The probability of Positive Test! - USMLE Forums
 USMLE Forums         Your Reliable USMLE Online Community     Members     Posts
 Home USMLE Articles USMLE News USMLE Polls USMLE Books USMLE Apps
 USMLE Forums The probability of Positive Test!
 Register FAQs Members List Search Today's Posts Mark Forums Read

 USMLE Step 1 Forum USMLE Step 1 Discussion Forum: Let's talk about anything related to USMLE Step 1 exam

#1
10-16-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 203 Threads: 32 Thanked 124 Times in 50 Posts Reputation: 134
The probability of Positive Test!

Medical clinic wants to assess population for CHF. measurement of BNP to be able to detect CHF with sensitivity 90% and specificity 90%. 1 in 10 in the population have CHF. Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value

A 0.09
B 0.10
C 0.50
D 0.81
E 0.90

Found this questiona bit confusing will try getting the answer. mean while any guesses and explanations!!

#2
10-16-2012
 USMLE Forums Addict Steps History: Not yet Posts: 126 Threads: 17 Thanked 42 Times in 33 Posts Reputation: 52

CHF with an abnormal BNP value implies true positive ? ? Think the answer is 0.9 ???
#3
10-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,186 Threads: 119 Thanked 815 Times in 434 Posts Reputation: 825

i'm thinking E. 0.90.......i'm not sure though(i'm bad at stats)

CHF prevalence----1 in 10 --0.1

likelihood ratio is used to assess whether the disease condition exist in a patient(remember doing a question from uworld)..

positive likelihood ratio= sensitivity/1-specificity
= 0.90/1-0.90 ==> 9

so with the prevalence of 0.1 in population for CHF, the probability he is diseased would be----9 * 0.1===>0.9

thanks
 The above post was thanked by: alice (10-16-2012)
#4
10-16-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 203 Threads: 32 Thanked 124 Times in 50 Posts Reputation: 134

this question is super tricky found it on kaplan q bank simulated test. keep posting in your answers , i will post the right answer in sometime.

@venky2600 & @alice - i too got the same answer
#5
10-16-2012
 USMLE Forums Guru Steps History: Not yet Posts: 375 Threads: 17 Thanked 131 Times in 81 Posts Reputation: 147

B. 0.10 ??????
#6
10-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,186 Threads: 119 Thanked 815 Times in 434 Posts Reputation: 825

Quote:
 Originally Posted by nishaa this question is super tricky found it on kaplan q bank simulated test. keep posting in your answers , i will post the right answer in sometime. @venky2600 & @alice - i too got the same answer

ohh so i may be thinking it as A) 0.09

coz with such gr8 sensitivity and specificity test(with 9 likelihood ratio---means the result is likely accurate) should show results very accurately

as population has 1 in 10 (0.1)----so, the probability of disease with this accurate test must be less than 0.1---so thinking of 0.09(just thinking logically)

thanks
#7
10-16-2012
 USMLE Forums Guru Steps History: Not yet Posts: 414 Threads: 30 Thanked 86 Times in 69 Posts Reputation: 96

c)
by the way i m confused by languge of question they are asking PPV?
__________________
There is no shortcut to SUCCESS ,and there is no excuse for FALIURE - by Kabutar111
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
#8
10-16-2012
 USMLE Forums Addict Steps History: Not yet Posts: 126 Threads: 17 Thanked 42 Times in 33 Posts Reputation: 52

Phew!! .. Biostats .. : waiting for the answer ...
#9
10-16-2012
 USMLE Forums Guru Steps History: 1+CK+CS Posts: 331 Threads: 39 Thanked 308 Times in 117 Posts Reputation: 318

I would choose C), because the question asks what is the PPV (probability that a person with a positive test result - abnormal BNP - actually has the disease CHF)

So, if you make a 2/2 table you will get TP=9, FP=9, FN=1, TN=81, because the prevalence of the disease is 10 % (meaning that 10 people out of 100 really have the disease). So the PPV=TP/TP+FP ->9/18 = 0.5
 The above post was thanked by: nishaa (10-16-2012)
#10
10-16-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 557 Times in 318 Posts Reputation: 567

A 0.09
#11
10-16-2012
 USMLE Forums Master Steps History: Step 1 Only Posts: 864 Threads: 62 Thanked 370 Times in 224 Posts Reputation: 380

Quote:
 Originally Posted by nishaa Medical clinic wants to assess population for CHF. measurement of BNP to be able to detect CHF with sensitivity 90% and specificity 90%. 1 in 10 in the population have CHF. Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value A 0.09 B 0.10 C 0.50 D 0.81 E 0.90 Found this questiona bit confusing will try getting the answer. mean while any guesses and explanations!!
So it would be E. 0.90

