Ask Marilyn. BY MARILYN V os A particularly interesting and important question t
ID: 3239134 • Letter: A
Question
Ask Marilyn. BY MARILYN V os A particularly interesting and important question today is that of testing for drugs. Suppose it is assumed that about 5% of the general population uses drugs. You employ a test that is 95% accurate, which we'll say means that if the individual is a user, the test will be positive 95% of the time, and if the individual is a nonuser, the test will be negative 95% of the time. A person is selected at random and given the test. It's positive. What does such a result suggest? Would you conclude that the individual is highly likely to be a drug-user? -Charles Feinstein, Ph.D., Santa Clara University, Santa Clara, Calif.Explanation / Answer
a)
100%T test negative
a)prevalence of test posiitve=T test positive/total *100
=100/200*100
=50%
b)sensitivity=
c)
d)
Positive Predictive Value:
A/(A + B) × 100
=95/(95+5)*100=95
negative predictive value:
D/(D+C)*100=95/(95+5)*100=95
using bayies theorem:
positive predictive value=sensitivity*prevalence/sensitivity*prevalence+(1-specificity)*(1-prevalence)
=0.95*0.50/0.95*0.50+(1-0.95)*(1-0.50)
=0.475/0.475+(0.05*0.50)
=0.95
negative predictive value:
npv=specificity*(1-prevalence)/(1-sensitivity)*prevalence+specificiy*(1-prevalence)
=0.95*0.50/0.05*0.50+0.95*0.50
=0.95
e)
positive predictive value=sensitivity*prevalence/sensitivity*prevalence+(1-specificity)*(1-prevalence)
=0.95*0.01/0.95*0.01+(0.05*0.50)
=0.0095/0.0095+0.025
=0.0095/0.0345
=0.2753
negative predictive value:
npv=specificity*(1-prevalence)/(1-sensitivity)*prevalence+specificiy*(1-prevalence)
=0.95*0.99/0.05*0.01+0.95*0.99
=0.9405/0.9405+0.0005
=0.9405/0.941
=0.9994
user non user total positive 95%(A) (B)5% 100%T test positive negative (C)5% 95%(D)100%T test negative
100%T user 100% T non user total 200%Related Questions
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