Survival analysis

Survival analysis

Dr HAR ASHISH JINDALJR

Contents

• Survival• Need for survival analysis• Survival analysis• Life table/ Actuarial • Kaplan Meier product limit method• Log rank test• Mantel Hanzel method• Cox proportional hazard model• Take home message

Survival

• It is the probability of remaining alive for a specific length of time.

• point of interest : prognosis of disease e.g.– 5 year survivale.g. 5 year survival for AML is 0.19, indicate 19% of patients with AML will survive for 5 years after diagnosis

Survival

• In simple terms survival (S) is mathematically given by the formula;S = A-D/A

A = number of newly diagnosed patients under observation

D= number of deaths observed in a specified period

e.g For 2 year survival: S= A-D/A= 6-1/6 =5/6 = .83=83%

e.g For 5 year survival: S= A-D/A

Censoring

• Subjects are said to be censored– if they are lost to follow up – drop out of the study, – if the study ends before they die or have an outcome of

interest.

• They are counted as alive or disease-free for the time they were enrolled in the study.

• In simple words, some important information required to make a calculation is not available to us. i.e. censored.

Types of censoring

Three Types of Censoring

Right censoring Left censoring Interval censoring

Right Censoring

• Right censoring is the most common of concern. • It means that we are not certain what happened to

people after some point in time.• This happens when some people cannot be

followed the entire time because they died or were lost to follow-up or withdrew from the study.

• Left censoring is when we are not certain what happened to people before some point in time.

• Commonest example is when people already have the disease of interest when the study starts.

Left Censoring

• Interval censoring is when we know that something happened in an interval (i.e. not before starting time and not after ending time of the study ), but do not know exactly when in the interval it happened.

• For example, we know that the patient was well at time of start of the study and was diagnosed with disease at time of end of the study, so when did the disease actually begin?

• All we know is the interval.

Interval Censoring

FOR 5 YEAR SURVIVAL

B AND E Survived 5 years S=6- 2/6=4/6=0.67=67%

B and E did not survive for full 5

years .S=6-4/6= 2/6= 0.33=33%

Conclusion: since the observations are censored , it is not possible to know how long will subject survive . Hence the need for Special techniques to

account such censored observations

2 possibilities

Need for survival analysis

• Investigators frequently must analyze data before all patients have died; otherwise, it may be many years before they know which treatment is better.

• Survival analysis gives patients credit for how long they have been in the study, even if the outcome has not yet occurred.

• The Kaplan–Meier procedure is the most commonly used method to illustrate survival curves.

• Life table or actuarial methods were developed to show survival curves; although surpassed by Kaplan–Meier curves.

15

What is survival analysis?

• Statistical methods for analyzing longitudinal data on the occurrence of events.

• Events may include death, injury, onset of illness, recovery from illness (binary variables) or transition above or below the clinical threshold of a meaningful continuous variable (e.g. CD4 counts).

• Accommodates data from randomized clinical trial or cohort study design.

Randomized Clinical Trial (RCT)

Target population

Intervention

Control

Disease

Disease-free

Disease

Disease-free

Timeline

Random assignment

Disease-free, at-risk cohort

Target population

Treatment

Control

Cured

Not cured

Cured

Not cured

TIME

Random assignment

Patient population


Timeline

Target population

Treatment

Control

Dead

Alive

Dead

Alive

TIME

Random assignment

Patient population


Timeline

Cohort study (prospective/retrospective)

Target population

Exposed

Unexposed

Disease

Disease-free

Disease

Disease-free

TIME

Disease-free cohort

Timeline

20

Estimate time-to-event for a group of individuals, such as time until second heart-attack for a group of MI patients.

To compare time-to-event between two or more groups, such as treated vs. placebo MI patients in a randomized controlled trial.

To assess the relationship of co-variables to time-to-event, such as: does weight, insulin resistance, or cholesterol influence survival time of MI patients?

