Lisa is worried about her health. Her doctor tells her not to worry too much as there is only a 1 in 1,000 chance that women of her age has the dreaded Econitus virus. Nevertheless, Lisa remains anxious about this possibility and decides to obtain a test that can detect Econitus. The test is moderately accurate: When someone has Econitus it delivers a positive result 86% of the time. But there is, however, a small 'false positive' rate: 5% of people produce a positive result despite not having Econitus. Lisa takes the test and obtains a positive result. What are the chances that she has Econitus?
0-20 percent chance
21-40 percent chance
41-60 percent chance
61-80 percent chance
81-100 percent chance
The correct answer is 1.7%!
If you answered 86% then you fell into the common trap of ignoring 'base rates'...
If 1,000 women like Lisa take the test, 999 will not have Econitus. But the false positive result means 50 will be told they have Econitus. Therefore there is only 1.7% chance that Lisa has Econitus (trust me!).
What this demonstrates is that people ignore background information in favour of more salient information about a specific case.
Hopefully this will be useful the next time you are given statistics by a doctor...
I know this isnt the point, but how did you get to 1.7%? This is making my brain hurt. My understanding is 50.86 people per 1000 will get a positive diagnosis, with only 1 person having it? Therefore the chance is near enough 2%? (1.97 to be snooty).
ReplyDeleteSure - because there is a chance that none of the 50 people actually have it it is less 2%. Acoording to Bazerman (2009) the correct calculation is: 0.86/(0.86+49.95)=1.7
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