ESTIMATING SURVIVAL PROBABILITY
KAPLAN-MAYER'S ESTIMATE OF SURVIAL FUNCTION
Now we are interested in computing the estimate of the survival function S(t).
S(t) = P(T > t | T < t)
T = Survival Time
t = Observed survival time
S(t) is the probability of a subject surving past time t given that the subject has already survived upto time t.
In this problem the event of interest is attrition and time is time in years until attrition. This is a right censored problem. Although some of the employee had left the company but some of them are still working. We will use the following R code to compute the estimate of the survival probabilities for each time point t (in years).
R Code:
#loading the survival package
library(survival)
#attaching the variables
attach(HR)
#fitting the survival model using survfit taking Attrition == 1 (KM survival probabilities)
HR.surv <- survfit(Surv(YearsAtCompany, Attrition) ~ 1, conf.type="none")
summary(HR.surv) #summary of the HR.surv
Interpretation:
From the above summary table of survival analysis using “Kaplan Mayer “we can see that when time =0 it means that when a person had joined recently in the company his probability of surviving past the 0th year in the company is 0.98 which is very high as we expect. With the increase in time of staying in the company the probability of surviving in that company decreases as we can see from the monotonically decreasing in survival curve.
For example: when t = 3, S(t) = 0.76, t = 5, S(t)= 0.54 which shows that there was a gradient fall in the probability rate with the increase in time which shows that there is a high fall in the survival rate of the employee as the Years at companies increases.
When t = 5 we can see that there is 50% chances of the employee staying or leaving the company. We can also see that when the time was between 21 to 37 there was stability in the probability of the surviving time of the employee in the company. So from the above table we conclude that as the time increases the probability of surviving time of the employee’s decreases.