Conditional Probability and Screening Tests

1. Conditional Probability and Screening Tests Clinical topics: Mammography and Pap smear (screening for cancer)

2. 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.

3. 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)

4. To assess your cumulative risk: http://bcra.nci.nih.gov/brc/

5. 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.

6. 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.

7. 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.

8. Progression to cervical cancer

9. 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

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

11. 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?

12. Key Concepts -Independence -Conditional probability -Sensitivity -Specificity -Positive Predictive Value

13. 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

14. 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…

15. 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♀)

16. 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!

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

18. 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%

19. 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.

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

21. 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%

22. 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%!

23. Part II: Epidemiology is the study of patterns of diseases in populations.

24. 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.

25. 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.”

26. 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

27. 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”

28. Introduction to Case-Control studies

29. 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

30. Case-Control Studies Disease (Cases) Exposed in past Not exposed Target population Exposed No Disease (Controls) Not Exposed

31. 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.

32. 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).