Class Notes (839,590)
Anthropology (1,602)
ANTC67H3 (65)
Lecture

# Tutorial 5 .docx

5 Pages
137 Views

Department
Anthropology
Course Code
ANTC67H3
Professor
Larry Sawchuk

This preview shows pages 1 and half of page 2. Sign up to view the full 5 pages of the document.
Description
ANTC67 Tutorial #5 ODDS RATIOS: In class lesson 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. Diseased Not Diseased TOAL (Leukemia) Exposure (electromagnetic a 28 b 12 40 fields) No Exposure (no electromagnetic c 92 d 118 210 fields) 120 130 250 * This is always the set-up for the table, do not switch up things, you will get the wrong answer if not in this format OR = 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? H o There is no relationship between the exposure and developing leukemia; OR = 1 H = OR not equal to 1. A 3. We use Mantel-Haenszel chi-square test to assess statistical significance of odds ratios. What is the decision rule for accepting or rejecting the null hypothesis? Hint: the critical value for a chi- square test with an alpha of 0.05 is X = 3.84. When p<0.05, our CV (critical value) = 3.84 Fail rejectoH , if X < 3.84 Reject Hoif X = (or greater than) 3.84 1 ANTC67 Tutorial #5 4. Calculate the test statistic using the Chi-square formula. Each cell’s value must be at least 5 in order to apply the Chi square formula. 2 MH chi-square= (│ad-bc│-N/2) *N C1*C2*R1*R2 = (( )( ) ( )( )| ) = 8.21 C = column R = row 5. What is your conclusion based on your calculations? Reject H oecause the chi square value is greater than our critical value of 3.84. Part II 1. Calculate the 95% confidence interval for the odds ratio. 95% CI for OR= e^[(ln(OR)) + 1.96SE(ln(OR))] SE(ln(OR))] = √ = e^[(ln(OR)) + 1.96SE(ln(OR))] = e^[(ln(OR)) + 1.96 √ = e^[(ln(2.99)) + 1.96√ = e^(1.096 1.96(0.372)) = e^(1.096 1.96(0.372)) 2 ANSWERS: 1.44, 6.20 2. Interpret the odds ratio including the confidence interval. Lower limit 1.44 Upper Limit 6.20 2 ANTC67 Tutorial #5 2.99 fits within 95% confidence interval. We are 95% confident that the OR falls b/w 1.44 and 6.20 The odds of leukemia in those that are exposed to electromagnetic fields are approximately three times the odds of those who are not exposed to electromagnetic fields.
More Less

Only pages 1 and half of page 2 are available for preview. Some parts have been intentionally blurred.

Unlock Document

Unlock to view full version

Unlock Document
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.