Transcript Lecture 3

Lecture 3
Survival analysis
The slang
• Current life table
– What is the life time now?
• Cohort life table
– What is the life time of this group?
Problem
• Do patients survive longer after treatment
A than after treatment B?
• Possible solutions:
– ANOVA on mean survival time?
– ANOVA on median survival time?
• Survival analysis
– Aka. Actuarial / life table analysis
The example
Person-year of observation
• In total: 15122 days ~ 41.4y
• 11 patients died: 11/41.4y = 0.266 y-1
26.6 death/100y
• 1000 patients in 1 y
• or
• 100 patients in 10y
Mortality rates
• 11 of 25 patients died
• 11/25 = 44%
• When is the analysis done?
1-year survival rate
• 6 patients dies the first year
• 25 patients started
• 24%
1-year survival rate
•
•
•
•
3 patients less than 1 year
6/(25-3) = 27%
Patient 7
24% -27%
Actuarial / life table anelysis
• Patients on dialysis
• 180 day pariods
Actuarial / life table anelysis
• Ni number of patients
starting a given period
Actuarial / life table anelysis
• Wi number of patients that
have not yet been in the
study long enough to
finish this period
Actuarial / life table anelysis
• Number exposed to risk:
• ni – wi/2
• Assuming that patients
withdraw in the middle of
the period on average.
Actuarial / life table anelysis
• di : Number of patients
that died
Actuarial / life table anelysis
• qi = di/(ni – wi/2)
• Proportion of patients
terminating in the period
Actuarial / life table anelysis
• p i = 1 - qi
• Proportion of patients
surviving
Actuarial / life table anelysis
• Si = pi pi-1 ...pi-N
• Cumulative proportion of
surviving
• Conditional probability
Survival curves
• How long will a
dialysis patient
survive?
A few concerns
• Withdrawals are assumed to happen at
time midpoints
– Kaplan Meiers method will fix this
• The probability of surviving a period is
treated as if it is independent of survival of
other periods.
Kaplan-Meier product limit method
• Changes in the survival curve are
calculated when an event occurs
• Withdrawals are ignored as events
Kaplan-Meier
• Simple example with only
4 ”death-events”.
Confidence of survival curves
• Standard Error (SE)
SE ( Si )  Si
di
 n n  d 
i
i
i
• Assumed to be
normally distributed
• 95% confidence is
1.96*SE
Confidence interval of the KaplanMeier method
SE ( Si )  Si
di
 n n  d 
i
i
i
The Kaplan-Meier method
Actuarial and Kaplan-Meier
Survival curves
• Actuarial
• 0.6117
• Kaplan-Meier
• 0.6154
Comparing survival curves
Comparing survival curves
• Two principle methods
– Logrank statistics
– Mantel Haenszel chi-square statistics
Logrank statistics
# deaths  # group
E
# total
2  
O  E 
E
2
Logrank statistics
Hazard ratio
O E
Hazard ratio  1 1  4.13
O2 E2
Mantel Haenszel chi-square
statistics
Mantel Haenszel chi-square
statistics
OR 
  a  d / n
 b  c / n
E (ai ) 
  ai   E  ai  

Mantel  Haenszel  
V  ai 
Row _ total  Col _ total
Grand _ total
2
V (ai ) 
(a  b)(b  d )(a  b)(c  d )
n2 (n  1)
Hazard function
H   log( Si )
H
d
 f  c