Introduction to Survival Analysis With Stata Seminar

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Statistical Computing SeminarsSurvival Analysis with Stata

Background for Survival AnalysisThe UIS dataExploring the data: Univariate AnalysesModel BuildingInteractionsProportionality AssumptionGraphing Survival Functions from stcox commandGoodness of Fit of the Final Model

The Stata program on which the seminar is based.The UIS_small data file for the seminar.

Background for Survival AnalysisThe goal of this seminar is to give a brief introduction to the topic of survival analysis. We will be using a smaller and slightly modified version of the UISdata set from the book "Applied Survival Analysis" by Hosmer and Lemeshow. We strongly encourage everyone who is interested in learning survivalanalysis to read this text as it is a very good and thorough introduction to the topic.

Survival analysis is just another name for time to event analysis. The term survival analysis is predominately used in biomedical sciences where theinterest is in observing time to death either of patients or of laboratory animals. Time to event analysis has also been used widely in the social scienceswhere interest is on analyzing time to events such as job changes, marriage, birth of children and so forth. The engineering sciences have also contributedto the development of survival analysis which is called "reliability analysis" or "failure time analysis" in this field since the main focus is in modeling the timeit takes for machines or electronic components to break down. The developments from these diverse fields have for the most part been consolidated intothe field of "survival analysis". For more background please refer to the excellent discussion in Chapter 1 of Event History Analysis by Paul Allison.

There are certain aspects of survival analysis data, such as censoring and non-normality, that generate great difficulty when trying to analyze the datausing traditional statistical models such as multiple linear regression. The non-normality aspect of the data violates the normality assumption of mostcommonly used statistical model such as regression or ANOVA, etc. A censored observation is defined as an observation with incomplete information. There are four different types of censoring possible: right truncation, left truncation, right censoring and left censoring. We will focus exclusively on rightcensoring for a number of reasons. Most data used in analyses have only right censoring. Furthermore, right censoring is the most easily understood ofall the four types of censoring and if a researcher can understand the concept of right censoring thoroughly it becomes much easier to understand theother three types. When an observation is right censored it means that the information is incomplete because the subject did not have an event during thetime that the subject was part of the study. The point of survival analysis is to follow subjects over time and observe at which point in time they experiencethe event of interest. It often happens that the study does not span enough time in order to observe the event for all the subjects in the study. This could bedue to a number of reasons. Perhaps subjects drop out of the study for reasons unrelated to the study (i.e. patients moving to another area and leaving noforwarding address). The common feature of all of these examples is that if the subject had been able to stay in the study then it would have been possibleto observe the time of the event eventually.

It is important to understand the difference between calendar time and time in the study. It is very common for subjects to enter the study continuouslythroughout the length of the study. This situation is reflected in the first graph where we can see the staggered entry of four subjects. Red dots denoteintervals in which the event occurred, whereas intervals without red dots signify censoring. It would appear that subject 4 dropped out after only a shorttime (hit by a bus, very tragic) and that subject 3 did not experience an event by the time the study ended but if the study had gone on longer (had morefunding) we would have known the time when this subject would have experienced an event.

clearinput subj tp censored str11 datestr1 1 0 "1 jan 1990"1 2 0 "1 mar 1991"2 1 1 "1 feb 1990"2 2 1 "1 feb 1991"3 1 1 "1 jun 1990"3 2 1 "31 dec 1991"4 1 0 "1 sep 1990"4 2 0 "1 apr 1991"end

gen date = date(datestr, "DMY")format date %dmytwoway (line subj date, connect(L))(scatter subj date if censored==0), /// ylabel(1 2 3 4) legend(order (2 "censored"))

gen time =0 if tp==1replace time= (date-date[_n-1])/30.5 if tp==2twoway (line subj time, connect(L))(scatter subj time if censored==0), /// ylabel(1 2 3 4) legend(order (2 "censored")) xlabel(0 8 12 14 19 24)

>stat >stata >seminars stata_survival

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http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#Background

