Lecture 20

Comparing groupsCox PHM

Comparing two or more samples

Anova type approach

where τ is the largest time for which all groups have at least one subject at risk

Data can be right-censored for the tests we will discuss

tthH

tthththH

jA

K

somefor different is )( of oneleast at :

allfor ,)()()(: 210

Notation

t1<t2<…tD be distinct death times in all samples being compared

At time ti, let dij be the number of deaths in group j out of Yij individuals at risk. (j=1,..,K)

Define

K

j iji

K

j iji

YY

dd

1

1

Log-Rank Test Rationale

Comparisons of the estimated hazard rate of the jth population under the null and alternative hypotheses

If the null is true, the pooled estimate of h(t) should be an estimator for hj(t)

Applying the Test

for j = 1,…,K

If all Zj(τ)’s are close to zero, then little evidence to reject the null.

D

i i

i

ij

ijj Y

d

Y

dZ

1

)(

Others?

LOTS! Gehan test Fleming-Harrington

Not all available in all software Worth trying a few in each situation

to compare inferences

2+ samples

Let’s look at a prostate cancer dataset

Prostate cancer clinical trial 3 trt groups (doce Q3, doce weekly, Q3

mitoxantrone) 5 PSA doubling times categories outcome: overall survival

TAX327 results

0 10 20 30 40 50 60 70

0.0

0.2

0.4

0.6

0.8

1.0

time (months)

S(t

)

Q3 DoceWeekly DoceMitox

R: survdiff

################################## test for differences by trt grp

plot(survfit(st~trt), mark.time=F, col=c(1,2,3))

test1 <- survdiff(st~trt)test2 <- survdiff(st~factor(trt, exclude=3))test3 <- survdiff(st[trt<3]~trt[trt<3])

PSADT categories

0 10 20 30 40 50 60 70

0.0

0.2

0.4

0.6

0.8

1.0

01234

R: survdiff

table(psadt)plot(survfit(st~psadt), mark.time=F, col=1:5, lwd=rep(2,5))legend(50,1,as.character(0:4), lty=rep(1,5), col=1:5,

lwd=rep(2,5))

test1 <- survdiff(st~psadt)

test2 <- survdiff(st[psadt<3 & psadt>0]~psadt[psadt<3 & psadt>0])

test3 <- survdiff(st[psadt>2]~psadt[psadt>2])

Caveat

Note that we are interested in the average difference (consider log-rank specifically)

What if hazards ‘cross’? Could have significant difference

prior to some t, and another significant difference after t: but, what if direction differs?

What about all those differences in our prostate cancer KM curves?

Not much evidence of crossing if there isnt overlap, then tests will

be somewhat consistent log-rank: most appropriate for

‘proportional hazards’

Example

Klein & Moeschberger 1.4 Kidney infection data Two groups:

patients with percutaneous placement of catheters (N=76)

patients with surgical placement of catheters (N=43)

Kaplan-Meier curves

0 5 10 15 20 25

0.0

0.2

0.4

0.6

0.8

1.0

Time to infection (months)

S(t

)

PercutaneousSurgical

Log-rank

Comparisons

p

0.110.960.530.240.260.002

0.24

0.002

0.002

0.004

Why such large differences?

Notice the differences!

Situation of varying inferences Need to be sure that you are testing what

you think you are testing Check:

look at hazards? do they cross?

Problem: estimating hazards is messy and imprecise recall: h(t)= derivative H(t)

Misconception

Survival curves crossing telling about appropriateness of log-rank

Not true: survivals crossing depends on censoring and study

length what if they will cross but t range isnt sufficient?