Objectives of survival analysis

Censored observations

Kaplan- meier or

Actuarial

Cox Proportional Hazard Model

Scale of measurement of

dependent variable

Censored observations

nominal

numerical

Life table

History of life table

• John Graunt developed a life table in 1662 based on London’s bills of mortality, but he engaged in a great deal of guess work because age at death was unrecorded and because London’s population was growing in an un-quantified manner due to migration.

HISTORY OF THE LIFE TABLE

Edmund Halley (1656 – 1742) - ‘An estimate of the Degree of the Mortality of Mankind drawn from the curious Table of the Births and Funerals at the city of Breslau’.

Life table/ Actuarial methods

Actuary means “someone collection and interpretation of numerical data (especially someone who uses statistics to calculate insurance premiums)”

Known as the Cutler–Ederer method (1958) in the medical literature

Widely used for descriptive and analytical purposes in demography, public health, epidemiology, population geography, biology and many other branches of sci ence.

Describe the extent to which a generation of people dies off with age.

Life table

A special type of analysis which takes into account the life history of a hypothetical group or cohort of people that decreases gradually by death till all members of the group died.

A special measure not only for mortality but also for other vital events like reproduction, chances of survival etc.

Uses & Applications

The probability of surviving any particular year of ageRemaining life expectancy for people at different agesMoreover, can be used to assess:

At the age of 5, to find number of children likely to enter primary school.

At the age of 15, to find number of women entering fertile period.

At age of 18, to find number of persons become eligible for voting.

Computation of net reproduction rates.Helps to project population estimates by age & sex. To estimate the number likely to die after joining service

till retirement, helping in budgeting for payment towards risk or pension.

Uses & Applications

If we want to construct a life table

showing survival & death in a cohort of 150000 babies

STEPS

1. These 1,50,000 babies born at same time were subjected to those mortality influences at various ages that influence population at certain period of time.

2. On the basis of mortality rates operating, we can estimate what number would be alive at first birthday by applying mortality rates during first year on them.

3. By applying mortality rates of second year on numbers of babies surviving at the end of the first year, we estimate number who would survive at the end of second year.

4. Similarly for other ages by applying mortality rates of selected year follow them till all members of cohort die.

5. These number of survivors at various ages form the basic data set out in a life table.

6. From these numbers we can calculate the average life time a person can expect to live after any age.

Steps

Table 1. Life table of a birth cohort

1 2 3 4 5 6 7 8

AgeX

Living at age x

(lx)

Dying b/w x & x+1 (dx)

Mortality rate (qx)

Survival rate (px)

Living b/w x & x+1 (Lx)

Living above

age x (Tx)

Life expectancy

at x (ex0)

0 142759 27124 .19000 .81000 129197 4638611 32.49

1 115635 7472 .06462 .93538 111899 4509414 39.00

2 108163 3144 .02907 .97093 106591 4397525 40.66

3 . . . . . . .

Life Table consist of 8 columns

FIRST COLUMN (x)• It gives the exact years of age starting from age

0,1,2,3…………………99.

SECOND COLUMN (lx)• lx is the of persons who are expected to attain exact age ‘x’ out of

number of births. • Thus the number 1,42,759 in lx column against ‘0’ year indicates the

number that begin their life together and are running first year of their life.

• Similarly figure 1,15,635 against 1 year indicates the number who have completed first year of life and running the second & so on.

