Survival analysis: Kaplan-Meier estimates and the log-rank test
The following example is from the recommended course text, “Primer of Biostatistics”
by Stanton Glantz, Chapter 12: How to analyze survival data.
Kaplan-Meier estimates: Survival after exposure to second-hand smoke
This study examines survival after exposure to second-hand smoke.
10 subjects are observed from the time of their first exposure to second-hand smoke until
the subject’s death, the end-of-study (EOS), or the patient is lost to follow up (FU). Table
1 below shows the outcomes for the 10 subjects.
Table 1. Survival outcomes for the 10 subjects
Subject Time in study (years) Outcome
A
7
Death
B
12
Death
C
7
Death
D
12
Alive at EOS
E
11
Lost to FU
F
8
Death
G
9
Death
H
6
Death
I
7
Alive at EOS
J
2
Death
Censored
No
No
No
Yes
Yes
No
No
No
Yes
No
The study ends after 12 years. Subjects enter the study at different times, and may be
alive at the end of the study.
Seven subjects (A, B, C, F, G, H, J) died during the study period, so these observations
are uncensored.
Two subjects (D and I) were alive at the end of the study, so these observations are
censored.
One subject (E) was lost to follow-up after 7 years in the study, so this observation is also
censored.
Estimating the survival curve
We want to estimate the survival function based on this sample of subjects.
Survivor function S(t):
S(t) = The probability that a subject survives longer than time t.
S (t ) 
Number of individual s surviving longer tha n time t
Total number of individual s in the population
The time at which half the population is alive and half is dead is called the median
survival time.
Because most studies run for at most a few years, we often do not know the length of
time every individual in the sample survives. Some observations are censored.
We need a method to estimate the survival function with censored data. To do this, we
compute the probability of surviving at each time we observe a death, based on the
number at risk (known to be alive) immediately before that death.
We list all the observations in the order of the time of death or censored observation, as
shown in Table 2. Uncensored observations (where the time of death is known) are listed
before censored observations. Censored observations are indicated with a “+”.
Table 2. Survival time, deaths, and number at risk.
Subject
Survival time, ti
J
H
A and C
I
F
G
E
B
D
2
6
7
7+
8
9
11+
12
12+
Number alive at
beginning of
interval, ni
10
9
8
5
4
2
-
Number of deaths at
end of interval, di
1
1
2
1
1
1
-
The first death (J) occurred at time 2.
The second death (H) occurred at time 6.
Two deaths (A and C) occurred at time 7
Subject I is censored at time 7.
etc.
We now estimate the probability of death within any time period, based on the number of
subjects that survive to the beginning of the time period (the number at risk).
Just before the first death (J) at time 2, there are 10 subjects alive. There is one death at
time two, so there are 10 – 1 = 9 survivors. Our best estimate of the probability of
surviving past time 2 if alive just before time 2 is
Fraction alive just before time 2 surviving past time 2 =
n 2  d 2 10  1 9


 0.9
n2
10
10
n2 is the number of individuals alive just before time 2
d2 is the number of deaths at time 2.
At the beginning of the time interval ending at time 2, 100 percent of the subjects were
alive, so we estimate the cumulative survival rate at time 2 is 1.00 * 0.9 = 0.9.
The same calculations apply at the next death, at t=6.
Fraction alive just before time 6 surviving past time 6 =
n6  d 6 9  1 8

  0.889
n6
9
9
At the beginning of the time interval ending at time 6, 90 percent of the subjects were
alive, so the estimated cumulative survival rate at time 2, S(2) is 0.9 * 0.889 = 0.8.
Table 3 shows these calculations for further time points.
Subject
Survival time, Number alive
ti
at beginning
of interval, ni
Number of
deaths at end
of interval, di
J
H
A and C
I
F
G
E
B
D
2
6
7
7+
8
9
11+
12
12+
1
1
2
1
1
1
-
10
9
8
5
4
2
-
Fraction
surviving
interval,
(ni – di)/ni
.9
.889
.75
Cumulative
survival
rate
.8
.75
.48
.36
.5
.18
.9
.8
.6
We have to do something different for the death at time 8, because one subject (I) was
censored at time 7, and is no longer in the at-risk group at time 8. We know that the
subject died sometime after time 7, but we don’t know exactly when.
The next death is subject F at time 8. Because of the censoring of subject I, who was last
observed at time 7, we don’t know if this subject was alive or dead at time 8. As a result,
we drop subject I from the calculation of the survival function.
Just before time 8, there are 5 subjects known to be alive (at risk), and 1 dies at time 8, so
we use the formula as before.
Fraction alive just before time 6 surviving past time 6 =
n8  d 8 5  1 4

