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical:

8. 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical:

9. 1. Construct Model Formal: Can we have an interaction term? Let’s look at the df df Drug = df Subject = df Drug*Subject = Df residual =

10. 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical:

11. 1. Construct Model Formal: Revised Model:

12. 2. Execute analysis lm1 <- lm(T~XS+XD, data=sleep) XSTXD 10.7A 2-1.6A 3-0.2A 4-1.2A 5-0.1A 63.4A 73.7A 80.8A 90.0A 102.0A 11.9B 20.8B 31.1B 40.1B 5-0.1B 64.4B 75.5B 81.6B 94.6B 103.4B R: multiple ways to model random effects Instead of lm: lmer{lme4} lme{nlme} use aov(), specifying Error(subject)

13. 2. Execute analysis 1.Compute 2.Compute mean per drug mean (T D=A )= 0.75 hs 3.Compute drug effect 4. Compute mean per subject mean(T S=1 )= 1.3 hs 5. Compute subject effect 6. Compute fits 7. Compute residualsresiduals = T - fits

14. 2. Execute analysis XSTXD β0β0 βDβD βSβS fitsres 10.7A1.54-0.79-0.240.51-0.19 2-1.6A1.54-0.79-1.94-1.190.41 3-0.2A1.54-0.79-1.09-0.34-0.14 4-1.2A1.54-0.79-2.09-1.34-0.14 5-0.1A1.54-0.79-1.64-0.89-0.79 11.9B1.540.79-0.242.090.19 20.8B1.540.79-1.940.39-0.41 31.1B1.540.79-1.091.240.14 40.1B1.540.79-2.090.240.14 5-0.1B1.540.79-1.640.690.79

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

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

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

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

19. 4.State the population and whether the sample is representative. Drugs set by experimental design  fixed effects We will infer only to those drugs Subjects, chosen at random. Hopefully from a larger population  random effects Population of all possible measurements of hours of extra sleep, given the mode of collection Infer to a population of subjects with characteristics similar to those in the study

20. 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. – Assume no interaction, i.e. effect of drug is consistent across subjects – Focus on drug effect

21. 6.State H A / H o pair, test statistic, distribution, tolerance for Type I error. Test Statistic Distribution of test statitstic Tolerance for Type I error

22. 7. ANOVA n = 20 SourcedfSSMSFp Subject Drug Res______ Total

23. 7. ANOVA n = 20 SourcedfSSMSFp Subject958.078 Drug112.482 Res___9___6.808 Total1977.37

24. 7. ANOVA n = 20 SourcedfSSMSFp Subject958.0786.453 Drug112.48212.48 Res___9___6.8080.7564 Total1977.37

25. 7. ANOVA n = 20 SourcedfSSMSFp Subject958.0786.453 Drug112.48212.4816.50.0028 Res___9___6.8080.7564 Total1977.37

26. 7. ANOVA n = 20 SourcedfSSMSFp Subject958.0786.453 Drug112.48212.4816.50.0028 Res___9___6.8080.7564 Total1977.37 SourcedfSSMSFp Drug112.48 3.46260.079 Res__18__64.8863.6048 Total1977.37 BUT we did this before  Ch 10.2 2 sample t-test STATISTICAL CONTROL r 2 = 0.91 r 2 = 0.16

27. 8. Decide whether to recompute p-value Slight deviation from normality n<30, p=0.0028 not near α  no need to recompute

28. 9.Declare decision about terms Only the fixed term was tested p=0.0028 < α =0.05 Reject H 0  extra sleep depends on drug administered We did a 2 way ANOVA, also known as a paired t-test. 1 random factor 1 fixed factor with 2 levels

29. 9.Declare decision about terms ABA-Bfitsres 0.71.91.21.58-0.38 -1.60.82.41.580.82 -0.21.11.31.58-0.28 -1.20.11.31.58-0.28 -0.1 0.01.58-1.58 3.44.41.01.58-0.58 3.75.51.81.580.22 0.81.60.81.58-0.78 0.04.6 1.583.02 2.03.41.41.58-0.18 Paired t-test: 1.Calculate difference within each random category 2.Test if the mean diff differs from zero p=0.0028

30. 10.Report and interpret parameters of biological interest Means per drug, not controlled for among subject variation SELCL (5%)UCL(95%) mean(T A )=0.75 hs0.5657-0.53 hs2.03 hs mean(T B )=2.33 hs0.63320.89 hs3.76 hs Confidence limits for the average difference, controlled for among subject variation SELCL (5%)UCL(95%) mean(T B -T A )=1.58 hs 0.3880.7 hs2.46 hs

32. Quizz 7 Good luck! Clock