Biostatistics Case Studies 2010

1. Biostatistics Case Studies 2010 Session 2: Survival Analysis Fundamentals Peter D. Christenson Biostatistician http://gcrc.labiomed.org/biostat

2. Question #1 243/347 = 70% Mortality 100%-20% = 80% Mortality

3. 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.

4. Kaplan-Meier: Cumulated Probabilities • Suppose 105, 92, and 46 (total 243) died in months 0-18, 18-36, and 36-54. Proportion surviving=(347-243)/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 242/347=0.70 surviving. • The 18-36 month interval has 139/231=0.60 surviving. • The 36-54 month interval has 42/88=0.48 surviving. • So, 54-month survival is (242/347)(139/231)(42/88)=0.20. • The real curve is made by creating a new interval whenever someone dies or completes follow-up (“censored”).

5. Question #2

6. Questions #3 and #4 81.2% 73.4%

7. Question #5 Taxol + mab Taxol 316/347=0.91 308/326=0.94 0.91 / 0.94 = 0.96

8. Question #5 27 RR1Yr = (1-0.50)/(1-0.27)=0.68 RR2Yr = (1-0.16)/(1-0.04)=0.88

9. Question #6 Hazard: “Sort-term” incidence at a specified time. E.g., events per 100,000 persons per day at 1 month. determines Hazard Prob of Survival 3 e-1(time) 1 e-3(time) Time Time Constant Hazard ↔ Exponential

10. Question #6 Heuristic: Often, HR for Group1 to Group2 ≈ Median Survival Time for Group 2 Median Survival Time for Group 1

11. Question #7 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)

12. Question #7 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 in Taxol+mab group for Proportional Hazards

13. Question #8 Case Non-Case 347 326 mab No mab Case = 1-Yr Progression For mab: Risk = Prob(Case) = 174/347 = 0.50 Odds = Prob(Case)/Prob(Non-Case) = 0.50/0.50 = 1.00 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

14. 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.

15. 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 Controls2 + 90 60 600 - 1040400 100 100 1000 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

16. Advantage of OR: Symmetry B Not B 347 326 A Not A Case = 1-Yr Progression 412 261 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