3.1
Hypothesis Testing
• One sample test
— A censored sample of size n from some population
— Want to test the hypothesis that the population hazard
rate is h0(t) for all t ≤ τ , i.e.,
H0 : h(t) = h0(t), 0 < t ≤ τ
— Typically, take τ to be the largest of the observed study
time
— Note that we have information on the failure times only
and the estimator of the hazard function at time ti is
ˆ i) = di
h(t
ni
Therefore, we compare the observed hazard and the
hazard under the null hypothesis
ˆ i) − h0(ti) = di − h0(ti)
h(t
ni
ˆ i) is the maximum likelihood estimator of h(ti),
Since h(t
it, asymptotically, is unbiased, i.e.,


di
E   = h0(ti)
ni
and the variance under H0:
3.2
Therefore, under H0,


h0(ti)(1 − h0(ti)) 
di
h0 (ti ),

∼N
ni
ni
asymptotically. Further, we have
di − Ei ∼ N (0, Ei)
where Ei = nih0(ti)
— If the data follow the distribution under H0, di − Ei
should be small. Therefore, large diﬀerences indicate
evidences against the null hypothesis.
— However we have k diﬀerences,
tj
t1
t2
...
tk
nj
n1
n2
...
nk
dj
d1
d2
...
dk
Ej
Diﬀ.
E1 d1 − E(t1)
E2 d2 − E(t2)
...
...
Ek dk − E(tk )
To test H0, so we sum them up
i=1
(di − Ei)
under H0, E(Z) = 0
— How about the variance of Z under H0
3.3
— Therefore, under H0,
Z
=
V ar(Z)
k
i=1 (di − Ei)
k E
i=1 i
∼ N(0, 1)
or, equivalently,
2
k
(d
−
E
)
i
i
i=1
k E
i=1 i
∼ χ21
— This is called the log-rank test
— Example: A study of the remission times in weeks of
21 patients:
6, 6, 6, 7, 10, 13, 16, 22, 23,
6+, 9+, 10+, 11+, 17+, 19+, 20+, 25+, 32+, 32+, 34+, 35+
Test the null hypothesis: H0 : h0(t) = 0.05
tj 6 7 10 13 16 22 23
nj
dj
Ej
3.4
Comparisons of Several Survivor Functions
— Suppose we have r groups of lifetimes and want to test
the equality among them
— Example: Pike (1966)
Table 1.1 Days to Vaginal Cancer Mortality in Rats
Group 1
Group 2
∗
Censored
143
216∗
142
233
344∗
164
220
156
233
188
227
162
239
188
230
198
240
190
234
204∗
261
192
244∗
205
280
206
246
232
280
209
265
232
296
213 216
304
233 233
296 323
3.5
— Suppose there are r independent censored samples
i.e., for sample i, we have
di1 failures
ti01 ≤ · · · ≤ ti0mi0 <
censored obs
ti1
di2 failures
≤ ti11 ≤ · · · ≤ ti1mi1 <
ti2
diki failures
≤ ··· <
tiki
≤ · · · ≤ tikmik
censored obs
— We want to test if all the survival functions are the same
— Under this null hypothesis, all these samples are coming
from the same distribution
— We can pool all samples together to form a pooled sample
— Note that the KM estimator for the survival function
only changes at the observed failures
— Therefore we only focus on the observed failures
— Let
t1 < t2 < · · · < tk
denote the failure times for the pooled sample
— Suppose dj failures occurs at tj and that nj study subjects are at risks just prior to tj (j = 1, . . . , k) in the
pooled sample
— Let dij and nij be the corresponding numbers in sample
i (i = 1, . . . , r)
— Then , at tj , we can form a 2 × r contingency table
Group failures survivors Sub-total
1
d1j
n1j − d1j
n1j
2
d2j
n2j − d2j
n2j
...
...
...
...
r
drj
nrj − drj
nrj
Total
dj
nj − dj
nj
3.6
— Under the null hypothesis (there is no diﬀerence among
groups), the conditional distribution of the d1j , . . . , drj
given dj is then the hypergeometric distribution
Pr {d1j from group 1, d2j from group 2, . . . , drj from group r|dj }





n
n
n
 1j   2j 
 rj 


...

