If we live with a deep sense of gratitude,
our life will be greatly embellished.
1
Categorical
Data Analysis
Chapter 10: Tests for
Matched Pairs
2
Meta Analysis
• Also known as stratified analysis
• Section 6.3.2: Cochran-Mantel-Haenszel test;
test for conditional independence
Situation: When another variable (strata Z) may
“pollute” the effect of a categorical explanatory
variable X on a categorical response Y
Goal: Study the effect of X on Y while controlling
the stratification variable Z without assuming a
model
3
Example: Respiratory
Improvement (SAS textbook, P. 46)
Center Treatment Yes
No
Total
1
Test
29
16
45
1
Placebo
14
31
45
43
47
90
Total
2
Test
37
9
45
2
Placebo
24
21
45
61
29
90
Total
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SAS Output
Summary Statistics for trtmnt by response
Controlling for center
Cochran-Mantel-Haenszel Statistics (Based on Table Scores)
Statistic
Alternative Hypothesis
DF
Value
Prob
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Nonzero Correlation
1
18.4106
<.0001
2
Row Mean Scores Differ
1
18.4106
<.0001
3
General Association
1
18.4106
<.0001
5
What to Do if Dependent
• (Section 6.3.5) When X and Y are NOT
conditionally independent given Z, we
would like to test for homogeneous
association
• (Section 6.3.6) If X, Y, Z have
homogeneous association, we would like
to estimate the common conditional odds
ratio for X, Y given Z
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SAS Output
Estimates of the Common Relative Risk (Row1/Row2)
Type of Study
Method
Value
95% Confidence Limits
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Case-Control
Mantel-Haenszel
4.0288
2.1057
7.7084
(Odds Ratio)
Logit
4.0286
2.1057
7.7072
Cohort
(Col1 Risk)
Mantel-Haenszel
Logit
1.7368
1.6760
1.3301
1.2943
2.2680
2.1703
Cohort
(Col2 Risk)
Mantel-Haenszel
Logit
0.4615
0.4738
0.3162
0.3264
0.6737
0.6877
Breslow-Day Test for
Homogeneity of the Odds Ratios
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Chi-Square
0.0002
DF
1
Pr > ChiSq
0.9900
Total Sample Size = 180
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Matched-pair Data
• Comparing categorical responses for two
“paired” samples
When either
• Each sample has the same subjects (or
say subjects are measured twice)
Or
• A natural pairing exists between each
subject in one sample and a subject from
the other sample (eg. Twins)
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Example: Rating for Prime Minister
Second Survey
First Survey
Approve
Disapprove
Approve
794
150
Disapprove
86
570
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Marginal Homogeneity
• The probabilities of “success” for
both samples are identical (The data
table shows “symmetry” across the
main diagonal)
• Eg. The probability of approve at the
first and 2nd surveys are identical
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Estimating Differences of
Proportions
• Sample estimate: P+1-P1+
• Standard error of P+1-P1+ (based on the
multinomial distribution of data):
p1 (1  p1 )  p1 (1  p1 )  2( p11 p22  p12 p21 )
n
• Asymptotical (1-a) confidence interval:
( p1  p1 )  Za / 2  SE( p1  p1 )
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McNemar Test (for 2x2 Tables
only)
• See SAS textbook Sec 3.7 (p. 40)
• Ho: marginal homogeneity
Ha: no marginal homogeneity
• A special case of C-M-H test; an
approximate test (when n*=n12+n21>10)
• Exact test (when n*=n12+n21<10)
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Level of Agreement: Kappa
Coefficient
• The larger the Kappa coefficient is;
the stronger the agreement is
• The difference between observed
agreement and that expected under
independence compared to the
maximum possible difference is
called Kappa coefficient
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SAS Output
McNemar's Test
Statistic (S)
DF
Asymptotic Pr > S
Exact
Pr >= S
1
<.0001
17.3559
3.716E-05
Simple Kappa Coefficient
Kappa
ASE
95% Lower Conf Limit
95% Upper Conf Limit
Sample Size = 1600
0.6996
0.0180
0.6644
0.7348
Level of agreement
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Chi-square Test for Square
Tables
Consider a IxI table
• Marginal homogeneity:
 i    i , i  1,...,I
• Symmetry: for all pairs of cells,  ij
Symmetry
  ji
=> marginal homogeneity
<=
15
Chi-square Test for Square Tables
Ho: symmetry
vs.
Ha: not symmetry
• Fitted values: ˆij  ˆ ji  (nij  n ji ) / 2
• Standardized Pearson residuals:
r ij  (nij  n ji ) / (nij  n ji )
• Pearson Chi-square Test statistic:
X 2   rij2
i j
X^2 follows approximately Chi-square with df =
16
I(I-1)/2
Example: Coffee Purchase
2nd purchase
1st
High
purchase point
High
point
Taster’s
Sanka
Nescafe
Brim
Taster’s
Sanka
Nescafe
Brim
93
17
44
7
10
9
46
11
0
9
17
11
155
9
12
6
4
9
15
2
10
4
12
2
27
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Example: Coffee Purchase
• X^2 = 20.4 and df is 5(5-1)/2=10
 lack of fit (reject Ho: symmetry)
 which pairs of cells cause the lack
of fit? Examine their standardized
Pearson residuals
 The pair (1,3) and (3,1)
contribute the most; other pairs are
fine (rij^2 is around 1 or less)
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