# Tag: drug testing

ABC News/Washington Times Investigation Leads To Action

32,000 to be notified? 32,000 Veterans, how many are probably suffering from PTSD/Depression already, apparently are in this program unearthed by ABC News and the Washington Times, 32,000!

Responding to an ABC News/Washington Times investigation, the Veterans Administration plans to inform 32,000 veterans that they are using a drug linked to suicide or violent behavior.

And we already know where the administration stood on this from the previous reports:

The Bush White House had initially defended the VA’s handling of the Chantix experiment.

Brought back to light in this one.

The report:

## BREAKING: ‘Disposable Heroes’:

Look I’m so F**KING PISSED OFF AGAIN, I really can’t wrap my mind around the words to describe this BULLS**T!!

I may, or may not, add more, or just let comments take charge.

## Answer to Monday Brain Teaser

Bayes’ theorem is useful in evaluating the result of drug tests. Suppose a certain drug test is 99% sensitive and 99% specific, that is, the test will correctly identify a drug user as testing positive 99% of the time, and will correctly identify a non-user as testing negative 99% of the time. This would seem to be a relatively accurate test, but Bayes’ theorem will reveal a potential flaw. Let’s assume a corporation decides to test its employees for opium use, and 0.5% of the employees use the drug. We want to know the probability that, given a positive drug test, an employee is actually a drug user. Let “D” be the event of being a drug user and “N” indicate being a non-user. Let “+” be the event of a positive drug test. We need to know the following:

* P(D), or the probability that the employee is a drug user, regardless of any other information. This is 0.005, since 0.5% of the employees are drug users. This is the prior probability of D.

* P(N), or the probability that the employee is not a drug user. This is 1 ? P(D), or 0.995.

* P(+|D), or the probability that the test is positive, given that the employee is a drug user. This is 0.99, since the test is 99% accurate.

* P(+|N), or the probability that the test is positive, given that the employee is not a drug user. This is 0.01, since the test will produce a false positive for 1% of non-users.

* P(+), or the probability of a positive test event, regardless of other information. This is 0.0149 or 1.49%, which is found by adding the probability that the test will produce a true positive result in the event of drug use (= 99% x 0.5% = 0.495%) plus the probability that the test will produce a false positive in the event of non-drug use (= 1% x 99.5% = 0.995%). This is the prior probability of +.

Given this information, we can compute the posterior probability P(D|+) of an employee who tested positive actually being a drug user:

P(D|+)  = (0.99 x 0.005)/(0.99 x 0.005)+(0.01 x 0.005)

P(D|+)  = 0.3322

Despite the high accuracy of the test, the probability that an employee who tested positive actually did use drugs is only about 33%, so it is actually more likely that the employee is not a drug user. The rarer the condition for which we are testing, the greater the percentage of positive tests that will be false positives.

## Monday Brain Teaser

0.5% of the employees of Company X use drugs.  Company X therefore decides to administer a drug test to all of its employees.  The drug test is 99% accurate and 99% specific (which means that 99% of all positives will be true positives, and 99% of all negatives will be true negatives).

What is the probability that any given positive test will be that of an actual drug user?