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### Chapter 12

H0: Pc(t) = PT(t) A. Mantel-Haenszel Test Ref: Mantel & Haenszel (1959) J Natl Cancer Inst Mantel (1966) Cancer Chemotherapy Reports

Survival Analysis

Survival Analysis Terminology

- Concerned about time to some event
- Event is often death
- Event may also be, for example
1. Cause specific death

2. Non-fatal event or death,

whichever comes first

death or hospitalization

death or MI

death or tumor recurrence

Survival Rates at Yearly Intervals

- 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

Beta-Blocker Heart Attack Trial

LIFE-TABLE CUMULATIVE MORTALITY CURVE

Discuss

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

T years

1

T years

2

Subject

T years

3

T years

4

0

T

2T

Time Since Start of Trial (T years)

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

o

Administrative

Censoring

1

Failure

2

*

•

Censoring

Loss to Follow-up

3

*

4

T

0

2T

Time Since Start of Trial (T years)

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

Administrative

Censoring

o

1

Failure

2

*

•

3

Censoring

Loss to Follow-up

4

*

T

0

Follow-up Time (T years)

Clinical Trial with Common Termination Date

Subject

o

1

2

*

•

3

•

4

o

•

5

*

•

•

6

•

•

7

•

*

8

•

9

o

o

•

10

o

*

•

11

o

o

0

T

2T

Trial

Terminated

Follow-up Time (T years)

Years of Cohort

Follow-Up Patients I II Total

Entered 100 100 200

1

Died 20 25 45

Entered 80 75 155

2

Died 20

Survived 60

- 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

- Ref: Berkson & Gage (1950) Proc of Mayo Clinic
- Cutler & Ederer (1958) JCD
- Elveback (1958) JASA
- Kaplan & Meier (1958) JASA
- - Note that we can express P(2 yr survival) as
- P(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

I1 I2 I3 I4 I5

t0 t1 t2 t3 t4 t5

Actuarial Estimate (2)

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

- 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)

Ii

ti-1 ti

Actuarial Estimate (3)

- 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 in

Interval Ii

- a. All deaths precede all losses
- b. All losses precede all deaths
- 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 we
- know the exact date of deaths and losses, the
- Kaplan‑Meier method is preferred.

- 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

Lifetable

At Number Number Adjusted Prop Prop. Surv. Up to

Interval 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

Kaplan-Meier Estimate (1)(JASA, 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

Kaplan-Meier Estimate (2)(JASA, 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——

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

titi = 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

- Variance of P(t)
Greenwood’s Formula

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

1. failure at t1 = 0.5 [.5, 1.5)

n1 = 20

d1 = 1

l1 = 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

^

^

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

2. failure at t2 = 1.5 n2 = n1 - d1 - l1

[1.5, 3.0) = 20 - 1 - 1

= 18

d2 = 2

l2 = 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

Kaplan-Meier Life Table for 20 Subjects

Followed for One Year

Interval Interval Time

Number of death nj djlj

[.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

nj: number of subjects alive at the beginning of the jth interval

dj: number of subjects who died during the jth interval

lj : 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

Kaplan-Meier Estimate

1.0

o

*

0.9

^

*

o

*

0.8

o

o

*

*

Estimated Survival Cure [P(t)]

0.7

o

o

*

o

o

0.6

o

o

*

o

o

o

0.5

0

4

6

8

10

12

2

Survival Time t (Months)

Comparison of Two 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

Trt1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12+'S

Kaplan-Meier Estimate for Treatment

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

^

1.0

o

*

TRT

0.9

*

^

*

o

*

0.8

o

o

*

*

Estimated Survival Cure [P(t)]

0.7

CONTROL

o

o

*

0.6

o

*

o

o

o

o

0.5

0

4

6

8

10

12

2

Survival Time t (Months)

Comparison of Two 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:

- - 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 death
- time, determining the number in each group who were at
- risk and number who died

Comparison of Two Survival Curves (1)

- 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 of
- deaths in the TRT group, under H0

Comparison of Survival Data for a Control Group and an Intervention Group Using the Mantel-Haenszel Procedure

Rank Event Intervention Control Total

Times

j tj aj + bj ajlj cj + dj cjlj aj + cj bj + dj

1 0.5 20 0 0 20 1 1 1 39

2 1.0 20 1 0 18 0 0 1 37

3 1.5 19 0 2 18 2 1 2 35

4 3.0 17 0 1 15 1 2 1 31

5 4.5 16 1 0 12 0 0 1 27

6 4.8 15 0 1 12 1 0 1 26

7 6.2 14 0 1 11 1 2 1 24

8 10.5 13 0 1 8 1 1 20

- aj + bj = number of subjects at risk in the intervention group prior to the death at time tj
- cj + cj = number of subjects at risk in the control group prior to the death at time tj
- aj = number of subjects in the intervention group who died at time tj
- cj = number of subjects in the control group who died at time tj
- lj = number of subjects who were lost or censored between time tj 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 tj