Consider: Survival curves cross hazards cross Hazards cross survivals may or may not cross

solution? test in regions of t prior to and after cross based on looking at hazards some tests allow for crossing (Yang and Prentice

2005)

Cox Propotional Hazards Model

Names Cox regression semi-parametric proportional hazards Proportional hazards model Multiplicative hazards model

When? 1972

Why? allows adjustment for covariates (continuous or

categorical) in a survival setting allows prediction of survival based on a set of

covariates Analogous to linear and logistic regression in

many ways

Cox PHM Notation

Data on n individuals: Tj : time on study for individual j

dj : event indicator for individual j

Zj : vector of covariates for individual j

More complicated: Zj(t) covariates can be time dependent they may change with time/age

Basic Model

)exp()(

)(

)()()|(

0

0

0

kk

Z

t

Zth

eth

ZcthZtht

)exp()()|( 10 iiii ZthZth

For a Cox model with just one covariate:

Comments on basic model

h0(t): arbitrary baseline hazard rate. notice that it varies by t

β: regression coefficient (vector) interpretation is a log hazard ratio

Semi-parametric form non-parametric baseline hazard parametric form assumed only for covariate

effects

Linear model formulation

Usual formulation Coding of covariates similar to linear

and logistic (and other generalized linear models)

kk

kk

Zth

Zth

ZthZth

)(

)|(log

)exp()()|(

0

0

Why “proportional”?

hazard ratio Does not depend on t (i.e., it is a constant

over time) But, it is proportional (constant

multiplicative factor) Also referred to (sometimes) as the

relative risk.

)(exp

exp)(

exp)(

)|(

)|(

21

20

10

2

1

ZZ

Zth

Zth

Zth

Zth

Simple example

one covariate: z = 1 for new treatment, z=0 for standard treatment

hazard ratio = exp(β) interpretation: exp(β) is the risk of having

the event in the new treatment group versus the standard treatment at any point in time.

Interpretation: at any point in time, the risk of the event in the new treatment group is exp(β) times the risk in the standard treatment group

Hazard Ratios

Assumption: “Proportional hazards” The risk does not depend on time. That is, “risk is constant over time” But that is still vague…..

Hypothetical Example: Assume hazard ratio is 0.5. Patients in new therapy group are at half the risk of death

as those in standard treatment, at any given point in time.

Hazard function= P(die at time t | survived to time t)

Hazard Ratios

Hazard Ratio = hazard function for Newhazard function for Std

Makes assumption that this ratio is constant over time.

Time

Ha

zard

Fu

nct

ion

0 10 20 30 40

0.0

0.2

0.4

0.6

0.8

Std TherapyNew Drug

Interpretation Again

For any fixed point in time, individuals in the new treatment group are at half the risk of death as the standard treatment group.

Hazard ratio is not always valid ….Nelson-Aalen cumulative hazard estimates, by group

analysis time0 10 20 30 40

0.00

1.00

2.00

3.00

4.00

group 0

group 1

Hazard Ratio = .71

Kaplan-Meier survival estimates, by group

analysis time0 10 20 30 40

0.00

0.25

0.50

0.75

1.00

group 0

group 1

Refresher of coding covariates

This should be nothing new Two kinds of ‘independent’ variables

quantitative qualitative

Quantitative are continuous need to determine scale

units transformation?

Qualitative are generally categorical ordered nominal coding affects the interpretation

Tests of the model

Testing that βk=0 for all k=1,..,p Three main tests

Chi-square/Wald test Likelihood ratio test score(s) test

All three have chi-square distribution with p degrees of freedom

Example: TAX327

Randomized clinical trial of men with hormone-refractory prostate cancer

three treatment arms (Q3 docetaxel, weekly docetaxel, and Q3 mitixantrone)

other covariates of interest: psa doubling time lymph node involvement liver metastases number of metastatic sites pain at baseline baseline psa tumor grade alkaline phosphatase hemoglobin performance status

Some of the covariates

0 10 30 50 70

0.0

0.4

0.8

time (months)

S(t

)

Performance Status

0 10 30 50 70

0.0

0.4

0.8

time (months)

S(t

)

PSADT

0 10 30 50 70

0.0

0.4

0.8

time (months)

S(t

)

