Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(This question relates to a problem I had at work a while ago, doing a little data mining at a car rental company. Names changed, of course.)

There was a flight of steps out the front of our building. It had a dodgy step on it, on which people often stub their toes.

I had records for everyone who works in the building, detailing how many times they climbed these steps and how many of these times they stubbed their toes on the dodgy step. There's a total of 3000 stair-climbing incidents and 1000 toe-stubbing incidents.

Jack climbed the steps 15 times and stubbed his toes 7 times, which is 2 more than you'd expect. What's the probability that this is just random, vs the probability that Joe is actually clumsy?

I'm pretty sure from half-remembered statistics 1 that its something to do with chi-squared, but beats me where to go from there.

...

Of course, we actually had several flights of steps, each with different rates of toe stubbing and instep bashing. How would I combine the stats from those to get a more accurate better likelihood of Joe being clumsy? We can assume that there's no systematic bias in respect of more clumsy people being inclined to use certain flights of steps.

share|improve this question
4  
This question is probably more appropriate for stats.stackexchange.com where you're likely to get more answers. –  Noah Snyder Aug 18 '10 at 18:18

3 Answers 3

A standard answer would be to assume that the number of times Joe stubs his toe out of 15 is a binomial random variable. If he's typical, then p = 1/3. Asking is it "random, or is Joe just clumsy," could be translated as asking how likely it is for someone to stub their toe at least 7 times out of 15 if the probability each time is only 1/3, that is, Prob(X>=7). A low probability indicates that the hypothesis that p = 1/3 should be rejected in favor of p > 1/3.

I computed this in Minitab, and found P(X>=7) = 0.20, that is, a 20% chance.

If you're a scientist, 20% isn't not strong enough evidence that Joe's clumsy. You need < 5%. If you're a judge, however, you might conclude that Joe is clumsy "beyond a reasonable doubt."

share|improve this answer
    
Thank you, but Oracle DBMS does not have minitab embedded in it, and I was more interested in the general question than in the particular instance of Joe. I'm dealing with potentially thousands of occurrences, so using the binomial distribution potentially will result in roundoff and overflow errors. –  paulmurray Aug 19 '10 at 4:41

Of course, it also depends on whether there's some sort of selection bias going on, i.e. did you choose to ask about Jack because he stubbed his toe particularly often? If so, then David's answer shows that even if everyone is equally clumsy we would expect 20% of people to stub their toes at least 7 times out of 15, so it's entirely reasonable to conclude that some people were just that unlucky.

share|improve this answer

if an issue is working with a binomial distribution with a large $n$, you could use a normal approximation to the binomial or perhaps a poisson approximation.

the normal approximation says that for $n$ large, the bin($n,p$) distribution is approximately N($np, np[1-p]$). the poisson approximation applies if $n$ is large and $p$ is small [rare occurrence, large number of trials] and approximates bin($n,p$) by Poi($np$).

from the numbers you cite, it sounds as tho the normal approximation would be a better bet.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.