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Old 08-05-2012
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Stats Calculate the Likelihood Ratio

The table below summarizes the results of a new test to identify patients with an acute myocardial infarction (AMI).
The test was performed on 400 subjects.

Acute Myocardial Infarction...... Present............ Absent................. Total


Test Result
...................Positive................ 120.................. 30....................... 150


..................Negative................. 80.................... 170 ................... 250




.......................Total............... 200..................... 200..................... 400



If one of the 400 subjects in the study is randomly chosen, what is the likelihood that an abnormal test indicates that the individual has an AMI?


A. 0.50

B. 0.60

C. 0.68

D. 0.80

E. 0.85
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Positive predictive value....C...0.68
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Explanation:

The correct answer is D. The likelihood of an abnormal test indicating disease, or the positive predictive value of
a test (PV+), is expressed with the following formula:

PV+ = TP/(TP + FP)

where a true positive (TP) is an abnormal test result in an individual with disease, while a false positive (FP) is an
abnormal test result in a healthy person. Note that as the FP rate decreases, the PV+ approaches 100%. The FP
rate reflects the test's specificity, or the chance that a test is normal in a healthy individual. As the FP rate of a
test decreases, the test's specificity approaches 100%. Therefore, a test with 100% specificity always has a PV+
of 100%. This underscores why tests with high specificity are used to confirm disease, since an abnormal test is
more likely to represent a TP rather than a FP.

From the table we can see the following: TP = 120, FN = 80, TN = 170, and FP = 30. The likelihood of an
abnormal test indicating an AMI is therefore:

PV+ = 120/(120 + 30) = 0.80

Thus there is an 80% chance that an abnormal test is a TP and a 20% chance that it is a FP.

Choice A is the prevalence of AMI in this study. Prevalence refers to the number of subjects with disease (TP +
FN) divided by the total number of subjects in the study. Hence, the prevalence of an AMI in this study is (120 +
80)/(120 + 80 + 170 + 30) = 0.50, or 50%. As expected, the higher the prevalence of disease, the greater the
PV+ , since the number of individuals with disease (TPs) is increased.

Choice B is the test's sensitivity [TP/(TP + FN)], which is calculated as follows: 120/(120 + 80) = 0.60, or 60%.
The test's sensitivity is too low to be useful as a screening test for an AMI.

Choice C is the negative predictive value (PV-) = [TN/(TN + FN)], which is calculated as follows: 170/(170 + 80) =
0.68, or 68%. This percentage indicates that there is a 68% chance that a normal test is a TN and a 32% chance
that it is a FN. The higher the prevalence of disease, the lower the PV-, since a greater number of FNs are
present.

Choice E is the test's specificity [TN/(TN + FP)], which is calculated as follows: 170/(170 + 30) = 0.85, or 85%.
Although this test's specificity is higher than its sensitivity, its specificity is not high enough for it to be useful as a
confirmatory test for an AMI.
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Old 08-05-2012
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seems am sleepy calculated the wrong thing...its 0.8....still PPV as i mentioned earlier....
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Old 08-06-2012
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Quote:
Originally Posted by Novobiocin View Post
The table below summarizes the results of a new test to identify patients with an acute myocardial infarction (AMI).
The test was performed on 400 subjects.

Acute Myocardial Infarction...... Present............ Absent................. Total


Test Result
...................Positive................ 120.................. 30....................... 150


..................Negative................. 80.................... 170 ................... 250




.......................Total............... 200..................... 200..................... 400



If one of the 400 subjects in the study is randomly chosen, what is the likelihood that an abnormal test indicates that the individual has an AMI?


A. 0.50

B. 0.60

C. 0.68

D. 0.80

E. 0.85
LR+ = sens / (1-spec) = 60 / 15 = 4
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Quote:
Originally Posted by Novobiocin View Post
The table below summarizes the results of a new test to identify patients with an acute myocardial infarction (AMI).
The test was performed on 400 subjects.

Acute Myocardial Infarction...... Present............ Absent................. Total


Test Result
...................Positive................ 120.................. 30....................... 150


..................Negative................. 80.................... 170 ................... 250




.......................Total............... 200..................... 200..................... 400



If one of the 400 subjects in the study is randomly chosen, what is the likelihood that an abnormal test indicates that the individual has an AMI?



D. 0.80

2X2 Statistics / Epidemiology Table (Punnett Square)
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Quote:
Originally Posted by cingulate.gyrus View Post
LR+ = sens / (1-spec) = 60 / 15 = 4
?????? please explain this for me
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Old 08-07-2012
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Quote:
Originally Posted by riya rai View Post
?????? please explain this for me
Sorry I was calculating The Likelihood Ratio . Seeing the word likelihood i jumped into conclusion of calculating likelihood ratio..

Sorry in this case we are supposed to calculate positive predictive value.

Likelihood ratio is the likelihood that a given test result would be expected in a patient with the target disorder compared to the likelihood that that same result would be expected in a patient without the target disorder.

Likelihood ratio for +ve result and -ve result are calculated by the formula.

LR+ = sensitivity / (1-specificity) = (a/(a+c)) / (b/(b+d))
LR- = (1-sensitivity) / specificity = (c/(a+c)) / (d/(b+d)
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