NBME 22 Answers : Block 3 : #21 = A screening test for cancer is developed and ...

NBME 22 Answers ↦

nbme22/Block 3/Question#21 (38.7 difficulty score)
A screening test for cancer is developed and ...
Data point labeled ‘A’🔍
tags:

+9
submitted by brise(31),

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sbryant6 SPin and SNout. Specificity in, sensitivity out. +5

+7
submitted by adisdiadochokinetic(66),

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owH nmya olppee d'dint ees thta ti swa eicsy-p1itcfi and edcpki E lkei em :(

jfny21 Here +1

eacv yesss! me too -.- +3

jinzo me also :( +1

+6
submitted by hello(257),

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wnKogin het +RL auelv = 10 esod otn pleh to loves ihts Q ebuscae itntsaemig werhe "0"1 uohdsl lfla no an isax is b.rryiatra o,lAs eht adta sotinp are intrceoosda -- htye vaeh na -elavuX adn a l-aeuvY ,(X )Y.

eTh awy to carphapo tsih Q si ot kown htat a ighh tyeficiipcs esanm atht a psivetoi letsur si y,evr evyr iylekl to eb a etur iipotves.

pepSsuo hatt hte sycteiiifcp is 09.9 -- tihs si 9%9 ifitp.cceisy ,Tneh ouy loko at teh rap.gh heT axXis- si c."tcy1ief-ispi" ,So osusppe hte best ttes ahs a peyicifitsc of .%99 ehT,n catcaulle icyfs epci1-it = 1 - .909 = 00..1 You dwluo nthe schoe het ipdtoatan atth orcoressnpd to gahniv an -eavX"u"l hatt si ostsecl ot hte .niroig nI itsh obrl,pme ti esocdrrnops ot atad onipt "".A

uYo otn'd vene nede to nkow a siipfcec ticifsipcye veaul ot eslvo sthi bolmpr.e lAl ouy dene ot do is asntnerdud htat if hte iyftspiceic is xetmeyler ihh,g uyo iwll deen ot dinf a titnopada that si lsceots ot eth iginor -- ta altse rfo hte lauve ni eht axs-iX ni eth dirneooact of het data ponti -- bueaesc eht isx-Xa ooeprrcsnsd to a lctncouaial fo it"1cs-pfiiy.ec"

link981 Excellent explanation but a minor typo. 1-0.99 = 0.01 not 0.10 :) +2

+4
submitted by alexb(43),

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The rosdw tsmo" eyl"vitceeff dsuecfno .em I uhthtgo ot mefysl ee"nv if ti has het hghstie yiceif,tcpsi sti' otn vyer icetefvfe fi is't otg purse wol inesisvtiyt -- nesic it llwi miss a olt of het teru v+s'e (figlani ot eulr ni cceran ofr eht enos atth get ime)ss.d" hatT wsa a styor I tdlo sfeylm ubtoa eithr sue fo het swdor ms"ot "elvefiftyec oll.

+0
submitted by amirmullick3(50),

Very easy.
Sensitivity; SnOUT.
Sensitivity is your Y axis, and the higher you go, the more sensitive, which means you can rule OUT cancer.
So A is the least sensitive.

Specificity; SpIN. Specific tests rule out cancer. If your specificity is 1, then 1-specificity would be 0, which is the 0 point of the X axis.

-2
submitted by seagull(1171),

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Thsi si a dikioehllo tro.ia L=R+ /spfnS1eceiS-

nyA vlaeu egterra atnh 01 epr( rsfit adi) eatdnicid suesnf"usel fo iagdncitos tets" whcih is cpbamleaor to PVP rgl(uni in a .d)z tnPio A"" si eht seslotc rmak to heewr 01 usohdl eb no hte Y xa.is

brise The question is asking what point would be the most likely to rule in cancer, and high specificity when positive rules in cancer. The highest specificity value is A, bc the the X axis shows (1-specificity)! +5

hello brise is correct. Knowing the LR+ value = 10 does not help in this situation because estimating where "10" should fall on an axis is arbitrary. The way to approach this Q is to know that a high specificity is will mean that a positive result is very very likely to be a true positive. In theory, suppose that the specificity was 0.99. This is 99% specificity. Then, you look at the graph. The X-axis is "1-specificity." So, suppose the best test has a specificity of 99%. Then, calculating 1-specificity = 1 - 0.99 = 0.1. You would then chose the datapoint that corresponds to having an "X-value" that is closest to the origin. In this problem, it corresponds to data point "A." +3