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  #1  
Old 08-24-2014
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Default Biostat question for the day

A test has just become available to detect a new strain of hepatitis virus. In early trials 1000 people were enrolled to undergo testing. Four hundred participants had a positive initial test, but confirmatory testing showed that only 200 people were truly infected. The remaining participants were negative by both experimental testing and confirmatory testing.
What is the specificity of this new test?

40%
50%
70%
75%
80%
90%
100%
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Old 08-24-2014
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TP= 200

FP= 200

TN= 800

Specificity= TN/FP+TN

800/200+800=80%
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Old 08-24-2014
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OK so the FN is Zero.

Also, for the new test TN is 600.

So, Specificity is 600/200+600=75%
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Old 08-24-2014
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I think it ll be 80%
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Quote:
Originally Posted by Dr.ann View Post
I think it ll be 80%
How did you reach that conclusion ?
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Old 08-24-2014
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Well i m not sure but
If total 1000 subject
400 r positive, out of that 200 TP. AND 200 FP
1000 minus 200 = 800 are TN

SO SPECIFICITY TN/TN+FP
800/800+200
800/1000= 0.8, 80%
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Old 08-24-2014
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You guys are doing very well
More answers please. Then I give final answer!
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Old 08-24-2014
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Quote:
Originally Posted by Novobiocin View Post
OK so the FN is Zero.

Also, for the new test TN is 600.

So, Specificity is 600/200+600=75%
I calculated by your way too but how did u assume FN is zero??
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Old 08-24-2014
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Well now i m confuse between both methods
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Old 08-24-2014
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600 were negative and also did not have disease. So, TN = 600

Out of the 400 who were positive, 200 were falsely positive. So, FP = 200

Specificity is out of all the disease free how many got a negative result or were truly negative.

All disease free = 800, True negative = 600

Specificity = 600/800 = 3/4 = 75%

Thanks for the question.
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Quote:
Originally Posted by Dr.ann View Post
I calculated by your way too but how did u assume FN is zero??
There is no need to calculate FN as it is not needed for the calculations.
Specificity by definition does not take into account false negatives.

Think of it this way to understand the concept:

Specificity (true negative rate) is defined as the proportion of negatives which truly does not have the disease (e.g. the percentage of healthy people who are correctly identified as not having the disease).
The last sentence is telling you that the remaining (1000-400=600) were negative by both the new test as well as the confirmatory test. This means that the TN has to be 600.
Also, since the total of all four has to be 1000 it leaves FN as Zero.
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  #12  
Old 08-24-2014
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Tp=200

fp=200

tn=1000-400=600

specificity=tn/tn+fp
=600/600+200
=600/800
=75%
plz tell the correct answer.!
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Old 08-25-2014
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Well, guys answer is 75% already explained by earlier posters!
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Old 09-30-2014
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600\800 = 75% is the correct answer
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Old 09-30-2014
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200. 200
0. 600



So specificity is 600/600+200x100 is 75%
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Old 10-03-2014
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By 2x2 table

Disease + -
Test result + 200(a) 200(b)
- 0 (c) 400(d)
As there were no individuals with negative results in screening test who had positive confirmatory test result so c=0

Specifiity=d/c+d 400/200+400=.75 or 75%
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TN=1000-400=600
FP=200
>specificity=TN/TN+FP
>600/600+200=0.75=75%
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