2. (50 points total) A 60-year old man who has never smoked cigarettes complains
ID: 3306394 • Letter: 2
Question
2. (50 points total) A 60-year old man who has never smoked cigarettes complains to a physician with symptoms consisting of a chronic cough and occasional breathlessness. The physician becomes concerned and orders the patient to be admitted to the hospital for a lung biopsy. Results of the lung biopsy are consistent with either lung cancer or sarcoidosis, a fairly common, benign nonfatal lung disease. Experience shows that of the patients referred to the hospital for a biopsy under these conditions, 25% are diagnosed with lung cancer while 75% are diagnosed with sarcoidosis. There are no other diagnoses under these conditions. Of those with lung cancer, 90% suffer from such symptoms as chronic coughs and occasional breathlessness. Similarly, 60% of those with sarcoidosis suffer from such symptoms. No others show symptoms. a. (25 points) Given that the man has such symptoms, what is the chance that he has lung cancer? b. (5 points) What is another name for the resulting probability in (a)? (6 points) Among those referred to the hospital for a lung biopsy under these conditions, what is the sensitivity of the 'symptom' test for lung cancer? c. d. (6 points) What is the prevalence of lung cancer for those who have been referred to the hospital for a lung biopsy under these conditions? e. (8 points) Calculate the: (i) false negative 'symptom' test for lung cancer; and (ii) false positive 'symptom' test for lung cancer, including providing probability statements for each for this example.Explanation / Answer
a)
Probability of cancer, P(C) = 0.25
Probability of sarcoidosis P(S) = 0.75
Probability of symptoms GIVEN Cancer P (sym/ C) = 0.9
Probability of sarcoidosis GIVEN Cancer P (sym/ S) = 0.6
From this calculate the probability of symptom AND cancer P( sym intersection C)
P (sym AND C) = P(sym/ C) * P (C)
= 0.9*0.25 = 0.225
P (sym AND S) = P(sym/ S) *P (S)
= 0.6*0.75 = 0.45
Construct the table as follows
Probability of symptoms = 0.225+0.45 = 0.675
Probability of no symptoms = 0.025+0.3 = 0.325
GIVEN a man has symptoms the probability of cancer is defined as P (C/ sym)
By Bayesian probability
P (A/ B) = P (B/A) * P (A) / P(B)
Therefore P (C/ sym) = P (sym/ C) * P (C)/ P(sym) = 0.9 * 0.25/0.675 =0.33
b) the probability of cancer GIVEN he has symptoms is called posterior probability
Bayesian proabability is defined as
Posterior probabilty P (A/ B) = Likelihood P (B/A) * Prior probability P (A) / Evidence P(B)
c) Probability of having symptoms GIVEN he has lung cancer is called sensitivity of the test i.e. P (sym/ C) which is 90%. That is anyone having a cancer will show the symptoms with 90% chance.
d)
For those who have referred to the biopsy the prevalence of cancer is the True positive which is the top left of the table.
P (sym AND C) i.e. 22.5%
Cancer(C) Sarcoidosis(S) Symptoms (sym) 0.225 0.45 No sympoms (nsym) 0.025 0.3 0.25 0.75Related Questions
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