Biostatistics 513 Spring Homework 1 Key solution. See the Appendix below for Stata code and output related to this assignment.

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1 Biostatistics 513 Spring 2010 Homework 1 Key solution * Question will be graded for points, the remaining question is credit/no credit. See the Appendix below for Stata code and output related to this assignment. 1. *(Hosmer & Lemeshow (1989)). Infant mortality and birth defect rates are elevated for low birth weight infants (defined as birth weight below 2500gms). A study undertaken in Massachusetts investigated the impact of various maternal characteristics on the risk of a low birth weight delivery. Information and data on this study are available in lowbwt description.pdf and lowbwt.txt respectively on the class website. Input these data into STATA and recode mothers pre pregnancy weight into five categories , , , and >150 lbs. [20 points] a) Construct a contingency table showing the frequencies and percentages of low birth weight infants according to these categories of maternal weight. [3] Table of observed counts and percentages of low birth weight infants across the five categories of maternal weight (1: , 2: , 3: , 4: , 5: >150 lbs): Maternal weight (categorized) Birth weight Normal (%) Low (%) Total Total b) Calculate and plot the estimated probability of low birthweight for these maternal weight categories. [3] Estimated probabilities of low birth weight across the five categories of maternal weight: Maternal weight Category Estimated probability of low bw Note the plot below is included for your interest. Stata code for a plot without the confidence intervals is provided in the Stata appendix.

2 Estimated probabilities of low birth weight infant across maternal weight categories Probabilities Maternal weight category c) Calculate the estimated odds ratios for low birthweight for maternal weight categories , , , lbs versus weight category >150 lbs. [2] First, we calculate the odds of low birth weight for each category of maternal weight: Mat. Weight Category Odds of low bw The odds ratios of categories 1-4 versus category 5 are then easily computed as following ratios: OR1 = 0.893/0.241 = 3.7 (OR of cat.1 versus 5, etc.) OR2 = 0.3/0.241 = 1.24 OR3 = 0.55/0.241 = 2.28 OR4 = 0.304/0.241 = 1.26 d) We are interested in assessing whether maternal pre pregnancy weight is associated with low birth weight. State the null hypothesis and the alternative hypothesis corresponding to i) an unordered test; ii) an ordered test. [4] The null hypothesis is H 0 : p 1 =p 2 = =p 5, where p i is the probability of low birth weight infant born to a mother with pre-pregnancy weight in the category i. i) The unordered alternative is Ha: at least one p i is not equal to the others ii) The ordered alternative is Ha: p1>=p2>= >=p5 (or <=), with strict inequality for at least one pair. e) Calculate the expected cell counts under the null hypotheses. [2] The expected cell counts under the null hypothesis are equal to

3 E ij = m i * n j / N, where m i is the marginal count for i-th row and n j is the marginal count for the j-th column. Maternal weight (cat.) Birth weight Total Normal Low Total f) Carry out an appropriate test for the unordered hypothesis in d). Provide the value of your test statistic and the appropriate reference distribution. State your conclusion. [3] The appropriate test for the unordered alternative hypothesis is the test of independence as we have a random sample of subjects (189 mother-baby pairs). The test statistic has a χ 2 (4) distribution under H 0 and in our case its value is With p-value = , we reject the hypothesis and conclude that there is evidence of an association between maternal pre-pregnancy weight and low birth weight of the infant. g) Carry out an appropriate test for the ordered hypothesis in d). Provide the value of your test statistic and the appropriate reference distribution. State your conclusion. [3] The appropriate test for the ordered alternative hypothesis is the score test for trend. The test statistic has a χ 2 (1) distribution under H 0 and in our case, with scaling of the weight categories 1,2,3,4,5, its value is With p- value = , we reject the hypothesis that the maternal pre-pregnancy weight is not associated with low birth weight and conclude that there statistically significant trend in the probability of low birth weight with increasing maternal weight. Note #1: This example shows that one can achieve greater power by designing a statistical procedure to test for a specific alternative (ordered probabilities, in this case) compared to a general alternative. Note #2: If we were to carry out the test with another maternal weight scaling (instead of the above values of 1,2,3,4,5), we find that the test statistic will depend on the choice of the values, or level of categories (lecture p.30). For example, if use actual weight values 100, 115, 125, 140 and 200 we find that we obtain a smaller test statistic =5.59 with p-value=0.0181, but our conclusion is unchanged. The choice of scaling should be made a priori based on what is meaningful for the particular context. 2. In a study of HIV infection among women entering the NYS prison system, 475 inmates were cross-classified with respect to HIV seropositivity and their history of intravenous drug use (Smith et al., 1991, AJPH Supplement). Use either direct calculations or STATA to answer the following questions (Note: If using Stata, compare the cci and csi commands. Which is more appropriate?) HIV+ HIV- Total IVDU Yes IVDU No Total a) Test the null hypothesis that there is no association between IVDU and HIV serostatus. State the hypotheses, and write a sentence (or two) that explains the results to a general medical audience. Ho: P(HIV IVDU) = P(HIV not IVDU) or Ho: HIV and IVDU are independent,

