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If we live with a deep sense of gratitude, our If we live with a deep sense of gratitude, our life will be greatly embellished.life will be greatly embellished.

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Categorical Categorical Data AnalysisData Analysis

Chapter 10: Tests for Chapter 10: Tests for Matched Pairs Matched Pairs

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Meta AnalysisMeta 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

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Example: Respiratory Example: Respiratory Improvement Improvement (SAS textbook, P. 46)(SAS textbook, P. 46)

Center Treatment Yes No Total

1 Test 29 16 45

1 Placebo 14 31 45

Total 43 47 90

2 Test 37 9 45

2 Placebo 24 21 45

Total 61 29 90

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SAS OutputSAS Output

Summary Statistics for trtmnt by response Controlling for center

Cochran-Mantel-Haenszel Statistics (Based on Table Scores)

Statistic Alternative Hypothesis DF Value Prob ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ 1 Nonzero Correlation 1 18.4106 <.0001 2 Row Mean Scores Differ 1 18.4106 <.0001 3 General Association 1 18.4106 <.0001

What to Do if DependentWhat 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 OutputSAS Output Estimates of the Common Relative Risk (Row1/Row2)

Type of Study Method Value 95% Confidence Limits ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Case-Control Mantel-Haenszel 4.0288 2.1057 7.7084 (Odds Ratio) Logit 4.0286 2.1057 7.7072

Cohort Mantel-Haenszel 1.7368 1.3301 2.2680 (Col1 Risk) Logit 1.6760 1.2943 2.1703

Cohort Mantel-Haenszel 0.4615 0.3162 0.6737 (Col2 Risk) Logit 0.4738 0.3264 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 DataMatched-pair Data• Comparing categorical responses for two

“paired” samplesWhen 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 MinisterExample: Rating for Prime Minister

Second Survey

First Survey Approve Disapprove

Approve 794 150

Disapprove 86 570

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Marginal HomogeneityMarginal 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 Estimating Differences of ProportionsProportions

• Sample estimate: P+1-P1+

• Standard error of P+1-P1+ (based on the multinomial distribution of data):

• Asymptotical (1- confidence interval:

n

pppppppp )(2)1()1( 211222111111

)()( 112/11 ppSEZpp

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McNemar Test (for 2x2 Tables McNemar Test (for 2x2 Tables only)only)

• See SAS textbook Sec 3.7 (p. 40)

• Ho: marginal homogeneityHa: 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)

Level of Agreement: Kappa Level of Agreement: Kappa CoefficientCoefficient

• 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 OutputSAS Output McNemar's Test Statistic (S) 17.3559 DF 1 Asymptotic Pr > S <.0001 Exact Pr >= S 3.716E-05

Simple Kappa Coefficient Kappa 0.6996 ASE 0.0180 95% Lower Conf Limit 0.6644 95% Upper Conf Limit 0.7348

Sample Size = 1600Level of agreement

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Chi-square Test for Square Chi-square Test for Square TablesTables

Consider a IxI table• Marginal homogeneity:

• Symmetry: for all pairs of cells,

Symmetry => marginal homogeneity

<=

Iiii ,...,1,

jiij

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Chi-square Test for Square TablesChi-square Test for Square Tables

Ho: symmetry vs. Ha: not symmetry

• Fitted values:

• Standardized Pearson residuals:

• Pearson Chi-square Test statistic:

X^2 follows approximately Chi-square with df = I(I-1)/2

2/)(ˆˆ jiijjiij nn

)(/)( jiijjiijij nnnnr

jiijrX 22

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Example: Coffee PurchaseExample: Coffee Purchase2nd purchase

1st purchase

High point

Taster’s Sanka Nescafe Brim

High point

93 17 44 7 10

Taster’s 9 46 11 0 9

Sanka 17 11 155 9 12

Nescafe 6 4 9 15 2

Brim 10 4 12 2 27

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Example: Coffee PurchaseExample: 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)