You are given the following data based on 60 lives at time 0 :

[math]j[/math]	[math]t_{(j)}[/math]	Deaths at [math]t_{(j)}[/math]	Exits in [math](t_{(j)}^{+},t_{(j+1)}^{-})[/math]	Entrants in [math](t_{(j)}^{+},t_{(j+1)}^{-})[/math]
0			0	0
1	5.3	1	8	1
2	8.6	1	6	7
3	13.2	2	7	7
4	16.1	1	6	5
5	21.0	1	6	4

Calculate the upper limit of the 80% linear confidence interval for [math]S(21.0)[/math] using the Kaplan Meier estimate and Greenwood's approximation.

• 0.872
• 0.893
• 0.915
• 0.944
• 0.968

Answer: D

[math]\hat{S}(21.0)=\frac{59}{60} \times \frac{59-8+1-1}{59-8+1} \times \frac{51-6+7-2}{51-6+7} \times \frac{50-7+7-1}{50-7+7} \times \frac{49-6+5-1}{49-6+5}=0.8899[/math]

[math]\operatorname{Var}[\hat{S}(21.0)] \approx 0.8899^{2}\left(\frac{1}{60 \times 59}+\frac{1}{52 \times 51}+\frac{2}{52 \times 50}+\frac{1}{50 \times 49}+\frac{1}{48 \times 47}\right)=0.00181[/math]

The upper limit of the [math]80 \%[/math] linear confidence interval is

[[math]] 0.8899+1.282(0.00181)^{0.5}=0.944 [[/math]]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.