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STATISTICS 542STATISTICS 542Intro to Clinical TrialsIntro to Clinical Trials

SURVIVAL ANALYSISSURVIVAL ANALYSIS

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Survival Analysis Survival Analysis TerminologyTerminology

• Concerned about time to some event

• Event is often death

• Event may also be, for example

1. Cause specific death2. Non-fatal event or death,

whichever comes firstdeath or hospitalizationdeath or MI

death or tumor recurrence

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0 1 2 3 4 50

0.25

0.5

0.75

1Group A Group B

YEARS•At 5 years, survival rates the same•Survival experience in Group A appears more favorable, considering 1 year, 2 year, 3 year and 4 year rates together

Survival Rates at Yearly Intervals

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Beta-Blocker Heart Attack TrialBeta-Blocker Heart Attack Trial

LIFE-TABLE CUMULATIVE MORTALITY CURVELIFE-TABLE CUMULATIVE MORTALITY CURVE

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Survival AnalysisSurvival AnalysisDiscuss

1. Estimation of survival curves

2. Comparison of survival curves

I. Estimation

• Simple Case– All patients entered at the same time and followed for

the same length of time– Survival curve is estimated at various time points by

(number of deaths)/(number of patients)– As intervals become smaller and number of patients

larger, a "smooth" survival curve may be plotted

• Typical Clinical Trial Setting

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• Each patient has T years of follow-up• Time for follow-up taking place may be different for each patient

T years

T years

T years

T years

T0 2T

4

3

2

1

Time Since Start of Trial (T years)

Su

bje

ct

Staggered Entry

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•Failure time is time from entry until the time of the event•Censoring means vital status of patient is not known beyond that point

o

*

*

•

AdministrativeCensoring

Failure

CensoringLoss to Follow-up

Time Since Start of Trial (T years)

T 2T0

4

3

2

1

Subject

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o

*

•

*

T0

4

3

2

1

Subject

AdministrativeCensoring

CensoringLoss to Follow-up

Failure

Follow-up Time (T years)

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Clinical Trial with Common Termination DateClinical Trial with Common Termination Date

*

*

*

1

2

3

4

5

6

7

8

9

10

11

0 T 2TFollow-up Time (T years) Trial

Terminated

Subject

*

••• •

•••••

o

o

•

o o

o

o o

•

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Years of Cohort

Follow-Up Patients I IITotalEntered 100 100 200

1Died 20 25 45

Entered 80 75 1552

Died 20

Survived 60

Reduced Sample Estimate (1)

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– Suppose we estimate the 1 year survival rate

a. P(1 yr) = 155/200 = .775

b. P(1 yr, cohort I) = 80/100 = .80

c. P(1 yr, cohort II) = 75/100 = .75

– Now estimate 2 year survival

Reduced sample estimate = 60/100 = 0.60

Estimate is based on cohort I only

Loss of information

Reduced Sample Estimate (2)

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Ref: Berkson & Gage (1950) Proc of Mayo ClinicCutler & Ederer (1958) JCDElveback (1958) JASAKaplan & Meier (1958) JASA

- Note that we can express P(2 yr survival asP(2 yrs) = P(2 yrs survival|survived 1st yr)

P(1st yr survival) = (60/80) (155/200) = (0.75) (0.775) = 0.58

• This estimate used all the available data

Actuarial EstimateActuarial Estimate (1)

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• In general, divide the follow-up time into a series of intervals

Actuarial EstimateActuarial Estimate (2)

• Let pi = prob of surviving Ii given patient alive at beginning of Ii (i.e. survived through Ii -1)

• Then prob of surviving through tk, P(tk)

iP PP P P P(t tP(Sk

1-ik321kk

...))

