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

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Part IVThe General Linear Model. Multiple
Explanatory VariablesChapter 13.3 Fixed
Random EffectsPaired t-test
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Overview of GLM

GLM
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GLM Paired t-test

• Two factors (2 explanatory variables on a nominal
  scale)
• One fixed (2 categories)
• The other random (many categories)

Random factor Remove var. among units ?
sensitive test
Fixed factor
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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?

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
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1. Construct Model

• Response variable Thours 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

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1. Construct Model

• Verbal Hours of extra sleep depends on drug.
• Graphical

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1. Construct Model

• Verbal Hours of extra sleep depends on drug.
• Graphical

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1. Construct Model

• Verbal Hours of extra sleep depends on drug.
• Graphical

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1. Construct Model

• Formal

Can we have an interaction term? Lets look at
the df dfDrug dfSubject
dfDrugSubject Dfresidual
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1. Construct Model

• Verbal Hours of extra sleep depends on drug.
• Graphical

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1. Construct Model

• Formal
• Revised Model

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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
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
lm1 lt- lm(TXSXD, datasleep)
R multiple ways to model random effects Instead
of lm lmerlme4 lmenlme use aov() ,
specifying Error(subject)
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2. Execute analysis

• Compute
• Compute mean per drug mean (TDA)
  0.75 hs
• Compute drug effect
• 4. Compute mean per subject mean(TS1) 1.3
  hs
• 5. Compute subject effect
• 6. Compute fits
• 7. Compute residuals residuals T - fits

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2. Execute analysis
XS T XD ß0 ßD ßS 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
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3. Evaluate model

• Straight line model ok?
• Errors homogeneous?
• Errors normal?
• Errors independent?

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3. Evaluate model
NA

• Straight line model ok?
• Errors homogeneous?
• Errors normal?
• Errors independent?

?
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3. Evaluate model
NA

• Straight line model ok?
• Errors homogeneous?
• Errors normal?
• Errors independent?

?
?
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3. Evaluate model
NA

• Straight line model ok?
• Errors homogeneous?
• Errors normal?
• Errors independent?

?
?
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• 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

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• Decide on mode of inference. Is hypothesis
  testing appropriate?
• 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

 
 
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• State HA / Ho pair, test statistic, distribution,
  tolerance for Type I error.
• Test Statistic
• Distribution of test statitstic
• Tolerance for Type I error

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7. ANOVA
n 20
Source df SS MS F p
Subject
Drug
Res ______ ______
Total
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7. ANOVA
n 20
Source df SS MS F p
Subject 9 58.078
Drug 1 12.482
Res ___9__ _6.808
Total 19 77.37
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7. ANOVA
n 20
Source df SS MS F p
Subject 9 58.078 6.453
Drug 1 12.482 12.48
Res ___9__ _6.808 0.7564
Total 19 77.37
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7. ANOVA
n 20
Source df SS MS F p
Subject 9 58.078 6.453
Drug 1 12.482 12.48 16.5 0.0028
Res ___9__ _6.808 0.7564
Total 19 77.37
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7. ANOVA
n 20
Source df SS MS F p
Subject 9 58.078 6.453
Drug 1 12.482 12.48 16.5 0.0028
Res ___9__ _6.808 0.7564
Total 19 77.37
r2 0.91
STATISTICAL CONTROL
BUT we did this before ? Ch 10.2 2 sample t-test
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
r2 0.16
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8. Decide whether to recompute p-valueSlight
deviation from normalitynlt30, p0.0028 not near
a ? no need to recompute
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• Declare decision about termsOnly the fixed term
  was tested p0.0028 lt a 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

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1. Declare decision about terms

• Paired t-test
• Calculate difference within each random category
• 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 1.58 -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
p0.0028
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• 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
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Quizz 7Good luck!
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