  0. 8
n8
5
5
At the beginning of the time interval ending at time 8, 60 percent of the subjects were
alive, so the estimate of the cumulative survival rate at time 8 is 0.6 * 0.8 = 0.48.
This method of estimating the survival curve is known as the Kaplan-Meier product-limit
estimate. The general formula for the Kaplan-Meier product-limit estimate of the survival
curve is
n d
Sˆ(t j )  ( n
i
i
)
i
where there are ni individuals alive just before time ti and di deaths occur at time ti.
The standard error of the Kaplan-Meier product-limit estimate at any time ti is estimated
using Greenwood’s formula:
S Sˆ (t )  Sˆ (t
j
j
)
di
i  di )
 n (n
i
The standard error may be used to calculate confidence intervals in the usual way.
Here’s an example from the text, Statistical Analysis of Medical Data Using SAS, by
Geoff Der and Brian Everitt.
data melanoma;
input weeks status$;
censor=status='alive';
cards;
12.8
dead
15.6
dead
24.0
alive
26.4
dead
29.2
dead
30.8
alive
39.2
dead
42.0
dead
58.4
alive
72.0
alive
77.2
dead
82.4
dead
87.2
alive
94.4
alive
97.2
alive
106.0
alive
114.8
alive
117.2
alive
140.0
alive
168.0
alive
;
proc lifetest data=melanoma plots=(s);
time weeks*censor(1);
run;
proc greplay igout=work.gseg nofs;
delete _all_;
run;
quit;
The LIFETEST Procedure
Product-Limit Survival Estimates
weeks
0.000
12.800
15.600
24.000*
26.400
29.200
30.800*
39.200
42.000
58.400*
72.000*
77.200
82.400
87.200*
94.400*
Survival
1.0000
0.9500
0.9000
.
0.8471
0.7941
.
0.7374
0.6807
.
.
0.6126
0.5445
.
.
Failure
0
0.0500
0.1000
.
0.1529
0.2059
.
0.2626
0.3193
.
.
0.3874
0.4555
.
.
Survival
Standard
Error
0
0.0487
0.0671
.
0.0814
0.0919
.
0.1014
0.1083
.
.
0.1169
0.1221
.
.
Number
Failed
Number
Left
0
1
2
2
3
4
4
5
6
6
6
7
8
8
8
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
97.200*
106.000*
114.800*
117.200*
140.000*
168.000*
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
8
8
8
8
8
8
5
4
3
2
1
0
NOTE: The marked survival times are censored observations.
E x a mp l e
21. 1:
HS V - 2
Ep i s o d e s
wi t h
g D2
Va c c i n e
1. 00
0. 75
0. 50
0. 25
0. 00
0
25
50
75
100
125
we e k s
Legend:
P r o d u c t - L i mi t
E s t i ma t e
Cu r v e
Ce n s o r e d
Ob s e r v a t i o n s
150
175
Comparing two survival curves: the logrank test
This description is from the course text, Glenn Walker, “Common Statistical Methods for
Clinical Research with SAS Examples”.
Usually, event times are not well-modeled using the normal distribution. The log-rank
test is a non-parametric test that does not require any assumptions about the distribution
of the event times.
If every patient were followed until the event occurrence, the event times could be
compared between two groups using the Wilcoxon rank-sum test. However, some
patients might drop out or complete the study before the event occurs. In such cases, the
data are censored. The log-rank test adjusts for this censoring.
The null hypothesis tested by the log-rank test is that of equal event time distributions
among groups. Equality of the distributions of event times implies similar risk-adjusted
event rates among groups not only for the clinical trial as a whole, but also for any
arbitrary time point during the trial. Rejection of the null hypothesis indicates that the
event rates differ among groups at one or more time points during the study.
In this example we will use two treatment groups, but the method extends readily to more
than two groups. Consider the following data-set.
Group 1
Patient ID
101
103
104
…
N1
Event time
Y11
Y12
Y13
Censored
Yes
Y1N1
Group 2
Patient ID
102
105
106
…
N2
Event time
Y21
Y22
Y23
Censored
Yes
Y1N2
The essence of the log-rank test is that, at each time when there is an event, we create a
contingency table (a 2x2 table) such as we use in a chi-square test. The entries in the 2*2
table is a count of the number of patients in each group who are event positive or event
negative.
At the time of an event, we have
1. The group that has the observed event(s)
2. The expected number of events in each treatment group.
We calculate a chi-square statistic by summing the differences between observed (O) and
expected (E) number of events at each time point, and dividing by the variance.
(Oi  Ei )2
 
Ei
i 1
g
2
The test statistic has an approximate chi-square distribution with g-1 degrees of freedom
where g is the number of groups (g=2 in the case of 2 treatment groups).
If one group has more events than the other group, we will get a large chi-square statistic.
Under the null hypothesis of no difference in event rates between the groups, we can
calculate the probability that we would get as extreme a chi-square statistic as we see in
the data. The calculations are shown in the texts by Glantz and Walker.