The UIS data
The goal of the UIS data is to model time until return to drug use for patients enrolled in two different
residential treatment programs that differed in length (treat=0 is the short program and treat=1 is the long
program). The patients were randomly assigned to two different sites (site=0 is site A and site=1 is site
B). The variable age indicates age at enrollment, herco indicates heroin or cocaine use in the past three months
(herco=1 indicates heroin and cocaine use, herco=2 indicates either heroin or cocaine use and herco=3
indicates neither heroin nor cocaine use) and ndrugtx indicates the number of previous drug treatments. The
variables time contains the time until return to drug use and the censor variable indicates whether the subject
returned to drug use (censor=1 indicates return to drug use and censor=0 otherwise).
Let’s load the dataset shared on your site and stset the data
use uis.dta, clear
gen id = ID
drop ID
stset time, failure(censor)
sts graph, na
Let’s look at the first 10 observations of the UIS data set. Note that subject 5 is censored and did not experience
an event while in the study. Also note that the coding for censor is rather counter-intuitive since the value 1
indicates an event and 0 indicates censoring. It would perhaps be more appropriate to call this variable “event”.
list id time censor age ndrugtx treat site herco in 1/10, nodisplay
Exploring the data: Univariate Analyses
In any data analysis it is always a great idea to do some univariate analysis before proceeding to more
complicated models. In survival analysis it is highly recommended to look at the Kaplan-Meier curves for all
the categorical predictors. This will provide insight into the shape of the survival function for each group and
give an idea of whether or not the groups are proportional (i.e. the survival functions are approximately
parallel). We also consider the tests of equality across strata to explore whether or not to include the predictor in
the final model. For the categorical variables we will use the log-rank test of equality across strata which is a
non-parametric test. For the continuous variables we will use a univariate Cox proportional hazard regression
which is a semi-parametric model. We will consider including the predictor if the test has a p-value of 0.2 –
0.25 or less. We are using this elimination scheme because all the predictors in the data set are variables that
could be relevant to the model. If the predictor has a p-value greater than 0.25 in a univariate analysis it is
highly unlikely that it will contribute anything to a model which includes other predictors.
The log-rank test of equality across strata for the predictor treat has a p-value of 0.0091, thus treat will be
included a potential candidate for the final model. From the graph we see that the survival function for each
group of treat are not perfectly parallel but separate except at the very beginning and at the very end. The
overlap at the very end should not cause too much concern because it is determined by only a very few number
of censored subjects out of a sample with 628 subjects. In general, the log-rank test places the more emphasis
on differences in the curves at larger time values. This is why we get such a small p-value even though the two
survival curves appear to be very close together for time less than 100 days.
sts test treat, logrank
sts graph, by(treat)
failure _d:
analysis time _t:
censor
time
Log-rank test for equality of survivor functions
|
Events
Events
treat | observed
expected
------+------------------------0
|
265
235.80
1
|
243
272.20
------+------------------------Total |
508
508.00
chi2(1) =
Pr>chi2 =
6.80
0.0091
The log-rank test of equality across strata for the predictor site has a p-value of 0.1240, thus site will be
included as a potential candidate for the final model because this p-value is still less than our cut-off of 0.2.
From the graph we see that the survival curves are not all that parallel and that there are two periods ( [0, 100]
and [200, 300] ) where the curves are very close together. This would explain the rather high p-value from the
log-rank test.
sts test site, logrank
sts graph, by(site)
failure _d:
analysis time _t:
censor
time
Log-rank test for equality of survivor functions
|
Events
Events
site | observed
expected
------+------------------------0
|
364
347.94
1
|
144
160.06
------+------------------------Total |
508
508.00
chi2(1) =
Pr>chi2 =
2.37
0.1240
The log-rank test of equality across strata for the predictor herco has a p-value of 0.1473, thus herco will be
included as potential candidate for the final model. From the graph we see that the three groups are not parallel
and that especially the groups herco=1 and herco=3 overlap for most of the graph. This lack of parallelism
could pose a problem when we include this predictor in the Cox proportional hazard model since one of the
assumptions is proportionality of the predictors.
