text: lecture27 notes
  
  Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )
  
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  Stat 312: Lecture 27 Testing for Independence Moo K. Chung [email protected] May 6, 2003 Concepts 1. Suppose that each individual in a population is classified with respect to two different factors into a two-way contingency table of size I × J . 2. The null hypothesis of interest would be test- ing if the two factors are independent: H 0 : p ij = p i p j for all i, j, where p i = J j =1 p ij and p j = J i =1 p ij . 3. Under H 0 , the expected number of element N ij = np ij = np i p j . Since the marginal probabilities are unknown, we estimate them by ˆ p i = n i /n , the sample proportion for cat- egory i for factor 1 and ˆ p j = n j /n , the sam- ple proportion for category j for factor 2. So N ij = n i n j /n . 4. Test statistic χ 2 = X i,j ( n ij - N ij ) 2 N ij χ 2 ( I - 1)( J - 1) . This is the test statistic that has been used in
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  Unformatted text preview: testing homogeneity! In-class problems Excercise 14.41. Test if the parental use of alcohol and drugs is independent of the student usage of marijuana for 445 college students. parent \ student never occasional regular neither 141 54 40 one 68 44 51 both 17 11 19 Self-study problems Example 14.14. Example. Show that the two hypotheses H : p ij = p i • p • j for all i, j and H : p 1 j = p 2 j = ··· = p Ij for all j are equivalent. Hence testing for ho-mogeneity is equivalent to testing independence. The End of Lectures. ra. ..ra. ..ra. .. Thank you all. In the next class, I will solve some sample final exams....
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  • Fall '04
  • Chung
  • Statistics, Null hypothesis, two-way contingency table, Moo K. Chung, nij − Nij
  
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