I1 I2 I3 I4 I5

t0 t1 t2 t3 t4 t5

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- Define the following

ni= number of subjects alive at beginning of Ii (i.e. at ti-1)di= number of deaths during interval Ii

li = number of losses during interval Ii (either administrative or lost to follow-up)

- We know only that di deaths and losses occurred inInterval Ii

Actuarial EstimateActuarial Estimate (3)

Ii

ti-1 ti

i

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a. All deaths precede all losses

b. All losses precede all deaths

c. Deaths and losses uniform, (1/2 deaths before 1/2 losses)

Actuarial Estimate/Cutler-Ederer

- Problem is that P(t) is a function of the interval choice.

- For some applications, we have no choice, but if weknow the exact date of deaths and losses, theKaplan‑Meier method is preferred.

ii

iii

n

dn

iP

2/

2/P * i

ii

iii

n

dn

i

ii

n

dn

iP

Estimation of PEstimation of Pii

542-08-#16

Actuarial Lifetime Method (1)Actuarial Lifetime Method (1)

• Used when exact times of death are not known

• Vital status is known at the end of an interval period (e.g. 6 months or 1 year)

• Assume losses uniform over the interval

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Lifetable

At Number Number Adjusted Prop Prop. Surv. Up toInterval Risk Died Lost No. At Risk Surviving End of Interval

(ni) (di) (li)

0-1 50 9 0 50 41/50-0.82 0.82

1-2 41 6 1 41-1/2=40.5 34.5/40.5=0.852 0.852 x 0.82=0.699

2-3 34 2 4 34-4/2=32 30/32=0.937 0.937 x 0.699=0.655

3-4 28 1 5 28-5/2=25.5 24.5/25.5=0.961 0.961 x 0.655=0.629

4-5 22 2 3 22-3/2=20.5 18.5/20.5=0.902 0.902 x 0.629=0.567

2i

iinn

ip

ipII

Actuarial Lifetime Method (2)Actuarial Lifetime Method (2)

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Actuarial Survival CurveActuarial Survival Curve100

80

60

40

20

0

X ___

X___X___

X___X___

X___

1 2 3 4 5

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Kaplan-Meier Estimate (1)Kaplan-Meier Estimate (1)((JASAJASA, 1958), 1958)

• Assumptions

1. "Exact" time of event is known

Failure = uncensored event

Loss = censored event

2. For a "tie", failure always before loss

3. Divide follow-up time into intervals such that

a. Each event defines left side of an interval

b. No interval has both deaths & losses

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Kaplan-Meier Estimate (2)Kaplan-Meier Estimate (2)((JASAJASA, 1958), 1958)

• Then

ni = # at risk just prior to death at ti

• Note if interval contains only losses, Pi = 1.0

• Because of this, we may combine intervals with only losses with the previous interval containing

only deaths, for convenience

X———o—o—o——

i

ii

i n

dnP

542-08-#21

Estimate of S(t) or P(t)Estimate of S(t) or P(t)

Suppose that for N patients, there are K distinct failure (death) times. The Kaplan-Meier estimate of survival curves becomes P(t)=P (Survival t)

K-M or Product Limit Estimate

ti t i = 1,2,…,k

where ni = ni-1 - li-1 - di-1

li-1 = # censored events since death at ti-1

di-1 = # deaths at ti-1

i

ii

n

dni

tP)(

)(ˆ

542-08-#22

Estimate of S(t) or P(t)Estimate of S(t) or P(t)

• Variance of P(t)

Greenwood’s Formula

tt

dnn

d

itPtPV

i

iii

i

)( )](ˆ[)](ˆ[ 2

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Example (see Table 14-2 in FFD)

Suppose we follow 20 patients and observe the event time, either failure (death) or censored (+), as

[0.5, 0.6+), [1.5, 1.5, 2.0+), [3.0, 3.5+, 4.0+), [4.8],

[6.2, 8.5+, 9.0+), [10.5, 12.0+ (7 pts)]

There are 6 distinct failure or death times

0.5, 1.5, 3.0, 4.8, 6.2, 10.5

KM Estimate (1)

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1. failure at t1 = 0.5 [.5, 1.5)n1 = 20d1 = 1l1 = 1 (i.e. 0.6+)

If t [.5, 1.5), p(t) = p1 = 0.95

V [ P(t1) ] = [.95]2 {1/20(19)} = 0.0024

9520

120

n

dnp

1

1 1

1.