sts test herco
sts graph, by(herco) noborder
failure _d:
analysis time _t:
censor
time
Log-rank test for equality of survivor functions
|
Events
Events
herco | observed
expected
------+------------------------1
|
228
242.14
2
|
100
84.19
3
|
180
181.67
------+------------------------Total |
508
508.00
chi2(2) =
Pr>chi2 =
3.83
0.1473
It is not feasible to calculate a Kaplan-Meier curve for the continuous predictors since there would be a curve
for each level of the predictor and a continuous predictor simply has too many different levels. Instead we
consider the Cox proportional hazard model with a single continuous predictor. Unfortunately it is not possibly
to produce a plot when using the stcox command. Instead we consider the Chi-squared test for ndrugtx which
has a p-value of 0.0003 thus ndrugtx is a potential candidate for the final model since the p-value is less than
our cut-off value of 0.2. We specify the option nohr to indicate that we do not want to see the hazard ratio
rather we want to look at the coefficients.
stcox ndrugtx, nohr
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
611
496
143002
Number of obs
=
611
LR chi2(1)
=
13.35
Log likelihood =
-2868.299
Prob > chi2
=
0.0003
-----------------------------------------------------------------------------_t |
_d |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------ndrugtx |
.029372
.0074979
3.92
0.000
.0146763
.0440676
------------------------------------------------------------------------------
In this model the Chi-squared test of age also has a p-value of less than 0.2 and so it is a potential candidate for
the final model.
stcox age, nohr
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
623
504
146816
Number of obs
=
623
LR chi2(1)
=
3.24
Log likelihood =
-2931.4929
Prob > chi2
=
0.0719
------------------------------------------------------------------------------
_t |
_d |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0128641
.0071888
-1.79
0.074
-.0269539
.0012256
------------------------------------------------------------------------------
Model Building
For our model building, we will first consider the model which will include all the predictors that had a p-value
of less than 0.2 – 0.25 in the univariate analyses which in this particular analysis means that we will include
every predictor in our model. The categorical predictor herco has three levels and therefore we will include this
predictor using dummy variable with the group herco=1 as the reference group. We can create these dummy
variables on the fly by using the xi command with stcox.
stcox age ndrugtx i.treat i.site i.herco, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2851.6989
= -2851.0884
= -2851.0863
= -2851.0863
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2851.0863
Number of obs
=
610
LR chi2(6)
Prob > chi2
=
=
34.94
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0237543
.0075611
-3.14
0.002
-.0385737
-.0089349
ndrugtx |
.034745
.0077538
4.48
0.000
.0195478
.0499422
1.treat | -.2540169
.091005
-2.79
0.005
-.4323834
-.0756504
1.site | -.1723881
.1020981
-1.69
0.091
-.3724966
.0277205
|
herco |
2 |
.2467753
.1227597
2.01
0.044
.0061706
.4873799
3 |
.125668
.1030729
1.22
0.223
-.0763513
.3276873
-----------------------------------------------------------------------------test 2.herco 3.herco
( 1)
( 2)
2.herco = 0
3.herco = 0
chi2( 2) =
Prob > chi2 =
4.36
0.1130
The predictor herco is clearly not significant and we will drop it from the final model. The predictor site is also
not significant but from prior research we know that this is a very important variable to have in the final model
and therefore we will not eliminate site from the model. So, the final model of main effects include: age,
ndrugtx, treat and site.
stcox age ndrugtx i.treat i.site, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2853.8641
= -2853.2393
= -2853.2371
= -2853.2371
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2853.2371
Number of obs
=
610
LR chi2(4)
Prob > chi2
=
=
30.64
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0221289
.0075108
-2.95
0.003
-.0368499
-.007408
ndrugtx |
.0350249
.0076676
4.57
0.000
.0199967
.050053
1.treat | -.2436784
.0905411
-2.69
0.007
-.4211358
-.0662211
1.site | -.1683325
.1004119
-1.68
0.094
-.3651362
.0284712
------------------------------------------------------------------------------
Interactions
Next we need to consider interactions. We do not have any prior knowledge of specific interactions that we
must include so we will consider all the possible interactions. Since our model is rather small this is manageable
but the ideal situation is when all model building, including interactions, are theory driven.
