Part IV - Questions 15-16 Data Analysis We will examine the association between current alcohol drinking (1= yes, 0 = no), current cigarette smoking (1 = yes, 0 = no), and current marijuana use (1= yes, 0 = no) in a substance use population. It is well-known that marijuana use is associated with alcoholic drinking and cigarette smoking. Hence, Mantel-Haenszel analysis was conducted to assess the association between current alcohol drinking and current cigarette smoking while taking into account current marijuana use. https://brightspace.nyu.edu/d21/Ims/quizzing/user/quiz_submissions_attempt.d21?isprv=&qi=222227&ai=1275479&isInPopup=0&cfql=0&fromQB=0&fromSubmis... 7/19 4/22/23, 10:48 AM : Final Exam NEW - Biostatistics for Public Health, Section 001 - NYU marijuana = no smoke alcohol 1 Total 20 15 35 45 53 98 Total | 65 58 133 marijuana = yes smoke alcohol 1 | Total 10 11 21 45 91 136 Total | 55 102 157 Based on the above cross-tabulation, the Stata output for the Mantel-Haenszel test is below.Based on the above cross-tabulation, the R output for the Mantel-Haenszel test is below. https://brightspace.nyu.edu/d21/Ims/quizzing/user/quiz_submissions_attempt.d21?isprv=&qi=222227&ai=1275479&isInPopup=0&cfql=0&fromQB=0&fromSubmis... 8/19 4/22/23, 10:48 AM : Final Exam NEW - Biostatistics for Public Health, Section 001 = NYU > Bres lowDayTest(x) Breslow-Day test on Homogeneity of Odds Ratios data: x X-squared = 0.065076, df = 1, p-value = 0.7986 > mantelhaen. test(Q15$alcohol , Q15$smoke , Q15$marijuana, correct=FALSE) Mantel-Haenszel chi-squared test without continuity correction data: Q15$alcohol and Q15$smoke and Q15$marijuana Mantel-Haenszel X-squared = 2.8772, df = 1, p-value = 0.08984 alternative hypothesis: true common odds ratio is not equal to 1 95 percent confidence interval: 0. 9207721 3. 0400141 sample estimates : common odds ratio 1. 673069 Question 15 1 / 1 point From the above cross-tabulation, we can calculate the odds ratio representing the association betw> Bres lowDayTest(x) Breslow-Day test on Homogeneity of Odds Ratios data: x X-squared = 0.065076, df = 1, p-value = 0.7986 > mantelhaen. test(Q15$alcohol , Q15$smoke , Q15$marijuana, correct=FALSE) Mantel-Haenszel chi-squared test without continuity correction data: Q15$alcohol and Q15$smoke and Q15$marijuana Mantel-Haenszel X-squared = 2.8772, df = 1, p-value = 0. 08984 alternative hypothesis: true common odds ratio is not equal to 1 95 percent confidence interval: 0. 9207721 3.0400141 sample estimates: common odds ratio 1. 673069 Question 15 1 / 1 point From the above cross-tabulation, we can calculate the odds ratio representing the association between current alcohol use and current cigarette smoking among those who currently use marijuana and those who do not currently use marijuana, respectively. The results of the Breslow-Day test on homogeneity of odds ratios indicate Among those who currently use marijuana, the odds of current alcohol drinking among those who currently smoke cigarettes is 1.84 times the odds of current alcohol drinking among those who do not currently smoke cigarettes. Among those who currently use marijuana, the probability of current alcohol drinking among those who currently smoke cigarettes is 1.84 times the probability of current alcohol drinking among those who do not currently smoke cigarettes. The odds ratios representing the association between current alcohol use and current cigarette smoking among those who currently use marijuana and those who do not currently use marijuana are not statistically different. The odds ratios representing the association between current alcohol use and current cigarette smoking among those who currently use marijuana and those who do not currently use marijuana are both greater than 1. Page 9 19Question 10 Which below is NOT a plausible value of an odds ratio? O 1.35 17.56 -0.20 0.05 Part II - Questions 11-12 https://brigBased on the above cross-tabulation, the Stata output for the Mantel-Haenszel test is below. cc alcohol smoke, by(marijuana) bd marijuana Odds ratio [95% conf. interval] M-H weight 1.57037 . 6727884 3. 706019 5. 075188 (exact) 1.838384 . 6447654 5. 151368 3. 152866 (exact) Crude 1. 846154 9833089 3.471573 (exact) M-H combined 1. 673069 9207721 3. 040014 Test of homogeneity (M-H) chi2(1) = 0.07 Pr>chi2 = 0.7987 Test of homogeneity (B-D) chi2(1) = 0. 07 Pr>chi2 = 0.7986 Test that combined odds ratio = 1: Mantel-Haenszel chi2(1) = 2. 88 Pr>chi2 = 0. 0898 Based on the above cross-tabulation, the R output for the Mantel-Haenszel test is below. https://brightspace.nyu.edu/d21/Ims/quizzing/user/quiz_submissions_attempt.d21?isprv=&qi=222227&ai=1275479&isInPopup=0&cfql=0&fromQB=0&fromSubmis. 4/22/23, 10:48 AM > Bres lowDayTest(x) : Final Exam NEW - Biostatistics for Public Health, Section 001 - NYU Breslow-Day test on Homogeneity of Odds Ratios data: x X-squared = 0.065076, df = 1, Page 8 19 +