Running head: PROBABILITY REVISION
NORTHWESTERN UNIVERSITY
Probability Revision using
2x2 Probability Revision
and Bayes Theorem
Formula
406-DL CDSS
Session 5 Assignment
Maria Almacen
2/9/2014
Submitted to Professor Robert Cundick
PROBABILITY REVISION
Given
Pretest (or prior) probability is the probability of a subject having a disease based on the
information available prior to the performance of a diagnostic test. Post-test (or posterior)
probability is the probability of the subject having a disease given the pretest probability along
with the added information about the result of the diagnostic test. Probability revision can be
termed as the method used to convert a pretest probability into a post-test probability, taking the
result of the test into account. Techniques for probability revision include the 2 x 2 table, Bayes'
formula, chance tree inversion, the odds-likelihood form of Bayes', and a nomogram.
It has been observed that patients with undifferentiated aseptic meningitis may be infected with
Lyme disease. This is especially true if they live in an area with a high prevalence for Lyme. For
purposes of this exercise, let us say that the incidence of Lyme disease among Connecticut
residents presenting with aseptic meningitis is 13 in 100. Further, let us assume that a certain
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 Connecticut resident presenting with aseptic meningitis has Lyme
disease given a positive result for the test.
b) The probability that the person does not have Lyme disease given a negative test result.
Be sure and show all of the steps you took in calculating the result. Emphasis is on effort to
complete the exercise. Eighty percent credit will be given if your answer is incorrect but your
method and reasoning are shown in detail. Formatting Microsoft Word 1997–2003.
PROBABILITY REVISION
Answer
Part A. Explanation of 2x2 Probability Revisions
The data given by this homework and used in the 2x2 probability revision are as follows:
Pretest Probability is 13 in 100
Sensitivity or TPR is 81%
Specificity or TNR is 96%
T+
TTotal
LD+
LD-
Total
81% TPR/Sensitivity
(1053/1300)
4% FPR (348/8700)
85%P(T+) (81+4)
1053
348
1401
75% PVP/Positive
Probability (1053/1401)
19% FNR (247/1300)
25% (348/1401)
100% (75+25)
96% TNR/Specificity
(8352/8700)
115% (19+96)
247
8352
8599
3% Post negative Test
Probability(247/8599)
97% PVN (8352/8599)
100% (3+97)
1300
8700
10000
13% Pretest Probability
of Disease P(D+) or the
Prevalence in population
of interest
87% P(D-) (100001300) or (8700 /10000)
100% Total sample
cohort
Legend:
LD+
=
Lyme disease positive
LD-
=
Lyme disease negative
T+
=
Test positive
T-
=
Test negative
TPR
=
Tre positive rate
FPR
=
False positive rate
PVP
=
Predictive value positive of test
FNR
=
False negative ratio
PROBABILITY REVISION
TNR
=
True negative rate
P (D+) =
Pretest probability of disease
P (D-) =
Pretest probability of non-disease
P (T+) =
Ratio of positive test
PVN =
Predictive value negative
The steps of computing the 2x2 Probability Revision are as follows:
1. A table was created for the Test Results (+/-) and for Disease Results (D+/D-).
2. Since 13 in 100 or 13% is the probability of Lyme Disease (LD), a hypothetical cohort
population was increased by 100 times more. Therefore there is 1300 or 13% prevalence
of Lyme disease (LD) in a 10,000 hypothetical population of interest (Pretest Probability
of Disease).
3. 1300 was plotted under LD Positive total also known as Pretest Probability of Disease (P
(D+)) or Prevalence of disease, while 10,000 (increased by 100 times more for more
accurate computation) was plotted in the total area as given in the homework.
1300 or 13% (1300 divided by 10,000) is the Pretest Probability of Disease or P (D+).
4. 10,000 (Total) minus 1300 (P (D+)) equals 8700 which is also 87% Pretest Probability of
Non-disease (P (D-)).
5. The Sensitivity or True Positive Ratio (TPR) was plotted in the Test Positive and LD
Positive as 1053. This is because 1300 (total LD Positive) was multiplied to 81% (given
in homework) equals 1053.
6. The False Negative Ratio (FNR) was calculated as 1300 minus 1053 equals 247 or 19%
(247 divided by 1300). (Formula of 1-TPR).
7. The Specificity or True Negative Ratio (TNR) was calculated as 96% (given by
homework) multiplied to 8700 (Total no LD) is 8352.
8. The False Positive Ratio (FPR) was calculated as 8700 minus 8352 equals 348 or 4%
(348 divided 8700). (Formula of 1-TNR).
9. The Ratio of Negative Test (P (T-) was calculated as 8599 (247 plus 8352).
10. The Ratio of Positive Test (P (T+)) was calculated as 1401 (1053 plus 348)).
Answer to Question 1. The probability that a Connecticut resident presenting with aseptic
meningitis has Lyme disease (LD) given a positive result for the test is 75% or 0.75 probability
(1053 divided 1401). 75% is the Post Positive Test Probability that a Connecticut resident
presenting with aseptic meningitis has Lyme disease.
PROBABILITY REVISION
Answer to Question 2. The probability that the Connecticut resident does not have Lyme
disease given a negative test result is 97%. This is the Total Negative Test Probability of Nondisease (97%).
Part B. Using Bayes Formula in application to given scenario
1. Using Bayes formula to find the probability of a Connecticut resident who presented
with aseptic meningitis has Lyme disease (LD) is presented as follows:
P (D+/T+) = P (T+, D+) / P (+)
0.75 = 1053/1401
75% is the probability that a Connecticut resident who presented with aseptic meningitis
has Lyme disease.
2. Using Bayes formula to find the probability of a Connecticut resident does not have
Lyme disease given a negative test result is as follows:
P (D-/T-) = P (T-, D-) / P (T-)
0.97 = 8352/8599
97% is the probability that a Connecticut resident does not have Lyme disease given a
negative test result.