KM Estimate (2)KM Estimate (2)

^

^

542-08-#25

Data [0.5, 0.6+), [1.5, 1.5, 2.0+), 3.0 etc.

2. failure at t2 = 1.5 n2 = n1 - d1 - 1

[1.5, 3.0) = 20 - 1 - 1= 18

d2 = 22 = 1 (i.e. 2.0+)

If t [1.5, 3.0), then P(t) = (0.95)(0.89) = 0.84

V [P(t2)] = [0.84]2 { 1/20(19) + 2/18(18-2) } = 0.0068

89.18

218

2

22

2

n

dnp

KM Estimate (4)KM Estimate (4)

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Life Table 14.2Kaplan-Meier Life Table for 20 Subjects

Followed for One Year

Interval Interval Time

Number of death nj dj j

[.5,1.5) 1 .5 20 1 1 0.95 0.95 0.0024

[1.5,3.0) 2 1.5 18 2 1 0.89 0.84 0.0068

[3.0,4.8) 3 3.0 15 1 2 0.93 0.79 0.0089

[4.8,6.2) 4 4.8 12 1 0 0.92 0.72 0.0114

[6.2,10.5) 5 6.2 11 1 2 0.91 0.66 0.0135

[10.5, ) 6 10.5 8 1 7* 0.88 0.58 0.0164

jp̂ )(p̂

jt )](p̂V[

jt

nj : number of subjects alive at the beginning of the jth intervaldj : number of subjects who died during the jth intervalj : number of subjects who were lost or censored during the jth interval : estimate for pj, the probability of surviving the jth interval

given that the subject has survived the previous intervals : estimated survival curve : variance of

* Censored due to termination of study

jp̂

(t)pj

ˆ

(t)]pV[j

ˆ )(ˆ tP

542-08-#27

0

0.5

0.6

0.7

0.8

0.9

1.0

2 4 6 8 10 12

*

* *

*

*

*

*

Survival Time t (Months)

Est

imat

ed S

urvi

val C

ure

[P(t

)]Survival Curve

Kaplan-Meier Estimate

^

o

oo o

o o

oooo

ooo

542-08-#28

Comparison of Comparison of Two Survival CurvesTwo Survival Curves

• Assume that we now have a treatment group and a control group and we wish to make a comparison between their survival experience

• 20 patients in each group

(all patients censored at 12 months)

Control 0.5, 0.6+, 1.5, 1.5, 2.0+, 3.0, 3.5+, 4.0+,

4.8, 6.2, 8.5+, 9.0+, 10.5, 12+'s

Trt 1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12+'S

542-08-#29

1. t1 = 1.0 n1 = 20 p1 = 20 - 1 = 0.95

d1 = 1 20

l1 = 3

p(t)= .95

2. t2 = 4.5 n2 = 20 - 1 - 3 p2 = 16 - 1 =0 .94

= 16 16

d2 = 1

0024.0)]t(P̂V[1

89.0)94.0)(95.0()t(P̂ 0530.)](ˆ 2

tPV[

Kaplan-Meier Estimate for Kaplan-Meier Estimate for TreatmentTreatment

^

542-08-#30

Kaplan-Meier EstimateKaplan-Meier Estimate

0

0.5

0.6

0.7

0.8

0.9

1.0

2 4 6 8 10 12

*

* *

*

*

*

Survival Time t (Months)

Est

imat

ed S

urvi

val C

ure

[P(t

)]^

TRT

CONTROL

o

* *

o o

ooooo

oo o

542-08-#31

Comparison of Comparison of Two Survival CurvesTwo Survival Curves

• Comparison of Point Estimates

– Suppose at some time t* we want to compare PC(t*) for the control and PT(t*) for treatment

– The statistic

has approximately, a normal distribution under H0

– Example:

2/1]*)(ˆ*)(ˆ[

*)(ˆ*)(ˆ

tPVtPV

tPtPZ

CT

CT

)(ˆ)(ˆ 6P vs6P CompareCT

3211290

170

0114000530

7208921

..

