STAT 405 - BIOSTATISTICS
Handout 8 – Bayes Rule for Screening Tests
The information in this handout focuses on Sections 3.7-3.9 of your text.
BAYES RULE
Suppose S = B 1  B 2  B n where P(Bi) > 0 for i = 1, 2, … n and B i  B j   for i ≠ j.
Then, for any event A of S with nonzero probability,
P(B i | A) 
P(B i  A)

P(A)
P(B i )P(A | B i )
n
 P(B )P(A | B )
j
.
j
j 1
Simple version:
Example: Obstetrics (Similar to exercises 3.48 – 3.50 of your text)
The following data are derived from the 1973 Final Natality Statistics Report issued by
the National Center for Health Statistics. These data are pertinent to live births only.
Suppose that infants are classified as low birthweight if they have a birthweight ≤ 2500 g
and as normal birthweight if they have a birthweight ≥ 2501 g. Suppose that infants are
also classified by length of gestation.
Length of Gestation P(Gestation) P(Low Birthweight | Gestation)
< 20 weeks
.0004
.940
20-27 weeks
.0059
.813
28-36 weeks
.0855
.379
> 36 weeks
.9082
.035
Questions:
1. Find the probability of having a low birthweight infant.
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2. Find the probability of having a length of gestation greater than 36 weeks given
that a child is low birthweight.
BAYES RULE AND SCREENING TESTS
Bayes’ Rule is used in the medical field and epidemiology to calculate the probability that
an individual has a disease, given that they test positive on a screening test.
Example: Down syndrome is a variable combination of congenital malformations
caused by trisomy 21. It is the most commonly recognized genetic cause of mental
retardation, with an estimated prevalence of 1 case per 773 live births in the United
States. Because of the morbidity associated with Down syndrome, screening and
diagnostic testing for this condition are offered as optional components of prenatal care.
Prenatal diagnosis of trisomy 21 allows parents the choice of continuing or terminating
an affected pregnancy. Many studies have been conducted looking at the effectiveness of
screening methods used to identify “likely” Down syndrome cases. One of these tests
called “triple test” or “triple screen” is described below:
Alpha-fetoprotein (AFP), unconjugated estriol and human chorionic gonadotropin (hCG) are the serum
markers most widely used to screen for Down syndrome. This combination is known as the "triple test"
or "triple screen." AFP is produced in the yolk sac and fetal liver. Unconjugated estriol and hCG are
produced by the placenta. The maternal serum levels of each of these proteins and of steroid hormones
vary with the gestational age of the pregnancy. With trisomy 21, second-trimester maternal serum levels
of AFP and unconjugated estriol are about 25 percent lower than normal levels and maternal serum hCG
is approximately two times higher than the normal hCG level.
One study looking at the effectiveness of the “triple test” produced the following results:
Triple Test Result
Test Positive ( T  )
Test Negative ( T  )
Column Totals
Down Syndrome Status
Down Syndrome
No Down

(D )
Syndrome ( D  )
87
203
31
3869
118
4072
Row
Totals
290
3900
4190
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Define the following events:
D+ = has Down syndrome
T+ = tests positive
D- = does not have Down syndrome
T- = tests negative
How well does this study suggest the “triple test” performs? We can use the following
conditional probabilities to help answer this question.
The sensitivity is the probability that the test result is positive given that the person has
the disease, or P(T+|D+).
The specificity is the probability that the test result is negative given that the person
does not have the disease, or P(T-|D-).
Questions:
Triple Test Result
Test Positive ( T  )
Test Negative ( T  )
Column Totals
Down Syndrome Status
Down Syndrome
No Down

