Question

The department of health of a certain state estimates a 10% rate of HIV

The Department of Health of a ce a state estimates a 10% ate of HN for the at sk population and ล 0.3% rate for the general population. Tests for HM are 95% accurate in detect ng bath true negatives and 20,000 people frorn the general papulation resuts in the follawing table. Use the table below to complete parts (a) through (e) ue pasitives Random selection of 5000 at risk people and Test Positive Test tive Test Psitive Tes 53 987 nfected Not Infected O uimoe ine to.ar number of irue negaives Dy tne totai numoer cr pauenms. O D. Divide the total number of false positives by the total number of patients. 25 277 223 18,953 b. Consider a patient in the "at risk population. Of those with HIV, what percentage test postive? Of those who test positive, what percentage have Hiv? Explain why these two percentages are different Of the patients in the "at risk population with HIV. Type an integer or decimal rounded to the nearest tenth as needed.) % est positve. Of the patients in the "at isk population who test positive, % have HM Why are these two percentages dierent? A O B. O C. O D. The percentages are different because people who test posit we don't always have HIV and people who have HIV don't always test positive. The percentages are different because thie people are in two different categories. The percentages are different because there are people being accounted for that don't have HIV in the second calculation. The percentages are different because the first test includes everyone who tested positive. c. Suppose a patient in the "at risk" category tests positive for the disease. As a doctor using this table, how would you describe the patient's chance af actually having the disease? Compare this figure to the averall rate of the disease in the "at risk category A patient in the-at risk" category who tests positive has a □% chance of having the disease which is Type an integer or decimal raunded to the nearest tenth as needed.) d. Consider a patient in the general population. Of those with HIV, what percentage test positive? Of those who test positive, what percentage have HIV? Explain why these two peroentages are different. ▼| the overall ''at nsk' incidence rate of 10%. Of the patients in he general population with H % test positive Of the patients in the (Type an integer or decimal rounded to the nearest tenth as needed.) neral population who test positive % have HIV Why are these two percentages diferent? Ο A. The percentages are different because there are people being accounted for that don't have HIV in the second calculation. O B. The percentages are different because peaple who test positive don't always have HIV and peaple who have HIV don't always test positive O C. The percentages are different because the first test includes everyone who tested positive. O D. The percentages are different because the pecple are in two different categories. e. Suppose a patient in the general population tests positive for the disease. As a doctor using this table, haw would you describe the patient's chance of actually having the disease? Compare this figure with the overall incidence rate of the discase The chance of the patient having HIV is | %, compared to the overal incidence rato of 0.3%. Type an integer or decimal rounded to the nearest tenth as needed.)

Answer 1

at risk	test positive	test negative	total
HIV	475	25	500
HIV free	223	4277	4500
total	698	4302	5000

b) of those with HIV,postive = 475/500=   0.9500   =   95%
          
of those with test positive, HIV are = 475/698=   0.6805   or    68.10%

option a) is correct

c)

P(disease|test positive) = P(disease and positive)/P(test positve)=   =475/698=   0.6805 or 68.05%

A patient in the at risk category who test postive has a 68.1% chance of having disease which is greater than the overall at risk incidence rate of 10%  
  
d)

gen. population	test positive	test negative	total
HIV	53	7	60
HIV free	987	18953	19940
total	1040	18960	20000

of those with HIV,% of test positive = 53/60*100=   88.3%
  
of those test positive,% of having HIV=53/1040*100=   5.1%
option b) is correct          
          
e) P(disease| positive) = P(disease and positive)/P(positive) = 5.1%

the chance of patient having HIV is 5.1% ,compared to overall incidence rate of 0.3%