4 H a : P(HIV IVDU) P(HIV not IVDU). This is a cross-sectional sample so we are talking about disease prevalence, not incidence, and use the chi-square test of independence (lecture p.60). We see that the chi-square statistic for testing this hypothesis is 78.0 and the p-value for this test is very small (p < 0.001). Therefore, we conclude that the risk of HIV infection in this population of women is associated with a history of IVDU use. b) Give a point estimate and 95% confidence interval for the difference in HIV risk in the two groups and write a sentence (or two) that explains this interval to a general medical audience. The risk of HIV infection in women in this sample who do not report a history of IVDU use is 29/339 = 0.08 while the risk of HIV infection among women in this sample who report a history of IVDU use is 59/136 = Therefore, the estimated difference in risk (the increased risk among women reporting a history of IVDU use compared to those who do not report a history of IVDU use) is A 95% confidence interval for the true difference in prevalence between these populations is (0.26, 0.44) (lecture p.61, 64). A summary might be: Among incarcerated women who report a history of IV drug use, we find that 43% are HIV seropositive while among women who do not report a history of IV drug use, we find that 8% are HIV seropositive. The increased prevalence associated with IV drug use as estimated by the risk difference is 35% (95% CI: 26%, 44%). These results suggest a relatively high prevalence of HIV among incarcerated women that do not report IV drug use (8%), and a statistically significant increase in seropositivity among the women that do report IV drug use. c) Give a point estimate and 95% confidence interval for the relative risk of HIV, comparing IV drug using women to women that are not IV drug users, and write a sentence (or two) that explains this interval to a general medical audience. As discussed above, the prevalence of HIV infection in women in this sample who do not report a history of IVDU use is 29/339 = 0.08 while the risk of HIV infection among women in this sample who report a history of IVDU use is 59/136 = Therefore, the estimated relative risk is 0.43/0.08 = 5.1 with a 95% confidence interval of (3.4, 7.5). Thus, the prevalence of HIV infection in women reporting a history of IVDU is 5.1 times higher than the prevalence of HIV infection in women who do not report a history of IVDU use (lecture p.61, 64). d) Give a point estimate and 95% confidence interval for the odds ratio, comparing IV drug using women to women that are not IV drug users, and write a sentence (or two) that explains this interval to a general medical audience. The estimated odds of HIV seropositivity among women who report IV drug use was (59/136)/(1-59/136))=0.766 while among women who did not report IV drug use, the odds was (29/339)/(1-29/339))= The association between HIV seropositivity and IV drug use can be summarized by the odds ratio of 8.2 (95% confidence interval 4.9, 13.6) that compares the odds of seropositivity among IDVU women relative to the odds of seropositivity among non-ivdu women. These results suggest a more than an 8-fold increase in the odds of being seropositive among women that report IV drug use relative to those who do not. e) Based on these summaries, what do you conclude? These results suggest a relatively high prevalence of HIV seropositivity among incarcerated women that do not report IV drug use (8%), but an additional 35% are seropositive among the women that do report IV drug use. We further conclude that the risk of HIV seropositivity is 5-fold greater among incarcerated women using IV drugs as

5 compared to those not using IV drugs and that odds of HIV seropositivity is 8-fold greater among incarcerated women exposed to IV drug use as compared to those not exposed to IV drug use. Note: Since HIV seropositivity has a prevalence of 8% among the unexposed, the OR overestimates the RR if used as an approximation of the RR (lecture p.54). Appendix Problem 1. infile id lbwt age lwt race smoke ptl hyper urirr pvft bwt using.../lowbwt.txt (maternal weight categorized): gen wtcat=1 if lwt<111 replace wtcat=2 if lwt>110 & lwt<=120 replace wtcat=3 if lwt>120 & lwt<=130 replace wtcat=4 if lwt>130 & lwt<=150 replace wtcat=5 if lwt>150 (a) tabulate lbwt wtcat, col wtcat lbwt Total Total (b) mean(lbwt), over(wtcat) Over Mean Std. Err. [95% Conf. Interval] lbwt graph dot (mean) lbwt, over(wtcat) (c) tabodds lbwt wtcat wtcat cases controls odds [95% Conf. Interval]

6 (e) tabulate lbwt wtcat, expected wtcat lbwt Total Total (f) + (g) tabodds lbwt wtcat Test of homogeneity (equal odds): chi2(4) = Pr>chi2 = Score test for trend of odds: chi2(1) = 6.74 Pr>chi2 = * with an alternative scaling for weight: gen wtcat2=100 if lwt<111 replace wtcat2=115 if lwt>110 & lwt<=120 replace wtcat2=125 if lwt>120 & lwt<=130 replace wtcat2=140 if lwt>130 & lwt<=150 replace wtcat2=200 if lwt>150 tabodds lbwt wtcat2 Test of homogeneity (equal odds): chi2(4) = Pr>chi2 = Score test for trend of odds: chi2(1) = 5.59 Pr>chi2 = Problem 2. Note: The csi command with the or option provided the information necessary to answer the question. The csi command is appropriate for cohort or cross-sectional studies. The cci command is for case-control studies and only gives odds ratios.. csi , or Exposed Unexposed Total Cases Noncases Total Risk Point estimate [95% Conf. Interval] Risk difference Risk ratio Attr. frac. ex Attr. frac. pop Odds ratio (Cornfield) chi2(1) = Pr>chi2 =

Contingency Tables Summer 2017 Summer Institutes 187

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