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TUT6 ANSWER.doc

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Department
Anthropology
Course
ANTC67H3
Professor
Larry Sawchuk
Semester
Winter

Description
ANTC67 ODDS RATIOS In a case control study, researchers examined the association between exposure to electromagnetic fields and the development of childhood leukemia. The researchers were able to recruit a total of 250 participants. Of the 120 children with leukemia, 28 reported to have exposure to electromagnetic fields. In the control group, 12 had reported to have exposure to electromagnetic fields. Part I 1. Set up a 2 X 2 table with appropriate labels and calculate the odds ratio. Leukemia No leukemia Electro- 28 12 40 magnetic No Electro 92 118 210 120 130 250 O.R = 2.99 Cannot conclude anything right now so we need to set up our hypothesis 2. We want to test if this difference is statistically different. Set up the hypotheses for this test. Hint: what is the null and alternative hypothesis? Ho  OR = 1 H1  OR is not equal to 1 (there is some sort of relationship between the exposure and the disease) 3. Is the Chi Square test suitable for this data? Each cell’s expected frequency must be at least 5 in order to apply the Chi square formula. Calculate the expected frequencies in each cell using the following formula: Expected cell frequency = (row total * column total)/n Leukemia No leukemia Electro- 28 (19.2) 12 (20.8) 40 magnetic No Electro 92 (100.8) 118 (109.2) 210 120 130 250 What is the difference between Chi Sq and MH Chi Sq? Sample size. 4. What is the decision rule for accepting or rejecting the null hypothesis? Hint: the critical value for a test with an alpha of 0.05 is X = 3.84. X^2 = sum of (O-E)^2 / E = 5. Calculate the test statistic using the Chi-square formula and the expected frequencies. Formula: X = ∑ (Observed – Expected) 2 Expected Exposure/disease Observed Expected (O-E) (O-E) 2 (O E) 2 SUM E +/+ 28 19.2 8.8 77.44/19.2 +/- 12 20.8 -8.8 77.44/20.8 -/+ 92 100.8 -8.8 77.44/100.8 -/- 118 109.2 8.8 77.44/109.2 Sum: 4.03 3.72 0.77 0.71 X^2 = 9.23 What does this mean? Critical value is 3.84 when P is 0.05 and d.f. is 1 6. What is your conclusion based on your calculations? Reject null hypothesis CV tells us whether we accept Null or reject. If it is greater 3.84 we reject null hypothesis and if it is less than 3.85 we reject null. This means we reject null hypothesis – there is significant relationship between electromagnetic exposure and leukemia Interpreting odds ratio Now that we rejected, we can say that the odds of leukemia that were exposed to EM are approx. 2.99 (3) times in odds than the odds of those who are not exposed to EM. Part II 1. Calculate the 95% confidence interval for the odds ratio. Formula: e^(ln(OR) + 1.96SE(ln(OR)) 2. Interpret the odds ratio including the confidence interval. ANTC67 ODDS RATIOS A case control study was conducted by University researches to examine the relationship bet
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