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Contingency Table Analysis Mary Whiteside, Ph.D.

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

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

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

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.

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.

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

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

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)

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.

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

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

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

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

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

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

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

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)

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

Examples Heart attack Outcomes by Anticoagulant Treatment Admission Decisions by Gender

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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