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  #1  
Old 02-06-2012
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Q. The blood supply used by a local hospital for transfusions during surgical procedures is discovered to be infected with hepatitis C. To protect the blood supply from future contamination, all new donors are required to be screened for hepatitis C. The test used for this screening has a sensitivity of 95%, and a specificity of 90%. If this test is used on a sample of donors in which 10% are known to have hepatitis C, what is the chance that a donor who tests negative is actually free from the disease? .
A. About 99%
B. About 95%
C. About 90%
D. About 85%
E. About 50%
F. About 45%
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  #2  
Old 02-06-2012
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ans.............D.......
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  #3  
Old 02-06-2012
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It will be negative predictie value for the test B. 95%
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  #4  
Old 02-06-2012
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Quote:
Originally Posted by mohitkmc View Post
It will be negative predictie value for the test B. 95%
hey how did you calculate NPV ??

yes you are right its NPV , but could not get it , so thought that sensitivity is the best test to predict a person without a disease and since a few false negative will be present which i take as 5% (100 - 95 % of sensitivity ) so 90% - 5% = 85% ......

thanks ..
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Old 02-07-2012
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u r doing right in thinking
FN=1-sensitivity
1-0.95=0.05
FP=1-specifity
1-0.9=0.1
then u have 0.1 people with the dis so u left with 0.9 without dis (fp+TN) so ur TN is )0.8
now just apply NPV=TN\FN+TN
0.8\0.8+0.05
u ll get around 95%
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  #6  
Old 02-07-2012
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A is the answer
Took me 3 minutes for the calculation
Just pray questions like this dont come in the real test
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Old 02-07-2012
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Quote:
Originally Posted by jinni View Post
A is the answer
Took me 3 minutes for the calculation
Just pray questions like this dont come in the real test
how did you do it ????????
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  #8  
Old 02-07-2012
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Sensitivity =TP/TP+FN => 95/100 = 10/10+FN

so FN= 10/19
TN +FN = 90 => TN = 90 - 10/19 => TN = 1700/19
NPV = TN/TN+FN
=1700/19/90
1700/1710
=99%
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Old 02-07-2012
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[QUOTE=jinni;92799]Sensitivity =TP/TP+FN => 95/100 = 10/10+FN

so FN= 10/19
TN +FN = 90 => TN = 90 - 10/19 => TN = 1700/19
NPV = TN/TN+FN
=1700/19/90
1700/1710
=99%[/ Hi u can't assume the 0.1 is a TP value The q is saying the disease people not people who have a test positive .
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[QUOTE=Amenah;92888]
Quote:
Originally Posted by jinni View Post
Sensitivity =TP/TP+FN => 95/100 = 10/10+FN

so FN= 10/19
TN +FN = 90 => TN = 90 - 10/19 => TN = 1700/19
NPV = TN/TN+FN
=1700/19/90
1700/1710
=99%[/ Hi u can't assume the 0.1 is a TP value The q is saying the disease people not people who have a test positive .
yes Jinni its 10 % of ppl who have Hep C it contains both TP and FN (a+c) , thats correct , the answer is 95 %
the only difference i found in Amenah your and my calculation id that TN is 0.9 not 0.8 I put the value of FP = 0.1 in the equation d/b+d = specificity and got the ans as 0.9 .. still the final ans doesnot change as 0.9/0.9+.05 and 0.8/0.8 + 0.05 both equal to 95% ....
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Thanks Amenah and Hitman. You both are correct.

But i again calculated and the answer is 99%.Please correct me.
Take the sample to be 100. So

95/100=TP/10
TP = 9.5

NOW FN = 10-9.5 = .5
TN= 100-10=90
NPV = TN/TN+FN
NPV = 90/90+.5 = 90/90.5 = 99%
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  #12  
Old 02-08-2012
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The answer is A.
To answer this question you must construct a 2 x 2 table. Start with a sample of 1,000 because it is a nice, round number and, because screening tests are about ratios, it does not matter what sample size you use. The Table is then completed by computing 10% of 1,000 for the total number of diseased in the marginal (100), taking 95% of this 100 and 90% of the remaining 900 non-diseased part of the sample. With the table complete, Negative predictive value can be computed using the given numbers. The complete table and the negative predictive value calculation are presented below: Negative predictive value = 810/815 or just over 99%. Note that you should be able to estimate that without even actually completing the calculation. To complete the full set of calculations: Note: the test has excellent negative predictive value, but a positive predictive value just slightly better than chance (50%)
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  #13  
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Quote:
Originally Posted by jinni View Post
Thanks Amenah and Hitman. You both are correct.

But i again calculated and the answer is 99%.Please correct me.
Take the sample to be 100. So

95/100=TP/10
TP = 9.5

NOW FN = 10-9.5 = .5
TN= 100-10=90
NPV = TN/TN+FN
NPV = 90/90+.5 = 90/90.5 = 99%
wow ! now it looks so simple , nice one , should have taken a sample , i guess i made a mistake by taking FP = 1 - specificity , otherwise i did the same calculation upto finding FN then i found FP from the above formula and placed it in d/b+d then found d ie TN as 0.9 ..... but have made a mistake in decimal point , should have taken a sample ... thanks ..
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Old 02-08-2012
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Default

\%
Quote:
Originally Posted by jinni View Post
Thanks Amenah and Hitman. You both are correct.

But i again calculated and the answer is 99%.Please correct me.
Take the sample to be 100. So

95/100=TP/10
TP = 9.5

NOW FN = 10-9.5 = .5
TN= 100-10=90
NPV = TN/TN+FN
NPV = 90/90+.5 = 90/90.5 = 99%
dear jinni,
however the answer is 99%,but it finally comes from 81/81.5 not from 90/90.5.the problem with your calculation is TN#100-10.why?
assume there are 100 people(no matter whatever you take the sample size though).now 10 people are diseased but 90 are nondiseased,not necessarily TN.in other words,this doesnt give us TN=90,rather tells us TN+FP=90.
in the other hand,TN=9FP(can be extracted from specifity).now,we know two things about the TN&FP relationship:
a.TN=9FP
b.TN+FP=90
from these two===>TN=81 FP=9

the complete table assuming n=100 would be :
TP=9.5 FN=0.5 FP=9 TN=81
NPV=0.993
PPV=0.513
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