THIRD COLUMN (dx)• It gives the number of persons among ‘lx’ who die before reaching

‘x+1’dx=lx-(lx+1)

• In this life table corresponding to ‘x = 0’ dx=142759-115635=27124

• If ‘x = 1’ then dx=115635-108163=7472

FOURTH COLUMN (qx)• It is the mortality rate to which population groups would be exposed, but it

is not the same as the age specific death rates obtained from death registration records.

qx= dx/lx.• If in above life table, x=0 then q0=27124/142759=.19000• Similarly for x=1, q1=7472/115635=.06462

FIFTH COLUMN (Px)• It is the probability that a person of precise age x will survive till his next

b’day• Since a person must either live or die in a particular year of life so

qx+px=1So px=1-qx

• For x=0, p0=1-0.19000=0.81000• Similarly x=1, p1=1-0.06462=0.93538

SIXTH COLUMN (Lx)• It is number of years lived in aggregate by cohort of lx persons

between ages x & x+1Lx=lx-1/2dx

• For age 2, Lx=108163-1/2×3144=106591• Lx is based on assumption that deaths are evenly distributed

throughout the year.• Lx can also be calculated as: Lx = lx+(lx+1)

2

SEVENTH COLUMN (Tx)• It is number of years lived by group from age x until all of them die. • Thus Tx=(Lx)+(Lx+1)+(Lx+2)+……………Ln. OR• Thus for the above table T0=129197+111899+106591+……..+9+5+2=

4638611z

EIGHT COLUMN (ex0)

• It measures average numbers of years a person of a given age x can be expected to live under the prevailing mortality conditions.

• The expectation of life at age x is obtained by• ex

0=Tx/lx• For x=2, e2

0 = 4397525 / 108163 = 40.66• For x=95, e95

0 = 32 / 21 = 1.5

Agex

Living at age x

(lx)

Dying b/w x & x+1 (dx)

Mortality rate (qx)

Survival rate (px)

Living b/w x & x+1 (Lx)

Living > age x

(Tx)

Life expectancy

at x (ex0)

1 2 3 4 5 6 7 8

0 142759 27124 .19000 .81000 129197 4638611 32.49

1 115635 7472 .06462 .93538 111899 4509414 39.00

2 108163 3144 .02907 .97093 106591 4397525 40.66

3 105019 3254 .03098 .96902 103392 4290934 40.86

4 102006 2006 .01967 .98033 101121 4187542 41.05

5 100000 1710 .01710 .98290 99145 4086420 40.86

. . . . . . . .

. . . . . . . .

. . . . . . . .

95 21 9 .40957 .59043 16 32 1.52

96 12 6 .42932 .57068 9 16 1.34

97 6 3 .44964 .55036 5 7 1.17

98 3 1 .47046 .52954 2 2 0.64

99 1 1 .49176 .50823 … … …

Modified life table

1. For survival in different treatment regimens

2. Arrange the the 13 patients on etoposide plus cisplatin(treatment arm =1) according to length of time they had no progression of their disease.

3. Features of the intervals:1. arbitrary

2. should be selected with minimum censored observations

Life table for sample of 13 patient treated with etoposide with cisplatin

Life Table Survival Variable: Progression-Free Survival

ni wi diqi = di/[ni-

(wi/2)] pi = 1–qi si = pipi–1pi-

2…p1

Interval Start Time

No. entering Interval

No. withdrawn

du. Interval

No. exposed to risk

No. of terminal events

Propn terminating

Propn surviving

Cumul Propn Surv at End

0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021

Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–

91.

No. of pts (13) began the study, so n1 is 13

1 patient's disease progressed, referred

to as a terminal event (d1)

2 patients are referred to as withdrawals (w1).

• The actuarial method assumes that patients withdraw randomly throughout the interval; therefore, on the average, they withdraw halfway through the time represented by the interval.

• In a sense, this method gives patients who withdraw credit for being in the study for half of the period.

Assumption:



ni wi di qi = di/[ni-(wi/2)]

pi = 1–qi si = pipi–1pi-2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021


91.

One-half of the number of patients withdrawing is subtracted from the number beginning the interval, so the EXPOSED TO RISK during the period, 13 – (½ × 2), or 12 in first interval.





2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021


91.

The proportion terminating (q1 = d1/[n1-(w1/2]) is 1/12 = 0.0833.

. Life table for sample of 13 patient treated with etoposide with cisplatin




2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021


91.

