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Contingency Table Analysis Mary Whiteside, Ph.D.

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Overview Hypotheses of equal proportions Hypotheses of independence Exact distributions and Fishers test The Chi squared approximation Median test Measures of dependence The Chi squared goodness-of-fit test Cochrans test

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Contingency Table Examples Countries - religion by government States – dominant political party by geographic region Mutual funds - style by family Companies - industry by location of headquarters

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More examples - Countries - government by GDP categories States - divorce laws by divorce rate categories Mutual funds - family by Morning Star rankings Companies - industry by price earnings ratio category

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Statistical Inference hypothesis of equal proportions H 0 : all probabilities (estimated by proportions, relative frequencies) in the same column are equal, H 1 :at least two of the probabilities in the same column are not equal Here, for an r x c contingency table, r populations are sampled with fixed row totals, n 1, n 2, … n r.

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Hypothesis of independence H 0 : no association i.e. row and column variable are independent, H 1 : an association, i.e. row and column variable are not independent Here, one populations is sampled with sample size N. Row totals are random variables.

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Exact distribution for 2 x 2 tables: hypothesis of equal proportions; n 1 = n 2 = 2 20 20 20 02 02 20 02 02 20 11 02 11

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Fishers Exact Test For 2 x 2 tables assuming fixed row and column totals r, N-r, c, N-c: Test statistic = x, the frequency of cell 11 Probability = hyper-geometric probability of x successes in a sample of size r from a population of size N with c successes

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Large sample approximation for either test Chi squared = Observed - Expected] 2 /Expected Observed frequency for cell ij comes from cross-tabulation of data Expected frequency for cell ij = Probability Cell ij * N Degrees of freedom (r-1)*(c-1)

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Computing Cell Probabilities Assumes independence or equal probabilities (the null hypothesis) Probability Cell ij = Probability Row i * Probability Column j = (R i /N) * (C j /N) Expected frequency ij = (R/N)*(C/N)*N = R*C/N.

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Distribution of the Sum Chi Square with (r-1)*(c-1) degrees of freedom Assumes Observed - Expected] 2 /Expected is standard normal squared

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Implies Observed - Expected] / Square root[Expected ] is standard normal Implies and Observed is a Poisson RV Poisson is approximately normal if > 5, traditional guideline Conovers relaxed guideline page 201

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Measures of Strength: Categorical Variables Phi 2x2 Cramer's V for rxc Pearson's Contingency Coefficient Tschuprow's T

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Measures of Strength: Ordinal Variables Lambda A.. Rows dependent Lambda B.. Columns dependent Symmetric Lambda Kendall's tau-B Kendall's tau-C Gamma

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Steps of Statistical Analysis Significance - Strength 1- Test for significance of the observed association 2 - If significant, measure the strength of the association

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Consider the correlation coefficient a measure of association (linear relationship between two quantitative variables ) significant but not strong significant and strong not significant but strong not significant and not strong

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r and Prob (p-value) r =.20 p-value <.05 r =.90 p-value <.05 r =.90 p-value >.05 r =.20 p-value >.05

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Concepts Predictive associations must be both significant and strong In a particular application, an association may be important even if it is not predictive (I.e. strong)

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More concepts Highly significant, weak associations result from large samples Insignificant strong associations result from small samples - they may prove to be either predictive or weak with larger samples

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Examples Heart attack Outcomes by Anticoagulant Treatment Admission Decisions by Gender

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Summary Is there an association? –Investigate with Chi square p-value If so, how strong is it? –Select the appropriate measure of strength of association Where does it occur? –Examine cell contributions

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