Available data:
P(appendicitis) = 0.16
P(Perforated at time 0|appendicitis) = 0.1875
P(Perforated at time 6|appendicitis) = 0.24
P(Sx worse at time 6|Perforated at time 6|appendicitis) = 0.84
P(Sx same at time 6|Perforated at time 6|appendicitis) = 0.16
P(Sx worse at time 6|Not perforated at time 6|appendicitis) = 0.8
P(Sx same at time 6|Not perforated at time 6|appendicitis) = 0.2
P(Sx same at time 6|No appendicitis) = 0.39
P(Sx better at time 6|No appendicitis) = 0.61
a. Perforation of appendix
If the appendix is perforated at time 0, it will always be perforated thereafter. Thus, 0.24 - 0.1875 = 0.0525 is the new occurence of perforation. This occured among patients without perforation at time 0, therefore, 0.0525 / (1 - 0.1875) = 0.0646 is the answer.
b. Chance tree
c. Probability of perforation at time 0
Prevalence of appendicitis * probability of perforation at time 0 = 0.16 * 0.1875 = 0.03.
d. Probability of perforation at time 6
Prevalence of appendicitis * probability of perforation at time 6 = 0.16 * 0.24 = 0.0384.
e. Probability of symptom outcome at time 6
better <- (1 - 0.16) * 0.61
same <- (1 - 0.16) * 0.39 + 0.16 * (0.1875 * 0.16 + (1 - 0.1875) *
(0.0646 * 0.16 + (1 - 0.0646) * 0.2))
worse <- 0.16 * (0.1875 * 0.84 + (1 - 0.1875) * (0.0646 * 0.84 +
(1 - 0.0646) * 0.8))
list(worse = worse, same = same, better = better)
## $worse
## [1] 0.1295
##
## $same
## [1] 0.3581
##
## $better
## [1] 0.5124
##
f. Probability of perforation conditional on symptom outcome
The values calculated in question e will be denominators.
P(perforation|Sx better) = P(Sx better AND perforation) / P(better)
Better: 0 / 0.5124 = 0
P(perforation|Sx same) = P(Sx same AND perforation) / P(same)
Same: (0.16 * (0.1875 * 0.16 + (1 - 0.1875) * 0.0646 * 0.16)) / 0.3581 = 0.0172.
P(perforation|Sx worse) = P(Sx worse AND perforation) / P(worse)
Worse: (0.16 * (0.1875 * 0.84 + (1 - 0.1875) * 0.0646 * 0.84)) / 0.1295 = 0.249.
g. Probability of appendicitis conditional on symptom outcome
P(appendicitis|Sx better) = P(Sx better AND appendicitis) / P(better)
Better: 0 / 0.5124 = 0
P(appendicitis|Sx same) = P(Sx same AND appendicitis) / P(same)
Same: (0.16 * (0.1875 * 0.16 + (1 - 0.1875) * (0.0646 * 0.16 + (1 - 0.0646) * 0.2))) / 0.3581 = 0.0851.
P(appendicitis|Sx worse) = P(Sx worse AND appendicitis) / P(worse)
Worse: (0.16 * (0.1875 * 0.84 + (1 - 0.1875) * (0.0646 * 0.84 + (1 - 0.0646) * 0.8))) / 0.1295 = 1.
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