Post on 28-Dec-2015
Part IVThe General Linear Model.
Multiple Explanatory Variables
Chapter 13.3 Fixed *Random Effects
Paired t-test
Overview of GLM
GLM
Regression
ANOVA
ANCOVA
One-Way ANOVA
Two-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
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
• 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
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 3.4Data are means
1. Construct ModelResponse 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 ModelVerbal: Hours of extra sleep depends on drug.Graphical:
1. Construct ModelVerbal: Hours of extra sleep depends on drug.Graphical:
1. Construct ModelVerbal: Hours of extra sleep depends on drug.Graphical:
1. Construct ModelFormal:
Can we have an interaction term?
Let’s look at the df
dfDrug =
dfSubject =
dfDrug*Subject =
Dfresidual =
1. Construct ModelVerbal: Hours of extra sleep depends on drug.Graphical:
1. Construct ModelFormal:Revised Model:
2. Execute analysis
lm1 <- lm(T~XS+XD, data=sleep)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
3 1.1 B
4 0.1 B
5 -0.1 B
6 4.4 B
7 5.5 B
8 1.6 B
9 4.6 B
10 3.4 B
R: multiple ways to model random effectsInstead of lm:
lmer{lme4}lme{nlme}use aov() , specifying
Error(subject)
2. Execute analysis
1. Compute
2. Compute mean per drug mean (TD=A)= 0.75 hs
3. Compute drug effect
4. Compute mean per subject mean(TS=1)= 1.3 hs
5. Compute subject effect
6. Compute fits
7. Compute residuals residuals = T - fits
hs54.1ˆ0
hshsAD 79.0)54.175.0(ˆ
hshsS 24.0)54.13.1(ˆ1
SDfits ˆˆˆ0
hshshs SAD 24.0ˆ;79.0ˆ;54.1ˆ10
2. Execute analysis
XS T XD β0 βD βS fits res
1 0.7 A1.54
-0.79
-0.24
0.51-
0.19
2 -1.6 A1.54
-0.79
-1.94
-1.19
0.41
3 -0.2 A1.54
-0.79
-1.09
-0.34
-0.14
4 -1.2 A1.54
-0.79
-2.09
-1.34
-0.14
5 -0.1 A1.54
-0.79
-1.64
-0.89
-0.79
1 1.9 B1.54
0.79-
0.242.09 0.19
2 0.8 B1.54
0.79-
1.940.39
-0.41
3 1.1 B1.54
0.79-
1.091.24 0.14
4 0.1 B1.54
0.79-
2.090.24 0.14
5 -0.1 B1.54
0.79-
1.640.69 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?
NA
3. Evaluate model
□ Straight line model ok?
□ Errors homogeneous?
□ Errors normal?
□ Errors independent?
NA
3. Evaluate model
□ Straight line model ok?
□ Errors homogeneous?
□ Errors normal?
□ Errors independent?
NA
4. State the population and whether the sample is representative.
Drugs set by experimental design fixed effectsWe 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 n = 20
Source df SS MS F p
Subject
Drug
Res______
______
Total
7. ANOVA n = 20
Source df SS MS F p
Subject 958.078
Drug 112.482
Res___9__
_6.808
Total 19 77.37
7. ANOVA n = 20
Source df SS MS F p
Subject 958.078
6.453
Drug 112.482
12.48
Res___9__
_6.808
0.7564
Total 19 77.37
7. ANOVA n = 20
Source df SS MS F p
Subject 958.078
6.453
Drug 112.482
12.48 16.5 0.0028
Res___9__
_6.808
0.7564
Total 19 77.37
7. ANOVA n = 20Source df SS MS F p
Subject 958.078
6.453
Drug 112.482
12.48 16.5 0.0028
Res___9__
_6.808
0.7564
Total 19 77.37
Source df SS MS F p
Drug 1 12.48 12.48 3.4626 0.079
Res__18__
64.886
3.6048
Total 19 77.37
BUT we did this before Ch 10.2 2 sample t-test
STATISTICAL CONTROL
r2 = 0.91
r2 = 0.16
8. Decide whether to recompute p-value
Slight deviation from normalityn<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 =0.05Reject H0 extra sleep depends on drug administered
We did a 2 way ANOVA, also known as a paired t-test. 1 random factor1 fixed factor with 2 levels
9. Declare decision about terms
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 1.58 -0.28
-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
0.0 4.6 4.6 1.58 3.02
2.0 3.4 1.4 1.58 -0.18
Paired t-test:
1. Calculate difference within each random category
2. Test if the mean diff differs from zero
p=0.0028
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
SE LCL (5%) UCL(95%)
mean(TB-TA)=1.58 hs
0.388 0.7 hs 2.46 hs