Problem #2 A recent study to investigate congestive heart failure (CHF) enrolled

Question

Problem #2 A recent study to investigate congestive heart failure (CHF) enrolled 259 patients. Among other variables collected, the researchers checked if the patient had a Gallop (S3) murmur. The researchers are interested to see if the presence of a murmur is an indication of CHF. That is, consider the murmur to be the "test" (presence of the murmur is a positive test; no murmur is a negative test) for the detection of CHF, the "disease". Here is the data: No CHF CHF 10 Gallop (S3) murmur No gallop murmur 215 STAT:3510 -Homework 4 Problem #2 (continued a) What would be considered a positive "test" result in this situation'? b) What is the sensitivity of using a gallop murmur indication for CHF? c) What is the specificity of using a gallop murmur indication for CHF? d) What is the probability of a false negative using gallop murmur as an indication for CHF? e) What is the probability of a false positive using gallop murmur as an indication for CHF? f) What is the probability that a person exhibiting a gallop murmur has CHF?

Explanation / Answer

Ans:

a)Probability of having Gallop murmur will be a positive test.

P(gallop murmur)=P(positive)=P(gallop murmur/CHF)*P(CHF)+P(gallop murmur/No CHF)*P(No CHF)

b)

Sensitivity=P(positive/disease)=P(Gallop murmur/CHF)

=P(Gallop murmur and CHF)/P(CHF)=(10/259)/(41/259)=10/41=0.2439

c)P(specificity)=P(negative/No disease)=P(no gallop mumur/no CHF)=215/218=0.9862

d)P(fasle negative)=P(no gallop murmur/CHF)=P(no Gallop mumur and CHF)/P(CHF)=31/41=0.7561

e)P(false positive)=P(gallop murmur/No CHF)=P(gallop murmur and No CHF)/P(No CHF)=3/218=0.0138

f)P(CHF/gallop murmur)=P(gallop murmur and CHF)/P(gallop murmur)

=10/13=0.7692

CHF No CHF Total Gallop murmur 10 3 13 No gallop murmur 31 215 246 Total 41 218 259 Joint probability distribution CHF No CHF Total Gallop murmur 0.0386 0.0116 0.0502 No gallop murmur 0.1197 0.8301 0.9498 Total 0.1583 0.8417 1.0000

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