Thanks,
that helped!
Yours,
Kay
Zitat von Gene Leynes :
I think that people are afraid to say "You can't do that in R"...
But I think the real answer is: you can't do that in R.
Although, it is helpful to understand Jeff's reply. I hadn't fully
realized why this particular problem occurs bef
successfully by call
install.packages("E:/R/R-2.13.0/library/RGoogleDocs_0.5-0.tar.gz",
repos=NULL, type="source")
thanks and sorry for my clumsiness,
kay
Zitat von Uwe Ligges :
On 05.09.2011 14:52, Kay Cecil Cichini wrote:
internet connection exists - dont' know why, but i
ll from the source that i downloaded to:
"E:/R/R-2.13.0/library/RCurl_0.91-0.tar.gz"
i tried to apply the manual for R Installation and Administration but
admittedly doesn't grasp it - any help how to achieve this would be
greatly
appreciated.
kay
Zitat von Uwe Ligges :
hi,
i'd like to install the package "RGoogleDocs ".
i downloaded to path "E:/R/R-2.13.0/library/RCurl_0.91-0.tar.gz"
i run R from an usb-stick and can't get the install.packages() prompt
to run correctly - can anyone help with this?
thanks,
kay
sessionInfo()
R version 2.13.0 (2011-04-13
hello,
yes, thanks a lot - i noticed relevel() beeing very convinient for this
purpose.
having an authority at hand may i kindly ask, if you could reinsure me
that the contrasts below are set up correctly, supposing i want to test
the earlier mentioned hypotheses simultanously.
thanks,
kay
the only possible value for T since
no other outcome is possible. Hence Prob(T==1) = 1 whether the coin
is fair or not. It is not possible for such data to discriminate
between a fair and an unfair coin.
And, as explained above, a P-value of 1 cannot prove that the
null hypothesis is true. All th
i test the null that the coin is fair (p(succ) = p(fail) = 0.5) with
one trail and get a p-value of 1. actually i want to proof the
alternative H that the estimate is different from 0.5, what certainly
can not be aproven here. but in reverse the p-value of 1 says that i
can 100% sure that t
felix,
thanks a lot for the hint!
i actually found another way by setting up a panel function by which i
can control every single panel with panel.number(). maybe there is
more efficient coding - i don't know. i also alternated tickmarks and
tick-labeling by panel-rows, which is nicer, but
exactly -
thanks a lot, richard!
kay
Zitat von "RICHARD M. HEIBERGER" :
Kay,
doe this do what you want?
dotplot(y1+y2 ~ facs$Treatment|facs$Sites,
outer=TRUE,
scales = list(x = list(rot = 90, tck=c(1,0))),
ylab=c("y1", "y2"),
xlab=c("Site 1", "Site 2"),
Hallo List,
I'm trying to implemement a restricted permutation scheme in permutest(). More
precisely I have dependence in my data that should be allowed for in the
permutation - I simulated the problem in the example of the vegan documentation
p.24:
library(vegan)
data(varespec)
## Bray-Curtis
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