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This is a kaplan 2009 lecture notes question. They have given the answer but haven't taken the time to explain. I just don't get it. Help me understand this plz.
I'm so bad at anything remotely MATH....

Kaplan book psychiatry and biostats, page 130, question # 34:

Q. 55 yr old man complains of urinary frequency. exam shows a 1 cm nodule on his prostate. physician order PSA. It comes out to be 7ng/ml ( while anything above 4ng/ml is abnormal). Using this standard , this test has a sensitivity of 80% and a specificity of 90%. recent studies show that 10% of men f this age have prostate CA.
what is your best estimate of the likelihood that this man actually has prostate CA?

A. 13%
B. 25%
C. 36%
D. 47%
E. 58%

answer is D. 47% ( I already know the answer, I just need you to explain this answer. I just don't get it at all, the concept is not clear)


this q has been explained before in the privious posts and in kaplan the q is explained on the video so maybe thats way they dont explain it anymore in the notes
u must make the 2 by 2 table
10% is the prevalence of the disease
always start with it when u make the table and assume u have a total of 1000 people (to make ur calculations easier but u can also put 100 or 10)

disease + disease -

test + 80 TP 90 FP

test - 20 FN 810 TN

100 900 1000
80 from Se 80% goes into TP box
90 from Sp 90 goes into FP box
the rest u just substract to fill the empty boxes
PPV is 80/170=47%


hmm..I still don't get it....coz if specificity is 90%--> then don't u have to put the 90 into the TN box?
like TP 80 FP 10
FN 20 TN 90



dont forget ur first choice to have 1000 total
so after u put 80 in TP box will look like this

tp 80 fp
fn 20 tn

T 100 900 1000

now put Sp knowing FP+TN is 900
u must put 90 in FP to have that 900 total
this is the tricky part i know

  #5 the concept. Thanks benstar!nod




wow!! thanks!!


I dont know the current status of the person asked the question...possibly it is too late to answer...but the simplest way of answering the question is stated below---
You should use Bayes Theorem to solve this problem...
the question ask to find out the PPV (Positive Predictive Value) of the test which is estimated by the following formula--
PPV=(Sensitivity x Prevalence)/[(Sensitivity x Prevalence)+(1-Specificity) x (1-Prevalence)] = (0.8 x .01)/[(0.8 x 0.1) + (1-0.9) x (1-0.1)] = 0.47
I hope this answers the problem. Further queries are always welcome. Cheers!

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