Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Stat 312: Lecture 24 Chi-square Goodness-of-fit Test Moo K. Chung [email protected] April 24, 2003 Concepts 1. The null hypothesis of interest is H 0 : 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 goodness-of-fit test can be used to test whether sample data follows a particular dis- tribution. In-class 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 0 : p A = 0 . 34 , p B = 0 . 15 , p AB = 0 . 23 , p O = 0 . 28 .
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