.

]..[

../

Z

542-08-#32

• Comparison of Overall Survival Curve

H0: Pc(t) = PT(t)

A. Mantel-Haenszel Test

Ref: Mantel & Haenszel (1959) J Natl Cancer InstMantel (1966) Cancer Chemotherapy Reports

- Mantel and Haenszel (1959) showed that a series of 2 x 2

tables could be combined into a summary statistic(Note also: Cochran (1954) Biometrics)

- Mantel (1966) applied this procedure to the comparison of

two survival curves

- Basic idea is to form a 2 x 2 table at each distinct deathtime, determining the number in each group who were

atrisk and number who died

542-08-#33

Suppose we have K distinct times for a death occurring

ti i = 1,2, .., K. For each death time,

Died At Risk at ti Alive (prior to ti)

Treatment ai bi ai + bi

Control ci di ci + di

ai + ci bi + di Ni

• Consider ai, the observed number ofdeaths in the TRT group, under H0

Comparison of Comparison of Two Survival Curves (1)Two Survival Curves (1)

542-08-#34

E(ai) = (ai + bi)(ai + ci)/Ni

Mantel-Haenszel Statistic

1)(NN

)d(c)b(a)db)(c(a )V(a

i

2

i

iiiiiiii

i

21iii χ~)V(a Σ/)]}a(E[a{MH 2

N(0,1) ~ MHZ

Comparison of Comparison of Two Survival Curves(2)Two Survival Curves(2)

542-08-#35

Table 14.3Table 14.3Comparison of Survival Data for a Control Group and an Comparison of Survival Data for a Control Group and an

Intervention Group Using the Mantel-Haenszel ProcedureIntervention Group Using the Mantel-Haenszel ProcedureRank Event Intervention Control Total

Times

j tj aj + bj aj j cj + dj cj j aj + cj bj + dj

1 0.5 200 0 201 1 1 39

2 1.0 201 0 180 0 1 37

3 1.5 190 2 182 1 2 35

4 3.0 170 1 151 2 1 31

5 4.5 161 0 120 0 1 27

6 4.8 150 1 121 0 1 26

7 6.2 140 1 111 2 1 24

8 10.5 130 1 8 1 1 20

aj + bj = number of subjects at risk in the intervention group prior to the death at time t j

cj + cj = number of subjects at risk in the control group prior to the death at time t j aj = number of subjects in the intervention group who died at time t j

cj = number of subjects in the control group who died at time t j

j = number of subjects who were lost or censored between time t j and time tj+1

aj + cj = number of subjects in both groups who died at time tj

bj + dj = number of subjects in both groups who are at risk minus the number who died at time t j

542-08-#36

• Operationally

1. Rank event times for both groups combined2. For each failure, form the 2 x 2 table

a. Number at risk (ai + bi, ci + di)b. Number of deaths (ai, ci)c. Losses (lTi, lCi)

• Example (See table 14-3 FFD) - Use previous data set

Trt: 1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12.0+'s

Control: 0.5, 0.6+, 1.5, 1.5, 2.0+, 3.0, 3.5+, 4.0+, 4.8, 6.2, 8.5+, 9.0+, 10.5, 12.0+'s

Mantel-Haenszel TestMantel-Haenszel Test

542-08-#37

1. Ranked Failure Times - Both groups combined

0.5, 1.0, 1.5, 3.0, 4.5, 4.8, 6.2, 10.5 C T C C T C C C

8 distinct times for death (k = 8)

