If you look at the documentation for ar, you will find the following
description of aic:
aic The differences in AIC between each model and the best-fitting
model
So this means that your model order is 5 (look at the pacf).
If you want the actual AIC of your model: (I call your datafi
I am trying to work out a bootstrapped Tobit regression model. I get the
coefficients all right, but they all have standard error zero. And I am
unable to figure out why. I know the coefficients are correct because that's
what I get when do a Tobit (without bootstrapping). Here's my code:
# Bootst
Hello,
I am trying to run the code for quadratic discriminant analysis with the
Spambase data set following the code example in the book "R in a Nutshell"
by Joseph Adler (page 443).
The line of code:
> spam.qda <- qda(formula=is_spam~.,
data=spambase.trainin
I know this is not really an R question - it is a query about a recent book
on R ("R In Action") by Robert Kabacoff, (Manning Publications 2011).
There are many references to interesting topics in R in the book, BUT, I do
not find a bibliography/list of references in the book!
Does anybody know i
This is in version 2.10.1
> sqrt(3)
[1] 1.732051
> 6/7
[1] 0.8571429
> options(digits=3)
> sqrt(3)
[1] 1.73
> 6/7
[1] 0.857
> sqrt(7)
[1] 2.65
> 7/9
[1] 0.778
Ravi
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Can anyone give me links to reviews/comparisons of R with Amos for SEM? I
have found some but they are a little old (2009).
Ravi
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I am trying to reproduce the output of a time series example in Koop's book
"Analysis of Financial Data". Koop does the example in Excel and I used the
ts function followed by the lm function.
I am unable to get the exact coefficients that Koop gives - my coefficients
are slightly different.
After
Of course! Works fine now.
Thanks, Martyn and Jim...
Ravi
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I know this is a very elementary question... I could not find a solution
looking at old posts.
I am unable to access a variable outside the scope of a for loop, even when
the variable was defined before the loop:
haar <- function() {
a = c(1.4560773, 2.3752412, 0.9798882, 3.0909252, 2.3986487, 1.8
Hello Forum,
I have data for two groups of subjects for a ratio variable (cortisol
level) assessed three times during the day (at 0600, 0900 and 2100 hours).
So I have three means for each group. I have been asked to do a "slope
analysis" to compare the two groups.
According to my understandin
I do not know if this is what you want, but take a look at the "External
links" at the bottom of the Wikipedia page:
http://en.wikipedia.org/wiki/Theil_index
Ravi
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actor(group), data=mydata)
gives the right answers. But
> aov(score~group, data=mydata)
also produces an ANOVA table, with incorrect entries. My question is: what
exactly is R doing when I did not specify that "group" was a factor?
Ravi Kulkarni
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Yes, there does not seem to be a method for comparing two ROC curves in ROCR.
But I found a discussion in the R-archives which may be useful:
http://tolstoy.newcastle.edu.au/R/help/06/03/23667.html
You should also see the replies to that post.
I am also interested in comparing
The ROCR package has methods to compute AUC and related methods. You might
want to check it out.
Ravi
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R-h
Use R Commander to do this. R Commander is a R package that offers a GUI for
R. It can be downloaded like any other R package.
If you use R Commander, there is a menu option where you specify that you
want to read an Excel file (you can also read text, SPSS, Minitab, Stata,
Access... files). It i
quot;optcosine".
The boundary kernel apparently improves the estimate in a neighborhood of
zero.
Thanks,
Ravi Kulkarni
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riables in the model? Is that what is causing the difference?
Thanks,
Ravi Kulkarni
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How do I use the "extra" information that two of my predictors are ordinal?
(I did not know I could do that.)
Thanks,
Ravi
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a numerical coding scheme for the
categorical variables as required by some statistical software packages,
with some sort of numeric dummy-variable coding?
I am using:
glm(binvar~x+y+z, family=binomial(link="logit"))
Thanks,
Ravi Kulkarni
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ce for explicit examples of ANOVA in R:
http://www.personality-project.org/R/r.anova.html
Ravi Kulkarni
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I have tried looking for the source code for the pt() function in
https://svn.r-project.org/R/trunk/src/library/stats/
and am unable to find it there. Can someone please tell me where to find it?
Thanks,
Ravi Kulkarni
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It's probably binary data - which implies that you can only "read" it with
the application that created it. What is the filename extension?
Ravi
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Hello,
Since I initiated this discussion some days ago, I discovered a paper that
may be of interest:
ANOVA for unbalanced data: Use Type II instead of Type III sums
of squares
by ØYVIND LANGSRUD
Statistics and Computing 13: 163–167, 2003
Ravi
s what the various "Types" of ANOVA correspond to? I mean that I
think of my ANOVA as a regression model (a General Linear Model) and the
various ways of entering predictors as the various ANOVA "Types".
Hope that makes sense...
Ravi Kulkarni
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Apologies. I should have checked the archives first... I did not know that it
was so widely discussed...
Ravi
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Hello,
I believe the aov() function in R uses a "Type-I sum-of-squares" by
default as against "Type-III".
This is relevant for me because I am trying to understand ANOVA in R using
my knowledge of ANOVA in SPSS. I can only reproduce the results of an ANOVA
done using R through SPSS if I specif
Giving different colors to plots of different levels of a factor can
be pleasing. For a boxplot, for example, the following produces a nice
plot:
boxplot(len~supp*dose, data=ToothGrowth, col=c(“orange”,”gray”))
If I try to specify (using the lattice package)
histogram(~len|supp*factor(dose), da
I forgot to reply to Peter Ehler's question: I am using Levene's test in the
"car" package.
Ravi
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I think I have an answer: SPSS uses absolute deviations from the _mean_ in
Levene's test.
(See calculation in
http://www.uvm.edu/~dhowell/gradstat/psych340/Lectures/Anova/anova2.html)
R uses absolute deviations from the _median_ (R help).
So the difference.
Ravi
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Hello,
I notice that when I do Levene's test to test equality of variances across
levels of a factor, I get different answers in R and SPSS 16.
e.g.: For the chickwts data, in R, levene.test(weight, feed) gives
F=0.7493, p=0.5896.
SPSS 16 gives F=0.987, p=0.432
Why this difference? Which on
Thanks! Exactly what I wanted.
Ravi
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Hello,
I have data for an ANOVA where the between-subjects factor has three
levels. How do I run a test of normality (using shapiro.test) on each
of the levels of the factor for the dependent variable separately
without creating extra datasets?
Thanks,
Ravi
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