Re: [R] Don´t know what test i have to use

2011-01-13 Thread gaiarrido
Thanks for the advice, I will use next years. Till know i´ve just got data for 3 independent months and one of the months it´s the joining for all the summer because of the small sample size, so, I suppose, I can't use it in the way you say. - Mario Garrido Escudero PhD student Dpto. de

[R] Don´t know what test i have to use

2011-01-12 Thread gaiarrido
Hello, I´m starting with my PhD and I have to stop because i got a little knowledge in R and statistics. I´ve got a model of this kind: binary response variable: prevalence of infection (0/1) 3 categorical independent variables: sex, month and name of the area I was trying with a full model

Re: [R] Don´t know what test i have to use

2011-01-12 Thread David Winsemius
On Jan 12, 2011, at 12:51 PM, gaiarrido wrote: Hello, I´m starting with my PhD and I have to stop because i got a little knowledge in R and statistics. I´ve got a model of this kind: binary response variable: prevalence of infection (0/1) 3 categorical independent variables: sex, month and

Re: [R] Don´t know what test i have to use

2011-01-12 Thread Joshua Wiley
Hi, That is basically correct. You can specify the link as logit (see my example), but that is the default so you do not strictly need to in this case. II would encourage you to keep your variables (prevalencia, edad, sexo, mes) stored in a data frame, in which case you would add the data =

Re: [R] Don´t know what test i have to use

2011-01-12 Thread gaiarrido
Thanks very much both. I´m starting playing with it, i was a little afaid because it was part of my job, but now i've found it very funny. Josh, I've got just data for 3 representatives months, and it's not a priori rejectable that could be differences in the ratio of changes along the months

Re: [R] Don´t know what test i have to use

2011-01-12 Thread Bert Gunter
... But I would think that month should be treated as a cyclical quantity, not as a factor with 12 independent levels, e.g. by transforming month to sin( 2*pi*monthNumber/12) . This assumes 1 year periodicity, which might not be right, of course. Time series methods could obviously be relevant