Biostatistics Case Studies 2008

Peter D. Christenson

Biostatistician

http://gcrc.labiomed.org/biostat

Session 1:

Survival Analysis Fundamentals

Question #1

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 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 243/347=0.70 surviving.• The 18-36 month interval has 139/232=0.60 surviving.• The 36-54 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

Question #9

27

RR1Yr = (1-0.50)/(1-0.27)=0.68

RR2Yr = (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

Time

Prob of Survival

Time

Hazard

1

3

e-1(time)

e-3(time)

Hazard Ratio (Red to Blue) = 1/3 =

Blue to Red Ratio of Survival Times

1 2 3 6

Cox Regression: Any hazard pattern for one group; other group is proportional.

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 #13

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

Question #14

mab

No mab

174 173

238 88

Case Non-Case

Case = 1-Yr Progression

For mab:

Risk = Prob(Case) = 174/347 = 0.50

Odds = Prob(Case)/Prob(Non-Case) = 174/173 = 1.00

347

326

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

174 173

238 88

B Not B

Case = 1-Yr Progression

347

326

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

412 261

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 - 10 40 400 100 100 1000Ratio 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