Research has shown that speci c biochemical markers are found exclusively in the

Question

Research has shown that speci c biochemical markers are found exclusively in the
breath of patients with lung cancer. However, no lab test can currently distinguish the breath
of lung cancer patients from that of other subjects. An experiment trained dogs to distinguish
between the breath of lung cancer patients and control (healthy) subjects. After the training, the
dogs were tested in a new double blind experiment (neither the dogs nor the researchers were given
any clues as to the condition of the subjects). Here are the results for a sample of 1286 specimens:
Dog result Control Subject Cancer Subject Total

  

Find the probability that the dog gave a positive result GIVEN that the subject actually had
lung cancer. Round your answer to three decimal places. This is called the test's sensitivity.
(b) Find the probability that the dog gave a negative result GIVEN that the subject did NOT
have lung cancer (so was in the control group). Round your answer to three decimal places.
This is called the test's specificity.
(c) Now suppose the rate of lung cancer in the general population is 0:5% (Note: this is much
lower than the rate in the table above, so we're no longer using the table). Also assume that
the sensitivity and specificity you found in (a) and (b) are accurate in the general population
as well. If a randomly chosen person from the general population is given the dog test, nd
the probability of a positive test result. (Do NOT round o until you nish the calculation,
then round to four decimal places.)
(d) What is the probability that a randomly chosen person from the population who tests positive,
actually HAS lung cancer. (Do NOT round o until you finish the calculation, then round to
four decimal places.)

I need help with C and B. FOR A I got 0.983 and B 0.994

I tried to do C and got 0.412 and D 0.542 but I know thats wrong. please help me with C and D and do it step by step. Thank you guys

Dog result control subject cancer subject Total negative 708 10 718 positive 4 564 568 total 712 574 1268

Explanation / Answer

A)

564/574=0.982578

B)

708/712=0.9944

The probability (Having disease and tested positive)(D and P)=0.005*0.982578=0.0049

The probability (Having disease and tested negative)(D and N)=0.005*0.01742=0.0000871           

The probability (Not having disease and tested positive)(NHD and P)=0.995*0.0056=0.005572    

The probability (Not having disease and tested negative)(NHD and N)=0.995*0.9944=0.9894

C) I am not sure of what the question is exactly because some of the letters are being cut but I guess, it is asking for the below:-

The probability (Having disease and tested positive)(D and P)=0.005*0.982578=0.0049

D) 0.0049/0.0049+0.005572=0.4679

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