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#data

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#exploring

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#building

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#interactions

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#proportionality

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#graphs

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/default.htm#fit

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/survival.do

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/uis.dta

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival/uis.dta

http://www.ats.ucla.edu/stat/stata/examples/asa

http://www.ats.ucla.edu/stat/books/default.htm

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http://www.ats.ucla.edu/stat/stata

http://www.ats.ucla.edu/stat/stata/seminars

http://www.ats.ucla.edu/stat/stata/seminars/stata_survival

The other important concept in survival analysis is the hazard rate. From looking at data with discrete time (time measured in large intervals such asmonth, years or even decades) we can get an intuitive idea of the hazard rate. For discrete time the hazard rate is the probability that an individual willexperience an event at time t while that individual is at risk for having an event. Thus, the hazard rate is really just the unobserved rate at which eventsoccur. If the hazard rate is constant over time and it was equal to 1.5 for example this would mean that one would expect 1.5 events to occur in a timeinterval that is one unit long. Furthermore, if a person had a hazard rate of 1.2 at time t and a second person had a hazard rate of 2.4 at time t then itwould be correct to say that the second person's risk of an event would be two times greater at time t. It is important to realize that the hazard rate is anun-observed variable yet it controls both the occurrence and the timing of the events. It is the fundamental dependent variable in survival analysis.

Another important aspect of the hazard function is to understand how the shape of the hazard function will influence the other variables of interest such asthe survival function. The first graph below illustrates a hazard function with a 'bathtub shape'. This graph is depicting the hazard function for the survival oforgan transplant patients. At time equal to zero they are having the transplant and since this is a very dangerous operation they have a very high hazard (agreat chance of dying). The first 10 days after the operation are also very dangerous with a high chance of the patient dying but the danger is less thanduring the actual operation and hence the hazard is decrease during this period. If the patient has survived past day 10 then they are in very good shapeand have a very little chance of dying in the following 6 months. After 6 months the patients begin to experience deterioration and the chances of dyingincrease again and therefore the hazard function starts to increase. After one year almost all patients are dead and hence the very high hazard functionwhich will continue to increase.

The hazard function may not seem like an exciting variable to model but other indicators of interest, such as the survival function, are derived from thehazard rate. Once we have modeled the hazard rate we can easily obtain these other functions of interest. To summarize, it is important to understandthe concept of the hazard function and to understand the shape of the hazard function.

An example of a hazard function for heart transplant patients.

We are generally unable to generate the hazard function instead we usually look at the cumulative hazard curve.

use http://www.ats.ucla.edu/stat/data/uis.dta, cleargen id = IDdrop IDstset time, failure(censor)sts graph, na

The UIS dataThe 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 issite B). The variable age indicates age at enrollment, herco indicates heroin or cocaine use in the past three months (herco=1 indicates heroin andcocaine use, herco=2 indicates either heroin or cocaine use and herco=3 indicates neither heroin nor cocaine use) and ndrugtx indicates the number ofprevious drug treatments. The variables time contains the time until return to drug use and the censor variable indicates whether the subject returned todrug use (censor=1 indicates return to drug use and censor=0 otherwise). 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 notethat the coding for censor is rather counter-intuitive since the value 1 indicates an event and 0 indicates censoring. It would perhaps be more appropriateto call this variable "event".

list id time censor age ndrugtx treat site herco in 1/10, nodisplay

id time censor age ndrugtx treat site herco 1. 1 188 1 39 1 1 0 3 2. 2 26 1 33 8 1 0 3 3. 3 207 1 33 3 1 0 2 4. 4 144 1 32 1 0 0 3 5. 5 551 0 24 5 1 0 2 6. 6 32 1 30 1 1 0 1 7. 7 459 1 39 34 1 0 3 8. 8 22 1 27 2 1 0 3 9. 9 210 1 40 3 1 0 2 10. 10 184 1 36 7 1 0 2

