You can find notes on Bayesian probability in three posts from last term you can find through this link. Here are a few more practice problems, with answers in the comments.
A) A trait shows up in 20% of the population and the test has a 2% error rate. Find p(error given test positive) and p(error, given test negative).
B) A trait shows up in 10% of the population and the test has a 1% error rate. Find p(error given test positive) and p(error, given test negative).
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2 comments:
I agree with all of the steps taken, but it should be a total of 892 for test -. this comes from subtracting 900-9=891.
You are right, thanks. Here's the correction.
A) A trait shows up in 20% of the population and the test has a 2% error rate. Find p(error given test positive) and p(error, given test negative).
20% = 1/5 and 2% = 1/50 so the grand total is 250.
_______don't have__have
test +_________________
test -_________________
totals_________________ 250
20% of 250 is 50, and 250-50 = 200.
_______don't have__have
test +_________________
test -_________________
totals______200_____50_ 250
1/50*50 = 1, so there will be 1 error in have trait, but test negative. Since there will be 5 errors total, there is a 4 in the don't have but test positive slot.
_______don't have__have
test +________4________
test -_______________1_
totals______200_____50_ 250
We can fill in the rest using degrees of freedom.
_______don't have__have
test +________4_____49_ 53
test -______196______1_ 197
totals______200_____50_ 250
p(error, given test positive) = 4/53 ~= 7.5%
p(error given test negative) = 1/197 ~= 0.5%
B) A trait shows up in 10% of the population and the test has a 1% error rate. Find p(error given test positive) and p(error, given test negative).
10% = 1/10 and 1% = 1/100 so the grand total is 1000.
_______don't have__have
test +_________________
test -_________________
totals_________________ 1000
10% of 1000 is 100, and 100-100 = 900.
_______don't have__have
test +_________________
test -_________________
totals____900______100_ 1000
1/100*100 = 1, so there will be 1 error in have trait, but test negative. Since there will be 10 errors total, there is a 9 in the don't have but test positive slot.
_______don't have__have
test +______9__________
test -_______________1_
totals____900______100_ 1000
We can fill in the rest using degrees of freedom.
_______don't have__have
test +______9_______99_ 108
test -____891________1_ 892
totals____900______100_ 1000
p(error, given test positive) = 9/108 ~= 8.3%
p(error given test negative) = 1/892 ~= 0.1%
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