2. At t1 = 0.5 (k = 1) [.5, .6+, 1.0)

T: a1 + b1 = 20 a1 = 0 lT1 = 0 c1 + d1 = 20 c1 = 1 lC1 = 1 1 loss @ .6+

D A R

T 0 20 20

C 1 19 20

1 39 40

E(a1)= 1•20/40 = 0.5

V(a1) = 1•39 • 20 • 20 402 •39

542-08-#38

3. At t2 = 1.0 (k = 2) [1.0, 1.5)

T: a2 + b2 = (a1 + b1) - a1 - lT1 a2 = 1.0= 20 - 0 - 0

= 20 lT2 = 0

C. c2 + d2 = (c1 + d1) - c1 - lC1 c2 = 0 = 20 - 1 - 1

= 18 lC2 = 0

so

D A R

T 1 19 20

C 0 18 18

1 37 38

E(a2)= 1•20 38V(a2) = 1•37 • 20 • 18

382 •37

542-08-#39

Eight 2x2 Tables Corresponding to the Event TimesEight 2x2 Tables Corresponding to the Event TimesUsed in the Mantel-Haenszel Statistic in Survival Used in the Mantel-Haenszel Statistic in Survival

Comparison of Treatment (T) and Control (C) GroupsComparison of Treatment (T) and Control (C) Groups

1. (0.5 mo.)* D† A‡ R§ 5. (4.5 mo.)* D A RT 0 20 20 T 1 15 16C 1 19 20 C 0 12 12

1 39 40 1 27 28

2. (1.0 mo) D A R 6. (4.8 mo.) D A RT 1 19 20 T 0 15 15C 0 18 18 C 1 11 12

1 37 38 1 26 27

3. (1.5 mo.) D A R 7. (6.2 mo.) D A RT 0 19 19 T 0 14 14C 2 16 18 C 1 10 11

2 35 37 1 24 25

4. (3.0 mo.) D A R 8. (10.5 mo.) D A RT 0 17 17 T 0 13 13C 1 14 15 C 1 7 8

1 31 32 1 20 21

* Number in parentheses indicates time, tj, of a death in either group† Number of subjects who died at time tj

‡ Number of subjects who are alive between time tj and time tj+1

§ Number of subjects who were at risk before the death at time tj R=D+A)

542-08-#40

Compute MH StatisticsCompute MH Statistics

Recall K = 1 K = 2 K = 3t1 = 0.5 t2 = 1.0 t3 = 1.5

D A0 20 201 19 201 39 40

D A1 19 200 18 181 37 38

D A0 19 192 16 182 35 37

a. ai = 2 (only two treatment deaths)

b. E(ai ) = 20(1)/40 + 20(1)/38 + 19(2)/37 + . . .= 4.89

c. V(ai) =

= 2.22d. MH = (2 - 4.89)2/2.22 = 3.76 or ZMH =

K

i 1

K

i 1

K

i 1

...)(

))()((

)(

))()((

3738

1820371

3940

202039122

941.MH

542-08-#41

B. Gehan Test (Wilcoxon)Ref: Gehan, Biometrika (1965)

Mantel, Biometrics (1966)Gehan (1965) first proposed a modified Wilcoxon rankstatistic for survival data with censoring. Mantel (1967) showed asimpler computational version of Gehan’s proposed test.

1. Combine all observations XT’s and XC’s into a single sample Y1, Y2, . . ., YNC + NT

2. Define Uij wherei = 1, NC + NT j = 1, NC + NT

-1 Yi < Yj and death at Yi Uij = 1 Yi > Yj and death at Yj

0 elsewhere

3. Define Ui

i = 1, … , NC + NT ji

ij

NTNC

1ji

U U

542-08-#42

Note:

Ui = {number of observed times definitely less than i}{number of observed times definitely greater}

4. Define W = Ui (controls)

5. V[W] = NCNT

Variance due to Mantel

6.

