Conditional Probability and Screening Tests

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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.

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### 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 pre-disease) 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.

Her chance of dying from any cause is 670/1000=67%

Given an 80-year old woman, her chance of getting breast cancer in her next decade of life is: 11/1000=.011

Or 1.1%

Cumulative risk of disease

To assess your cumulative risk:

http://bcra.nci.nih.gov/brc/

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.
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.
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 pre-cancerous lesions prevents cancer from ever developing.
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
Thought Question 1
• A 54-year old woman has an abnormal mammogram; what is the chance that she has breast cancer?

Guesses?

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?

Key Concepts

-Independence

-Conditional probability

-Sensitivity

-Specificity

-Positive Predictive Value

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

Child’s outcome

Father’s allele

Mother’s allele

P(DD)=.5*.5=.25

P(♀D=.5)

P(♂D=.5)

P(♀d=.5)

P(Dd)=.5*.5=.25

P(♀D=.5)

P(dD)=.5*.5=.25

P(♂d=.5)

P(dd)=.5*.5=.25

______________

1.0

P(♀d=.5)

Using a probability tree…
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♀)

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!

Thought Question 1
• A 54-year old woman has an abnormal mammogram; what is the chance that she has breast cancer?

sensitivity

True positives

P (+, test +)=.0027

P(test +)=.90

P(BC+)=.003

P(test -) = .10

P(+, test -)=.0003

False positives

P(test +) = .11

P(-, test +)=.10967

P(BC-)=.997

P(test -) = .89

P(-, test -) = .88733

______________

1.0

specificity

Marginal probabilities of breast cancer….(prevalence among all 54-year olds)

Hypothetical example: Mammography

P(BC/test+)=.0027/(.0027+.10967)=2.4%

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.

Thought Question 2
• A 35-year old woman has an abnormal pap smear; what is the chance that she has HSIL or cervical cancer?

sensitivity

True positives

P (+, test +)=.0044

P(test +)=.88

P(BC+)=.005

P(test -) = .12

P(+, test -)=.0006

False positives

P(test +) = .11

P(-, test +)=.10945

P(BC-)=.995

P(test -) = .89

P(-, test -) = .88555

______________

1.0

specificity

Marginal probability of cervical cancer/HSIL….(prevalence among all 35-year olds)

Hypothetical example: Pap Test

P(CC or HSIL/test+)=.0044/(.0044+.10945)=3.8%

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%!

### 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.”
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

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”

### Introduction to Case-Control 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

(Cases)

Exposed in past

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).