Download Part III The General Linear Model. Multiple Explanatory Variables

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!
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