Kaplan-Meier Estimation &Log-Rank Test Survival of Ventilated and Control Flies (Old Falmouth Line 107) R.Pearl and S.L. Parker (1922). “Experimental Studies.

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Kaplan-Meier Estimation &Log-Rank Test Survival of Ventilated and Control Flies (Old Falmouth Line 107) R.Pearl and S.L. Parker (1922). “Experimental Studies on the Duration of Life. V. On the Influence of Certain Environmental Factors on Duration of Life in Drosophila,” The American Naturalist, Vol. 56, #646, pp

Data Description 946 Flies are Venilated on Day Flies are not Ventilated on Day 1 (Controls) Counts of Survivors and Non-Survivors Reported every 6 Days Subsequently (Days 7,13,…,85) Goal 1: Estimate the Survival Function for the two groups: S(t) = P(T > t) where T is the (random) Survival time of a fly Goal 2: Test whether the (population) Survival Functions differ for the 2 Groups

Data VentDie and CntlDie represent numbers dying in period just prior to this day

Kaplan-Meier Estimation - Notation For each fly we observe the pair (t i,  i ) where t i is the i th fly’s time to death or censoring and  i is a censoring indicator (1=censored, 0=not censored (actual death)) Observed failure times: t (1) < … <t (k) –Number of failures at t (i)  d i –Subjects censored at t (i) treated as if censored between t (i) and t (i+1). Number censored in [t (i),t (i+1) )  m i –Subjects at risk Prior to t (i)  n i = (d i +m i )+…+(d k +m k ) (this removes all who have died or censor prior to t (i) ) Note: For this dataset, there is no censoring (all flies were observed to die)

Kaplan-Meier Estimates

Note: For the ventilated flies, 13 died at day 7 (actually between days 1 and 7), out of 946 that were at risk prior to that period: 13/946= At day 13, 15/933=.0161 was the proportion dying The Survival function at day 7 is obtained as ( )= At day 13, the survival function is ( )( ) =.9704

Log-Rank Test Used to test whether two (or more) survival functions are equal Involves obtaining the “expected” number of deaths in (say) the treatment group at time t (i) if the hazard functions for the two groups were equal: –Let n 1i and n 2i be the numbers at risk just prior to t (i) for the 2 groups, with total at risk  n i = n 1i + n 2i –Let d 1i and d 2i be the numbers dying at t (i) for the 2 groups, with total deaths  d i = d 1i + d 2i –Then the expected deaths for group 1 is: e 1i =d i (n 1i /n i ) which represents the total deaths at t (i) times the fraction of the total at risk that are in group 1

Log-Rank Test

Log-Rank Test (Fly Data) Both tests are highly significant