NBME 23 80% (NBME Answers)

Retired NBME 23 Answers

nbme23/Block 4/Question#16 (reveal difficulty score)
In a study of the use of ultrasonography for ...
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submitted by ∗yb_26(316)

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abnormal test result means that test detects cancer =>

35 of 50 men with prostate cancer have abnormal test result => n of pts with cancer = 50. Test shows cancer in 35 men => TP=35 => we can calculate FN = 50-35 = 15
20 of 100 men without prostate cancer have abnormal test results => FP =20 => we can calculate TN = 100-20=80
now we can calculate specificity = TN/(TN+FP) = 80/100 = 0.8 (in % will be 80%)

here is my 4/4 table: [https://www.reddit.com/r/usmlestep1/comments/ccul3w/biostat_question_from_nbme23/]

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smc213 Exactly what I did! +

smc213 I googled the meaning of abnormal test results just to make sure. A positive test is one in which the result of the test is abnormal; a negative test is one in which the test's result is normal. +4

This guy on youtube does a killer job on explaining all these problems. Just does problem after problem on 2x2 tables and a few other topics. The link to part 1 is added, part two you can find on the side panel I'm sure. Part 1

+1/- paulkarr(73)

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submitted by ∗meningitis(643)

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The question is confusing because a TRUE POSITIVE test result is finding detecting cancer by US (goes to show you we gotta look at what the test is looking for).

An ABNORMAL test detecting PrCa means its NOT A POSITIVE: ie: 35 out of 50 were False Positives

An ABNORMAL test WITHOUT PrCa means its NOT A NEGATIVE: 20 out of 100 WITHOUT PrCa were FALSE NEGATIVE.

TN = 80, FP = 15. Specificity: TN/(TN+FP); 80/(80+15) = 84%. And since it says "best represents" then 84% is closest to 80%.

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submitted by ∗lfsuarez(160)

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20 of the 100 men without prostate cancer have abnormal test results.

Specificity = FP/FP+TN = 20/100 = 0.8 = 80%

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seagull almost. 100/120 = 83% roughly 80% +

amirmullick3 Not sure what lfsuarez and seagull above mean. Here is my explanation. Specificity = TN/(TN+FP). This test gave 20 false positives out of 100 people, and only 15 true negatives out of 50 men. Specificity also equals 1-FPrate, and here the FP rate seems 20% so 100%-20%=80%. +4

yb_26 abnormal test result means pt has cancer => TP = 35, FN = 15 (50-35), FP=20, TN =80 (100-20) => specificity = TN/(TN+FP) = 80/100 = 0.8 (in % will be 80%) true negatives are 80 out of 100, not 15 out of 50 +2

bulgaine If you replace the values from the question in the table of page 257 of FA 2019, yb_26 explanation is correct. Abnormal test = patient has cancer = test + Question says 35/50 men with prostate cancer (so all 50 have cancer) only 35 have abnormal test results, meaning that TP=35 (disease + test +) and FN= 15 (disease + test - because they do have cancer but the test was not abnormal for them ). 20/100 men without prostate cancer have abnormal test results meaning all 100 DONT have cancer but 20 show that they have cancer when its not true so FP=20 (disease - test +) and TN =80 (disease - test -) +

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