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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

Conditional Probability and Screening Tests

Clinical topics:

Mammography and Pap smear

(screening for cancer)

what factors determine the effectiveness of screening
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.
cumulative risk of disease

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

(from your course reader)

to assess your cumulative risk
To assess your cumulative risk:

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

mammography
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.
mammography6
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
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 smear9
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
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
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
Key Concepts

-Independence

-Conditional probability

-Sensitivity

-Specificity

-Positive Predictive Value

independence
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
using a probability tree

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…
independence15
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
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 117
Thought Question 1
  • A 54-year old woman has an abnormal mammogram; what is the chance that she has breast cancer?
hypothetical example mammography

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 119
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 220
Thought Question 2
  • A 35-year old woman has an abnormal pap smear; what is the chance that she has HSIL or cervical cancer?
hypothetical example pap test

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 222
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

Part II:

Epidemiology is the study of patterns of diseases in populations.

assumptions and aims of epidemiologic studies
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
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
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
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”

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 studies30
Case-Control Studies

Disease

(Cases)

Exposed in past

Not exposed

Target population

Exposed

No Disease

(Controls)

Not Exposed

case control studies in history
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
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).