Question

A pilot study was designed to evaluate the potential efficacy of a program designed to reduce prison recidivism amongst inmates who have a documented long-term history of drug and/or alcohol problems. A sample of 11 prisoners was followed for up to 24 months after their most recent release from prison. Six of the inmates returned to prison at 3, 7 9, 11, 14 and 21 months respectively. Five of the inmates had not returned to prison as of the last time they were last contacted which was at 4, 8, 16, 24, and 24 months respectively.

Use the Kaplan Meier approach to estimate the survival curve for this set of inmates (which tracks the proportion who have not yet returned to prison over time).
It will be helpful to construct a table like the ones appearing in lecture 5: however, all you will need to report in the quiz generator are certain quantities from this table for specific
times.

- Suppose you treated the censored observations as events: what would your estimate for S(11) be?

a. 45%

b. 83%

c. 33%

d. 57%

Answer 1

Answer:

j	tj	nj	dj	(dj/nj(xi))	1-(dj/nj(xi))	[1-(dj/nj(xi))]
1	3	11	1	1/11=0.09090	10/11=0.9090	0.9090
2	7	9	1	1/9=0.11111	8/9=0.8888	0.8081
3	9	7	1	1/7=0.14285	6/7=0.8571	0.6934
4	11	6	1	1/6=0.16666	5/6=0.8333	0.5778
5	14	5	1	1/5=0.2	4/5=0.800	0.4635
6	21	3	1	1/3=0.33333	2/3=0.666	0.31

where,

• j=jth risk
• tj=time for the jth risk
• nj=number of people avaliable at time "j"
• dj=number of people returned toprison
• one diservation is censored at time 4

then,

nj at time 7 =11-1-1

nj at time 7 =11-2

nj at time 7 =9

if one is consored at 8,so nj at 9

nj at 9=9-1-1

nj at 9=9-2

nj at 9=7

frome the above table

[1-(dj/nj(xi))]=sj

then

s(t)=0 for 0t3,

91% for 3t7

81% for 7t9

69% for 9t11

57% for 11t14

46% for 14t21

31%. for 21t24

• s(11)=57%