Displaying 20 results from an estimated 800 matches similar to: "MANOVA"
2009 Sep 21
2
Combine vectors in order to form matrixes with combn
Hello!
I've a problem with the combn function and a set of vector. I
would like to make a simple combination where, instead of scalars, i
would like to combine vector, in order to form matrixes.
In other
words, i have nineteen 6-items vectors (for example coef1-coef19), that
i would like to combine in n!/k!(n-k)! 6x6 matrixes.
I tried with a
code like this
mma <-
2006 Feb 16
2
MANOVA: how do I read off within and between Sum-of-Squares info from the manova result?
Hi all,
I am experimenting the function "manova" in R.
I tried it on a few data sets, but I did not understand the result:
I used "summary(manova_result)"
and "summary(manova_result, test='Wilks')"
and they gave a bunch of numbers...
But I need the Sum-of-Squares of BETWEEN and WITHIN matrices...
How do I read off from the R's manova results?
Any
2004 May 24
2
Manova and specifying the model
Hi,
I would like to conduct a MANOVA. I know that there 's the manova() funciton and the summary.manova() function to get the appropriate summary of test statistics.
I just don't manage to specify my model in the manova() call. How to specify a model with multiple responses and one explanatory factor?
If I type:
2007 Feb 22
1
MANOVA usage
Hello,
I had a couple questions about manova modeling in R.
I have calculated a manova model, and generated a summary.manova output
using both the Wilks test and Pillai test.
The output is essentially the same, except that the Wilks lambda = 1 -
Pillai. Is this normal? (The output from both is appended below.)
My other question is about the use of MANOVA. If I have one variable which
has a
2013 May 03
1
MANOVA summary.manova(m) :" residuals have rank"
Dear All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location)
Zx <- Z[c(4:504)]
Zxm <- as.matrix(Z)
m<- manova(Zxm~Season*location, data=Z)
when I do summary.aov, I get respond for each species but summary.manova
summary.manova(m) :" residuals have rank" 24<501.
What can it be the reason for this error
2003 Nov 22
3
summary.manova and rank deficiency
Hi all,
I have received the following error from summary.manova:
Error in summary.manova(manova.test, test = "Pillai") :
residuals have rank 36 < 64
The data is simulated data for 64 variables. The design is a 2*2 factorial with 10 replicates per treatment. Looking at the code for summary.manova, the error involves a problem with qr(). Does anyone have a suggestion as to how to
2004 Feb 15
1
manova() with a data frame
I'm trying to learn to use manova(), and don't understand why none of
the following work:
> data(iris)
> fit <- manova(~ Species, data=iris)
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
incompatible dimensions
> fit <- manova(iris[,1:4] ~ Species, data=iris)
Error in model.frame(formula, rownames, variables, varnames, extras,
2001 Nov 08
2
Manova in R vs. SAS
While I was helping a SAS-using friend with an analysis I noticed some
differences in the multivariate test statistics, approximate F statistics,
and p-values in the manova function using R and proc GLM using SAS. The
univariate coefficients are identical. Is there a reason to expect R and SAS
to give different results?
Thanks,
Bill Kristan.
2006 Mar 30
2
Unbalanced Manova
Dear all,
I need to do a Manova but I have an unbalanced design. I have
morphological measurements similar to the iris dataset, but I don't have
the same number of measurements for all species. Does anyone know a
procedure to do Manova with this kind of input in R?
Thank you very much,
Naiara.
--------------------------------------------
Naiara S. Pinto
Ecology, Evolution and Behavior
1
2012 Mar 19
1
car/MANOVA question
Dear colleagues,
I had a question wrt the car package. How do I evaluate whether a
simpler multivariate regression model is adequate?
For instance, I do the following:
ami <- read.table(file =
"http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat",
col.names=c("TCAD", "drug", "gender", "antidepressant","PR",
2006 Nov 09
2
Repeated Measures MANOVA in R
Can R do a repeated measures MANOVA and tell what dimensionality the statistical variance occupies?
I have been using MATLAB and SPSS to do my statistics. MATLAB can do ANOVAs and MANOVAs. When it performs a MANOVA, it returns a
parameter d that estimates the dimensionality in which the means lie. It also returns a vector of p-values, where each p_n tests
the null hypothesis that the mean
2012 Aug 25
2
Standard deviation from MANOVA??
Hi,
I have problem getting the standard deviation from the manova output.
I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1
+ x2 + x3, data=mydata) .
I tried to get the predicted values and their standard deviation by using:
predict(myfit, type="response", se.fit=TRUE)
But the problem is that I don't get the standard deviation values, I only
2008 Jul 15
2
extracting elements from print object of Manova()
Hi there,
Does anyone know how to extract elements from the table returned by Manova()?
Using the univariate equivalent, Anova(), it's easy:
a.an<-Anova(lm(y~x1*x2))
a.an$F
This will return a vector of the F-values in order of the terms of the model.
However, a similar application using Manova():
m.an<-Manova(lm(Y~x1~x2))
m.an$F
Returns NULL. So does any attempt at calling the
2003 Jun 10
1
Bootstraping with MANOVA
Hi,
Does anyone know what the error message mean?
> Boot2.Pillai <- function(x, ind) {
+ x <- as.matrix(x[,2:ncol(x)])
+ boot.x <- as.factor(x[ind, 1])
+ boot.man <- manova(x ~ boot.x)
+ summary(manova(boot.man))[[4]][[3]]
+ }
>
> man.res <- manova(as.matrix(pl.nosite) ~
+ as.factor(plankton.new[,1]))$residuals
> boot2.plank <-
2003 Nov 20
4
p value in MANOVA
Dear R users,
Can anyone tell me how to get the p value out of the output from
summary.manova?
I tried all the methods I can think of, but failed.
Many thanks
Yu-Kang
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2008 Aug 13
1
summary.manova rank deficiency error + data
Dear R-users;
Previously I posted a question about the problem of rank deficiency in
summary.manova. As somebody suggested, I'm attaching a small part of
the data set.
#***************************************************
"test" <-
structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)),
levels = c("1", "2", "3"),
class =
2004 Jan 05
1
MANOVA power, degrees of freedom, and RAO's paradox
Hi,
I have a nested unbalanced data set of four correlated variables. When I
do univariate analyses, my factor of interest is significant or
marginally significant with all of the variables. Small effect size but
always in the same direction. If I do a MANOVA instead (because the
variables are not independent!) then my factor is far from being
significant. How does that come about?
I have
2011 Jun 21
1
Stepwise Manova
Hello all,
I have a question on manova in R:
I'm using the function "manova()" from the stats package.
Is there anything like a stepwise (backward or forward) manova in R (like there is for regression and anova).
When I enter:
step(Model1, data=Mydata)
R returns the message:
Error in drop1.mlm(fit, scope$drop, scale = scale, trace = trace, k = k, :
no 'drop1'
2003 Jun 07
1
Extracting Numbers from MANOVA output
Hi,
Suppose I have:
> summary(manova(plank.man))
Df Pillai approx F num Df den Df Pr(>F)
plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 ***
Residuals 58
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
My understanding is the MANOVA summary returns a list.
2007 Apr 05
1
MANOVA with repeated measurements
Hello,
I have a question regarding performing manova. I have an experiment where I want to measure 10 output variables for 3 different measurement methods. Since each of the methods requires some user interaction, I would also like to include repeated measures for each of the output variables to include intraobserver variability in the design.
How can I perform such a repeated-measures