The proportion surviving (p1 = 1-q1) is 1 – 0.0833 = 0.9167

because we are still in the first period, the cumulative survival is 0.9167





2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021


91.

four patients withdraw w2 = 4

one's disease

progressed, so d2 = 1

At the beginning of the second interval, only 10 patients remain.n2=10





2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021


91.

the proportion terminating (q2 = d2/[n2-(w2/2]) during 2nd interval is 1/[10 – (4/2)] = 1/8, or 0.1250.

the proportion with no progression is 1 – 0.1250, or 0.8750





2…p1

Interval Start Time


No. withdrawn

du. Interval

No. exposed to risk


Propn terminating

Propn surviving


0.0 13.0 2.0 12.0 1.0 0.0833 0.9167 0.9167

3.0 10.0 4.0 8.0 1.0 0.1250 0.8750 0.8021

6.0 5.0 4.0 3.0 0.0 0.0000 1.0000 0.8021

9.0 1.0 1.0 0.5 0.0 0.0000 1.0000 0.8021

Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–9

Rule from probability theory:

P(A&B)=P(A)*P(B) if A and B independent

the cumulative proportion of surv =p1*p2= 0.0.9167 × 0.8750=

0.8021

• Like Cancer treatment, life table of survivorship after any treatment as treatment of cancer by irradiation or drugs or after operation, such as of cancer cervix or breast can be prepared & made use in probabilities of survival at beginning or at any point of time.

• More recently, survival can be enquired after-– Heart operation like bypass, angioplasty, ballooning,

stenting, heart transplantation.– Kidney, lung, liver & other organ transplantation.

Life table

• This computation procedure continues until the table is completed.

• pi = the probability of surviving interval i only; to survive interval i, a patient must have survived all previous intervals as well.

• The probability of survival at one period is treated as though it is independent of the probability of survival at others

• Thus, pi is an example of a conditional probability because the probability of surviving interval i is dependent, or conditional, on surviving until that point.

• This is called survival function.

Life Table

Limitation

• The assumption that all withdrawals during a given interval occur, on average, at the midpoint of the interval.

• This assumption is of less consequence when short time intervals are analyzed; however, considerable bias can occur :

• if the intervals are large, • if many withdrawals occur, &• if withdrawals do not occur midway in the

interval. • The Kaplan–Meier method overcomes this problem.• .

Kaplan Meier product limit method

46

Kaplan-Meier Product limit method

• Similar to actuarial analysis except time since entry in the study is not divided into intervals for analysis.

• Survival is estimated each time a patient has an event.• Withdrawals are ignored• It gives exact survival times in comparison to

actuarial because it does not group survival time into intervals

47

Introduction to Kaplan-Meier

• Non-parametric estimate of the survival function.• Commonly used to describe survivorship of study

population/s.• Commonly used to compare two study populations.• Intuitive graphical presentation.

Beginning of study End of study Time in months

Subject B

Subject A

Subject C

Subject D

Subject E

Survival Data (right-censored)

1. subject E dies at 4 months

X

0 12

100%

Time in months

Corresponding Kaplan-Meier Curve

Probability of surviving to 4 months is 100% = 5/5

Fraction surviving this death = 4/5

Subject E dies at 4 months

4


Subject B

Subject A

Subject C

Subject D

Subject E

Survival Data

2. subject A drops out after 6 months


X

3. subject C dies at 7 monthsX

100%

Time in months


subject C dies at 7 months

Fraction surviving this death = 2/3

74


Subject B

Subject A

Subject C

Subject D

Subject E

Survival Data

2. subject A drops out after 6 months

4. Subjects B and D survive for the whole year-long study period


X

3. subject C dies at 7 monthsX

12

100%

Time in months


Rule from probability theory:

P(A&B)=P(A)*P(B) if A and B independent

In kaplan meier : intervals are defined by failures(2 intervals leading to failures here).

