Hi again,
Two things, I named the data frame SR as shown in the model.
The other is for those who may wish to answer the OP. The mediafire
website is loaded with intrusive ads and perhaps malware.
Jim
On Thu, Oct 11, 2018 at 9:02 AM Jim Lemon wrote:
>
> Hi Tranh,
> I'm not sure why you are
Hi Tranh,
I'm not sure why you are converting your variables to factors, and I
think the model you want is:
lm(KIC~temperature+AC+AV+Thickness+temperature:AC+
temperature:AV+temperature:thickness+AC:AV+
AC:thickness+AV:thickness,SR)
Note the colons (:) rather than asterisks (*) for the
We cannot read your message. Should post pure text, not html. Hm, my phone
now may post html, must try to stop. Your R code not legible. It seems to
be output? Lines all run together. I tried find articles you mention, but
"not found" resulted.
You should use aov() for fitting, then get post hoc
Thank you all for your **very good** answers:
Using aovp(..., perm="Exact") seems to be the way to go for small datasets,
and also I should definitely try ?kruskal.test.
Juan
[[alternative HTML version deleted]]
__
R-help@r-project.org
> This package uses a modified version of aov() function, which uses
> Permutation Tests
>
> I obtain different p-values for each run!
Could that be because you are defaulting to perm="Prob"?
I am not familiar with the package, but the manual is informative.
You may have missed something when
Juan,
Your question might be borderline for this list, as it ultimately rather seems
a stats question coming in R disguise.
Anyway, the short answer is that you *expect* to get a different p value from a
permutation test unless you are able to do all possible permutation and
therefore use
Dear Juan
I do not use the package but if it does permutation tests it presumably
uses random numbers and since you are not setting the seed you would get
different values for each run.
Michael
On 03/09/2018 16:17, Juan Telleria Ruiz de Aguirre wrote:
Dear R users,
I have the following
> I am trying to decide between using a multiple linear regression or a linear
> mixed effects model for my data:
> ...
> but I keep getting the following error code:
>
> Error in anova.lmlist (object, ...):
>
> models were not all fitted to the same size of dataset
anova is defaulting to
Dear Pamela,
I'm afraid that this question seems confused. Type III sums of squares are
computed for a linear model -- it's the *linear model* that you'd check, in the
usual manner, for outliers, etc., and this has nothing to do with how the ANOVA
table for the model is computed.
Best,
John
On 27 Dec 2014, at 02:50 , Richard M. Heiberger r...@temple.edu wrote:
Kristi,
The easiest way to do what you are looking for is the aovSufficient function
in the HH package.
## install.packages(HH) ## if you don't have it yet
library(HH)
B.aov - aovSufficient(mean ~ site, data=B,
This is a statistical question primarily and, as such, is off topic
here. Either consult a local statistical expert or post to a
statistical site like stats.stackexchange.com .
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is
You could at least say that no, that is patently wrong! There are ways to
reconstruct an ANOVA from means, sd, and group sizes, but this isn't it. In
fact, the group sizes are not even used in the code.
I agree about the need for a statistical expert. Fundamental misunderstandings
seem to be
Kristi,
The easiest way to do what you are looking for is the aovSufficient function
in the HH package.
## install.packages(HH) ## if you don't have it yet
library(HH)
B.aov - aovSufficient(mean ~ site, data=B, sd=B$SE, weights=c(3,3,3,3))
summary(B.aov)
You must have the sample size for each
Hi Stephane,
This is the well known result of limitted floating point precision (e.g.,
http://www.validlab.com/goldberg/addendum.html). Using a test of
approximate rather than exact equality shows R yields the correct answer:
nperm - 1
Fperm - replicate(n=nperm, anova(lm(sample(Y) ~ F,
Beware of the trap of listening to people with no knowledge of basic
numerical methods!
It really is basic that the results of floating-point computer
calculations depends on the order in which they are done (and the
compiler can change the order). Using == on such calculations is warned
I pressed enter to soon.
Again
Hi all,
I have troubles with doing an anova. So I have the following variables: a
variable group with two levels, a continuous variable trait and within each
group we have 12 organisms (clone) and for each clone we have 5 replicates.