Mantel-Haenszel Test Intervention Group Using the Mantel-Haenszel Procedure

- Operationally
- 1. Rank event times for both groups combined
- 2. 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

1. Intervention Group Using the Mantel-Haenszel ProcedureRanked 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

E(a Intervention Group Using the Mantel-Haenszel Procedure2)= 1•20

38

V(a2) = 1•37 • 20 • 18

382 •37

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

Eight 2x2 Tables Corresponding to the Event Times Intervention Group Using the Mantel-Haenszel ProcedureUsed in the Mantel-Haenszel Statistic in Survival Comparison of Treatment (T) and Control (C) Groups

1. (0.5 mo.)* D† A‡ R§ 5. (4.5 mo.)* D A R

T 0 20 20 T 1 15 16

C 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 R

T 1 19 20 T 0 15 15

C 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 R

T 0 19 19 T 0 14 14

C 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 R

T 0 17 17 T 0 13 13

C 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)

Compute MH Statistics Intervention Group Using the Mantel-Haenszel Procedure

Recall K = 1 K = 2 K = 3

t1 = 0.5 t2 = 1.0 t3 = 1.5

D A

0 20 20

1 19 20

1 39 40

D A

1 19 20

0 18 18

1 37 38

D A

0 19 19

2 16 18

2 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) =

= 2.22

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

or ZMH =

B. Gehan Test (Wilcoxon) Intervention Group Using the Mantel-Haenszel Procedure

Ref: Gehan, Biometrika (1965)

Mantel, Biometrics (1966)

Gehan (1965) first proposed a modified Wilcoxon rank

statistic for survival data with censoring. Mantel (1967) showed a

simpler 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 Uijwhere i = 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

Gehan Test Intervention Group Using the Mantel-Haenszel Procedure

- Note:
- Ui = {number of observed times definitely less than i}
- {number of observed times definitely greater}
- 4. Define W = S Ui (controls)
- 5. V[W] = NCNT
- Variance due to Mantel
- 6.
- Example (Table 14-5 FFD)
- Using previous data set, rank all observations

The Gehan Statistics, G Intervention Group Using the Mantel-Haenszel Procedurei involves

the scores Ui and is defined as

G = W2/V(W)

where W = Ui (Uis in control group only)

and

Example of Gehan Statistics Scores U Intervention Group Using the Mantel-Haenszel Procedurei for Intervention and Control (C) Groups

Observation Ranked Definitely Definitely = Ui

i Observed Time Group Less More

1 0.5 C 0 39 -39

2 (0.6)* C 1 0 1

3 1.0 I 1 37 -36

4 1.5 C 2 35 -33

5 1.5 C 2 35 -33

6 (1.6) I 4 0 4

7 (2.0) C 4 0 4

8 (2.4) I 4 0 4

9 3.0 C 4 31 -27

10 (3.5) C 5 0 5

11 (4.0) C 5 0 5

12 (4.2) I 5 0 5

13 4.5 I 5 27 -22

14 4.8 C 6 26 -20

15 (5.8) I 7 0 7

16 6.2 C 7 24 -17

17 (7.0) I 8 0 8

18 (8.5) C 8 0 8

19 (9.0) C 8 0 8

20 10.5 C 8 20 -12

21 (11.0) I 9 0 9

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

*Censored observations

Gehan Test Intervention Group Using the Mantel-Haenszel Procedure

- 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

Cox Proportional Hazards Model Intervention Group Using the Mantel-Haenszel Procedure

Ref: Cox (1972) Journal of the Royal Statistical Association

- Recall simple exponential
S(t) = e-lt

- More complicated
If l(s) = l, get simple model

- Adjust for covariates
- Cox PHM
l(t,x) =l0(t) ebx

Cox Proportional Hazards Model Intervention Group Using the Mantel-Haenszel Procedure

- So
- S(t1,X) =
- =
- =
- Estimate regression coefficients (non-linear estimation) b, SE(b)
- Example
- x1 = 1 Trt
- 2 Control
- x2 = Covariate 1
- indicator of treatment effect, adjusted for x2, x3 , . . .
- If no covariates, except for treatment group (x1),
- PHM = logrank

Survival Analysis Summary Intervention Group Using the Mantel-Haenszel Procedure

- 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

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