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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Fri Jun 23, 2006 11:37 pm Post subject:
HIV antibody testing



Hello, sci.math readers.
I have added the following to the Wikipedia and would like to work out
P(AB) = P(A*B)/P(B)
for the SparkNotes example. I use * to represent intersection above.
I am responsible for the second, fourth, and final paragraphs.
This comes from the section on debunking HIV myths.

==[[HIV testHIV antibody test]]ing is unreliable==
[[Diagnosis]] of [[infection]] using [[antibody]] testing is one of the
bestestablished concepts in [[medicine]]. HIV antibody tests exceed
the performance of most other infectious disease tests in both
sensitivity (the ability of the screening test to give a positive
finding when the person tested truly has the disease ) and specificity
(the ability of the test to give a negative finding when the subjects
tested are free of the disease under study). All current HIV antibody
tests have sensitivity and specificity in excess of 96% (except the
HIVTEK G by Sorin Biomedica) and are therefore extremely reliable
([http://www.who.int/diagnostics_laboratory/publications/en/HIV_Report15.pdf,
WHO, 2004]).
Having the antibodies is different from actually having the virus
because HIV antibodies are formed from exposure to HIV test vaccines
which stimulate the body's response to HIV infection but do not fully
simulate HIV.
Progress in testing methodology has enabled detection of viral genetic
material, antigens and the virus itself in body fluids and cells. While
not widely used for routine testing due to high cost and requirements
in laboratory equipment, these direct testing techniques have confirmed
the validity of the antibody tests (Jackson et al., 1990; Busch et al.,
1991; Silvester et al., 1995; Urassa et al., 1999; Nkengasong et al.,
1999; Samdal et al., 1996).
Bayesian statistics can be used to argue that HIV or other testing is
less reliable than it might seem to be. If HIV antibody testing is 5%
false negative and 5% false positive, and HIV antibodies are present in
0.6% of all people, then Baye's rule of inverse probability can be used
to show that if a person selected at random tests positive, the
probability that they actually have the antibodies is only 10%
(SparkNotesLLC, SparkCharts(tm) series, "Statistics", Barnes and Noble,
2002).
Positive HIV antibody tests are usually followed up by retests and
tests for antigens, viral genetic material and the virus itself,
providing confirmation of actual infection.

Doug Goncz
Replikon Research
Seven Corners, VA 220440394 

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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Sat Jun 24, 2006 11:17 am Post subject:
Re: HIV antibody testing



Doug Goncz (I) wrote:
Quote:  Hello, sci.math readers.
I have added the following to the Wikipedia and would like to work out
P(AB) = P(A*B)/P(B)
for the SparkNotes example. I use * to represent intersection above.
(snip)
Bayesian statistics can be used to argue that HIV or other testing is
less reliable than it might seem to be. If HIV antibody testing is 5%
false negative and 5% false positive, and HIV antibodies are present in
0.6% of all people, then Baye's rule of inverse probability can be used
to show that if a person selected at random tests positive, the
probability that they actually have the antibodies is only 10%
(SparkNotesLLC, SparkCharts(tm) series, "Statistics", Barnes and Noble,
2002).

"if a person selected at random tests postive..."
Hm.
B1= Postive Test
A1 = Has Antibodies
Do mutually exclusive and exhaustive probabilities sum to one?
A*B B1 B2
A1 ,95 .05 1
A2 .05 .95 1
1 1
Like that?
P(A1B1) = P(A1*B1)/P(B1) or
P(AB) = P(A*B) / ( P(AB1)*P(B1) + P(AB2)*P(B2) ) or what?
Or like this?
A*B B1 B2
A1 .95x .05y 0.006
A2 .05x .95y 0.994 (Ivory Soap?)
x y 1
Looks a hell of a lot like SuDoKu to me.
Linear equations (too many of 'em?):
1: 0.95x + 0.05y = 0.006
2: 0.05x + 0.95y = 0.994
3: x + y = 1 (oh, I see, this is 1: + 2:)
That's solvable.
2*19:
3: 0.95x + 18.05y = 17.892
31:
4: 0.00x + 18y = 17.886
solve 4:
5: y = 17.886/18 = 0.99367
solve 3:
6: x = 0.00633
coefficients:
0.95 x = 0.006
I still don't see 0.01, but there is some hope. We were talking P(A*B).
P(A1B1) = P(A1*B1)/P(B1) or
P(A1B1) = 0.006 / 0.0633...
P(A1B1) = 0.0950
Darn close to 10%.
Coefficients:
0.05x = 0.0003165
0.95y = 0.9439865
0.05y = 0.0496835
Matrix:
A*B B1 B2
A1 .0950 .0497
A2 .0003 .9439
Probability of a testpostive person with the antibodies: about 9.5%
Probability of a testnegative person with no antibodies: about 4.9%
Probability of a testnegative person with no antibodies: about 94%
Probability of a testpostive person with the antibodies: about 0.03%,
which makes sense, since antibodies are rare and testing is rare.
Or what? What's the conclusion?
Wikipedians want to know.
Doug 

