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Introduction to Biostatistics II

- Survival analysis

Survival analysis

- The outcome is survival.
- In general this is, the outcome is time to an

event - e.g., response
- failure
- death
- pregnancy
- infection

Survival curves for three population groups

USA life table 1979-1981

Survival curve for the US population, 1979-1981

Hemophiliac data example

Estimates of the survival curve

- Consider the probability that an individual

younger than 40 years of age in the previous data

set will die at time t6 months after initiation

of observation. - This individual must have survived up to the

six-month point and then expired a short time

after that, at time point t?, where ? symbolizes

a very small unit of time. How will the

probability of a death at six months be

calculated? - Recall the definition of the conditional

probability of an event B given an event A - which results in the multiplicative law of

probability

Estimates of the survival curve (contd)

- Considering two events ASurvive up to time t

and B Failed at time t? (i.e., shortly after

t). - The event S(t?)Survived up to time t? is
- i.e., the probability of surviving up to time t?

is equal to the probability of surviving up to t

times the probability of failing at t? given

survival up to t.

The life table method

- The life-table method of estimation of the

survival curve works as follows - Splits the time scale into J time intervals of

the type tj-1-tj for j1,?,J - The number of people dying in each interval is dj
- The number of people alive at the beginning of

the interval (number at risk) is rj

Derivation of the life table method

- To derive the life table estimate of the survival

distribution we need to estimate the following

quantities - Conditional probability of dying at interval j

given survival up to j - P(BjAj)?qt dj/rj
- Thus, probability of survival up to j
- The life-table estimate of the survival

distribution is constructed as follows

Life table of under-40 hemophiliac data

This subfile contains 12 observations

Life Table Survival Variable survival

Number Number Number Number

Cumul Intrvl Entrng Wdrawn Exposd of

Propn Propn Propn Proba- Start this

During to Termnl Termi- Sur- Surv

bility Hazard Time Intrvl Intrvl Risk

Events nating viving at End Densty

Rate ------ ------ ------ ------ ------

------ ------ ------ ------ ------ .0

12.0 .0 12.0 2.0 .1667 .8333

.8333 .0333 .0364 5.0 10.0 .0

10.0 3.0 .3000 .7000 .5833 .0500

.0706 10.0 7.0 .0 7.0 1.0

.1429 .8571 .5000 .0167 .0308 15.0

6.0 .0 6.0 3.0 .5000 .5000

.2500 .0500 .1333 20.0 3.0 .0

3.0 .0 .0000 1.0000 .2500 .0000

.0000 25.0 3.0 .0 3.0 1.0

.3333 .6667 .1667 .0167 .0800 30.0

2.0 .0 2.0 2.0 1.0000 .0000

.0000 These calculations

for the last interval are meaningless. The

median survival time for these data is 15.00

Life table estimate of the survival distribution

The Kaplan-Meier method

- The K-M method differs from the life-table method

in that it separates the time spectrum according

to failure times (instead of fixed-width

intervals). - The first interval is (0 2) (2 is the time of

the first failure) when 1/12 individuals failed

(died) so 11/12 survived. The survival estimate

at t2 is, S(2)11/120.9167. - The second interval is (2 3) (the second

failure happens at t3) when 1/11 individuals

fails. The survival estimate at t3 is

S(3)(11/12)(10/11)(10/12)0.8333, since to

survive up to t3 you must survive up to t2 and

(given that you survived up to t2) then survive

beyond t3. - And so on

The product-limit Method

- Nothing happens except at the time of failure.

Survival Analysis for survival Time

Status Cumulative Standard

Cumulative Number

Survival Error Events

Remaining 2 Selected .9167

.0798 1 11 3

Selected .8333 .1076

2 10 6 Selected

3 9

6 Selected .6667 .1361

4 8 7 Selected

.5833 .1423 5

7 10 Selected .5000

.1443 6 6 15

Selected

7 5 15 Selected

.3333 .1361 8 4

16 Selected .2500 .1250

9 3 27 Selected

.1667 .1076 10

2 30 Selected .0833

.0798 11 1 32

Selected .0000 .0000

12 0 Number of Cases 12

Censored 0 ( .00) Events 12

Survival Time Standard Error 95

Confidence Interval Mean 14

3 ( 8, 20 ) Median

10 5 ( 1,

19 ) Percentiles

25.00 50.00 75.00

Value 16.00 10.00

6.00 Standard Error 9.00 4.62

2.45

Kaplan-Meier estimate of the survival distribution

- This plot is the Kaplan-Meier estimate of the

hemophiliac-patient survival distribution

corresponding to the previous output.

Censoring

- When failure has not been observed, then the only

information from the data is that the failure

time is no less than the time of the last

available observation (e.g., clinical visit).

This is easily incorporated into the estimation

procedure. - For example, consider the following data where

subjects 2 and 6 completed observation without

failure at months 3 and 10 (censor0 means

censoring)

Life table method in the presence of censoring

- To carry out the life-table estimate of the

survival distribution, when data include censored

observations, we include the number of censored

observations in interval j. - cj is the number of censored observations in

interval j - Since we do not know when exactly the censoring

occurred we have the following options for

calculating the number of individuals surviving

up to j - at the beginning of the interval (so the number

at risk at the beginning of interval j is

r'jrj-cj) - at the end of the interval (so the number at risk

is r'jrj) - at the middle of the interval (assuming that

censoring happens uniformly through the interval,

so r'jrj-cj/2). - The latter case is called the actuarial estimator

of survival.

