By the results of Section 3.2.6 we have that the standardized Kaplan-Meier estimator is approximately standard normally distributed, for given value of t, where is the variance estimator (3.27)....

By the results of Section 3.2.6 we have that the standardized Kaplan-Meier estimator

is approximately standard normally distributed, for given value of t, where

is the variance estimator (3.27). [Alternatively we may estimate the variance by Greenwood’s formula (3.28).]

a) Let

be the pth fractile of the survival distribution (cf. Section 3.2.3), and consider testing the hypothesis

versus the alternative


Explain why we get a test with level α if we reject the hypothesis if

b) A 100(1 − α)% confidence interval for

is given as all values

that are not rejected by the test in question a). Explain why this interval may be read directly from the lower and upper confidence limits (3.29) as illustrated in.

For ease of presentation we consider the situation without transformations in this exercise. However, the arguments in questions a) and b) carry over with only minor modifications to the situation in which we use a transformation of the Kaplan-Meier estimator.