Two by Two Table Assignment
Session 2
Subrata Behera
Northwestern University
Decision Support Systems and Healthcare – MMI 406
1. Two by Two Table Assignment
In a hypothetical elementary school class, 35% of the students are obese and 40% watch at
least four hours of television each day. Twenty-five percent of the students are both obese
and among the excessive television watchers. Draw a 2 x 2 table and fill in the missing rows
and columns to show the percentages of students who watch excessive television but are not
obese, who don’t watch television excessively but are obese, and who are not obese and do
not watch excessive television.
If you were to randomly select a child from the class and find that they are among the
group that watches television excessively, what is the probability that the child is also
obese?
Be sure and show how you arrived at your answer. Emphasis is on effort made to complete
the exercise. 80% credit will be given if answer incorrect but method and reasoning shown
in detail.
Answer:
As per the book the approach would be to use the strategy used for finding out the probabilities
of a certain disease.
O+
O-
35%
65%
25%
15%
T+
40%
TP
FP
10%
Sensitivity = TP /(TP + FN)
50%
T60%
FN
TN
Sensitivity = TN /(TN + FP)
The various variables used in the 2x2 table above are as follows:




O+ = Students who have disease (obesity) = 35% (provided in the question)
O- = Students who have disease no disease (obesity) = 1 – O+ = 1-0.35 = 65%
T+ = Students who watch excessive television = 40% (provided in the question)
T- = Students who don’t watch excessive television = 1 – T+ = 1 - 0.40 = 60%
Calculations:

TP = True Positives
o TP = 25% (provided in the question) - Students who are both obese and watch
excessive television.

FN = False Negatives.
o FN = (O+) – TP = 0.35 – 0.25 = 0.10 or 10% = Students who are obese and don’t
watch excessive television.

FP = False Positives.
o FP = (T+) - TP = 0.40 – 0.25 = 0.15 or 15% = Students who are not obese but watch
excessive television.

TN = True Negatives.
o TN = (O-) - FP = 0.65 – 0.15 = 0.50 or 50% = Students who are neither obese not
watch excessive television.
Sensitivity (is the proportion of patients with disease who test positive) = TP / (TP + FN) = .25 /
(.25 + .10) = 0.71 = 71 % (Thomas G. Tape)
Specificity = (is the proportion of patients without disease who test negative) TN / (TN + FP) = .50
/ (.50+.15) = 0.77 = 77% (Thomas G. Tape)
If you were to randomly select a child from the class and find that they are among the group
that watches television excessively, what is the probability that the child is also obese?
Students who are obese and watch excessive television = 25
Students who watch excessive television = 40
So P (Students who are obese and watch excessive television / Students who watch excessive
television) = P (25/40) = 0.625 = 62.5 %
Reference:
Hunink, H. & Glasziou, P. (2009). Decision making in health and medicine: Integrating evidence
and values (7th printing). Cambridge, England: Cambridge University Press.
Thomas G. Tape, M. " Interpreting Diagnostic Tests." Retrieved April 14, 2013, from
http://gim.unmc.edu/dxtests/Default.htm.