Transcript Angio_talk - Stanford University

Conditional Probability
and Screening Tests
Clinical topics:
Mammography and Pap smear
(screening for cancer)
What factors determine the
effectiveness of screening?

The prevalence (risk) of disease.
 The effectiveness of screening in preventing
illness or death.





Is the test any good at detecting disease/precursor
(sensitivity of the test)?
Is the test detecting a clinically relevant condition?
Is there anything we can do if disease (or predisease) is detected (cures, treatments)?
Does detecting and treating disease at an earlier
stage really result in a better outcome?
The risks of screening, such as false positives
and radiation.
2
Cumulative risk of disease
Her chance of
dying from any
Given an 80-year
cause is
old woman, her
670/1000=67%
chance of getting
breast cancer in
her next decade
of life is:
11/1000=.011
Or 1.1%
(from your course reader)
3
To assess your cumulative risk:
http://bcra.nci.nih.gov/brc/
4
Mammography

Mammography utilizes ionizing radiation to image breast
tissue.
 The examination is performed by compressing the breast
firmly between a plastic plate and an x-ray cassette that
contains special x-ray film.
 Mammography can identify breast cancers too small to
detect on physical examination, and can also find ductal
carcinoma in situ (DCIS), a noninvasive condition that
may or may not progress to cancer.
 Early detection and treatment of breast cancer (before
metastasis) can improve a woman’s chances of survival.
 Studies show that, among 50-69 year-old women,
screening results in 20-35% reductions in mortality from
breast cancer.
5
Mammography

Controversy exists over the efficacy of
mammography in reducing mortality from breast
cancer in 40-49 year old women.
 Mammography has a high rate of false positive
tests that cause anxiety and necessitate further
costly diagnostic procedures.
 Mammography exposes a woman to some
radiation, which may slightly increase the risk of
mutations in breast tissue.
6
Pap Smear




In a pap test, a sample of cells from a woman’s cervix is
collected and spread on a microscope slide. The cells
are examined under a microscope in order to look for
pre-malignant (before-cancer) or malignant (cancer)
changes.
Cellular changes are almost always a result of infection
with high-risk strains of human papillomavirus (HPV), a
common sexually transmitted infection.
Most cellular abnormalities are not cancer and will
regress spontaneously, but a small percent will progress
to cancer.
It takes an average of 10 years for HPV infection to lead
to invasive cancer; early detection and removal of precancerous lesions prevents cancer from ever
developing.
7
Progression to cervical cancer
8
Pap smear



