Help please Please show work Thank you Suppose a certain dru

Help please, Please show work. Thank you!

Suppose a certain drug test is 98% sensitive. that is. the test will correctly identify a drug user as testing positive 98% of the time. and 94% specific. that is. the test will correctly identify a non-user as testing negative 94% of the time. Suppose a corporation decides to test its employees for drug use. and that only 0.8% of the employees actually use the drug. What is the probability that. given a positive drug test. an employee is actually a drug user? Let D stand for being a drug user and N indicate being a non-user. Let be the event of a positive drug test and be the event of a negative drug test.

Solution

Given

Prevalence =0.8%

let we consider with sample size of 100000, prevalence =0.8% of 100000, ( 800 out of 100000)

P( D+) =0.008

Therefore, P( N) = P( D-) = 0.992 (99200 out of 100000)

Sensitivity = 98%

Therefore P( T+   and D+) = 0.98   ( 800*0.98=784)

Specificity = 94%

Therefore P( T-   and D-) = 0.94*99200=93248

P( D+/T+) = sensitivity*Prevalence / ( sensitivity*Prevalence+(1-specificity)(1-prevalence)

P( T+/D- )= Specificity*(1-Prevalence) / ( (1-sensitivity)*Prevalence + specificity(1-prevalence))

User

D+

D-

total

Test

T+

784

5952

6736

T-

16

93248

93264

total

800

99200

100000

b) P( N) = P( D-) = 0.992

c). P(T+/D-) =5952/99200 =0.06

d). P( T+) =6736/100000=0.06736

e). P( D+/T+) =784/6736 = 0.116389

		User
		D+	D-	total
Test	T+	784	5952	6736
	T-	16	93248	93264
	total	800	99200	100000