Biostatistics Case Studies 2009 Peter D. Christenson Biostatistician Session 2: Survival Analysis Fundamentals

Question #1

Question #2 243/347 = 70% Mortality 100%-20% = 80% Mortality

Kaplan-Meier: Cumulated Probabilities We want the probability of surviving for 54 months. If all subjects were followed for 54 months, then this prob is the same as the proportion of subjects alive at that time. If some subjects were not followed for 54 months, then we cannot use the proportion because we don’t know the outcome for these subjects at 54 months, and hence the numerator. Denominator? We can divide the 54 months into intervals using the follow-up times as interval endpoints. Ns are different in these intervals. Then, find proportions surviving in each interval and cumulate by multiplying these proportions to get the survival probability.

Kaplan-Meier: Cumulated Probabilities Suppose 104, 93, and 46 (total 243) died in months 0-18, 18-36, and Proportion surviving=( )/347=0.30. Of 104 survivors: suppose 11 had 18 months F/U, 51 had 36 months F/U, 35 had 54 months, and 7 had >54 months. Then, the 0-18 month interval has 243/347=0.70 surviving. The month interval has 139/232=0.60 surviving. The month interval has 42/88=0.48 surviving. So, 54-month survival is (243/347)(139/232)(42/88)=0.20. The real curve is made by creating a new interval whenever someone dies or completes follow-up (“censored”).

Question #3

Questions #4 and #5 81.2% 73.4%

Question #8

Question #9

27 RR 1Yr = (1-0.50)/(1-0.27)=0.68 RR 2Yr = (1-0.16)/(1-0.04)=0.88

Question #10 Even more basic, why bother with “hazards”, since we have already solved the problem of comparing groups with survival times?

Question #10 Hazard: “Sort-term” incidence at a specified time. E.g., events per 100,000 persons per day at 1 month. Time Prob of Survival Time Hazard 1 3 e -1(time) e -3(time) Constant Hazard ↔ Exponential determines

Question #10 Heuristic: Often, HR for Group1 to Group2 ≈ Median Survival Time for Group 2 Median Survival Time for Group 1

Question #11 For convex curves like these, the hazard ratio is approximately the ratio of survival times for any survival (y-axis). HR = 6/12=0.50 HR = 12/18=0.67 HR = 24/30=0.80 So this figure “obviously” violates proportional hazards. The authors used an interaction to resolve this violation (bottom of p 2671)

Question #11 For convex curves like these, the hazard ratio is approximately the ratio of survival times for any survival (y-axis). HR = 6/12=0.50 HR = 12/18=0.67 HR = 24/30=0.80 So this figure “obviously” violates proportional hazards. Needed for Proportional Hazards

Question #13 The circled p=0.02 verifies what seems clear in Fig 3 for subjects >65.

Question #14 mab No mab Case Non-Case Case = 1-Yr Progression For mab: Risk = Prob(Case) = 174/347 = 0.50 Odds = Prob(Case)/Prob(Non-Case) = 174/173 = RR = (174/347)/(238/326) = 0.50/0.73 = 0.68 OR = (174/173)/(238/ 88) = 1.00/2.70 = 0.37 → Effect by OR almost twice RR

When is Odds Ratio ≈ Relative Risk ? Odds = Prob(Case)/Prob(Non-Case) ≈ Risk = Prob(Case), if Prob(Non-Case) is close to 1. So, Odds Ratio ≈ Relative Risk in case-control studies of a rare disease.

Advantage of OR: Symmetry A Not A B Not B Case = 1-Yr Progression RR of A on B = (174/347)/(238/326) = 0.50/0.73 = 0.68 RR of B on A = (174/412)/(173/261) = 0.42/0.67 = 0.64 OR of A on B = (174/173)/(238/ 88) = (174x88)/(173x238) = (174/238)/(173/ 88) = OR of B on A

Odds Ratio in Case-Control Studies In case-control studies, cannot measure RR, or risk of outcome, due to separate control selection: Risk Factor Cases Controls1 Controls Ratio of (90/150) (90/690) Percents /(10/50) /(10/410) = 3.0 = 5.3 Odds [(90/150)/(60/150)] [(90/690)/(600/690)] Ratio /[(10/50)/(40/50)] /[(10/410)/(400/410)] = 6.0 = 6.0