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Unformatted text preview: Lecture 26 26.1 Test of independence. In this lecture we will consider the situation when data comes from the sample space X that consists of pairs of two features and each feature has a finite number of categories or, simply, X = { ( i, j ) : i = 1 , . . . , a, j = 1 , . . . , b } . If we have an i.i.d. sample X 1 , . . . , X n with some distribution on X then each X i is a pair ( X 1 i , X 2 i ) where X 1 i can take a different values and X 2 i can take b different values. Let N ij be a count of all observations equal to ( i, j ) , i.e. with first feature equal to i and second feature equal to j, as shown in table below. Table 26.1: Contingency table. Feature 2 Feature 1 1 2 ··· b 1 N 11 N 12 ··· N 1 b 2 N 21 N 22 ··· N 2 b . . . . . . . . . . . . . . . a N a 1 N a 2 ··· N ab We would like to test the independence of two features which means that ( X = ( i, j )) = ( X 1 = i ) ( X 2 = j ) . In we introduce the notations ( X = ( i, j )) = θ ij , ( X 1 = i ) = p i and ( X 2 = j ) = q j , 103 LECTURE 26. 104 then we want to test that for all i and j we have θ ij = p i q j . Therefore, our hypotheses can be formulated as follows: H 1 : θ ij = p i q j for some ( p 1 , . . . , p, ....
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 Spring '09
 DmitryPanchenko
 Statistics, ChiSquare Test, Maximum likelihood, Chisquare distribution, Pearson's chisquare test, Likelihoodratio test

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