Exploring the data: Univariate AnalysesIn 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 highlyrecommended to look at the Kaplan-Meier curves for all the categorical predictors. This will provide insight into the shape of the survival function for eachgroup 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 ofequality 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 ofequality across strata which is a non-parametric test. For the continuous variables we will use a univariate Cox proportional hazard regression which is asemi-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 schemebecause 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 univariateanalysis 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 finalmodel. 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 atthe 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 subjectsout 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 whywe 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, logranksts graph, by(treat)

failure _d: censor analysis time _t: time

Log-rank test for equality of survivor functions

| Events Eventstreat | observed expected------+-------------------------0 | 265 235.801 | 243 272.20------+-------------------------Total | 508 508.00

chi2(1) = 6.80 Pr>chi2 = 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 finalmodel 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 twoperiods ( [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, logranksts graph, by(site)



| Events Eventssite | observed expected------+-------------------------0 | 364 347.941 | 144 160.06------+-------------------------Total | 508 508.00

chi2(1) = 2.37 Pr>chi2 = 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 finalmodel. 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 isproportionality of the predictors.

sts test hercosts graph, by(herco) noborder



| Events Eventsherco | observed expected------+-------------------------1 | 228 242.142 | 100 84.193 | 180 181.67------+-------------------------Total | 508 508.00

chi2(2) = 3.83 Pr>chi2 = 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 acontinuous 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 nohrto 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 = 611 Number of obs = 611No. of failures = 496Time at risk = 143002 LR chi2(1) = 13.35Log 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


No. of subjects = 623 Number of obs = 623No. of failures = 504Time at risk = 146816 LR chi2(1) = 3.24Log 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 BuildingFor 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 univariateanalyses which in this particular analysis means that we will include every predictor in our model. The categorical predictor herco has three levels andtherefore we will include this predictor using dummy variable with the group herco=1 as the reference group. We can create these dummy variables on thefly by using the xi command with stcox.

stcox age ndrugtx i.treat i.site i.herco, nohr


Iteration 0: log likelihood = -2868.555Iteration 1: log likelihood = -2851.6989Iteration 2: log likelihood = -2851.0884Iteration 3: log likelihood = -2851.0863Refining estimates:Iteration 0: log likelihood = -2851.0863


No. of subjects = 610 Number of obs = 610No. of failures = 495Time at risk = 142994 LR chi2(6) = 34.94Log likelihood = -2851.0863 Prob > chi2 = 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.herco = 0 ( 2) 3.herco = 0

chi2( 2) = 4.36 Prob > chi2 = 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 weknow 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 maineffects include: age, ndrugtx, treat and site.

stcox age ndrugtx i.treat i.site, nohr





------------------------------------------------------------------------------ _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------------------------------------------------------------------------------

InteractionsNext we need to consider interactions. We do not have any prior knowledge of specific interactions that we must include so we will consider all thepossible interactions. Since our model is rather small this is manageable but the ideal situation is when all model building, including interactions, are theorydriven.

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


Iteration 0: log likelihood = -2868.555Iteration 1: log likelihood = -2854.6056Iteration 2: log likelihood = -2851.8845Iteration 3: log likelihood = -2851.8195Iteration 4: log likelihood = -2851.8195Refining estimates:Iteration 0: log likelihood = -2851.8195



------------------------------------------------------------------------------ _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





------------------------------------------------------------------------------ _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





------------------------------------------------------------------------------ _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


Iteration 0: log likelihood = -2868.555Iteration 1: log likelihood = -2854.1019Iteration 2: log likelihood = -2853.0275Iteration 3: log likelihood = -2853.0174Iteration 4: log likelihood = -2853.0174Refining estimates:

Iteration 0: log likelihood = -2853.0174



------------------------------------------------------------------------------ _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





------------------------------------------------------------------------------ _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





------------------------------------------------------------------------------ _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 turnsout 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 interactionusing the lrtest command since the models are nested. The significant lrtest indicates that we reject the null hypothesis that the two models fit the dataequally 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






estimates store m1

stcox age ndrugtx i.treat i.site, nohr





------------------------------------------------------------------------------ _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 LR chi2(1) = 4.69(Assumption: . nested in m1) Prob > chi2 = 0.0303

The final model and interpretation of the hazard ratios.

stcox age ndrugtx i.treat i.site c.age#i.site





------------------------------------------------------------------------------ _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 oneunit, 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 allother 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 usingHazard 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 andhere they are listed below just for convenience.

