This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 crossclassification table . Also called I × J table or in brief Iby J table. For example, 2 × 3 contingency table. Table 2.1: CrossClassification 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 ....
View
Full
Document
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.
 Spring '11
 DeyuanLi
 Probability

Click to edit the document details