The interaction term of age with ndrugtx is not significant and will not be included in the model.
stcox age ndrugtx i.treat i.site c.age#c.ndrug, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Iteration 4:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
=
=
=
=
=
-2868.555
-2854.6056
-2851.8845
-2851.8195
-2851.8195
= -2851.8195
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2851.8195
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
33.47
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0110172
.0100068
-1.10
0.271
-.0306302
.0085959
ndrugtx |
.1054144
.0419532
2.51
0.012
.0231875
.1876412
1.treat | -.2352811
.0906447
-2.60
0.009
-.4129416
-.0576207
1.site | -.1746173
.1004498
-1.74
0.082
-.3714953
.0222607
|
c.age#|
c.ndrugtx | -.0020967
.0012469
-1.68
0.093
-.0045406
.0003472
------------------------------------------------------------------------------
The interaction age and treat is not significant and will not be included in the model.
stcox age ndrugtx i.treat i.site c.age#i.treat, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2852.845
= -2852.1654
= -2852.1631
= -2852.1631
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2852.1631
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
32.78
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0114621
.0103995
-1.10
0.270
-.0318448
.0089205
ndrugtx |
.0357659
.0077155
4.64
0.000
.0206437
.050888
1.treat |
.4483383
.4809163
0.93
0.351
-.4942403
1.390917
1.site | -.1492698
.1010768
-1.48
0.140
-.3473766
.048837
|
treat#c.age |
1 |
-.021469
.0146588
-1.46
0.143
-.0501996
.0072616
------------------------------------------------------------------------------
The interaction age anf site is significant and will be included in the model.
stcox age ndrugtx i.treat i.site c.age#i.site, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2851.487
= -2850.8935
= -2850.8915
= -2850.8915
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
610
495
142994
Number of obs
=
610
LR chi2(5)
=
35.33
Log likelihood
=
-2850.8915
Prob > chi2
=
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0336943
.0092913
-3.63
0.000
-.051905
-.0154837
ndrugtx |
.0364537
.0077012
4.73
0.000
.0213597
.0515478
1.treat | -.2674113
.0912282
-2.93
0.003
-.4462153
-.0886073
1.site | -1.245928
.5087349
-2.45
0.014
-2.24303
-.2488262
|
site#c.age |
1 |
.0337728
.0155087
2.18
0.029
.0033764
.0641693
------------------------------------------------------------------------------
The interaction drug anf treat is not significant and will be not included in the model.
stcox age ndrugtx i.treat i.site c.ndrug#i.treat, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Iteration 4:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
=
=
=
=
=
-2868.555
-2854.1019
-2853.0275
-2853.0174
-2853.0174
= -2853.0174
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
610
495
142994
=
-2853.0174
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
31.08
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0220158
.0075029
-2.93
0.003
-.0367212
-.0073105
ndrugtx |
.0404798
.011066
3.66
0.000
.0187909
.0621686
1.treat | -.1949252
.1166714
-1.67
0.095
-.423597
.0337465
1.site | -.1708522
.1004592
-1.70
0.089
-.3677487
.0260442
|
treat#|
c.ndrugtx |
1 | -.0099061
.0149405
-0.66
0.507
-.0391889
.0193767
------------------------------------------------------------------------------
The interaction drug and site is not significant and will not be included in the model.
stcox age ndrugtx i.treat i.site c.ndrug#i.site, nohr
failure _d:
analysis time _t:
Iteration
Iteration
Iteration
Iteration
0:
1:
2:
3:
log
log
log
log
censor
time
likelihood
likelihood
likelihood
likelihood
= -2868.555
= -2853.9255
= -2853.1789
= -2853.1746
Iteration 4:
log likelihood = -2853.1746
Refining estimates:
Iteration 0:
log likelihood = -2853.1746
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2853.1746
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
30.76
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0222734
.0075266
-2.96
0.003
-.0370251
-.0075216
ndrugtx |
.0366438
.0088665
4.13
0.000
.0192658
.0540218
1.treat | -.2454197
.0906816
-2.71
0.007
-.4231524
-.067687
1.site | -.1417165
.1253391
-1.13
0.258
-.3873766
.1039435
|
site#|
c.ndrugtx |
1 | -.0059702
.0169939
-0.35
0.725
-.0392776
.0273373
------------------------------------------------------------------------------
The interaction treat and site is not significant and will not be included in the model.
stcox age ndrugtx i.treat i.site i.treat#i.site, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2852.4136
= -2851.8662
= -2851.8645
= -2851.8645
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2851.8645
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
33.38
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0238596
.007638
-3.12
0.002
-.0388297
-.0088895
ndrugtx |
.0361507
.0077457
4.67
0.000
.0209694
.0513321
1.treat | -.3404088
.107682
-3.16
0.002
-.5514616
-.129356
1.site | -.3238557
.139417
-2.32
0.020
-.597108
-.0506033
|
treat#site |
1 1 |
.3335144
.2009322
1.66
0.097
-.0603054
.7273342
------------------------------------------------------------------------------
The final model including interaction. Now we can see why it was important to include site in our model as
prior research had suggested because it turns out that site is involved in the only significant interaction in the
model. We can compare the model with the interaction to the model without the interaction using the lrtest
command since the models are nested. The significant lrtest indicates that we reject the null hypothesis that the
two models fit the data equally well and conclude that the bigger model with the interaction fits the data better
than the smaller model which did not include the interaction.