So we want to see if for the
Hi Lynn,
Please read the posting guide, this document:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and don't post in HTML. (See your own email copied below for why.)
Sarah
On Thu, Aug 21, 2014 at 6:08 AM, Lynn Govaert lynn.gova...@gmail.com wrote:
Hi
... and you do get P values; you just don't understand what you're
looking at, which is more or less what you said.
This is not a statistical help site. You should seek local statistical
consulting advice or post on a statistical help site (caveat emptor!)
like stats.stackexchange.com, not here.
## Manuel,
## Please look at the maiz example in ?mmc in the HH package.
install.packages(HH) ## if necessary
library(HH)
?mmc
## After the full maiz example, then you need one more command
maiz.proj - proj(maiz.aov)
maiz.proj
## I think you are looking for
maiz.proj$Within[, Residuals]
##
Although your queries certainly intersect R, they are primarily about
statistical modeling, which is OT for this list. Your issues also
appear to be complex. I would therefore suggest that you eschew remote
Internet advice and consult a local statistical expert for help.
Cheers,
Bert
Bert Gunter
Bert,
My queries are directly related to R:
1. Can the R package frm can be used to compare nested models. If so, how.
2. Are there alternative R packages to perform ANOVAs on a dependent variable
that is a proportion with significant mass one one extreme?
Also, my statistical problem is well
Hi
You have only single value for each participant in each gruppo and AFAIK you
can not do statistic on single value.
You can check differences among participants
fit-lm(valor~participantes, data=df2)
fit-lm(valor~participantes, data=df2)
anova(fit)
Analysis of Variance Table
Response:
I think you start by doing your homework:
1. Read An Inroduction to R (ships with R) or other R online
tutorial. There are many good ones.
2. Use R's Help system:
?aov
?lm
?anova
there will be relevant links in these docs that you should follow,
especially to the use of formulas for model
Dear Catalin,
Have a look at the plyr package.
library(plyr)
dlply(
eg,
.(Exp),
function(x) {
aov(masa.uscat.tr~Clona,data=x)
}
)
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
But the ANOVA results is incorrect when I use
aov.out = aov(Response~-1+x1*x2*x3).
First, this doesn't work at all. Your variables are in upper case and your data
environment is not specified, so you must have done something else. I can get
an answer using something like
On 05/20/2013 10:52 AM, Luis Fernando García Hernández wrote:
Dear All,
This is a data relating leg shaking on differenntre treatments. I reused
several individuals so I want to know 1) if there are significative
differences on shaking per treatment and 2) if the reused individuals
presented
I would say that the OP should seek local statistical help, as he
appears to be out of his statistical depth. This would appear to be a
mixed effects models-type setup, but a local statistical expert would
be much better able to judge what the goals of the study were and what
sort of approach,
Jochen,
a) this is an English-spoken mailing list; other languages are not encouraged
nor will they typically generate a lot of replies...
b) your code is fine, so this is not an R-issue; you are rather stuck with some
of the stats background -- you might want to see a friendly local
Dear Claudia,
Your question has been posed on many previous occasions.
The (short) answer has always been the same: have a look at the Anova function
in the car package but before doing that, get a copy of John Fox's Applied
Regression Analysis and Generalized Linear Models book.
Best,
José
On 02/23/2013 08:55 PM, Robert Zimbardo wrote:
I have several linear models on the same data:
m1 - lm(y ~ poly(x,1))
m2 - lm(y ~ poly(x,2))
m3 - lm(y ~ poly(x,3))
What I don't understand is why
anova(m1, m2, m3, test=F)
- yields the same RSS and SS values, but a different p-value from
On Feb 23, 2013, at 10:06 , Rolf Turner wrote:
What am I missing?
A basic understanding of the theory of linear models. This really has little
to do with R. Go and read a good intro to linear modelling.
Insofar as your question has anything to do with R:
When you do
A (late) update to this question:
On Fri Aug 17 07:33:29, Henrik Singmann wrote:
Hi Diego,
I am struggeling with this question also for some time and there does
not seem to be an easy and general solution to this problem. At least
I
haven't found one.
However, if you have just one
Hi,
You can use the car package and choose Type-III test.
