Unformatted text preview: Stat 312: Lecture 24 Chisquare Goodnessoffit Test Moo K. Chung [email protected] April 24, 2003 Concepts 1. The null hypothesis of interest is H : p 1 = c 1 , ··· , p k = c k , H 1 : p j 6 = c j for some j. 2. The test statistic χ 2 = ( N 1 EN 1 ) 2 EN 1 + ··· ( N k EN k ) 2 EN k . Assuming EN j ≥ 5 for every j , χ 2 ∼ χ 2 k 1 . Proof. When k = 2 (binomial data), it is easy to prove. Note that χ 2 = ( N 1 np 1 ) 2 np 1 (1 p 1 ) . We have shown that for large n , N 1 /n p 1 p p 1 (1 p 1 ) /n ∼ N (0 , 1) 3. χ 2 goodnessoffit test can be used to test whether sample data follows a particular dis tribution. Inclass problems Example 1. There are four blood types (ABO sys tem). It is believed that 34, 15, 23 and 28 % of peo ple have blood type A, B, AB and O respectively. The blood samples of 100 students were collected: A(12), B(56), AB(2), O(30). Test if the collected blood sample contradicts the previous belief. Solu tion. The null hypothesis would be H...
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This note was uploaded on 01/31/2008 for the course STAT 312 taught by Professor Chung during the Spring '04 term at University of Wisconsin.
 Spring '04
 Chung
 Statistics

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