Thanks for your insight Mike

Thanks for taking the time to publish this comment about your and Xiaoming's conversation. I wish I had read this before the midterm (both in class and take home). Now that we have handed in the midterm, I was wondering what you all did to figure out the OR and 95%CI for Question #3, relating to the OR for a 53 year old with 155 mmHG compared to a 47 year old with SBP 113. I racked my brain trying to figure this out...sticking with the model (not using an interaction term) but bringing in all these variables. Anyone care to share?

Good things to know for the test ... perhaps?

I was talking with Xiaoming after class and he helped me out with a few concepts. These might be useful to know for the test.

1) (Looking at Table 7.2 on pg 235) If you want to compare the RR of CHD for a women with 71-80 mm Hg compared to a man with 101-110 mm Hg, then you divide the two numbers. 10.9 / 2.43 = 4.49 . This method also works when adding interaction terms (Table 7.3). You can not just divide the numbers to get the 95% C.I. for the RR because you have to take into account the covariance. Xiaoming said to use the “lincom” command in STATA to figure this out.

2) (Looking at the STATA output on top of pg 249) If you want to compare the relative risk for CHD of a 75 year old compared to a 45 year old (30 year difference) then you take the coefficient of age (1.04863) and raise it to the 30th power.

Mike

Aaron, HW6 #4

xi: stcox i.pd if pd~=1

*Risk = 6.045 95% CI = 3.658 – 9.989

xi: stcox i.pd if pd~=2

*Risk=2.018 95% CI = 1.327 – 3.069

Mike: Chapter 6, #5

There were only a few lines of codes for this problem.

summarize entage if pd==0
summarize entage if pd==1
summarize entage if pd==2
oneway entage pd, bonferroni tabulate