Also, have a look at the package easyanova.
Regards,
José
José Iparraguirre
Chief Economist
Age UK
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of humming-bird
Sent: 27
Now my question: Am I allowed to use these functions given
that my data is unbalanced?
Unusually, Yes, assuming all the other requisite assumptions are reasonably
well satisfied. One-way ANOVA interpretation is not much affected by imbalance
because (among other things) with only one factor
Hi Sachin,
You may find this tutorial useful:
http://goanna.cs.rmit.edu.au/~fscholer/anova.php
And you'll need the car package; but become yourself familiar with Type I, II
and III sums of squares models before running the Anova; the tutorial explains
these in detail.
Hope it helps.
José
I want to test whether the MEAN of two different variables,
(and different number of observations) are the same. I am
trying to use the anova test but it doesn't seem to like that
the number of observations are different:
a=c(1:5)
b=c(1:3)
aov_test=aov(a~b)
Error in
Thank you for the replies.
I am actually trying to gain p-values and f values, and tried the below
script but unsuccessful.
1) I have read in another forum to use the package lmer but apparently it
does not exist.
2) Then I tried the pvals.fnc but it is not a function.
3) I read also that a
Hello,
You're thinking of ?aov. anova() does _not_ have a formula interface, it
would be
anova(lm(HSuccess ~ Veg, data = data.to.analyze))
or
aov(HSuccess ~ Veg, data = data.to.analyze)
Hope this helps,
Rui Barradas
Em 05-10-2012 09:27, Jhope escreveu:
Hi R-listers,
I am trying to do an
Dear Jean,
On Fri, 5 Oct 2012 01:27:55 -0700 (PDT)
Jhope jeanwaij...@gmail.com wrote:
Hi R-listers,
I am trying to do an ANOVA for the following scatterplot and received the
following error:
library(car)
scatterplot(HSuccess ~ Veg,
data = data.to.analyze,
whether the str() I got matches with yours.
Have a great day!
A.K.
- Original Message -
From: Landi ent-ar...@gmx.de
To: r-help@r-project.org
Cc:
Sent: Friday, September 28, 2012 10:35 AM
Subject: Re: [R] Anova and tukey-grouping
Hello !
Thanks for your advice. I tried
typohne.csv http://r.789695.n4.nabble.com/file/n4644635/typohne.csv
Hello arun kirshna,
I checked str() but I do not see any mistake.
Find attached a subset of my data (did you mean this with dput() ?)
Typ= treatment
abun= abundance
div= diversity, number of species
best wishes.
--
View
Hello again !
It worked exactly the same way as yours!
I'm kind of astonished, because in my opinion our skripts are the same
(obviously not...).
Whatever! It works :-D
Tomorrow (as where I live it's already evening...sunday) will be a great
day, indeed, and I can go on with analysis !
I really
HI,
I guess there is a mistake in your code. You should have used typ instead of
abun as abun is the dependent variable.
summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks))
myresults - TukeyHSD(fm1, tension, ordered = TRUE)
library(agricolae)
HSD.test(fm1,wool,group=TRUE)
Hello !
Thanks for your advice. I tried it, but the output is the same:
HSD.test(anova.typabunmit, typ, group=TRUE)
Name: typ
ds.typabunmit$typ
I don't get the values...!?!?
--
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, 2012 10:35 AM
Subject: Re: [R] Anova and tukey-grouping
Hello !
Thanks for your advice. I tried it, but the output is the same:
HSD.test(anova.typabunmit, typ, group=TRUE)
Name: typ
ds.typabunmit$typ
I don't get the values...!?!?
--
View this message in context:
http://r.789695.n4
Please check if your independent variables or Xs are independent.
If they are not that can affect the sequential decomposition.
--
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Sent from the R help mailing list
Check out:
http://rtutorialseries.blogspot.com/2011/02/r-tutorial-series-two-way-repeated.html
On 8/15/2012 11:32 AM, Diego Bucci wrote:
Hi,
I performed an ANOVA repeated measures but I still can't find any good news
regarding the possibility to perform multiple comparisons.
Can anyone help
Hi Diego,
I am struggeling with this question also for some time and there does
not seem to be an easy and general solution to this problem. At least I
haven't found one.