Derivation of the life-table method

- To calculate the life table method for the period

between 5 and 10 (interval j1) months in our

example we proceed as follows - There is one failure and one censored observation

in the first interval (i.e., between 0 and 5

months). Assuming that the censoring happened at

the midpoint of the interval (actuarial survival)

the (effective) number at risk is

r'1(r1-c1/2)11.5. - Thus, ?q11/11.50.0870, so
- S(1)?q10.9130
- For the second interval (j2, time between 5 and

10 months) we have that three failures occurred

with no censoring thus after removing the first

failure and censored observation r'2r210 and

?q23/100.3000, so

Analysis via the life-table method

- Life Table
- Survival Variable survival
- Number Number Number Number

Cumul - Intrvl Entrng Wdrawn Exposd of Propn

Propn Propn Proba- - Start this During to Termnl Termi-

Sur- Surv bility Hazard - Time Intrvl Intrvl Risk Events nating

viving at End Densty Rate - ------ ------ ------ ------ ------ ------

------ ------ ------ ------ - .0 12.0 1.0 11.5 1.0 .0870

.9130 .9130 .0174 .0182 - 5.0 10.0 .0 10.0 3.0 .3000

.7000 .6391 .0548 .0706 - 10.0 7.0 1.0 6.5 .0 .0000

1.0000 .6391 .0000 .0000 - 15.0 6.0 .0 6.0 3.0 .5000

.5000 .3196 .0639 .1333 - 20.0 3.0 .0 3.0 .0 .0000

1.0000 .3196 .0000 .0000 - 25.0 3.0 .0 3.0 1.0 .3333

.6667 .2130 .0213 .0800 - 30.0 2.0 .0 2.0 2.0 1.0000

.0000 .0000 - These calculations for the last interval

are meaningless. - The median survival time for these data is

17.18

Life table estimate of the survival distribution

in the presence of censoring

K-M estimate in the presence of censoring

- Consider how censoring is handled in the K-M

procedure - In the first interval (time 0-2) one out of 12

individuals fails at 2 months so that the

estimate of survival at t2 is - No one fails at t3 months (second interval).
- At t6 months two total subjects have failed out

of the remaining ten (since one subject was

censored at 3 months and is no longer part of the

at-risk sample at six months), so (1q6 0.2000)

is the probability of failure at t6 months. The

estimate of the survival distribution is - S(6) S(2)(1-1q6) 0.9167(1-0.2000)0.7333
- So, censored observations are present up to the

interval where they are censored and disappear

after that.

Kaplan-meier estimate with censored observations

- Survival Analysis for survival
- Time Status Cumulative Standard

Cumulative Number - Survival Error

Events Remaining - 2 1.00 .9167 .0798

1 11 - 3 .00

1 10 - 6 1.00

2 9 - 6 1.00 .7333 .1324

3 8 - 7 1.00 .6417 .1441

4 7 - 10 .00

4 6 - 15 1.00

5 5 - 15 1.00 .4278 .1565

6 4 - 16 1.00 .3208 .1495

7 3 - 27 1.00 .2139 .1325

8 2 - 30 1.00 .1069 .1005

9 1 - 32 1.00 .0000 .0000

10 0

The K-M plot with censoring

- The K-M estimate of the survival distribution in

the presence of censoring is as shown in the

figure.

Testing

- Consider the survival curves of hemophiliacs

contracting AIDS above 40 years of age and before

40 years of age.

Survival distribution of gt40 year-olds

- Survival Analysis for survival
- Factor age gt40
- Time Status Cumulative Standard

Cumulative Number - Survival Error

Events Remaining - 1 1.00

1 8 - 1 1.00

2 7 - 1 1.00

3 6 - 1 1.00 .5556 .1656

4 5 - 2 1.00 .4444 .1656

5 4 - 3 1.00

6 3 - 3 1.00 .2222 .1386

7 2 - 9 1.00 .1111 .1048

8 1 - 22 1.00 .0000 .0000

9 0 - Number of Cases 9 Censored 0 (

.00) Events 9

Comparing two survival distributions

The log-rank test

- The log-rank test evaluates the null hypothesis
- H0 Slt40(t) Sgt40(t) versus the alternative
- H0 Slt40(t) ? Sgt40(t)
- the test is based on the statistic
- where, for each failure time j and group i1,2,
- , where dj is the number of deaths, Y(t)

is the number at risk (alive) at time t and Y1(t)

and Y2(t) are the total number at risk in group 1

and 2 respectively and - and are the

total numbers of expected and observed deaths.

The log-rank test with SPSS

- The SPSS output for the log-rank test is as

follows - since p0.006lt0.05, there is a statistically

significant difference in survival between the

two groups. - Since the log-rank test is two-sided, we must

check the median survival time to see the

direction of the difference (here it is the

younger lt40 year-old patients).

Test Statistics for Equality of Survival

Distributions for age Statistic

df Significance Log Rank

7.61 1 .0058

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