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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Sat Jun 24, 2006 3:58 pm Post subject:
Re: HIV antibody testing



Doug Goncz (I) wrote:
Quote:  A*B B1 B2
A1 .0950 .0497
A2 .0003 .9439
Probability of a testpostive person with the antibodies: about 9.5%
Probability of a testnegative person with no antibodies: about 4.9%
Probability of a testnegative person with no antibodies: about 94%
Probability of a testpostive person with the antibodies: about 0.03%,
which makes sense, since antibodies are rare and testing is rare.
Or what? What's the conclusion?
Wikipedians want to know.

No, I think that table should be labled AB, and you'd need separate
table for BA, right?
And the text is wrong.
Probability of a person having antibodies, given a positive test: about
9.5%.
Right?
Doug 

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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Sun Jun 25, 2006 4:18 pm Post subject:
Re: HIV antibody testing



Doug Goncz (I) wrote:
Quote:  Doug Goncz (I) wrote:
A*B B1 B2
A1 .0950 .0497
A2 .0003 .9439
Probability of a testpostive person with the antibodies: about 9.5%
Probability of a testnegative person with no antibodies: about 4.9%
Probability of a testnegative person with no antibodies: about 94%
Probability of a testpostive person with the antibodies: about 0.03%,
which makes sense, since antibodies are rare and testing is rare.
Or what? What's the conclusion?
Wikipedians want to know.
No, I think that table should be labled AB, and you'd need separate
table for BA, right?
And the text is wrong.
Probability of a person having antibodies, given a positive test: about
9.5%.
Right?
Doug
Not all right. The table is P(AB), the probability of A given B. 
One more table is needed: P(BA).
P(BA) (B test, A antibodies)
this is the given data, which affects the table with 0.95x in the
corner.
P(BA) B1 B2
A1 0.95 0.05 1
A2 0.05 0.95 1
1 1 z
I doubt that z=2. 0<=P(anything)<=1
I am not sure z has meaning. It is clear to me that x and y are P(B1)
and P(B2) in the first table. The problem is to find P(B1) and P(B2)
such that P(A1*B1) + P(A1*B2) = .006 and P(A2*B1) + P(A2*B2) = .994,
that is, such that P(A*B)=1
P(AB) B1 B2
A1 .0950 .0497 .1447
A2 .0003 .9439 .9442
.0953 .9936 w
Hm. Should these rows and colums sum to 1?
There is some matrix math to be learned in there.
Mathcad gives different values for x and y. x = 0.049; y = 1.049. But
then these can't be P(B1) and P(B2). That makes a certain amount of
sense, as nowhere in the problem is the incidence of testing mentioned.
But if P(A*B) = 1, then x + y = 1. Hm.
What is quoted in the SparkNote is P(A1B1) = 10%.
Doug 

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DGoncz@aol.com science forum Guru Wannabe
Joined: 25 Oct 2005
Posts: 122

Posted: Wed Jun 28, 2006 11:49 pm Post subject:
Re: HIV antibody testing



OK, I think I understand a little more Bayesian probability now,
and how to get Google to use fixedpitch in a post.
What a mess the previous posts are!
You start with (exhaustive?) probabilities (rows and columns sum to
one)
A: Antibodies
T: Test
We are given the conditinal probability P(TA),
the selectivity and specificity of the test,
given assumed antibody status; 95% each.
P(TA) T+ T Sum
A+ 0.95 0.05 1
A 0.05 0.95 1
1 1 n/a
Why n/a rather than 2?
There might be 3 cases in A and 2 in T. That sum doesn't matter in this
matrix.
We are given the marginial probability P(A+) = 0.006,
and convert to joint probabilities
(row and column sums sum to one).
Sumof P(A&T) = 1, now and forever.
P(T*A) = P(A*T) = P(TA) * P(A)
P(A&T) T+ T Sum
A+ 0.0057 0.0003 0.006
A 0.0497 0.9443 0.994
0.0554 0.9446 1
We compute the desired P(AT):
P(A+T+) = P(A+&T+) / P(T+)
= 0.0057 / 0.0554
~= 0.10 (~0.10288..)
And we now know what I think is called the power of the test, beta:
P(AT) T+ T Sum
A+ 0.10 0.0003 n/a
A 0.897 0.9996 n/a
n/a n/a n/a
So if a person at random (this could mean you) has a negative HIV
antibody test, the probability of being HIV negative is 0.9996, etc.,
given the assumptions.
Doug Goncz
Replikon Research
Seven Corners, VA 220440394 

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