Probability Revision. MED INF 406 Assignment 5. Golkonda, Jyothi 11/4/2012

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1 Probability Revision MED INF 406 Assignment 5 Golkonda, Jyothi 11/4/2012

2 Problem Statement Assume that the incidence for Lyme disease in the state of Connecticut is 78 cases per 100,000. A diagnostic test for the disease has a sensitivity of 81% and a specificity of 96%. Select two different probability revision techniques and use both to calculate: a) The probability that a person being tested in Connecticut has Lyme disease given a positive result for the test. b) The probability that the person has Lyme disease given a negative test result. 2 x 2 table Analysis True Positive A patient has a disease and the test result was positive False Positive A patient does not have a disease but the test result is positive False Negative A patient has a disease but the test result was negative True Negative A patient does not have a disease and the test result was negative 78 cases per 100,000 had Lyme disease. Hence percent of patients with Lyme disease = (78/100000) * 100 = 0.078% =( TP + FN) Percent of patients who do not have Lyme disease = = % = (TN + FP) Sensitivity = 81% = 0.81 = TP/(TP+FN) TP/(TP+FN) = 0.81 TP/0.078 = 0.81 TP = 0.81 * = 0.063% TP + FN = 0.078% FN = FN = = 0.015% Specificity = 96% = 0.96 = TN /(TN+FP) TN /(TN+FP) = 0.96 TN / = 0.96 TN = 0.96 * = % FP = = 3.997%

3 (Positive test) 4.06% Percent with Lyme disease (Disease) 0.078% 0.063% True positive (TP) Percent without Lyme disease (No Disease) % 3.997% False positive (FP) (Negative test) 95.94% 0.015% False Negative (FN) % True Negative (TN) Positive predictive value (Probability that the person actually has Lyme disease when the test result is positive) = TP/(TP + FP) = 0.063/( ) = or 1.55 % Negative predictive value (Probability that the person is disease free when the test result is negative) = TN/(TN + FN) = /( ) = or 99.98% Probability that the person has Lyme disease given a negative test result = FN/(TN+FN) = 0.015/( ) = 0.015/95.94 = = % 2x2 Alternate calculations 78 cases out of 100,000 had Lyme disease. Number of cases who did not have Lyme disease = 100, = 99,922 Sensitivity of diagnostic test = 81% i.e. 81% of 78 cases will have a positive test result = 78 * (81/100) = 63 = True positive The rest of the 78 cases will show negative. False negative = = 15 Specificity of the diagnostic test = 96% i.e. 96 % of 99,922 cases will have negative test result = * (96/100) = = True negative The rest of the cases will show positive when there is no disease.

4 False positive = = 3997 Lyme disease (D+) No Lyme disease (D-) Total Positive test (T+) 63 True positive (TP) 3997 False positive (FP) 4060 Negative test (T-) False Negative (FN) True Negative (TN) Total 78 99, ,000 Positive predictive value (Probability that the person actually has Lyme disease when the test result is positive) = TP/(TP + FP) = 63/( ) = 63/4060 = or 1.55 % Probability that the person has Lyme disease given a negative test result = FN/(TN+FN) = 15/( ) = 15/95940 = = % Bayes' formula Postpositive test probability = Sensitivity * pretest probability (Sensitivity * pretest probability) + ((1-specificity) * (1-pretest probability)) Sensitivity = 81% = 0.81 Pretest probability = 78/ = Specificity = 96% = 0.96

5 Postpositive test probability = probability that the person has Lyme disease when the test result is positive = (0.81 * )/ ((0.81 * ) + ((1-0.96) *( ))) = /( ) = / = or 1.55% Postnegative test probability = Probability that the person has Lyme disease when the test result is negative = (1-Sensitivity) * pretest probability ((1-Sensitivity) * pretest probability) + (specificity * (1-pretest probability)) = (1-0.81) * /(((1-0.81) * ) + (0.96 * ( )) = 0.19 * /((0.19 * ) + (0.96 * )) = /( ) = / = = 0.015% Decision Tree 78 out of 100,000 cases have LymeDisease In the Decision tree the initial node has the total population which is 100,000 The two chance nodes are cases with Lyme disease which is 78 Cases without Lyme disease will be 100, = 99,922 Hence the probability for each node is and respectively The sensitivity of the test is 81% The probability of the branch with positive test results for the chance node with Lyme disease is 0.81 The probability of the branch with negative test results = = 0.19 The specificity of the test is 96% The probability of the branch with a negative test result on the chance node with no Lyme disease is 0.96 The probability of the branch with positive test result = = 0.04 True Positives (TP) = 78 * 0.81 = 63 False Negative (FN) = 78 * 0.19 = 15 False Positive (FP) = * 0.04 = 3997 True Negative (TN) = * 0.96 = 95925

6 Lyme disease , No Lyme disease Positive Test Negative Test 0.19 Positive Test Negative Test 0.96 True positive = 78 *0.81 = 63 False Negative = 15 False Positive = 3997 True Negative = Positive predictive value (Probability that the person actually has Lyme disease when the test result is positive) = TP/(TP + FP) = 63/( ) = 63/4060 = or 1.55 % Probability that the person has Lyme disease given a negative test result = FN/(TN+FN) = 15/( ) = 15/95940 = = % The odds-likelihood form of Bayes' Likelihood Ratio = Sensitivity/ (1 specificity) = 0.81/(1-0.96) = Pretest odds = P/(1 P) Pretest probability = 78/100,000 = Pretest odds = /( ) = / = Posttest odds = * = Converting odds to probability = O/ (1 + O) = /( ) = / = = 1.55% Probability that the person actually has Lyme disease when the test result is positive = 1.55% Probability that the person has Lyme disease given a negative test result Likelihood Ratio = (1- Sensitivity)/ (specificity) = 0.19/(0.96) = Pretest odds = P/(1 P)

7 Pretest probability = 78/100,000 = Pretest odds = /( ) = / = Posttest odds = * = Converting odds to probability = O/ (1 + O) = /( ) = / = = 0.015% Probability that the person has Lyme disease given a negative test result = 0.015%