stcox, nohr




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 toexp(-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 subjectswithin 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 AssumptionOne of the main assumptions of the Cox proportional hazard model is proportionality. There are several methods for verifying that a model satisfies theassumption of proportionality and for more information on this topic please refer to our FAQ Tests of proportionality in SAS, Stata, SPLUS and R. We willcheck proportionality by including time-dependent covariates in the model by using the tvc and the texp options in the stcox command. Time dependentcovariates are interactions of the predictors and time. In this analysis we choose to use the interactions with log(time) because this is the most commonfunction of time used in time-dependent covariates but any function of time could be used. If a time-dependent covariate is significant this indicates aviolation 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 proportionalhazard.

stcox age ndrugtx i.treat i.site c.age#i.site, nohr tvc(age ndrugtx treat site) texp(ln(_t))




No. of subjects = 610 Number of obs = 610No. of failures = 495

http://www.ats.ucla.edu/stat/examples/asa/test_proportionality.htm

Time at risk = 142994 LR chi2(9) = 37.38Log 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 throughthe 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 ofproportionality 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 notsignificance (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. Ahorizontal line in the graphs is further indication that there is no violation of the proportionality assumption. The stphplot command uses log-log plots totest proportionality and if the lines in these plots are parallel then we have further indication that the predictors do not violate the proportionalityassumption.

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 onprior research.

quietly stcox age ndrugtx treat site c.age#i.site, schoenfeld(sch*) scaledsch(sca*)stphtest, detailstphtest, 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 non-proportional 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.

bysort treat: stcox age ndrugtx site c.age#i.site, nohr

-------------------------------------------------------------------------------------------------------------------------------------> treat = 0





------------------------------------------------------------------------------ _t | Coef. Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- age | -.0332994 .0139111 -2.39 0.017 -.0605646 -.0060341 ndrugtx | .0403077 .0115213 3.50 0.000 .0177263 .062889 site | -1.7505 .7047318 -2.48 0.013 -3.131749 -.3692513 | site#c.age | 1 | .0454033 .0213107 2.13 0.033 .0036351 .0871715------------------------------------------------------------------------------

-------------------------------------------------------------------------------------------------------------------------------------> treat = 1


Iteration 0: log likelihood = -1214.6484Iteration 1: log likelihood = -1206.8412Iteration 2: log likelihood = -1206.4699Iteration 3: log likelihood = -1206.4683

Refining estimates:Iteration 0: log likelihood = -1206.4683



------------------------------------------------------------------------------ _t | Coef. Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- age | -.0383652 .0126867 -3.02 0.002 -.0632308 -.0134997 ndrugtx | .0363666 .0105992 3.43 0.001 .0155926 .0571406 site | -.4327386 .7562894 -0.57 0.567 -1.915039 1.049561 | site#c.age | 1 | .0139769 .0232429 0.60 0.548 -.0315784 .0595322------------------------------------------------------------------------------

The parameter estimates are almost the same for each level of treat which further indicates that treat really is proportional. If treat were truly violating theassumption of proportionality we would expect the estimates to differ. The estimates are also very similar to the estimates obtained from the modelincluding treat as a predictor.

stcox age ndrugtx treat site c.age#i.site, nohr





------------------------------------------------------------------------------ _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 treat | -.2674113 .0912282 -2.93 0.003 -.4462153 -.0886073 site | -1.245928 .5087349 -2.45 0.014 -2.24303 -.2488262 | site#c.age | 1 | .0337728 .0155087 2.18 0.029 .0033764 .0641693------------------------------------------------------------------------------

Graphing Survival FunctionsEach covariate pattern will have a different survival function. The default survival function is for the covariate pattern where each predictor is set equal tozero. 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 usefulto 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 covariatepattern where all predictors are set to zero. Then we raise the baseline survival function to the exponential to the linear combination of the coefficients andthe 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 thesurvival functions of different groups. In the following example we generate a graph with the survival functions for the two treatment groups where all thesubjects 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 ModelWe 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 onthe covariate vector has an exponential distribution with a hazard rate of one. This translates into fitting the model using the stcox command andspecifying 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, asthe time variable. We then use the sts generate command to create the Nelson-Aalen cumulative hazard function. Finally, we graph the Nelson-Aalencumulative hazard function and the cs variable so that we can compare the hazard function to the diagonal line. If the hazard function follows the 45degree 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 = naline 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

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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 finalmodel fits the data very well.

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