Displaying 20 results from an estimated 5000 matches similar to: "MANOVA"
2018 Apr 18
1
Problem with regression line
Hi Anne,
I would suggest to change the linear model to lm(BloodPressure~Age), as
this model makes more sense in biological means (you would assume that
age influences pressure, not vice versa) and also obeys the statistical
assumption of weak exogeneity, that age can be measured without error,
at least compared to error-prone bp measures.
Cheers
Am 18.04.2018 um 16:07 schrieb Gerrit Eichner:
2018 Apr 18
0
Problem with regression line
Hi, Anne,
assign Age and Bloodpressure in the correct order
to the axes in your call to plot as in:
plot(y = Age, x = BloodPressure)
abline(SimpleLinearReg1)
Hth -- Gerrit
---------------------------------------------------------------------
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eichner at math.uni-giessen.de Justus-Liebig-University Giessen
Tel:
2018 Apr 18
3
Problem with regression line
Hello,
I am trying to graph a regression line using the followings:
Age <- c(39, 47, 45, 47, 65, 46, 67, 42, 67, 56, 64, 56, 59, 34, 42, 48, 45,
17, 20, 19, 36, 50, 39, 21, 44, 53, 63, 29, 25, 69)
BloodPressure <- c(144, 220, 138, 145, 162, 142, 170, 124, 158, 154, 162,
150, 140, 110, 128, 130, 135, 114, 116, 124, 136, 142, 120, 120, 160, 158,
144, 130, 125, 175)
SimpleLinearReg1=lm(Age ~
2006 Jul 05
2
Colinearity Function in R
Is there a colinearty function implemented in R? I
have tried help.search("colinearity") and
help.search("collinearity") and have searched for
"colinearity" and "collinearity" on
http://www.rpad.org/Rpad/Rpad-refcard.pdf but with no
success.
Many thanks in advance,
Peter Lauren.
2004 Jan 13
2
Manova for repeated measures
Hi everyone,
I'm posting again, since I haven't got an answer (yet :( ).
According to R help, manova does not support the inclusion of the Error()
term in the formula call. I have repeated measures data for two dependent
variables,
so this means I can't account for subject variance in time?. Any lights?
Thanks in advance,
Rodrigo Abt,
Department of Economic Studies,
SII, Chile.
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
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,
2018 Apr 20
1
Further questions
Hi R folks,
In my previous post I forgot to mention that I was new to R. I was really grateful for your quick help. I have two further questions:
1) In the graph of a regression line I would like to show one specific residual yi obs - yi pred (let's take the person whose residual is 76). How do I add a bracket to this vertical distance and name it? I'am getting stuck after the
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
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
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
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]))$residuals
> boot2.plank <-
2012 Mar 19
1
car/MANOVA question
Dear colleagues,
I had a question wrt the car package. How do I evaluate whether a
simpler multivariate regression model is adequate?
For instance, I do the following:
ami <- read.table(file =
"http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat",
col.names=c("TCAD", "drug", "gender", "antidepressant","PR",
2012 Aug 25
2
Standard deviation from MANOVA??
Hi,
I have problem getting the standard deviation from the manova output.
I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1
+ x2 + x3, data=mydata) .
I tried to get the predicted values and their standard deviation by using:
predict(myfit, type="response", se.fit=TRUE)
But the problem is that I don't get the standard deviation values, I only
2008 Aug 13
1
summary.manova rank deficiency error + data
Dear R-users;
Previously I posted a question about the problem of rank deficiency in
summary.manova. As somebody suggested, I'm attaching a small part of
the data set.
#***************************************************
"test" <-
structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)),
levels = c("1", "2", "3"),
class =
2008 Jul 15
2
extracting elements from print object of Manova()
Hi there,
Does anyone know how to extract elements from the table returned by Manova()?
Using the univariate equivalent, Anova(), it's easy:
a.an<-Anova(lm(y~x1*x2))
a.an$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$F
Returns NULL. So does any attempt at calling the
2013 May 03
1
MANOVA summary.manova(m) :" residuals have rank"
Dear All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location)
Zx <- Z[c(4:504)]
Zxm <- as.matrix(Z)
m<- manova(Zxm~Season*location, data=Z)
when I do summary.aov, I get respond for each species but summary.manova
summary.manova(m) :" residuals have rank" 24<501.
What can it be the reason for this error
2011 Jun 21
1
Stepwise Manova
Hello all,
I have a question on manova in R:
I'm using the function "manova()" from the stats package.
Is there anything like a stepwise (backward or forward) manova in R (like there is for regression and anova).
When I enter:
step(Model1, data=Mydata)
R returns the message:
Error in drop1.mlm(fit, scope$drop, scale = scale, trace = trace, k = k, :
no 'drop1'
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