KaplanMeier Estimation

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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. 385-405

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Data Description

• 946 Flies are Venilated on Day 1
• 931 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 gt t) where T is the
  (random) Survival time of a fly
• Goal 2 Test whether the (population) Survival
  Functions differ for the 2 Groups

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Data
VentDie and CntlDie represent numbers dying in
period just prior to this day
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Kaplan-Meier Estimation - Notation

• For each fly we observe the pair (ti, di) where
  ti is the ith flys time to death or censoring
  and di is a censoring indicator (1censored,
  0not censored (actual death))
• Observed failure times t(1) lt ltt(k)
• Number of failures at t(i) ? di
• Subjects censored at t(i) treated as if censored
  between t(i) and t(i1). Number censored in
  t(i),t(i1)) ? mi
• Subjects at risk Prior to t(i) ? ni
  (dimi)(dkmk) (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)

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Kaplan-Meier Estimates
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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 13/946.0137.
At day 13, 15/933.0161 was the proportion
dying The Survival function at day 7 is obtained
as (1-.0137).9863. At day 13, the survival
function is (1-.0137)(1-.0161) .9704
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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 n1i and n2i be the numbers at risk just prior
  to t(i) for the 2 groups, with total at risk ? ni
  n1i n2i
• Let d1i and d2i be the numbers dying at t(i) for
  the 2 groups, with total deaths ? di d1i d2i
• Then the expected deaths for group 1 is
  e1idi(n1i/ni) which represents the total deaths
  at t(i) times the fraction of the total at risk
  that are in group 1

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