stcox age ndrugtx i.treat i.site c.age#i.site, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2851.487
= -2850.8935
= -2850.8915
= -2850.8915
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2850.8915
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
35.33
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0336943
.0092913
-3.63
0.000
-.051905
-.0154837
ndrugtx |
.0364537
.0077012
4.73
0.000
.0213597
.0515478
1.treat | -.2674113
.0912282
-2.93
0.003
-.4462153
-.0886073
1.site | -1.245928
.5087349
-2.45
0.014
-2.24303
-.2488262
|
site#c.age |
1 |
.0337728
.0155087
2.18
0.029
.0033764
.0641693
-----------------------------------------------------------------------------estimates store m1
stcox age ndrugtx i.treat i.site, nohr
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2853.8641
= -2853.2393
= -2853.2371
= -2853.2371
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2853.2371
Number of obs
=
610
LR chi2(4)
Prob > chi2
=
=
30.64
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+----------------------------------------------------------------
age | -.0221289
.0075108
-2.95
0.003
-.0368499
-.007408
ndrugtx |
.0350249
.0076676
4.57
0.000
.0199967
.050053
1.treat | -.2436784
.0905411
-2.69
0.007
-.4211358
-.0662211
1.site | -.1683325
.1004119
-1.68
0.094
-.3651362
.0284712
-----------------------------------------------------------------------------lrtest . m1
Likelihood-ratio test
(Assumption: . nested in m1)
LR chi2(1) =
Prob > chi2 =
4.69
0.0303
The final model and interpretation of the hazard ratios.
stcox age ndrugtx i.treat i.site c.age#i.site
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2868.555
= -2851.487
= -2850.8935
= -2850.8915
= -2850.8915
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2850.8915
Number of obs
=
610
LR chi2(5)
Prob > chi2
=
=
35.33
0.0000
-----------------------------------------------------------------------------_t | Haz. Ratio
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age |
.966867
.0089835
-3.63
0.000
.9494191
.9846355
ndrugtx |
1.037126
.0079871
4.73
0.000
1.021589
1.052899
1.treat |
.7653582
.0698223
-2.93
0.003
.6400459
.9152049
1.site |
.2876737
.1463497
-2.45
0.014
.1061364
.7797155
|
site#c.age |
1 |
1.03435
.0160414
2.18
0.029
1.003382
1.066273
------------------------------------------------------------------------------
From looking at the hazard ratios (also called relative risks) the model indicates that as the number of previous
drug treatment (ndrugtx) increases by one unit, and all other variables are held constant, the rate of relapse
increases by 3.7%. If the treatment length is altered from short to long, while holding all other variables
constant, the rate of relapse decreases by (100% – 76.5%) = 23.5%. As treatment is moved from site A to site B
and age is equal to zero, and all other variables are held constant, the rate of relapse decreases by (100% –
28.8%) = 71.2%. These results are all based on the output using Hazard ratios. To discuss the variables that are
involved in an interaction term, such as age and site in our model, we need to use the raw coefficients and here
they are listed below just for convenience.
stcox, nohr
Cox regression -- Breslow method for ties
No. of subjects =
610
Number of obs
=
610
No. of failures =
Time at risk
=
Log likelihood
=
495
142994
LR chi2(5)
Prob > chi2
-2850.8915
=
=
35.33
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0336943
.0092913
-3.63
0.000
-.051905
-.0154837
ndrugtx |
.0364537
.0077012
4.73
0.000
.0213597
.0515478
1.treat | -.2674113
.0912282
-2.93
0.003
-.4462153
-.0886073
1.site | -1.245928
.5087349
-2.45
0.014
-2.24303
-.2488262
|
site#c.age |
1 |
.0337728
.0155087
2.18
0.029
.0033764
.0641693
------------------------------------------------------------------------------
Comparing 2 subjects within site A (site=0), an increase in age of 5 years while all other variables are held
constant yields a hazard ratio equal to exp(-0.03369*5) = .84497351. Thus, the rate of relapse is decreased by
(100% – 84.5%) = 15.5% with an increase of 5 years in age. Comparing 2 subjects within site B, an increase in
age of 5 years while holding all other variables constant, yields a hazard ratio equal to exp(-0.03369*5 +
0.03377*5) = 1.0004. Thus, the rate of relapse stays fairly flat for subjects at site B since 1.0004 if so close to
1.
