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#1




Estimating the likelihood of cancer?
This question was in Kaplan CK Qbank and I cannot figure out the explanation, if any one can explain it to me plesae
55 year old man visit his PCP with a complaint of urinary frequency. Examination shows a 1 cm nodule on his prostate. The physician orders a PSA. By common standards, a PSA level greater than 4 is considered abnormal. Using this standard, the test has a sensitivity of 80% and specificity of 90%. A recently published article found that in a cross sectional study, 10% of men of this age have prostate cancer. The result on the patients PSA is 7. What is your best estimate of likelihood that this man has cancer.? a. 13% b. 25% c. 36% d. 47% e. 58% f. 69% g. 72% h. 81% 
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svmle (07072012) 
#2




72%
???? 
#3




nah the answer is D but i want to know how to get to it



#4




The question is asking you to calculate the positive predictive value of the test. I think that the best way to calculate PPV is to use a 2x2 square, but some people prefer to memorize the formula.
Let's say you have 100 patients and 10 of them have prostate cancer, so 90 of them are healthy. If you work backwards from sensitivity and specificity, you can fill out this 2x2 chart: .......................Disease present.......................Disease absent Positive test ............A..................................... ......B Negative test............C..................................... ......D Total.......................10......................... ..................90 Sensitivity = % of patients with the disease who have a positive test You have 10 patients with the disease, so if the sensitivity is 80%, then you have 8 patients in Box A. Specificity = % of patients without the disease who have a negative test You have 90 patients without the disease, so if the specificity is 90%, then you have 81 patients in Box D. So: .......................Disease present.......................Disease absent Positive test ............8..................................... ......B Negative test............C..................................... ......81 Total.......................10......................... ..................90 And then you can work out the remaining boxes B and C with simple subtraction: (10 patients with disease)  (8 patients with disease and positive test) = 2 patients with disease and negative test (90 patients with no disease)  (81 patients with no disease and negative test) = 9 patients with no disease and positive test .......................Disease present.......................Disease absent Positive test ............8..................................... ......9 Negative test............2..................................... ......81 Total.......................10......................... ..................90 So, back to the question, which asks you to calculate the percentage chance of your patient having the disease if he has a positive test. Based on our table, 17 patients have a positive test and 8 of them have disease. Therefore, the PPV of the test is 8/17 = 47%. 
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datson (08162016), greentrees (11192016), Krazy (10212013), mbbs2010 (07062012), svmle (07072012) 
#5




i can give it a try.
What they seem to be asking is 'what is the probability that the positive test is actually positive" in other words the PPV Now we have to construct the 2 X 2 table.. Putting in the values from the question The table is as DISEASED NON DISEASED POSITIVE TEST..........8................................... ..9 (>4 PSA) NEGATIVE TEST..........2................................... .81 (4< PSA) TOTAL.....................10...................... .............90 TOTAL 100 PPV = 8/8+9 = 8/17 = 0.47 so the probability that the test is actually positive and the chance the patient actually has prostate ca is 47 %.. Correct me if i am wrong anywhere...
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#6




wow! i was looking for that for so long!!
please quick question  Does that quotation work only if you select 90 and 10 for total 100 patients? because i tried different number and its not 47 at the end....( 
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BiostatisticsEpidemiology, Step2Questions 
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