P(surviving intervals 1 and 2)=P(surviving interval 1)*P(surviving interval 2)

Product limit estimate of survival = P(surviving interval 1/at-risk up to failure 1) * P(surviving interval 2/at-risk up to failure 2) = 4/5 * 2/3= .5333

0

The probability of surviving in the entire year, taking into account censoring = (4/5) (2/3) = 53%

Example :kaplan–Meier survival curve in detail for patients on etoposide plus

cisplatinEvent Time(T)

Number at Risk ni

Number of Events di

Mortality qi = di/ni

Survival pi = 1 - qi

Cumulative Survival S = pip(i-1)…p2p1

1.0 13 1 0.076 0.9231 0.9231

2.4 12

2.8 11

3.1 10

3.7 9

4.4 8 1 0.1250 0.8750 0.8077

4.6 7

4.7 6

6.5 5

7.1 4

8.0 3

8.1 2

12.0 1

• One patient's disease progressed at 1 month and another at 4.4 months, and they are listed under the column “Number of Events.”

• Then, each time an event or outcome occurs, the mortality, survival, and cumulative survival are calculated in the same manner as with the life table method.

• In this method first step is to list the times when a death or drop out occurs, as in the column “Event Time”.

Contd…

• If the table is published in an article, it is often formatted in an abbreviated form, such as in Table 5.

Kaplan–Meier survival curve in abbreviated form for patients on etoposide plus cisplatinEvent Time

(T)Number at

Risk ni

Number of Events

di

Mortality qi = di/ni

Survival pi = 1 - qi

Cumulative Survival S = pip(i-1)…p2p1

1.0 13 1 0.076 0.9231 0.9231

4.4 8 1 0.1250 0.8750 0.8077

..

..

..

12.0

Kaplan meir survival curve for patients on etoposide & cisplatin(Source: Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–91.)

2.0 4.0 6.0 8.0 10.0 12.0

57

Limitations of Kaplan-Meier

• Requires nominal predictors only• Doesn’t control for covariates

Cox progressive hazard model solves these problems

Kaplan meir survival curve with 95 % confidence limits for patients on irinotecan & cisplatin(Source: Source: Noda K, Nishiwaki Y, Kawahara M, Negoro S, Sugiura T, Yokoyama A, et al: Irinotecan plus cisplatin compared with etoposide plus cisplatin for extensive small-cell lung cancer. N Engl J Med 2002; 346: 85–91.)

Comparison between 2 survival curve

• Don’t make judgments simply on the basis of the amount of separation between two lines

Comparison between 2 survival curve

• For comparison if no censored observations occur, the Wilcoxon rank sum test introduced, is appropriate for comparing the ranks of survival time.

• If some observations are censored, methods may be used to compare survival curves.– the Logrank statistic – the Mantel–Haenszel chi-square statistic.

Logrank test

• The log rank statistic is one of the most commonly used methods to learn if two curves are significantly different.

• This method also known as Mantel-logrank statistics or Cox-Mantel-logrank statistics

• The logrank test compares the number of observed deaths in each group with the number of deaths that would be expected based on the number of deaths in the combined groups that is, if group membership did not matter.

Hazard ratio

• The logrank statistic calculates the hazard ratio• It is estimated by O1st group/E1st group

divided by O2nd group/E2group• The hazard ratio is interpreted in a similar

manner as the odds ratio• Using the hazard ratio assumes that the hazard

or risk of death is the same throughout the time of the study.

Mantel– Haenszel chi test

• Another method for comparing survival distributions is an estimate of the odds ratio developed by Mantel and Haenszel that follows (approximately) a chi-square distribution with 1 degree of freedom.

• The Mantel– Haenszel test combines a series of 2 × 2 tables formed at different survival times into an overall test of significance of the survival curves.