Proportionality Assumption
One of the main assumptions of the Cox proportional hazard model is proportionality. There are several
methods for verifying that a model satisfies the assumption of proportionality. We will check proportionality
by including time-dependent covariates in the model by using the tvc and the texp options in the stcox
command. Time dependent covariates are interactions of the predictors and time. In this analysis we choose to
use the interactions with log(time) because this is the most common function of time used in time-dependent
covariates but any function of time could be used. If a time-dependent covariate is significant this indicates a
violation of the proportionality assumption for that specific predictor.
The conclusion is that all of the time-dependent variables are not significant either collectively or individually
thus supporting the assumption of proportional hazard.
stcox age ndrugtx i.treat i.site c.age#i.site, nohr tvc(age ndrugtx treat site)
texp(ln(_t))
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Iteration 4:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
=
=
=
=
=
-2868.555
-2850.4619
-2849.8647
-2849.8626
-2849.8626
= -2849.8626
Cox regression -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
610
495
142994
Number of obs
=
610
LR chi2(9)
=
37.38
Log likelihood
=
-2849.8626
Prob > chi2
=
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------main
|
age | -.0322788
.0340846
-0.95
0.344
-.0990834
.0345258
ndrugtx |
.0173789
.0321568
0.54
0.589
-.0456473
.0804052
1.treat | -.6671007
.4114915
-1.62
0.105
-1.473609
.1394078
1.site | -1.637207
.6801889
-2.41
0.016
-2.970353
-.3040617
|
site#c.age |
1 |
.033723
.015548
2.17
0.030
.0032495
.0641965
-------------+---------------------------------------------------------------tvc
|
age | -.0004057
.007119
-0.06
0.955
-.0143587
.0135473
ndrugtx |
.0042828
.0069637
0.62
0.539
-.0093658
.0179314
treat |
.0860457
.0863163
1.00
0.319
-.0831312
.2552226
site |
.084347
.0974399
0.87
0.387
-.1066317
.2753258
-----------------------------------------------------------------------------Note: variables in tvc equation interacted with ln(_t)
Another method of testing the proportionality assumption is by using the Schoenfeld and scaled Schoenfeld
residuals which must first be saved through the stcox command. In the stphtest command we test the
proportionality of the model as a whole and by using the detail option we get a test of proportionality for each
predictor. By using the plot option we can also obtain a graph of the scaled Schoenfeld assumption. If the tests
in the table are not significance (p-values over 0.05) then we can not reject proportionality and we assume that
we do not have a violation of the proportional assumption. A horizontal line in the graphs is further indication
that there is no violation of the proportionality assumption. The stphplot command uses log-log plots to test
proportionality and if the lines in these plots are parallel then we have further indication that the predictors do
not violate the proportionality assumption.
The predictor treat might warrant some closer examination since it does have a significant test and the curve in
the graph is not completely horizontal. The graph from the stphplot command does not have completely
parallel curves. However, we choose to leave treat in the model unaltered based on prior research.
quietly stcox age ndrugtx treat site c.age#i.site, schoenfeld(sch*) scaledsch(sca*)
stphtest, detail
stphtest, plot(age) msym(oh)
stphtest, plot(ndrugtx) msym(oh)
stphtest, plot(treat) msym(oh)
stphtest, plot(site) msym(oh)
stphtest, plot(c.age#1.site) msym(oh)
stphplot, by(treat) plot1(msym(oh)) plot2(msym(th))
stphplot, by(site) plot1(msym(oh)) plot2(msym(th))
drop sch1-sch5 sca1-sca5
Test of proportional hazards assumption
Time: Time
---------------------------------------------------------------|
rho
chi2
df
Prob>chi2
------------+--------------------------------------------------age
|
0.01210
0.07
1
0.7912
ndrugtx
|
0.05563
1.47
1
0.2260
treat
|
0.10598
5.61
1
0.0179
site
|
0.02336
0.25
1
0.6150
age_site
|
-0.01350
0.08
1
0.7722
------------+--------------------------------------------------global test |
8.27
5
0.1419
----------------------------------------------------------------
If one of the predictors were not proportional there are various solutions to consider. One solution is to include
the time-dependent variable for the non-proportional predictors. Another solution is to stratify on the nonproportional predictor. The following is an example of stratification on the predictor treat. Note that treat is no
longer included in the model statement instead it is specified in the strata statement.
