 # Lecture 3 Two Sample Comparison of Survival Distributions

```Lecture 3
Stat 255
V. Nguyen
Lecture 3
Two Sample Comparison of
Survival Distributions
Statistics 255 - Survival Analysis
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
Presented October 2, 2012
Vinh Nguyen
Department of Statistics
University of California, Irvine
3.1
Kaplan-Meier estimator
I
Show that the KM estimator is the nonparametric
maximum likelihood estimator of S(t). . .
Lecture 3
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
3.2
Uncensored data
1. Two sample t-test, H0 : µ1 = µ2 vs. HA : µ1 6= µ2
I
It could be that F1 6= F2 yet µ1 = µ2
I
Example: N (µ, σ12 ) vs. N (µ, σ22 )
2. Kolmogorov-Smirnov, H0 : F1 (x) = F2 (x) for all of x
vs. HA : F1 (x) 6= F2 (x) for some x
I
b1 (x) − F
b2 (x)| where Fˆ is the
Test statistic: supx |F
empirical distribution function
Lecture 3
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
3. Wilcoxon rank-sum test (Mann-Whitney),
H0 : F1 (x) = F2 (x) vs. HA : F1 (x) > F2 (x) for all x
4. Other tests . . .
3.3
Motivation
I
Recall: 6-MP vs. placebo leukemia trial
I
21 patients randomized to 6-MP treatment
I
21 patients randomized to placebo-control
I
Outcome: remission times (in weeks)
Lecture 3
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
3.4
Survival functions
Lecture 3
1.0
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
0.6
Comparing Two
Survival
Distributions
Logrank test
0.4
Ex: 6-MP Trial
0.0
0.2
Survival Probability
0.8
Control (N=21)
6−MP (N=21)
0
5
10
15
20
25
30
35
Study Time (weeks)
3.5
Comparison
I
Lecture 3
Question of interest: Is 6-MP associated with the time to
cancer remission?
Stat 255
V. Nguyen
One approach: compare survival at single time point t0
I
b 1 (t0 ) − S
b 2 (t0 )
Use the Kaplan-Meier estimator to obtain S
I
Use Greenwood’s variance estimator for the difference in
survival probabilities
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
I
Test:
H0 : S1 (t0 ) = S2 (t0 )
vs.
HA : S1 (t0 ) 6= S2 (t0 )
using the test statistic
Z =q
b 1 (t0 ) − S
b 2 (t0 )
S
b 1 (t0 )] + V
b 2 (t0 )]
b [S
b [S
V
.
∼H0 N (0, 1)
3.6
Comparing survival at a single time point
Lecture 3
Stat 255
V. Nguyen
Problems
1. Clinically, it may not be appropriate to compare
survival at a single time point
I
Example: 1-month survival for childhood leukemia
2. Inefficient since we throw away information on the
rest of the survival curves
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
Solution
I
Goal: compare the entire survival curves (up to the
maximum observed time)
3.7
Logrank test
Lecture 3
Stat 255
V. Nguyen
Setup
I
Let t1 < t2 < · · · < tD denote the set of distinct and
ordered observed survival times from the pooled sample
(both groups combined)
Kaplan-Meier
Estimator
I
Test
H0 : S1 (t) = S2 (t)
vs.
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
HA : S1 (t) = [S2 (t)]φ ,
or equivalently,
H0 : λ1 (t) = λ2 (t)
vs.
HA : λ1 (t) = φλ2 (t),
for all t, where φ > 0
I
HA : the hazard ratio φ is not 1
3.8
Conditional 2 × 2 table
I
At each failure time tk , k = 1, . . . , D, consider a 2 × 2 table
of the form
Group 1
Group 2
Total
I
Lecture 3
Failure
Yes
No
d1k y1k − d1k
d2k y2k − d2k
dk
yk − dk
Risk set size
y1k
y2k
yk
Conditional on the risk set at time tk , consider the
distribution of deaths in group 1 under H0 :
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
D1k ∼ Hypergeometric(yk , y1k , dk )
I
For failure time tk , compute the observed number of
deaths in group 1 and the expected number of deaths
under H0 :
O1k = d1k
E1k = y1k
(observed)
dk
yk
(expected)
Logrank test
I
3.9
Lecture 3
Stat 255
V. Nguyen
At each failure time, compute:
Uk = O1k − E1k
I
Kaplan-Meier
Estimator
Under H0 ,
E[Uk ] = 0
y1k y2k (yk − dk )dk
Var[Uk ] =
= Vk
yk2 (yk − 1)
I
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
The logrank statistic is given by
hP
D
k =1 (Ok
PD
− Ek )
k =1 Vk
i2
hP
D
k =1 Uk
= PD
k =1 Vk
i
.
∼ χ21
3.10
Example: 6-MP Trial
> library(KMsurv)
> data(drug6mp)
> drug6mp[1:5, ]
pair remstat t1 t2 relapse
1
1
1 1 10
1
2
2
2 22 7
1
3
3
2 3 32
0
4
4
2 12 23
1
5
5
2 8 22
1
>
> ##### Transform data to long form
> drug6mpLong <- with(drug6mp
+
, data.frame(time=c(t1, t2)
+
, irelapse=c(rep(1, length(t1)), relapse)
+
, sixmp=rep(c(0, 1), each=length(t1))))
> dim(drug6mpLong)
 42 3
> drug6mpLong[1:5, ]
time irelapse sixmp
1
1
1
0
2
22
1
0
3
3
1
0
4
12
1
0
5
8
1
0
Lecture 3
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
3.11
Example: 6-MP Trial
> library(survival)
> plot( survfit( Surv( time, irelapse ) ~ sixmp, data=drug6mpLong ),
+
lty=1:2, xlab="Study Time (weeks)", ylab="Survival Probability" )
> legend( 25, 1, lty=1:2, legend=c("Control (N=21)", "6-MP (N=21)"), bty="n" )
> ##
> #####
> ##### Logrank test
> #####
> ##
> library( survival )
> survdiff( Surv( time, irelapse ) ~ sixmp, data=drug6mpLong )
Call:
survdiff(formula = Surv(time, irelapse) ~ sixmp, data = drug6mpLong)
N Observed Expected (O-E)^2/E (O-E)^2/V
sixmp=0 21
21
10.7
9.77
16.8
sixmp=1 21
9
19.3
5.46
16.8
Chisq= 16.8
Lecture 3
Stat 255
V. Nguyen
Kaplan-Meier
Estimator
Comparing Two
Survival
Distributions
Logrank test
Ex: 6-MP Trial
on 1 degrees of freedom, p= 4.17e-05
3.12
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