• Example (Table 14-5 FFD)Using previous data set, rank all observations

Gehan TestGehan Test

NTNC

1i

N(NN(N

U

TCTC

2

i

)) 1

~V(W)

WZ

or

(0,1) N ~V(W)

WZ

2

1

2

2

G

G

542-08-#43

The Gehan Statistics, Gi involves the scores Ui and is defined as

G = W2/V(W)

where W = Ui (Uis in control group only)

and

iC NN

ii

iCiC

iC UNNNN

NNWV

1

2

1)(

))(()(

542-08-#44

Example of Gehan Statistics Scores UExample of Gehan Statistics Scores Uii for for

Intervention and Control (C) GroupsIntervention and Control (C) GroupsObservation Ranked Definitely Definitely = Ui

i Observed Time Group Less More

1 0.5 C 0 39 -392 (0.6)* C 1 0 13 1.0 I 1 37 -364 1.5 C 2 35 -335 1.5 C 2 35 -336 (1.6) I 4 0 47 (2.0) C 4 0 48 (2.4) I 4 0 49 3.0 C 4 31 -27

10 (3.5) C 5 0 511 (4.0) C 5 0 512 (4.2) I 5 0 513 4.5 I 5 27 -2214 4.8 C 6 26 -2015 (5.8) I 7 0 716 6.2 C 7 24 -1717 (7.0) I 8 0 818 (8.5) C 8 0 819 (9.0) C 8 0 820 10.5 C 8 20 -1221 (11.0) I 9 0 9

22-40 (12.0) 12I, 7C 9 0 9

*Censored observations

542-08-#45

Thus W = (-39) + (1) + (-36) + (-33) + (4) + . . . .

= -87

and V[W] = (20)(20) {(-39)2 +12 + (-36)2 + . . . } (40)(39)

= 2314.35

so

• Note MH and Gehan not equal

811352314

87.

.

GZ

Gehan TestGehan Test

542-08-#46

Cox Proportional Hazards ModelCox Proportional Hazards ModelRef: Cox (1972) Journal of the Royal Statistical Association

• Recall simple exponential

S(t) = e-t

• More complicated

If (s) = , get simple model

• Adjust for covariates

• Cox PHM

(t,x) =0(t) ex

})(exp{)( dsstS t 0

}),(exp{),( dsxsxtS t 0

542-08-#47

So

S(t1,X) =

=

=

• Estimate regression coefficients (non-linear estimation) , SE()

• Examplex1 = 1 Trt

2 Controlx2 = Covariate 1

indicator of treatment effect, adjusted for x2, x3 , . . .

• If no covariates, except for treatment group (x1),PHM = logrank

}exp ds(s)e{- xB'

0t

0

xB'00 e(s)ds ]e t[

xB'e

0(t)][S

,ˆ1

Cox Proportional Hazards ModelCox Proportional Hazards Model

542-08-#48

Homework ProblemHomework Problem

PatientNumber

Treatmenta Length ofObservationb Statusc

1 D 6 D2 P 16 A3 P 2 D4 D 1 D5 D 1 D6 P 4 A7 D 16 A8 D 3 A9 P 16 A

10 D 8 D11 P 16 A12 P 2 D13 D 4 D14 P 16 A15 P 5 A16 D 1 D17 D 10 A18 P 2 A19 P 2 A20 D 10 D21 P 16 A22 D 1 A23 P 16 A24 D 15 A

1. Kaplan-Meier2. Gehan-Wilcoxon3. Mantel-Haenszel

a D = drug; P = placebob In weeksc A = alive; D = dead

Source: P.B. Gregory (1974)

542-08-#49

Survival Analysis SummarySurvival Analysis Summary• Time to event methodology very useful in

multiple settings• Can estimate time to event probabilities or

survival curves• Methods can compare survival curves

– Can stratify for subgroups– Can adjust for baseline covariates using

regression model

• Need to plan for this in sample size estimation & overall design