Biostatistics Case Studies 2014 Youngju Pak Biostatistician Session 5: Survival Analysis Fundamentals

Question #1 Fourteen more subjects were assigned to one of the treatment groups, compared to the other (Figure 1). In most randomized studies there is a difference of at most a few subjects. Why is it so far off here?

Question #1 P2670, the second paragraph

Question #2 In Figure 2B, survival is 20% in the mab group at the end of the study, yet Figure 1 says that only 243/347=70% (or perhaps 6 more to give 72%) in that group died, so why isn’t survival 30%? 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 Time (Mon) Starting N N of LOST N at Risk N of Death N of Survival Rate /347 = /232 = /88 =.48 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. 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 The main positive conclusion is that adding mab therapy doubled median time until disease progression (Fig 2). Why are median, rather than mean, times always reported in survival analyses ?

Questions #4: Why time to event as an outcome? 81.2% 73.4%

Question # 5 The authors state that the effect (on disease progression) of adding mab therapy decreases with time. What should we look at? Is this apparent in Figure 2A? Does it matter? Why?

Question #6 Relative risk (RR) can not be used? At end of study, would RR be 316/347 divided by 308/326, equals 0.96 (from Fig 1)? Use Fig 2 to show that approximately RR=50/73=0.68 at 1-year and 84/96=0.88 at 2-years.

Question #6 27 RR 1Yr = (1-0.50)/(1-0.27)=0.68 RR 2Yr = (1-0.16)/(1-0.04)=0.88

Question # 7 Even more basic, why bother with “hazards”, since we have already solved the problem of comparing groups with survival times?

Question # 7 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 # 7 Heuristic: Often, HR for Group1 to Group2 ≈ Median Survival Time for Group 2 Median Survival Time for Group 1

Question # 8 What does the requirement of “proportional hazards” used in their (page 2670, paragraph 2, and in most paper’s) survival analyses mean? Sketch a figure like Fig 2A where this obviously does not occur. What could we do in that case, since the standard analysis needs it, and would therefore be wrong?

More on Question #8 The Cox PH model assumes that the hazards in the two groups remains constant over time. For example, drug effect on the hardy is proportionally the same as it is on the frail. This is difficult to see directly on Kaplan-Meier curves, but if they are curved downward, as in Fig 2A, then the ratio of the survival times at any survival probability is close to the hazard ratio, and should be constant for different y-values. This is obviously not the case in Fig 2A, so it is a good example where proportional hazards do not hold.

More on Question # 8 There are some plots (eg., Schonfeld Residual plots, Grambsch and Therneau test (1994) ) that can be used to assess this. The solution is to either use a more general Cox analysis that allows say one ratio over some time and another ratio over other times. Alternatively, as in this paper, an “interaction” can be used that puts a structured change in the hazard ratio over time.

More Question # 8 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)

More Question # 8 The circled p=0.02 verifies what seems clear in Fig 3 for subjects >65.

Q #9: RR vs. OR 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 Ratio ≈ Relative Risk, if Prob(Case) is close to 0. So, Odds Ratio ≈ Relative Risk in case-control studies of a rare disease. For most case-control studies, OR should be used but the odds should not interpreted as a risk or chance of case since odds is not a probability.

Question 10 : Too many subgroup analyses? The overall type I error rate (False positive) will be increased as multiple statistical tests are performed. – See Lagakos SW. The challenge of subgroup analyses—reporting without distorting. N Engl J Med 2006; 354(16): 1667–1669. The prevailing opinion is not to use p-values or adjust for multiple tests. –Bonferroni: if 10 tests, then p < 0.05/10=0.005 should be considered as statistically significant to maintain the overall type I error rate of 0.05 –There are several other methods such as Tukey –Most commercial statistical softwares have options to adjust for multiple tests.

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