• The Mantel–Haenszel statistic is very useful because it can be used to compare any distributions, not simply survival curves

Cox progressive hazard model

Why is called cox proportional hazard model

• Cox =scientist’s name(Sir David Roxbee Cox)– British statistician– In 1972 developed it.

• Uses hazard function

• covariates have a multiplicative or a proportional , effect on the probability of event

What does cox model do>

• It examines two pieces of information:– The amount of time since the event first happened

to a person– The person’s observations on the independent

variables.

Cox progressive hazard model

• Used to assess the simultaneous effect of several variables on length of survival.

• It allows the covariates(independent variables) in the regression equation to vary with time.

• Both numerical and nominal independent variables may be used in this model.

COX regression coefficient

• Determines relative risk or odd’s ratio associated with each independent variable and outcome variable, adjusted for the effect of all other variables .

Hazard function

• Opposite to survival function• Hazard function is the derivative of the survival

function over time h(t)=dS(t)/dt• instantaneous risk of event at time t (conditional

failure rate)

• It is the probability that a person will die in the next interval of time, given that he survived until the beginning of the interval.

Hazard function

• Hazard function given by

h(t,x1,x2…x5)=ƛ0 (t)eb1x1+b2x2+….b5x5

• ƛ0 is the baseline hazard at time t i.e. ƛ0(t)

• For any individual subject the hazard at time t is hi(t).

• hi(t) is linked to the baseline hazard h0(t) by

loge {hi(t)} = loge{ƛ0(t)} + β1X1 + β2X2 +……..+ βpXp

• where X1, X2 and Xp are variables associated with the subject

71

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Proportional hazards:

Hazard functions should be strictly parallel!

Produces covariate-adjusted hazard ratios!

Hazard for person j (eg a non-smoker)

Hazard for person i (eg a smoker)

Hazard ratio

72

The model: binary predictor

smoking

smoking

agesmoking

agesmoking

eHR

eet

et

th

thHR

smokingcancerlung

j

ismokingcancerlung

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This is the hazard ratio for smoking adjusted for age.

Table 2. Death rates for screenwriters who have won an academy award.* Values are percentages (95% confidence intervals) and are adjusted for the factor indicated Relative increase

in death rate for winners

Basic analysis 37 (10 to 70)

Adjusted analysis

Demographic:

Year of birth 32 (6 to 64)

Sex 36 (10 to 69)

Documented education 39 (12 to 73)

All three factors 33 (7 to 65)

Professional:

Film genre 37 (10 to 70)

Total films 39 (12 to 73)

Total four star films 40 (13 to 75)

Total nominations 43 (14 to 79)

Age at first film 36 (9 to 68)

Age at first nomination 32 (6 to 64)

All six factors 40 (11 to 76)

All nine factors 35 (7 to 70)

HR=1.37; interpretation: 37% higher incidence of death for winners compared with nominees

HR=1.35; interpretation: 35% higher incidence of death for winners compared with nominees even after adjusting for potential confounders

http://bmj.bmjjournals.com/cgi/content-nw/full/323/7327/1491/#TF2-150

Importance

• Provides the only valid method of predicting a time dependent outcome , and many health related outcomes related to time.

• Can be interpreted in relative risk or odds ratio• Gives survival curves with control of

confounding variables.• Can be used with multiple events for a subject.

Take Home Message

• survival analysis deals with situations where the outcome is dichotomous and is a function of time

• In survival data is transformed into censored and uncensored data

• all those who achieve the outcome of interest are uncensored” data

• those who do not achieve the outcome are “censored” data

Take Home Message

• The actuarial method adopts fixed class intervals which are most often year following the end of treatment given.

• The Kaplan-Meier method uses the next death, whenever it occurs, to define the end of the last class interval and the start of the new class interval.

• Log-Rank test used to compare 2 survival curves but does not control for confounding.

• Mantel henzel test can compare any curve not only survial curves

• For control for confounding use another test called as ‘Cox Proportional Hazards Regression.’

Thank you

Education

Survival analysis