However, if you have just one repeated-measures factor, use the solution
describe by me here:
-Original Message-
Say I have the following data:
a-data.frame(col1=c(rep(a,5),rep(b,7)),col2=runif(12))
a_aov-aov(a$col2~a$col1)
summary(aov)
Note that there are 5 observations for a and 7 for b, thus is
unbalanced. What would be the correct way of doing anova for
HI,
Check this link:
https://stat.ethz.ch/pipermail/r-help/2011-April/273858.html
A.K.
- Original Message -
From: Sachinthaka Abeywardana sachin.abeyward...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Monday, August 13, 2012 10:09 PM
Subject: [R] anova in unbalanced data
Hi all,
Dear Peter;
This is an exact duplicate of a question posted on SO. Cross-posting
is deprecated on Rhelp.
--
David.
On Jul 6, 2012, at 11:06 AM, mails wrote:
the study design of the data I have to analyse is simple. There is 1
control group (CTRL) and 2 different treatment groups (TREAT_1
Dear Peter,
Because your model is additive, type-II and type-III tests are identical,
and the t-tests for the linear and quadratic coefficients are interpretable.
I hope this helps,
John
John Fox
Sen. William McMaster Prof. of Social Statistics
On Jun 19, 2012, at 5:36 PM, James Johnson wrote:
Hi All,
I have a microarray dataset as follows:
expt1 expt2 expt3 expt4 expt 5
gene1val val val val val
gene2val val val valval
.
.
..
gene15000
Well that's that cleared up then. Thanks to all.
Chris B.
On 31/05/2012 17:51, Albyn Jones wrote:
No, both yield the same result: reject the null hypothesis,
which always corresponds to the restricted (smaller) model.
albyn
On Thu, May 31, 2012 at 12:47:30PM +0100, Chris Beeley wrote:
No, both yield the same result: reject the null hypothesis,
which always corresponds to the restricted (smaller) model.
albyn
On Thu, May 31, 2012 at 12:47:30PM +0100, Chris Beeley wrote:
Hello-
I understand that it's convention, when comparing two models using
the anova function
Dear Robert,
It is easier to use lm instead of aov if you want coefficients for each group.
Note that you can use rnorm vectorised.
set.seed(0)
N - 100 # sample size
MEAN - c(10, 20, 30, 40, 50)
VAR - c(20,20,1, 20, 20)
LABELS - factor(c(A, B, C, D, E))
# create a data frame with labels
df -
Hello Thierry,
thanks for your answer! There is one thing, however, that I don't
understand. The values labeled B in my data are generated with
1/20th the variance of the others, yet the standard error and
confidence intervals are the same for all levels of the factor. How
come?
Rob:
On Fri, May 4, 2012 at 9:18 AM, robgriffin247 rg.rfo...@hotmail.co.uk wrote:
Hi,
I need to create a data frame containing the results of a number of ANOVA's
but I'm having some trouble setting it up (some being enough for me to spend
3 days trying with no progress and be left staring in
The following constructs the data.frame that I think the original
poster asked for.
I don't understand the graph, so I didn't attempt it.
I agree with Bert that this might not make sense. Specifically, the distinction
between AB01 and AB02 is not modeled, and that is probably the critical
Gee Bert, thanks for the really helpful tip. But if you read my post properly
you'll note that I do know how ANOVA's work.
The anova of *G* /AB01 /would be some thing like: y=V, fixed=S, Random= L
L*S...
I didn't want to show a full model formula in case it led people do the
wrong path to
Thanks Richard, that works great on the test data,
I'll try it out on the full dataset now and let you know how it goes.
Thanks a lot!
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: Galen Sher galens...@gmail.com To: Brian S Cade ca...@usgs.gov
Date: 04/20/2012 09:59 AM Subject: Re: [R] ANOVA in quantreg - faulty
test for 'nesting'?
--
Thanks Brian. I think anova.glm() requires the user to specify the
appropriate distribution
Date: 04/20/2012 09:59 AM Subject: Re: [R] ANOVA in quantreg - faulty
test for 'nesting'?
