Sample of n subjects (where n is large) cross-classified according to disease status and screening test result are given below. Test Result Disease Positive(T) Negative(T) Total Present(D) 286 12 298 Absent(D) 18 284 302 Total 304 296 600 Assume P(D) = 0. 01. Denote the probabilities of false negative and false positive, respectively by FNR and FPR. Then the sensitivity, specificity, FNR, FPR, and predictive value negative of the screening test are, respectively, (Choose one from the following which is closest to the correct answer.)

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Sample of n subjects (where n is large) cross-classified according to disease status and screening test
result are given below.
Test Result
Disease Positive(T) Negative(T) Total
Present(D) 286
Absent(D) 18
Total
12
298
284
302
304
296
600
Assume P(D) = 0. 01. Denote the probabilities of false negative and false positive, respectively by
FNR and FPR. Then the sensitivity, specificity, FNR, FPR, and predictive value negative of the
screening test are, respectively, (Choose one from the following which is closest to the correct
answer.)
O a. 0.9597, 0.9404, 0.0596, 0.0403, 0.1399.
O d. 0.9404, 0.9597, 0.0403, 0.0596, 0.9996.
O b. 0.9404, 0.9597, 0.0596, 0.0403, 0.9996.
e. 0.9597, 0.9404, 0.0403, 0.0596, 0.9996.
O c. 0.9597, 0.9404, 0.0596, 0.0403, 0.9996.
Transcribed Image Text:Sample of n subjects (where n is large) cross-classified according to disease status and screening test result are given below. Test Result Disease Positive(T) Negative(T) Total Present(D) 286 Absent(D) 18 Total 12 298 284 302 304 296 600 Assume P(D) = 0. 01. Denote the probabilities of false negative and false positive, respectively by FNR and FPR. Then the sensitivity, specificity, FNR, FPR, and predictive value negative of the screening test are, respectively, (Choose one from the following which is closest to the correct answer.) O a. 0.9597, 0.9404, 0.0596, 0.0403, 0.1399. O d. 0.9404, 0.9597, 0.0403, 0.0596, 0.9996. O b. 0.9404, 0.9597, 0.0596, 0.0403, 0.9996. e. 0.9597, 0.9404, 0.0403, 0.0596, 0.9996. O c. 0.9597, 0.9404, 0.0596, 0.0403, 0.9996.
Expert Solution
Step 1: Basic idea
  1. Contingency Table (2x2 Table): You should be familiar with the structure of a contingency table, which is used to organize data for binary classification problems, such as medical screening tests. It typically includes four values: true positives (TP), true negatives (TN), false positives (FP), and false negatives (FN).

  2. True Positives (TP): These are cases where the screening test correctly identifies individuals who have the condition or disease.

  3. True Negatives (TN): These are cases where the screening test correctly identifies individuals who do not have the condition or disease.

  4. False Positives (FP): These are cases where the screening test incorrectly identifies individuals as having the condition or disease when they do not.

  5. False Negatives (FN): These are cases where the screening test incorrectly identifies individuals as not having the condition or disease when they actually do.

  6. Sensitivity: Sensitivity is a measure of how well a screening test correctly identifies individuals with the condition. It is calculated as TP / (TP + FN) and represents the true positive rate.

  7. Specificity: Specificity is a measure of how well a screening test correctly identifies individuals without the condition. It is calculated as TN / (TN + FP) and represents the true negative rate.

  8. False Negative Rate (FNR): FNR is the proportion of individuals with the condition who are incorrectly classified as not having it. It is calculated as FN / (TP + FN).


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