High Grade

0 10 30 50 70

0.0

0.4

0.8

time (months)

S(t

)

Number of Sites

Cox PHM approach

st <- Surv(survtime, died)attach(data, pos=2)

reg1 <- coxph(st ~ trtgrp)reg2 <- coxph(st ~ factor(trtgrp))summary(reg2)attributes(reg2)reg2$coefficientssummary(reg2)$coef

Results

> summary(reg2)Call:coxph(formula = st ~ factor(trtgrp))

n= 1006 coef exp(coef) se(coef) z pfactor(trtgrp)2 0.105 1.11 0.0882 1.19 0.2300factor(trtgrp)3 0.245 1.28 0.0863 2.84 0.0045

exp(coef) exp(-coef) lower .95 upper .95factor(trtgrp)2 1.11 0.900 0.935 1.32factor(trtgrp)3 1.28 0.783 1.079 1.51

Rsquare= 0.008 (max possible= 1 )Likelihood ratio test= 8.12 on 2 df, p=0.0173Wald test = 8.16 on 2 df, p=0.0169Score (logrank) test = 8.19 on 2 df, p=0.0167

Multiple regression In the published paper, the model included

all covariates included in previous list

Fitting it in R

reg3 <- coxph(st ~ factor(trtgrp) + liverny + numbersites + pain0c + pskar2c + proml + probs + highgrade + logpsa0 + logalkp0c + hemecenter + psadtmonthcat)

reg4 <- coxph(st ~ factor(trtgrp) + liverny + numbersites + pain0c + pskar2c + proml + probs + highgrade + logpsa0 + logalkp0c + hemecenter + factor(psadtmonthcat))

> reg3Call:coxph(formula = st ~ factor(trtgrp) + liverny + numbersites + pain0c + pskar2c + proml + probs + highgrade + logpsa0 + logalkp0c + hemecenter + psadtmonthcat)

coef exp(coef) se(coef) z pfactor(trtgrp)2 0.1230 1.131 0.1099 1.12 2.6e-01factor(trtgrp)3 0.3784 1.460 0.1070 3.54 4.0e-04liverny 0.4813 1.618 0.2168 2.22 2.6e-02numbersites 0.4757 1.609 0.1430 3.33 8.8e-04pain0c 0.3708 1.449 0.0925 4.01 6.1e-05pskar2c 0.3167 1.373 0.1339 2.37 1.8e-02proml 0.3132 1.368 0.1125 2.78 5.4e-03probs 0.2568 1.293 0.0991 2.59 9.5e-03highgrade 0.1703 1.186 0.0922 1.85 6.5e-02logpsa0 0.1549 1.168 0.0312 4.96 7.0e-07logalkp0c 0.2396 1.271 0.0483 4.96 7.0e-07hemecenter -0.1041 0.901 0.0351 -2.96 3.1e-03psadtmonthcat -0.0884 0.915 0.0430 -2.05 4.0e-02

Likelihood ratio test=205 on 13 df, p=0 n=641 (365 observations deleted due to missingness)

>

“Local” Tests

Testing individual coefficients But, more interestingly, testing sets of

coefficients Example:

testing the psa variables testing treatment group (3 categories)

Same as previous: Wald test Likelihood ratio Scores test

TAX327

reg5 <- coxph(st ~ liverny + numbersites + pain0c + pskar2c + proml + probs + highgrade + logpsa0 + logalkp0c + hemecenter + factor(psadtmonthcat))

lrt.trt <- 2*(reg4$loglik[2] - reg5$loglik[2])p.trt <- 1-pchisq(lrt.trt, 2)

#` to compare, you need to have the same datasetliverny1 <- ifelse(is.na(psadtmonthcat),NA,liverny)reg6 <- coxph(st ~ factor(trtgrp) + liverny1 + numbersites

+ pain0c + pskar2c + proml + probs + highgrade + logpsa0 + logalkp0c + hemecenter)

lrt.psadt <- 2*(reg4$loglik[2] - reg6$loglik[2])p.psadt <- 1-pchisq(lrt.psadt, 4)

proportional?