(D )
Syndrome ( D  )
87
203
31
3869
118
4072
Row
Totals
290
3900
4190
1. Find the sensitivity of the triple test.
2. Find the specificity of the triple test.
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Now, suppose an expectant parent has just been given the results of the “triple test” and
that they are positive for Down syndrome. Wouldn’t they be interested in the probability
that their unborn child actually has Down syndrome? To answer a question such as this,
we must find the following quantities:
The positive predictive value of a screening test is the probability that a person has
the disease given that the test is positive, or P(D+|T+).
The negative predictive value of a screening test is the probability that a person does
not have the disease given that the test is negative, or P(D-|T-).
Questions:
1. Find a formula for the positive predictive value of the triple test.
2. Find a formula for the negative predictive value of the triple test.
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Bayes’ rule requires that we have prior knowledge about the likelihood of having a child
with Down syndrome. This information is readily available from the U.S. from Centers
for Disease Control (http://www.msnbc.msn.com/id/10725449/).

About 1 in 2,000 fetuses carried by 20 year-old women are afflicted with Down
syndrome.

About 1 in 100 fetuses carried by 40 year-old women are afflicted with Down
syndrome.
Questions:
1. Find the positive predictive value of the triple test for 20 year-old women.
2. Find the negative predictive value of the triple test for 20 year-old women.
3. Find the positive predictive value of the triple test for 40 year-old women.
4. Find the negative predictive value of the triple test for 40 year-old women.
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Receiver Operating Characteristic (ROC) Curves
In situations where there the cutoff for a positive test result is controlled by a
continuous/ordinal variable, we can control the performance of screening test in terms
of sensitivity and specificity by moving the cutoff value. One way to investigate the
performance of a test is to examine a Receiver Operating Characteristic (ROC) Curve.
An ROC curve is a plot of the sensitivity of the test vs. (1 – specificity).
It can be shown that the area beneath an ROC curve represents the probability that when
given a pair of “normal” and “abnormal” patients, we correctly diagnose which is which
by simply comparing their test scores.
For example, if a high value indicates abnormality, then whichever subject has the higher
test score would be classified as the abnormal one. In the case of tie, it is assumed that a
coin is flipped to determine which is classified as abnormal.
In general, of two screening tests for the same disease, the test with the higher area
under the ROC curve is considered to be the better test, unless some particular level of
sensitivity or specificity is especially important in comparing the two tests.
Sketch the ROC curves for both the best and worst tests possible:
Best test possible
Worst test possible
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Example: HIV ELISA Tests
The enzyme-linked immuosorbent assay (ELISA) test was the main test used to screen
blood samples for antibodies to the HIV virus (rather than the virus itself) in 1985. It
gives a measured mean absorbance ratio for HIV (previously called HTLV) antibodies.
The table below gives the absorbance ratio values for 297 healthy blood donors and 88
HIV patients. Healthy donors tend to give low ratios, but some are quite high, partly
because the test also responds to some other types of antibody, such a human leucocyte
antigen or HLA. HIV patients tend to give high ratios, but a few give lower values
because they have not been able to mount a strong immune reaction.
To test this in practice we need a cutoff value so that those who fall below the value are
deemed to have tested negatively and those above to have tested positively. Any such
cutoff will naturally involve misclassifying some people without HIV as having a positive
HIV test (which will be a huge emotional shock), and some people with HIV as having a
negative HIV test (with consequences to their own health, the health of people around
them, and the integrity of the blood bank, etc.).
MAR (mean absorb. ratio)
<2
2 – 2.99
3 – 3.99
4 – 4.99
5 – 5.99
6 – 11.99
12+
Total
Healthy Donor
202
73
15
3
2
2
0
297
HIV Patients
0
2
7
7
15
36
21
88
Questions:
1. Compute the sensitivity and the specificity if an MAR < 2 leads to a negative test.
Test Result
Test Positive Test Negative
(D )
( D )
Row
Totals
Test Positive ( T  )
Test Negative ( T  )
Column Totals
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2. Fill in the following table:
MAR cutoff Sensitivity Specificity (1-Specificity)
0
<2
<3
<4
<5
<6
< 12
12 +
3. Sketch the ROC curve and compute the area beneath it.
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