Sent from my iPad using Tapatalk HD
__________________
One day i will infect these microbes and they will search for "Anti-MEbiotic"....
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
#12
10-16-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 203 Threads: 32 Thanked 124 Times in 50 Posts Reputation: 134
C

This is the explanation I got:
· Set up 2x2 table using a total of 100. Next put down as 10 people who have the disease, 90 who do not have the disease ( prevalence is 1 in 10 which is 10 in 100: prevalence of CHF = 10%)
· Multiply the sensitivity of 0.90 by 10 diseased persons to get TRUE POSITIVE. TP= 9
· Remaining 1 person is FALSE NEGATIVE. FN= 1
· Multiply specificity 0.90 by 90 non- diseased persons to get TRUE NEGATIVE TN=0.81
· Remaining 9 persons are FALSE POSITIVE FP= 9

disease present no disease

test positive 9(TP) 9(FP)
test negative 1(FN) 81(TN)

PPV using above table = TP/TP+FP= 9/18 =0.50

I found this exact explanation on kaplan test.. Hope this new (wierd) concept helps ya' all

@Teona congrats you nailed it!!!

Last edited by nishaa; 10-16-2012 at 09:43 PM.
 The above post was thanked by: venky2600 (10-17-2012)
#13
10-16-2012
 USMLE Forums Master Steps History: Step 1 Only Posts: 864 Threads: 62 Thanked 370 Times in 224 Posts Reputation: 380

Ya this Kaplan explanation is really hard to digest. That also with GERD of other informations. Hehhee

I thought this question with other view. Need other people suggestions
As question say
1 in 10 already affected. That means 10 out of 100 is already affected.
So, we have diseased 10 and non disease 90.
Now if we construct 2x2 table
Diseased. Non diseased
Test +. 9 9
Test - 1. 81
Total. 10. 100

So PPV= TP/TP+FP
= 9/9+9= 1/2= 0.5

Sent from my iPad using Tapatalk HD
__________________
One day i will infect these microbes and they will search for "Anti-MEbiotic"....
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
#14
10-17-2012
 USMLE Forums Guru Steps History: Not yet Posts: 414 Threads: 30 Thanked 86 Times in 69 Posts Reputation: 96

Quote:
 Originally Posted by nishaa Medical clinic wants to assess population for CHF. measurement of BNP to be able to detect CHF with sensitivity 90% and specificity 90%. 1 in 10 in the population have CHF. Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value A 0.09 B 0.10 C 0.50 D 0.81 E 0.90 Found this questiona bit confusing will try getting the answer. mean while any guesses and explanations!!
language of question was exactly same?
__________________
There is no shortcut to SUCCESS ,and there is no excuse for FALIURE - by Kabutar111
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
#15
10-17-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,338 Times in 347 Posts Reputation: 1348

PPV= 9/ 9 + 9 i.e. 50%
__________________
Step 1 - 244 [✔] Step 2 CK - 246 [✔] Step 2 CS [✔] Step 3 - 228 [✔] Match [EMORY SOM] YOG: 2013, 4 Months of USCE University Hospital

#16
10-17-2012
 USMLE Forums Scout Steps History: --- Posts: 34 Threads: 3 Thanked 21 Times in 15 Posts Reputation: 31

1- u make 2x2 table and assume the total as a big number minimum to be 1000.
2- You use to prevalence value giving to you to distribute it into the column of diseased people with TP= 90 FN=10 to match the 90 % sensitivity this will form total diseased as 100 which confirm the prevalence as = 100/1000 = 0.1
3- Now 2 cells left to fill by specificity data which = 90% so FP= 90 , TN=810
4- Now you have complete 2x2 table and all what's left to calculate the post-test probability (PPV)= TP/ ( TP+ FP )

PPV = 90 / ( 90 + 90 ) = 0.5
__________________
" The sign on the door to success says, 'PUSH'."
#17
10-17-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 203 Threads: 32 Thanked 124 Times in 50 Posts Reputation: 134

