# ansfin2001 - Final Exam, December 2001, ANSWERS Categorical...

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Final Exam, December 2001, ANSWERS Categorical Data Analysis for Epidemiologic Studies, CHL5407H 1a. No, the Pearson chi-square test is not an appropriate method of analysis to test for a diﬀerence in the proportion of learning disabled children comparing teachers to psychologists. The data provided in Table 1 are from a pair-matched study since each student was assessed twice. The Pearson chi-square test is more ap- propriately used to test for diﬀerences in proportions of two or more independent samples while McNemar’s chi-square statistic should be used to test for diﬀer- ences in two proportions for paired data. Moreover the expected cell counts in Table 1 are too small to support approximate statistical inferences constructed using the Pearson chi-square test. 1b. From Table 1 we see that teachers were far more likely to classify students as having a learning disability than were psychologists. For instance, the McNemar’s odds ratio of being classi±ed as having a learning disability, ˆ ψ MC =11 / 2=5 . 5 comparing teachers to psychologists. Another way of looking at the data is to note that 68% percent of teachers (100 × 15 / 22) but only 27% of psychologists (100 × 6 / 22) classi±ed children as having a learning disability. Since the same children are classi±ed twice - once by their homeroom teacher and once by the school psychologist - the proportions are almost certainly correlated. A method of analysis which takes this correlation into account should be used. We can use McNemar’s statistic to test H 0 : the proportion of students classi±ed as having a learning disability does not diﬀer at the population level between homeroom teachers and school psychologists vs. H A : the proportion of students classi±ed as having a learning disability does diﬀer at the population level between homeroom teachers and school psychologists. This, approximate, one degree of freedom chi-square test statistic uses only data from the discordant pairs. It is given by χ 2 MC = (11 - 2) 2 11 + 2 =6 . 23 ( p< 0 . 05) . Therefore, the much greater preference of teachers as compared to psychologists to classify students as having a learning disability is statistically signi±cant. This test is likely valid since there are more than 10 discordant pairs. Of course

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## ansfin2001 - Final Exam, December 2001, ANSWERS Categorical...

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