--
Thanks Brian. I think anova.glm() requires the user to specify the
appropriate distribution. In the example above, if I use either of the
following commands
anova(f1,f2
On Apr 17, 2012, at 1:03 PM, MatthewJudd wrote:
Hello I am having some trouble performing a simple ANOVA in R
I am working with 2001 census tract information regarding religion.
I am
quite a rookie at this whole R experiance, but I would like to
improve my
skills.
*I read in my data
On Mar 18, 2012, at 08:21 , shoreliner11 wrote:
I'm still relatively new to R but was wondering if anyone could help me force
R to compute the f-statistic etc using the the nested term rather than the
residual. In my particular case we were nesting a treatment effect by a
replicated tank
And a further problem with both approaches: is 11pm (23) really that
different from 1am?
On Thu, Dec 8, 2011 at 3:28 PM, Michael comtech@gmail.com wrote:
Hi all,
If we wanted to study the effect on the mean of the hourly data based on
the hours within a day...
and we wanted to do Anova
On Dec 8, 2011, at 3:28 PM, Michael wrote:
Hi all,
If we wanted to study the effect on the mean of the hourly data
based on
the hours within a day...
and we wanted to do Anova analysis...
We have two choices:
Who is we and how were these constraints imposed?
Please see below:
Why
On Nov 19, 2011, at 15:37 , Roberto wrote:
First of all, thank you for the reply to my topic.
Anova summary has shown no significative difference into Treatment (Pr
0.374), otherwise LSD Test has shown difference between 2 Treatment.
I suppose that's something strange.
It isn't. You
I didn't meant to rewrite textbook. I asked only for a clarification to my
doubt, because I have started to study statistic recently.
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Roberto,
Please be more specific. What are the differences between ANOVA and LSD that
give you concern?
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
First of all, thank you for the reply to my topic.
Anova summary has shown no significative difference into Treatment (Pr
0.374), otherwise LSD Test has shown difference between 2 Treatment.
I suppose that's something strange.
Roberto
--
View this message in context:
Hi David,
Apologies again and thankyou for your help, I've edited my original post to
clarify what I was asking. What I meant was that the factor had only 1
degrees of freedom when it should have had 2 (14 in total), so you're right
there were 14 but not in the right place.
In SPSS you select
Thanks bbolker, that's really helpful. I'll look out for the book too, it
could be helpful!
Yours,
Sam
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On Oct 9, 2011, at 11:10 AM, shardman wrote:
Hi,
I'm trying to analyse some data I have imported into R from a .csv
file but
when I carry out the aov command the results show only one degree of
freedom
when there should be 14. Does anyone know why? I'd really appreciate
some
help, the
Hi David,
Thanks for your message.
I can assure you this is not homework. I'm working on an ecology project and
am trying to analyse the results from the fieldwork. I don't want other
people to do the work for me I was just hoping someone might be able to spot
where I have made a mistake, I'm
On Oct 9, 2011, at 2:08 PM, shardman wrote:
Hi David,
Thanks for your message.
I can assure you this is not homework. I'm working on an ecology
project and
am trying to analyse the results from the fieldwork. I don't want
other
people to do the work for me I was just hoping someone might
shardman samuelhardman at hotmail.com writes:
Hi,
I'm trying to analyse some data I have imported into R from a .csv file but
when I carry out the aov command the results show only one degree of freedom
when there should be 14. Does anyone know why? I'd really appreciate some
help, the
Hi Daniel,
Thanks for your response, I don't completely follow. I also should have
split this into two questions.
Regarding my first question:
Daniel Malter wrote:
You are modeling Condition * Stimulus * Group as fully interacted fixed
effects... A simple random effect for the individual
You are modeling Condition * Stimulus * Group as fully interacted fixed
effects. I do not actually think that you would need a complex random effect
structure for this. A simple random effect for the individual might suffice.
You could try to model this with lmer (in the lme4 library) and inspect
HronnE wrote on 09/27/2011 06:27:38 AM:
Hi all
This is probably a simple problem but somehow I am having much trouble
with
finding a solution, so I seek your help!
I have a data-set with continuous response variables. The explanatory
variably is 4xpH treatments (so 8.08, 7.94, 7.81 and
Thank you very much for a quick reply!
