16. True or Fase Group A and Group D are mutually exchnive 17. True or False Gro
ID: 3057473 • Letter: 1
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16. True or Fase Group A and Group D are mutually exchnive 17. True or False Group C and Group D are mtualy eschaive a. 8 and D occur simutaneously-(99.05 emperature 5 101.53 b. Band D occur umaaneoaly . 198 5 temperature s 101 c Bor D or both occur) 990 s temperature s 101 S) d. In or o oro" or both ocor)·pa s s temperature s 101.9) Suppose th, probabilty of chid with chicken pox having a temperaturen Groep 8 h 25% and the probability of child wth chicken pox having temperature Group C is 10 5% 19. what., the probability of a chid having a temperature in Group D or Group a. 0675 0.25 d 0.32s a. 0.895 0.25 20. What is the probability of a child having a temperature not in Group C? d 075 A study relating smoking histony to several measures of cardiopulmonary disablity was recenty reported. The data below were presented relating the number of people with different disabilities according to cigarette-smoking status Ogarette-Smoking Status Non-Smoker 512 Ex-Smoker 611 15 319 13 15 1857 Shortness of Breath 10 13 Possible Infarction Total 650 350 21. What is the prevalence of ex-smokers in the sample? a. 0.256 b. 0.325 0.3055 d. 0.27 22. What is the prevalence of shortness of breath in the sample? a. 0.256 b. 0.9285 0.9481 d. 0.0275 23. What is the probability of a current smoker that smokesExplanation / Answer
For problems 16-20, important information regarding the groups A, B, C, and D is required. Please ask again with all the relevant information. Solving from problem 21 onwards.
21. Answer - b: The prevalance of ex-smokers in the sample is 650/2000 = 0.325.
22. Answer - d: The prevalance of shortness of breath in the sample is 55/2000 = 0.0275.
23. Answer - a: The probability of a current smoker that smokes < 15 g/day with a possible infraction is 5/2000 = 0.0025.
24. Answer - b: The probability of a non-smoker with angina = 12/2000 =0.006.
25. Answer - d: The probability of no disability given a current smoker who smokes > 15 g/day = 319/350 = 0.9114. This is a conditional probability to be found by looking at the values in the column of a current smoker > 15 g/day.
26. Answer - True: The probability of no disability and smoking > 15 g/day are dependent events. We can show this by calculating the conditional and unconditional probabilities and showing that they are not equal.
P(no disability) = 1857/2000 = 0.9285
P(no disability | > 15 g/day) = 319/350 = 0.9114
As the two values are not equal, the two events are dependent.
27. Answer - d: The prevalence of no CAD in this cohort is 95/125 = 0.76.
28. Answer - b: The predictive value positive of the exercise test (the probability of actually having a CAD given that the test was positive) is 24/32 = 0.75. This is a conditional probability on the exercise test positive row.
29. Answer - c: The predictive value negative of the exercise test (the probability of actually not having a CAD given that the test was negative) is 87/93 = 0.9355. This is a conditional probability on the exercise test negative row.
30. Answer - a: The sensitivity of the exercise test (the probability of a CAD condition correctly detected so by the test) is 24/30 = 0.8. This is a conditional probability on the CAD column,
31. Answer - d: The specificity of the exercise test (the probability of a non-CAD condition correctly detected so by the test) is 87/95 = 0.9158. This is a conditional probability on the non-CAD column,
32. Answer - True: The exercise test is not an acceptable alternative in diagnosing CAD.
33. Answer - We expect the new predictive value negative to improve there will be fewer cases of false negatives. However, the exact value will depend on how the distribution changes because of the lower prevalence.
Remaining questions not solved due to lack of time. Please ask again. Thanks.
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