d1j
d2j
drj


=
n
 j 


dj
— Why? (What is the hypergeometric distribution?)
3.7
— The conditional mean of dij is
wij = nij
dj
nj
— the conditional variance of dij is
(Vj )ii =
nij (nj − nij )dj (nj − dj )
n2j (nj − 1)
— The conditional covariance of dij and dlj is
(Vj )il = −
nij nlj dj (nj − dj )
n2j (nj − 1)
— the statistic
vj = (d1j − w1j , . . . , drj − wrj )
has (conditional) mean zero and variance matrix Vj
— Although the column vector vj has r components, we
need only the knowledge of r − 1 components
Redefine
vj∗ = (d1j − w1j , . . . , d(r−1)j − w(r−1)j )
and the corresponding covariance matrix is
Vj∗ = [(Vj )il ]i,l=1,2,...,r−1
— The log-rank statistic is
v=
k
1
vj∗
— If the k contingency tables were independent, the variance of the log-rank statistic would be
V = V1∗ + V2∗ + · · · Vk∗
3.8
— The log-rank test for the equality of the r survival
curves is
v V−1 − v
which is asymptotic χ2r−1 distributed
— the χ2r−1 statistic can be formed using any r−1 elements
of v and the corresponding (r − 1) × (r − 1) submatrix
of V
3.9
Pool two groups together
142(2)
204(2)∗
232(2)
246(1)
143(1)
205(2)
232(2)
261(2)
156(2)
206(1)
233(2)
265(1)
162(2)
209(1)
233(2)
280(2)
164(1)
213(1)
233(2)
280(2)
188(1)
216(1)
233(2)
296(2)
188(1)
216(1)∗
234(1)
296(2)
190(1)
220(1)
239(2)
304(1)
192(1)
227(1)
240(2)
323(2)
— Totally there are 29 distinct failures and hence 29 contingency tables
— just prior to t1 = 142, n1 = 40, d1 = 1, d11 = 0,
n11 = 19, d21 = 1, n21 = 21
Group failures survivors Subtotal
1
0
19
19
2
1
20
21
Subtotal
1
39
40
— Therefore, w11 = n11 ∗ d1/n1 = 19 ∗ 1/40, w21 = n21 ∗
d1/n1 = 21 ∗ 1/40 and
v1 = (0 − 0.475, 1 − 0.525) = (−0.475, 0.475)
198(2)
230(1)
244(1)∗
344(2)∗
3.10
— just prior to t2 = 143, n2 = 39, d2 = 1, d12 = 1,
n12 = 19, d22 = 0, n22 = 20
Group failures survivors Subtotal
1
1
18
19
2
0
20
20
Subtotal
1
38
39
— Therefore, w12 = n12 ∗ d2/n2 = 19 ∗ 1/39, w22 = n22 ∗
d2/n2 = 20 ∗ 1/39 and
v2 = (1 − 0.487, 0 − 0.513) = (0.513, −0.513)
— ···
— The log-rank statistic is 4.762 and variance is 7.263 and
hence the χ2 statistic is
4.762 ∗ 4.762
= 3.12
7.263
and
p − value = P r(χ21 > 3.12) = 0.0773
3.11
tj
142
143
156
162
164
188
190
192
198
205
206
209
213
216
220
227
230
232
233
234
239
240
246
261
265
280
296
304
323
dj
1
1
1
1
1
2
1
1
1
1
1
1
1
1
1
1
1
2
4
1
1
1
1
1
1
2
2
1
1
nj d1j
40 0
39 1
38 0
37 0
36 1
35 2
33 1
32 1
31 0
29 0
28 1
27 1
26 1
25 1
23 1
22 1
21 1
20 0
18 0
14 1
13 0
12 0
10 1
9 0
8 1
7 0
5 0
3 1
2 0
n1j d2j
19 1
19 0
18 1
18 1
18 0
17 0
15 0
14 0
13 1
13 1
13 0
12 0
11 0
10 0
8 0
7 0
6 0
5 2
5 4
5 0
4 1
4 1
3 0
2 1
2 0
1 2
1 2
1 0
0 1
n2j
21
20
20
19
18
18
18
18
18
16
15
15
15
15
15
15
15
15
13
9
9
8
7
7
6
6
4
2
2
3.12
Wilcoxon Test
— In log-rank test all contingency tables are treated equally
— Someone may think of large sample should carry more
information
— Proposed to use
k
j=1
nj vj
— The corresponding variance matrix is
V=
k
j=1
n2j Vj
— back to the example, the Wilcoxon statistic is 114 and
the variance is 4902.22291 and hence the χ21 statistic is
114 * 114 / 4902.22291 = 2.651 and
p − value = P r(χ21 > 2.651) = 0.1035