CT4 september 2016 question 10 ii

[Priyanka Malhotra]

Hi, when solving this question using the approach in examiner's report in April 2016 que 11 (v) and sept 2017 ques 6 (iii), the answer 1-33.2% using both methods given in mentioned diets, could someone please help me understand why the answer is not matching the solution in examiner's report i.e. 1 − 0.7219, thanks in advance!

Please see below solution using the method given in mentioned examiner's report of April 2016 and sept 2017 :

exp(0.15)/exp(-0.035)=exp(0.185)=1.20321 (A)

probability = Survival^A = (1-0.6)^1.20321= 0.3320

alternatively
int(0,3) -h(t)= ln(0.4)/exp(-0.035) ( A)
using A to calculate probability that the female is discharged
exp(A*exp(0.15))
=exp(-1.10250)
0.332

 

[Andrew Martin]

Hello

Unfortunately there is a bit of confusion in this question. The examiners gave credit to all reasonable interpretations.

If you interpret the question as asking for the hazard rate of a female heavy drinker who smokes and who is still in hospital after three days then the answer is 0.7219, which is part of the solution given in the examiners report.

If you interpret the question as asking for the probability of a female heavy drinker who smokes staying in the hospital for three days, then we need to interpret the given 0.6 as the probability (rather than hazard as is stated in the question) of a male moderate drinker who does not smoke leaving the hospital before three days are up. In which case the answer is 0.332 as you've calculated.

So overall I wouldn't too much about it.

Hope this helps.

Andy