Quote:
 Originally Posted by koolkiller88 Ya this Kaplan explanation is really hard to digest. That also with GERD of other informations. Hehhee I thought this question with other view. Need other people suggestions As question say 1 in 10 already affected. That means 10 out of 100 is already affected. So, we have diseased 10 and non disease 90. Now if we construct 2x2 table Diseased. Non diseased Test +. 9 9 Test - 1. 81 Total. 10. 100 So PPV= TP/TP+FP = 9/9+9= 1/2= 0.5 Sent from my iPad using Tapatalk HD

yep you are right the explanation too means the same its just too long and bit complicated. idea is to figure out the prevalence which is 10 in 100. then compute true positive from sensitivity (90% of 10)
then figure out true negative from specificity (90% of 90). rest is simple to compute ppv by making a table. i spent 2 hrs reading kaplan explanation finally got it !!
#18
10-17-2012
 USMLE Forums Addict Steps History: 1 + CS Posts: 106 Threads: 3 Thanked 40 Times in 32 Posts Reputation: 50

This question seems so hard for some reason!! I cant get my head around it?! And im done with the biostats part from uworld!
Do you think we'll get these sorta questions on the real exam? Cuz i better learn it well then...

Thanks for the question nishaa
#19
10-17-2012
 USMLE Forums Veteran Steps History: 1+CK+CS+3 Posts: 205 Threads: 6 Thanked 151 Times in 80 Posts Reputation: 161

Quote:
 Originally Posted by nishaa Medical clinic wants to assess population for CHF. measurement of BNP to be able to detect CHF with sensitivity 90% and specificity 90%. 1 in 10 in the population have CHF. Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value A 0.09 B 0.10 C 0.50 D 0.81 E 0.90 Found this questiona bit confusing will try getting the answer. mean while any guesses and explanations!!
I am having a hard time to translate " Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value" into PPV

as according to kaplan PPV : is the probabilty of a disease in a person WHO RECEIVES A POSITIVE TEST RESULT

please can some one throw more light on how question was precisely asking PPV????
 The above post was thanked by: Relid (10-17-2012)
#20
10-17-2012
 USMLE Forums Scout Steps History: --- Posts: 57 Threads: 8 Thanked 31 Times in 19 Posts Reputation: 41

Quote:
 Originally Posted by offpiste I am having a hard time to translate " Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value" into PPV as according to kaplan PPV : is the probabilty of a disease in a person WHO RECEIVES A POSITIVE TEST RESULT please can some one throw more light on how question was precisely asking PPV????
I totally agree!
When i first read the question i automatically recalled the multiplication rule. In this example, prevalence x TP result (maybe TP+FP???, because the test may falsely show abnormal BNP, but it's still abnormal BNP value!!!)... Exactly just what the q asks for, and not PPV. Can someone explain?

Really appreciate it!
 The above post was thanked by: offpiste (10-17-2012)
#21
10-21-2012
 USMLE Forums Scout Steps History: 1 + CS Posts: 82 Threads: 10 Thanked 76 Times in 31 Posts Reputation: 86
My Guess

Quote:
 Originally Posted by nishaa Medical clinic wants to assess population for CHF. measurement of BNP to be able to detect CHF with sensitivity 90% and specificity 90%. 1 in 10 in the population have CHF. Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value A 0.09 B 0.10 C 0.50 D 0.81 E 0.90 Found this questiona bit confusing will try getting the answer. mean while any guesses and explanations!!

I see other people's posting of the correct answer from the source but the interpretation of this question does not make sense to me. I did it this way:
1- the odds that a pt has CHF = 1/10
2- odds the pt will have a true positive= 9/10

1/10 x 9/10 = 9/100 =.09 A

Then I scrolled down and saw it is C, I understand how to get C but how does "Which is the approximate probability that the next patient who walks into the clinic has CHF with abnormal BNP value" mean they want you to get PPV based on the question. I think Kaplan worded this wrong or I am reading it wrong. If someone can shed some light on how that's what they were asking; do not want to be stuck on the exam answering the wrong question just because I read it wrong!! Thank you for any feedback!

 Tags Biostatistics-Epidemiology, Step-1-Questions

Message:
Options

Register Now

In order to be able to post messages on the USMLE Forums forums, you must first register.
User Name:
Medical School
Choose "---" if you don't want to tell. AMG for US & Canadian medical schools. IMG for all other medical schools.
 AMG IMG ---
USMLE Steps History
What steps finished! Example: 1+CK+CS+3 = Passed Step 1, Step 2 CK, Step 2 CS, and Step 3.

Choose "---" if you don't want to tell.

 Not yet Step 1 Only CK Only CS Only 1 + CK 1 + CS 1+CK+CS CK+CS 1+CK+CS+3 ---
Favorite USMLE Books
 What USMLE books you really think are useful. Leave blank if you don't want to tell.
Location
 Where you live. Leave blank if you don't want to tell.

Log-in

Human Verification

In order to verify that you are a human and not a spam bot, please enter the answer into the following box below based on the instructions contained in the graphic.