stcox age ndrugtx site c.age#i.site, nohr strata(treat)
failure _d:
analysis time _t:
censor
time
Iteration 0:
log likelihood
Iteration 1:
log likelihood
Iteration 2:
log likelihood
Iteration 3:
log likelihood
Refining estimates:
Iteration 0:
log likelihood
= -2525.8022
= -2511.524
= -2510.8632
= -2510.8606
= -2510.8606
Stratified Cox regr. -- Breslow method for ties
No. of subjects =
No. of failures =
Time at risk
=
Log likelihood
=
610
495
142994
-2510.8606
Number of obs
=
610
LR chi2(4)
Prob > chi2
=
=
29.88
0.0000
-----------------------------------------------------------------------------_t |
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
-------------+---------------------------------------------------------------age | -.0347463
.0092884
-3.74
0.000
-.0529512
-.0165413
ndrugtx |
.0363728
.0076961
4.73
0.000
.0212888
.0514569
site | -1.251301
.5085532
-2.46
0.014
-2.248047
-.2545555
|
site#c.age |
1 |
.0339858
.0155057
2.19
0.028
.0035951
.0643765
-----------------------------------------------------------------------------Stratified by treat
Graphing Survival Functions
Each covariate pattern will have a different survival function. The default survival function is for the covariate
pattern where each predictor is set equal to zero. However, for many predictors this value is not meaningful
because this value falls outside of the data such as age=0. It would be much more useful to specify an exact
covariate pattern and generate a survival function for subjects with that specific covariate pattern.
In the following example we want to graph the survival function for a subject who is 30 years old (age=30), has
had 5 prior drug treatments (ndrugtx=5), and is currently getting the long treatment (treat=1) at site A (site=0
and agesite=30*0 = 0). We first output the baseline survival function for the covariate pattern where all
predictors are set to zero. Then we raise the baseline survival function to the exponential to the linear
combination of the coefficients and the values of the covariates in the covariate pattern of interest. Thus, in this
particular instance the linear combination would be: -0.0336943*30+0.0364537*5 – 0.2674113*1 – 1.245928*0
– .0337728*0.
stcox age ndrugtx treat site c.age#i.site, nohr basesurv(surv0)
generate surv1 = surv0^exp( (-0.0336943*30+0.0364537*5 - 0.2674113))
line surv1 _t, sort ylab(0 .1 to 1) xlab(0 200 to 1200)
Looking at the survival function for one covariate pattern is sometimes not sufficient. It is often very useful to
have a graph where we can compare the survival functions of different groups. In the following example we
generate a graph with the survival functions for the two treatment groups where all the subjects are 30 years old
(age=30), have had 5 prior drug treatments (ndrugtx=5) and are currently being treated at site A (site=0 and
agesite=30*0=0). Thus, the two covariate patterns differ only in their values for treat.
generate surv2 = surv0^exp( (-0.0336943*30+0.0364537*5))
label variable surv1 "long treatment"
label variable surv2 "short treatment"
line surv1 surv2 _t, sort ylab(0 .1 to 1) xlab(0 200 to 1200)
drop surv0-surv2
Goodness of Fit of the Final Model
We can evaluate the fit of the model by using the Cox-Snell residuals. If the model fits the data well then the
true cumulative hazard function conditional on the covariate vector has an exponential distribution with a
hazard rate of one. This translates into fitting the model using the stcox command and specifying the mgale
option which will generate the martingale residuals. Then we use the predict command with the csnell option
to generate the Cox-Snell residuals for the model. We reset the data using the stset command specifying the
variable cs, the variable containing the Cox-Snell residuals, as the time variable. We then use the sts generate
command to create the Nelson-Aalen cumulative hazard function. Finally, we graph the Nelson-Aalen
cumulative hazard function and the cs variable so that we can compare the hazard function to the diagonal
line. If the hazard function follows the 45 degree line then we know that it approximately has an exponential
distribution with a hazard rate of one and that the model fits the data well.
quietly stcox age ndrugtx treat site c.age#i.site, nohr mgale(mg)
predict cs, csnell
stset cs, failure(censor)
sts generate H = na
line H cs cs, sort xlab(0 1 to 4) ylab(0 1 to 4)
drop mg
We see that the hazard function follows the 45 degree line very closely except for very large values of time. It is
very common for models with censored data to have some wiggling at large values of time and it is not
something which should cause much concern. Overall we would conclude that the final model fits the data very
well.