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Part IV The General Linear Model. Multiple Explanatory Variables Chapter 13.3 Fixed *Random Effects Paired t-test Overview of GLM Regression Simple regression Multiple regression ANOVA Two categories (t-test) GLM One-Way ANOVA Multiple categories - Fixed (e.g., treatment, age) - Random (e.g., subjects, litters) Two-Way ANOVA 2 fixed factors 1 fixed & 1 random (e.g., Paired t-test) Multi-Way ANOVA ANCOVA GLM: Paired t-test Two factors (2 explanatory variables on a nominal scale) One fixed (2 categories) The other random Fixed factor + (many categories) Random factor Remove var. among units → sensitive test GLM | Paired t-test • 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? Data are means Subject DrugA Drug B 1 0.7 1.9 2 -1.6 0.8 3 -0.2 1.1 4 -1.2 0.1 5 -0.1 -0.1 6 3.4 4.4 7 3.7 5.5 8 0.8 1.6 9 0.0 4.6 10 2.0 3.4 1. Construct Model Response variable: T=hours of extra sleep – ratio scale Explanatory variables: 1. Drug. XD = Drug A, Drug B. Nominal scale Fixed effect 2. Subject. XS = [1,2,…,10]. Nominal scale Random effect Mean value for each subject varies randomly and is not under the control of the investigator 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical: 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical: 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical: 1. Construct Model Formal: Can we have an interaction term? Let’s look at the df dfDrug = dfSubject = dfDrug*Subject = Dfresidual = 1. Construct Model Verbal: Hours of extra sleep depends on drug. Graphical: 1. Construct Model Formal: Revised Model: 2. Execute analysis XS T XD 1 0.7 A 2 -1.6 A 3 -0.2 A 4 -1.2 A 5 -0.1 A 6 3.4 A 7 3.7 A 8 0.8 A 9 0.0 A 10 2.0 A 1 1.9 B 2 0.8 B lm1 <- lm(T~XS+XD, data=sleep) R: multiple ways to model random effects Instead of lm: lmer{lme4} lme{nlme} use aov() , specifying Error(subject) 2. Execute analysis 1. Compute ˆ0 1.54 hs 2. Compute mean per drug 3. Compute drug effect mean (TD=A)= 0.75 hs ˆ D A (0.75 1.54)hs 0.79 hs 4. Compute mean per subject 5. Compute subject effect 6. Compute fits 7. Compute residuals mean(TS=1)= 1.3 hs ˆ S 1 (1.3 1.54)hs 0.24 hs fits ˆ0 ˆ D ˆS residuals = T - fits 2. Execute analysis ˆ0 1.54 hs ; ˆ D A 0.79 hs ; ˆ S 1 0.24 hs β0 βD βS XS T XD fits res 1 0.7 A 1.54 -0.79 -0.24 0.51 -0.19 2 -1.6 A 1.54 -0.79 -1.94 -1.19 0.41 3 -0.2 A 1.54 -0.79 -1.09 -0.34 -0.14 4 -1.2 A 1.54 -0.79 -2.09 -1.34 -0.14 5 -0.1 A 1.54 -0.79 -1.64 -0.89 -0.79 1 1.9 B 1.54 0.79 -0.24 2.09 0.19 2 0.8 B 1.54 0.79 -1.94 0.39 -0.41 3 1.1 B 1.54 0.79 -1.09 1.24 0.14 4 0.1 B 1.54 0.79 -2.09 0.24 0.14 5 -0.1 B 1.54 0.79 -1.64 0.69 0.79 3. Evaluate model □ Straight line model ok? □ Errors homogeneous? □ □ Errors normal? Errors independent? 3. Evaluate model □ Straight line model ok? □ Errors homogeneous? □ □ Errors normal? Errors independent? 3. Evaluate model □ Straight line model ok? □ Errors homogeneous? □ □ Errors normal? Errors independent? 3. Evaluate model □ Straight line model ok? □ Errors homogeneous? □ □ Errors normal? Errors independent? 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 5. Decide on mode of inference. Is hypothesis testing appropriate? 6. State HA / Ho 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 6. State HA / Ho pair, test statistic, distribution, tolerance for Type I error. Test Statistic Distribution of test statitstic Tolerance for Type I error 7. ANOVA Source n = 20 df SS Subject Drug Res Total ______ ______ MS F p 7. ANOVA n = 20 Source df SS Subject 9 58.078 Drug 1 12.482 Res Total ___9__ _6.808 19 77.37 MS F p 7. ANOVA n = 20 Source df Subject 9 58.078 6.453 Drug 1 12.482 12.48 Res Total SS MS ___9__ _6.808 0.7564 19 77.37 F p 7. ANOVA n = 20 Source df Subject 9 58.078 6.453 Drug 1 12.482 12.48 Res Total SS MS ___9__ _6.808 0.7564 19 77.37 F p 16.5 0.0028 7. ANOVA n = 20 Source df Subject 9 58.078 6.453 Drug 1 12.482 12.48 Res Total SS MS F p 16.5 0.0028 ___9__ _6.808 0.7564 19 r2 = 0.91 77.37 BUT we did this before Ch 10.2 2 sample t-test Source Drug Res Total df SS MS F p 1 12.48 12.48 3.4626 0.079 __18__ 64.886 3.6048 19 77.37 r2 = 0.16 8. Decide whether to recompute p-value Slight deviation from normality n<30, p=0.0028 not near α no need to recompute 9. Declare decision about terms Only the fixed term was tested p=0.0028 < α =0.05 Reject H0 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 9. Declare decision about terms Paired t-test: 1. Calculate difference within each random category 2. Test if the mean diff differs from zero A B A-B fits res 0.7 1.9 1.2 1.58 -0.38 -1.6 0.8 2.4 1.58 0.82 -0.2 1.1 1.3 -1.2 0.1 1.3 1.58 -0.28 -0.1 -0.1 0.0 1.58 -1.58 3.4 4.4 1.0 1.58 -0.58 3.7 5.5 1.8 1.58 0.22 0.8 1.6 0.8 1.58 -0.78 1.58 p=0.0028 -0.28 10.Report and interpret parameters of biological interest Means per drug, not controlled for among subject variation SE LCL (5%) UCL(95%) mean(TA)=0.75 hs 0.5657 -0.53 hs 2.03 hs mean(TB)=2.33 hs 0.6332 0.89 hs 3.76 hs Confidence limits for the average difference, controlled for among subject variation mean(TB-TA)=1.58 hs SE LCL (5%) UCL(95%) 0.388 0.7 hs 2.46 hs Quizz 7 Good luck! Clock