From | Roger Newson <[email protected]> |
To | [email protected] |
Subject | Re: st: RE: probability models |
Date | Mon, 07 Mar 2005 21:28:08 +0000 |
At 12:35 07/03/2005, Nick Cox wrote (in reply to Blau Blau):
If you are really confident that the 2 Poisson variables you are comparing really are Poisson, then you can use the fact that the conditional distribution of the first one, given their sum, is binomial, with total equal to their sum, and probability equal toFor the normal, you have -ttest- and -ttesti-. A comparison of two Poisson means is possible using -poisson-. I am not aware of an immediate equivalent. Nick [email protected]
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