Recent content by katxt

You may need a couple of brackets in there as well.

A p value of 0.1would mean that that there could well be a correlation, but that the evidence isn't strong enough to claim that there is. H0 can never be accepted as true (because it probably isn't). It just isn't rejected. You just don't know.

Right. You cannot prove statistically that something is zero. If you find no significant correlation, that doesn't mean the correlation is zero. "No significant correlation" is simply a face saving way of saying "We don't know if there is a correlation or not." You may be able to show that it is...

H0 is hardly a null hypothesis. What result would lead you to reject H0 here? Would p = 0.1 mean there was no correlation?

Sorry, B2 is =MOD(INT($A2/B$1),2)

These are the combinations nCr. You don't need them. They are just a check that all is well. Here is a sheet for five. You can expand it out to 10 and down to 1023. The yellow are the formulas to enter. All the rest is copy and fill. kat

OK Good luck. Actually, now that I have thought a little more about it, 1024 rows (or more) can probably be done in Excel in about 5 minutes. Just a few formulas and copying.

Have you looked for a pattern in these numbers for a clue?

Let's try the 5 event case. A B C D E can happen with probabilities a b c d e. There are 2^5 = 32 possibilities from 0 to 5 events happening. The probability of each possibility is the product of 5 other probabilities. The sum of these 32 possibilities will total 1. So, for instance, take B&C...

Just in case anyone is following this, I was looking for a function to generate Monte Carlo Poisson rates given an observation k. Given the "exact" CI expression I used mu = 1/2*CHIINV(k,2*(n+k)), where u is Uniform(0,1). This gives about 2k df for small values of u, and 2k+2 df for large u. It...

To get 1 and 2 only, you also have to get not 3, not 4 and not 5. Or have Interpreted the question wrong? I read it as there are 5 things that may or not happen. What is the probability that exactly 2 of them will happen, given the probability of each individual thing.

Did you include the ones which didn't occur? P1&2 = P1*P2*(1-P3)*(1-P4)*(1-P5)?

OK. I just thought it was common sense. Perhaps I'm a closet Bayesian.

Does "any 2" mean 2 or more? or exactly 2

Thanks. Good thought. I'm not too hot on Bayesian stuff, but a uniform prior seems to give (I think) more or less the same results as the likelihood function.