Example: Suppose a company plans to build a lead smelter in a community and the city council wishes to assess the health effects of the smelter. In particular, there is concern from previous literature that children living very close to the smelter will experience unusually high rates of lead poisoning in the first three years of life.Distance fromSmelterProjected Rateof LeadPoisoning% ofPopulation≤ 2km 50 / 100,000 5%> 2km but ≤5km20 / 100,000 15%> 5km 5 / 100,000 80%What is the overall probability that a child from this community will get lead poisoning?Let A = {child gets lead poisoning}B1 = {child lives ≤ 2km from smelter}B2 = {child lives > 2km but ≤ 5km from smelter}B3 = {child lives > 5km from smelter}Find Pr(A)Pr(A) = Pr(A∩B1) + Pr(A∩B2) + Pr(A∩B3)= Pr(A|B1)*Pr(B1) + Pr(A|B2) * Pr(B2) + Pr(A|B3)*Pr(B3)= [50/100,000]*0.05 + [20/100,000] * 0.15 + [5/100,000]*0.80= 0.000025 + 0.00003 + 0.00004= 0.000095Generalized Multiplication Law of ProbabilityIf A1, A2, …, Ak are a set of events,Pr(A1∩A2∩…Ak) = Pr(A1)*Pr(A2|A1) * Pr(A3|A2∩A1) * … * Pr(Ak|Ak-1∩…∩A2∩A1)Note: If A1, …, Ak are independent,Pr(A1∩A2∩…Ak) = Pr(A1) * Pr(A2) * … * Pr(Ak)Screening TestsDiseaseTest Present(B)Absent)B(TotalPositive(A)a b a + bNegative)A(c d c + dTotal a + c b + d na = number of subjects with disease and with a positive testDef: Predictive Value Positive (PV+)—probability that a person has a disease given the test is positivePV+ = Pr(B|A) = baaDef: Predictive Value Negative (PV-)—probabilitythat a person does not have a disease given thetest is negativedcd)A|BPr(PVDef: False Positive—test (A) is positive but true status is no disease )B(. This is the cell “b” in the tabledbb)B|Pr(APr(FP)Def: False Negative—test is negative but true status is disease is present. This is cell “c” in the tablecacB)|APr(Pr(FN)Def: Sensitivity—probability symptom is present given subject has a diseasecaaB)|Pr(ASENSDef: Specificity—probability symptom is not present given subject does not have diseasedbd)B|APr(SPECExample: Suppose we wish to study clinical depression and the CES-D screening test (symptom)depressionclinicalnoBDCESNormalAdepressionclinicalBDCESHighAClinical depressionCES-D Yes (B) No )B(TotalHigh (A) 90 100 190Normal )A(10 100 110Total 100 200 300FP = 100 people (cell “b”)FN = 10 people (cell “c”)0.909110100)A|BPr(PV0.47419090A)|Pr(BPV0.5200100)B|APr(SPEC0.910090B)|Pr(ASENSROC (Receiver Operating Characteristic) CurvesSeveral categories of response rather than 2 or if continuous, putting information into several categories with arbitrary cutpoints.Example: 109 computed tomography (CT) imagesclassified by a radiologist for subjects with possible neurological problems. Real disease status is known.True disease statusCT rating Abnormal Normal TotalNormal (1) 3 33 36Probably normal (2) 2 6 8Questionable (3) 2 6 8Probably abnormal (4)11 11 22Abnormal (5) 33 2 35Total 51 58 109No obvious cutpoint for saying subject is abnormal based on CT scan.Let’s compute Sensitivity and Specificity for the radiologist’s ratingsTest-positivecriteriaSensitivity Specificity1+ 1.00 0.002+ 0.94 0.573+ 0.90 0.674+ 0.86 0.785+ 0.65 0.976+ 0.00 1.000.67258633)D|2orPr(1yspecificit30.9025133112D)|Pr(3ysensitivit3We can display with a ROC curve—plot 1-Specificity on the X axis and Sensitivity on theY axis. Examine area under curve (need calculus). For the example, area = 0.89. Radiologist has an 89% probability of correctly distinguishing normal from abnormal based on CT scan. The higher the area, the better, particularly if comparing 2 or more tests.Prevalence: Probability of having a disease regardless of how long a person has had the disease. e.g. 20% of adults have high cholesterolIncidence: Probability that a person without disease develops disease in a specified time period. e.g. 0.6% of adult males ages 45-49 years will have a heart attack in the next six years.Extra ProblemThe relationship between physical fitness and cardiovascular disease mortality was recently studied in a group of railroad working men, ages 22-79. Data are presented in the table relating baseline exercise-test heart rate and coronary heart disease mortality.Exercise-testheart rate(beats/min)Coronary heart-disease mortality(20 years)(per100)% ofpopulation≤ 105 9.1 20106 – 115 8.7 30116 – 127 11.6 30> 127 13.2 40What is the overall probability of dying within 20 years from