Ch2-4 - Chapter 2. Describing Contingency Tables Deyuan Li...

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Unformatted text preview: Chapter 2. Describing Contingency Tables Deyuan Li School of Management, Fudan University Feb. 28, 2011 1 / 28 Outline • 2.1 Probability Structure for Contingency Tables • 2.2 Comparing Two Proportions • 2.3 Partial Association in Stratified 2 × 2 Tables • 2.4 Extensions for I × J Tables 2 / 28 2.1 Probability Structure for Contingency Tables 2.1.1 Contingency tables and their distributions X : categorical response variable with I categories; Y : categorical response variable with J categories. Construct a rectangle table with I rows and J columns, the cells contain frequency counts of the outcomes for a sample. This table is called contingency table (by Karl Pearson (1940)) or cross-classification table . Also called I × J table or in brief I-by- J table. For example, 2 × 3 contingency table. Table 2.1: Cross-Classification of Aspirin Use and Myocardial Infarction. Myocardial Infarction Fatal Attack Nonfatal Attack No Attack Placebo 18 171 10,845 Aspirin 5 99 10,933 3 / 28 Let π ij denote the probability of X is in category i an Y is in category j for i = 1 , 2 , ..., I and j = 1 , 2 , ..., J . Then { π ij } is the joint distribution of ( X , Y ), { π i + } is the marginal distribution of X , { π + j } is the marginal distribution of Y , where π i + = j π ij and π + j = i π ij . Of course, ∑ i π i + = ∑ j π + j = ∑ i ∑ j π ij = 1 . 0. 4 / 28 2.1.3 Independence of categorical variable Define π j | i = P ( Y is in category j | X is in category i ) = π ij /π i + . { π 1 | i , ..., π J | i } is the conditional distribution of Y at category i of X . X and Y are independent if π ij = π i + π + j for all i = 1 , 2 , ..., I and j = 1 , 2 , ..., J . Consequently, when X and Y are independent, π j | i = π ij /π i + = ( π i + π + j ) /π i + = π + j for all i , j . 5 / 28 2.1.4 Poisson and Multinomial sampling Poisson sampling model treats cell counts, denoted by { Y ij } , as independent Poisson variables with parameters { μ ij } . So the joint distribution of { Y ij } is P ( Y 11 = n 11 , ..., Y ij = n ij , ..., Y IJ = n IJ ) = i j exp( − μ ij ) μ n ij ij / n ij ! . Multinomial sampling : (1). total sample size n is fixed but the row and column totals are not. P ( Y ij = n ij , i = 1 , ..., I , j = 1 , ..., J ) = [ n ! / ( n 11 ! · · · n IJ !)] i j π n ij ij . Note ∑ i ∑ j π ij = 1 and ∑ i ∑ j n ij = n . This is the difference compared with Poisson sampling. 6 / 28 (2). rows total sample sizes ( n i = n i + ) are fixed ( Y is a response variable while X is an explanatory variable). P ( Y ij = n ij , j = 1 , ..., J ) = [ n i ! / ( n i 1 ! · · · n iJ !)] j π n ij j | i ....
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This note was uploaded on 11/20/2011 for the course ST 3241 taught by Professor Deyuanli during the Spring '11 term at Adams State University.

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Ch2-4 - Chapter 2. Describing Contingency Tables Deyuan Li...

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