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NBME 23 Answers

nbme23/Block 4/Question#16 (25.4 difficulty score)
In a study of the use of ultrasonography for ...
80%πŸ”
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 +6 
submitted by yb_26(243),
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arabomln stte urlste seanm that test edtetcs ccraen g;=&t

  • 53 of 05 nem ithw etroapts ncrace avhe abomlanr tets erustl =t;g& n of pts twhi caencr = 50. Tset shwso racnec in 53 nme =&t;g =3P5T =;t&g ew cna elaccautl FN = 0-355 = 15

  • 02 of 001 emn uthtoiw ertotaps ercacn ahev nmlaorab etst rstelus gt=&; FP 0=2 &;=gt ew nca aectalulc NT = 0=-820010

  • now we nac lelccuaat itiifpcseyc = NFTNP/+T() = 080/10 = 0.8 ni( % liwl eb %)08

erhe is my /44 :lebta [ im:biea3e_/snicso1m3msrt//c_sl_wmrd/.epwbtmtm/wrt2topscuooutf.nc/shneeqtutleow//d]

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. +3  

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

+/- paulkarr(54),


 +1 
submitted by meningitis(500),
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heT iustenoq si uscninofg bueeacs a TURE IOIVPSTE stet tulser si fngndii idetetcng cenrac by SU o(gse ot wsoh oyu ew oatgt olok ta thwa eht test is koionlg ro.)f

An OBNAMRAL sett etnticged aPrC enmas tis TON A PS:IIEVTO i:e 35 out fo 05 rwee lseaF oPtivessi

An ONLBMAAR sett TTIUWOH PaCr mesan ist TON A VAI:NEEGT 02 uto of 010 UWIOTTH PCra rewe FSLAE TAGNEIE.V

TN = 80, PF = icifp:iSc1 .ey5t TFTN/(P;N)+ /)500+188( = 8% nA4d. cesin it ssya t"ebs eernrp"esst tenh %48 is sstecol to %08.




 +0 
submitted by lfsuarez(141),
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20 fo hte 010 enm uowihtt psottrae acrcen aevh mbnarola ttes ss.urlte

petScicfiiy = PT/NFF+P = 100/02 = 8.0 = %80

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%. +3  
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 -) +