recall we are making strong assumption that we have proportional hazards for each covariate

we can investigate this to some extent via graphical displays

formal test of the assumption is also possible. can be used for quantitative or qualitative

variables depends strongly on N simply gives p-value

Approach for testing proportionality

there are residuals in Cox models “Schoenfield residuals” have similar interpretation

to residuals from linear regression recall in linear regression: no pattern between

residuals and covariates “rho” is estimated to be the correlation between

transformed survival time and the scaled Schoenfeld residuals

Test that the correlation is zero vs. non-zero A “good” model has correlation of zero

Testing proportionality in R

> reg6.z <- cox.zph(reg6)> > reg6.z rho chisq pfactor(trtgrp)2 0.00417 0.00917 0.9237factor(trtgrp)3 0.00456 0.01127 0.9154liverny1 -0.06709 2.44565 0.1179numbersites 0.01869 0.17896 0.6723pain0c -0.08400 3.69796 0.0545pskar2c -0.02822 0.41993 0.5170proml -0.05994 2.00080 0.1572probs -0.04290 1.00064 0.3172highgrade 0.01497 0.11942 0.7297logpsa0 -0.02583 0.29315 0.5882logalkp0c 0.00210 0.00210 0.9635hemecenter 0.06715 2.54094 0.1109GLOBAL NA 16.18273 0.1830

Stata Commands

insheet using "C:\....\nomogramdata.txt"

stset survtime, fa(died)

sts graph, cens(m)sts graph, by(trt) cens(m)

sts test trt xi: stcox i.trt

* diagnosticsstphplot, by(trt)stcoxkm, by(trt)stcoxkm, by(trt) separate

xi: stcox i.trt, schoenfeld(trtsch*) estat phtest

tab pain0cstphplot if pain0c<2, by(pain0c)stcoxkm if pain0c<2, by(pain0c) separatestcox pain0c if pain0c<2, schoenfeld(painsch*) estat phtest

Stata: Cox Regression. xi: stcox i.trti.trtgrp _Itrtgrp_1-3 (naturally coded; _Itrtgrp_1 omitted)

failure _d: died analysis time _t: survtime

Iteration 3: log likelihood = -4881.615Refining estimates:Iteration 0: log likelihood = -4881.615

Cox regression -- Breslow method for ties

No. of subjects = 1006 Number of obs = 1006No. of failures = 800Time at risk = 18886.96674 LR chi2(2) = 8.09Log likelihood = -4881.615 Prob > chi2 = 0.0175

------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- _Itrtgrp_2 | 1.111033 .0980152 1.19 0.233 .9346177 1.320748 _Itrtgrp_3 | 1.277127 .1101523 2.84 0.005 1.078495 1.512343------------------------------------------------------------------------------

-20

24

6-l

n[-ln

(Sur

viva

l Pro

bab

ility

)]

-2 0 2 4ln(analysis time)

trtgrp = 1 trtgrp = 2trtgrp = 3

0.0

00

.50

1.0

00

.00

0.5

01

.00

0 20 40 60 80

0 20 40 60 80

1 2

3

Observed: trtgrp = 1 Observed: trtgrp = 2Observed: trtgrp = 3 Predicted: trtgrp = 1

Predicted: trtgrp = 2 Predicted: trtgrp = 3

Su

rviv

al P

rob

abili

ty

analysis time

Graphs by trtgrp

* diagnosticsstphplot, by(trt)stcoxkm, by(trt)stcoxkm, by(trt) separate

Test of Proportionality

Test of proportional-hazards assumption

Time: Time ---------------------------------------------------------------- | chi2 df Prob>chi2 ------------+--------------------------------------------------- global test | 0.30 2 0.8616 ----------------------------------------------------------------

xi: stcox i.trt, schoenfeld(trtsch*) estat phtest