Widespread use of Pap test in the U.S. has been linked
to dramatic reductions in the country’s incidence of
cervical cancer.
However, cervical cancer is the leading cause of death
from cancer among women in developing countries,
because of lack of access to screening tests and
treatment.
However, as with mammography, false positives,
detection of abnormalities of unknown significance, or
detection of low-grade lesions that are not likely to
progress to cancer can cause anxiety and unnecessary
follow-up tests
9
Thought Question 1
 A 54-year
old woman has an abnormal
mammogram; what is the chance that she
has breast cancer?
Guesses?
10
Thought Question 2
 A 35-year
old woman has an abnormal
pap smear; what is the chance that she
has HSIL or cervical cancer?
Guesses?
11
Key Concepts
-Independence
-Conditional probability
-Sensitivity
-Specificity
-Positive Predictive Value
12
Independence
 Example:
if a mother and father both carry
one copy of a recessive disease-causing
mutation (d), there are three possible
outcomes for the child (the sample space):
 P(genotype=DD)=.25
 P(genotype=Dd)=.50
 P(genotype=dd)=.25
13
Using a probability tree…
Mother’s allele
Father’s allele
Child’s outcome
P(DD)=.5*.5=.25
P(♀D=.5)
P(♂D=.5)
P(♀d=.5)
P(Dd)=.5*.5=.25
P(♀D=.5)
P(♂d=.5)
P(♀d=.5)
P(dD)=.5*.5=.25
P(dd)=.5*.5=.25
______________
1.0
14
Independence
Formal definition: A and B are independent iff P(A&B)=P(A)*P(B)
The mother’s and father’s alleles are segregating independently.
P(♂D/♀D)=.5 and P(♂D/♀d)=.5
Note the “conditional probability”
Father’s gamete does not depend on the mother’s—does not depend on
which branch you start on.
Formally, P(DD)=.25=P(D♂)*P(D♀)
15
Characteristics of a diagnostic test
Sensitivity= Probability that, if you truly have
the disease, the diagnostic test will catch
it. P(test+/D)
Specificity=Probability that, if you truly do
not have the disease, the test will register
negative. P(test-/~D)
Note the conditional probabilities!
P(test+/~D)P(test+/D)); not
independent!
16
Thought Question 1
 A 54-year
old woman has an abnormal
mammogram; what is the chance that she
has breast cancer?
17
Hypothetical example:
Mammography
sensitivity
P(test +)=.90
P(BC+)=.003
P (+, test +)=.0027
True positives
P(test -) = .10
P(+, test -)=.0003
P(test +) = .11
P(-, test +)=.10967
P(BC-)=.997
False
positives
P(test -) = .89
P(-, test -) = .88733
______________
1.0
specificity
Marginal probabilities of breast cancer….(prevalence
among all 54-year olds)
P(BC/test+)=.0027/(.0027+.10967)=2.4%
18
Thought Question 1
The probability that she has cancer given
that she tested “abnormal” is just 2.4%!
This is called the “positive predictive value,”
or PPV.
The PPV depends on both the
characteristics of the test (sensitivity,
specificity) and the prevalence of the
disease.
19
Thought Question 2
 A 35-year
old woman has an abnormal
pap smear; what is the chance that she
has HSIL or cervical cancer?
20
Hypothetical example: Pap Test
sensitivity
P(test +)=.88
P(BC+)=.005
P (+, test +)=.0044
True positives
P(test -) = .12
P(+, test -)=.0006
P(test +) = .11
P(-, test +)=.10945
P(BC-)=.995
False
positives
P(test -) = .89
P(-, test -) = .88555
______________
1.0
specificity
Marginal probability of cervical
cancer/HSIL….(prevalence
among all 35-year olds)
P(CC or HSIL/test+)=.0044/(.0044+.10945)=3.8%
21
Thought Question 2
 A 35-year
old woman has an abnormal
pap smear; what is the chance that she
has HSIL or cervical cancer?
 Just
3.8%!
22
Part II:
Epidemiology is the study of
patterns of diseases in
populations.
Assumptions and aims of
epidemiologic studies
 1)
Disease does not occur at random but
is related to environmental and/or personal
characteristics.
 2) Causal and preventive factors for
disease can be identified.
 3) Knowledge of these factors can then be
used to improve health of populations.
Correlation studies
 Using
differences in the rate of diseases
between populations to gather clues as to
the cause of disease is called a
“correlation study” or an “ecologic study.”
25
Example: Patterns of disease
and cervical cancer
Initial examination of the patterns of cervical
cancer gave strong etiologic clues:
-high rates among prostitutes
-absence of cases among nuns
-higher rates among married and highest
rates among widowed women
 Suggested sexually transmitted cause
26
Problems with correlation
studies
 Hypothesis-generating,
not hypothesis-
testing
 Ecologic fallacy: Cannot infer causation;
association may not exist at the individual
level.
- Making observations of risk factor and
diseases status on individual subjects is
called “analytic epidemiology”
27
Introduction to CaseControl studies
Case-Control Studies
Sample on disease status and ask
retrospectively about exposures (for
rare diseases)

Marginal probabilities of exposure for cases
and controls are valid.
• Doesn’t require knowledge of the absolute risks
of disease
• For rare diseases, can approximate relative risk
Case-Control Studies
Disease
Exposed in
past
(Cases)
Not exposed
Target
population
Exposed
No Disease
(Controls)
Not Exposed
Case-Control Studies in History

In 1843, Guy compared occupations of men with
pulmonary consumption to those of men with
other diseases (Lilienfeld and Lilienfeld 1979).
 Case-control studies identified associations
between lip cancer and pipe smoking (Broders
1920), breast cancer and reproductive history
(Lane-Claypon 1926) and between oral cancer
and pipe smoking (Lombard and Doering 1928).
All rare diseases.
 Case-control studies identified an association
between smoking and lung cancer in the 1950’s.
 You read about two historical case-control
studies for homework.
Frequency Distributions
 Refer
to figure 4, p. 144 of “Bimodal age
distributions of mammary cancer.”
 Histograms, or frequency distributions, plot
the frequency of disease according to
categories of a predictor (such as age).
32