Ansmid - Midterm Exam February 2003 ANSWERS Categorical Data Analysis for Epidemiologic Studies CHL5407H 1a Adoption of a Pearson chi-square test

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Midterm Exam, February 2003, ANSWERS Categorical Data Analysis for Epidemiologic Studies, CHL5407H 1a. Adoption of a Pearson chi-square test is reasonable. Only the degrees of freedom are in question. The Pearson chi-square test should be calculated using two degrees of freedom since there are no observations, and hence no information, in the row reserved for subjects with severely impaired lung function. The expected number of observations in these two cells is zero. How could these cells even be used to calculate the Pearson chi-square test given by χ 2 P = ∑∑ ( O - E ) 2 /E ? The expected numbers of observations in all other cells are sufficiently large to support the use of a Pearson chi-square test with two degrees of freedom. 1b. Several different methods of analysis could be used. Here I will only review one of these. Several students discussed more than one method. Only the Frst of these was graded. Students are encouraged to try and answer the question which was asked. ±isher’s exact test is applicable to analysis of data from an r × c contingency table and so could be used to test for an association between smoking status and lung function. Adoption of this test would still require omitting the row reserved for subjects with severely impaired lung function. One advantage of an exact test is
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This note was uploaded on 11/20/2011 for the course STATISTICS ST3241 taught by Professor Manwai's during the Spring '11 term at National University of Singapore.

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Ansmid - Midterm Exam February 2003 ANSWERS Categorical Data Analysis for Epidemiologic Studies CHL5407H 1a Adoption of a Pearson chi-square test

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