hi all-
I am doing some research, have never used R before until today and need to
understand the following program for a project.
if some one could PLEASE help me understand this program ASAP i would
GREATLY appreciate it (any syntax/ statistic comments would be great)
PLEASE PLEASE HELP!! THANKYOU!!!
-on a side note, it seems to me that R doesnt include the pv, and it was
calculated seperatly, is this true?
fit=gee(foci~as.factor(time)*cond,id=exper,data=drt,family=poisson(link
"log"))
Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27
running glm to get initial regression estimate
(Intercept) as.factor(time)24
3.051177 -2.705675
condHypoxia as.factor(time)24:condHypoxia
-0.402259
1.429034> pv=2*(1-pnorm(abs(summary(fit)$coef[,5])))
> data.frame(summary(fit)$coef,pv)
Estimate Naive.S.E. Naive.z Robust.S.E.
Robust.z
(Intercept) 3.051177 0.02221052 137.37527 0.04897055
62.306363
as.factor(time)24 -2.705675 0.10890056 -24.84537 0.19987174
-13.537057
condHypoxia -0.402259 0.03907961 -10.29332 0.10661248
-3.773095
as.factor(time)24:condHypoxia 1.429034 0.12549576 11.38711 0.17867421
7.997988
pv
(Intercept) 0.000000e+00
as.factor(time)24 0.000000e+00
condHypoxia 1.612350e-04
as.factor(time)24:condHypoxia 1.332268e-15> ftable(table(drt$cond,drt$time,predict(fit)))
0.345501643340608 1.37227675004058 2.64891772174934
3.05117673373261
Oxia 0.5 0 0
0 485
24 315 0
0 0
Hypoxia 0.5 0 0
346 0
24 0 449
0 0> ## 3-th term gives the difference between the Hypoxia/Oxia at time=0.5
> ## the difference between Hypoxia/Oxia at time=24
> L=matrix(c(0,0,1,1),nrow=1)
> fit$coef[L==1]
condHypoxia as.factor(time)24:condHypoxia
-0.402259
1.429034> L%*%fit$coef
[,1]
[1,] 1.026775> wald.test(fit$robust.variance,fit$coef,L=L)
Wald test:
----------
Chi-squared test:
X2 = 23.8, df = 1, P(> X2) = 1.1e-06>
[[alternative HTML version deleted]]
On Jun 23, 2011, at 4:42 PM, elisheva corn <elishevacorn at gmail.com> wrote:> hi all- > > I am doing some research, have never used R before until today and need to > understand the following program for a project. > if some one could PLEASE help me understand this program ASAP i would > GREATLY appreciate it (any syntax/ statistic comments would be great) > > PLEASE PLEASE HELP!! THANKYOU!!! > -on a side note, it seems to me that R doesnt include the pv, and it was > calculated seperatly, is this true? > > > fit=gee(foci~as.factor(time)*cond,id=exper,data=drt,family=poisson(link > "log"))You apparently have count data (foci) which is measured repeatedly within exper, and you're interested in how foci changes with time and condition including their interaction. The code fits a generalized estimating equation (GEE) model, which can be an appropriate model for repeated measures data. See, for example, Diggle, Liang, Zeger & Heagerty for background.> Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27 > running glm to get initial regression estimate > (Intercept) as.factor(time)24 > 3.051177 -2.705675 > condHypoxia as.factor(time)24:condHypoxia > -0.402259 1.429034 >> pv=2*(1-pnorm(abs(summary(fit)$coef[,5]))) >> data.frame(summary(fit)$coef,pv)The gee package doesn't compute the value directly, though other functions like lm, glm and others do. What the code does is use the robust z statistic, which is the estimate/robust se, and relate it to the standard normal distribution.> Estimate Naive.S.E. Naive.z Robust.S.E. > Robust.z > (Intercept) 3.051177 0.02221052 137.37527 0.04897055 > 62.306363 > as.factor(time)24 -2.705675 0.10890056 -24.84537 0.19987174 > -13.537057 > condHypoxia -0.402259 0.03907961 -10.29332 0.10661248 > -3.773095 > as.factor(time)24:condHypoxia 1.429034 0.12549576 11.38711 0.17867421 > 7.997988 > pv > (Intercept) 0.000000e+00 > as.factor(time)24 0.000000e+00 > condHypoxia 1.612350e-04 > as.factor(time)24:condHypoxia 1.332268e-15 >> ftable(table(drt$cond,drt$time,predict(fit))) > 0.345501643340608 1.37227675004058 2.64891772174934 > 3.05117673373261 > > > Oxia 0.5 0 0 > 0 485 > 24 315 0 > 0 0 > Hypoxia 0.5 0 0 > 346 0 > 24 0 449 > 0 0 >> ## 3-th term gives the difference between the Hypoxia/Oxia at time=0.5 >> ## the difference between Hypoxia/Oxia at time=24 >> L=matrix(c(0,0,1,1),nrow=1) >> fit$coef[L==1] > condHypoxia as.factor(time)24:condHypoxia > -0.402259 1.429034 >> L%*%fit$coef > [,1] > [1,] 1.026775 >> wald.test(fit$robust.variance,fit$coef,L=L) > Wald test: > ---------- > > Chi-squared test: > X2 = 23.8, df = 1, P(> X2) = 1.1e-06 >> > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
I have taught myself R over the last few months and I learned quite a bit from reading the PDFs that accompany each package, articles on the R Journal (http://journal.r-project.org/) that cover similar problems to mine, and by asking specific questions about code to this list. I also search online for answers to questions similar to mine. Sometimes I even read other people's code that has been posted online because they often include explanations and I can learn new ways of coding things. I am still not proficient but I think that might come with years of practice. Good luck -- View this message in context: http://r.789695.n4.nabble.com/new-to-R-need-urgent-help-tp3621329p3622632.html Sent from the R help mailing list archive at Nabble.com.