I had not realized the degrees of freedom changed. It was my lack of
understanding of the aov function.
I will continue defining the pH as factor for the ANOVA's.
Cheers,
Hronn
-
--
Hrönn Egilsdóttir
PhD Student
Marine Research Institute
Hi:
You can try something like this: assuming the factor variables of
interest and the response variable are in a data frame named df,
ivset - c(comma separated vector of factor names)
myaovs - lapply(ivset, function(x) {
form - as.formula(substitute(yvar ~ foo, list(foo = as.name(x
It is not nearly as complicated as Dennis Murphy makes out.
Just do
aov(y ~ ., data=X)
where X is your data frame with one column (the response) name ``y''
and then any number of other columns which will then form the
predictors (which may be either numeric predictors or factors).
Where do I begin troubleshooting such a problem?
1. By examining your data to see whether you have entered it correctly.
2. By reading the posting Guide at the bottom of this and every
message and doing as it asks (posting a portion of your data and R
code so that we can see what you did).
3.
Please read the first two paragraphs of Details from ?formula
Ok, I just did.
So, If I understand this properly, the term Plot*Day would include both the
main effects of a and b and their second order interactions. So it could be
written Plot + Day + Plot:Day.
The term Plot:Day includes only
So what is the difference between a colon and an asterisk in this code? For
that matter what does the slash mean?
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David,
Please read the first two paragraphs of Details from ?formula
You can try out the formulas with the notation
expand.formula - function(f) colnames(attr(terms(f), factors))
expand.formula(~a+b)
[1] a b
expand.formula(~a:b)
[1] a:b
expand.formula(~a*b)
[1] a b a:b
On May 5, 2011, at 23:30 , Rovinpiper wrote:
Thanks slre,
I seem to be making some progress now.
Using a colon instead of an asterisk in the code really changes things. I
had been getting residual SS and MS of zero. Which is ridiculous. Now I get
much more plausible values.
Also,
Rovinpiper david.j.mee...@gmail.com 04/05/2011 22:43
So this seems to indicate that I have what I want. I have two
respiration data points at each plot on each day.
Yes; if you had only Plot+Day you'd have a completely balanced full
factorial ... for Plot and Day.
But I think I now see an
On May 5, 2011, at 2:31 PM, Asan Ramzan wrote:
Hello R-Help
How can i exctact and store the within group mean squared
difference from an
anova summary table into a varible.
In the absence of an example and code it's speculation, but something
along the lines of:
anova(fit)$`Mean Sq`
Thanks slre,
I seem to be making some progress now.
Using a colon instead of an asterisk in the code really changes things. I
had been getting residual SS and MS of zero. Which is ridiculous. Now I get
much more plausible values.
Also, When I used an asterisk instead of a colon It wouldn't
This response went to my email:
Without your data it's hard to say, but one possibility is that your
plots are nested within treatments instead of crossed, or that you have
something rather more cunning going on involving the Days. For example
if you had 8 days for six of your plots and another 8
And I responded as follows:
Hi,
Thanks for your advice. I tried using table() to check for missing
data. Here are the results:
table(Combined.Plot)
Combined.Plot
60m A1 B1 B3 B4 C5 C9 D2 D9 F60m F8 Q7
34 34 3434 343434 3434 34 34 34
Most likely your combined.trt is linearly dependent on the combined.plot
factor. Try
Anova.Trt.D.M.T.Pr.Model - aov(Combined.Rs ~ as.factor(Combined.Plot)
.
and see if combined.plot now has the 11 df you are anticipating.
Rich
On Tue, May 3, 2011 at 3:37 PM, Rovinpiper
Hi Richard,
Thanks for your advice.
I think that your suggestion is that I run the ANOVA with Combined.Plot as a
factor. I have tried that does not alleviate the problem.
Did I understand you properly?
Do you have another idea?
Thanks,
David
--
View this message in context:
On Fri, Mar 11, 2011 at 8:25 AM, Brian McLoone brianbmclo...@gmail.com wrote:
This is a follow-up to a query that was posted regarding some problems that
emerge when running anova analyses for cox models, posted by Mathias Gondan:
Matthias Gondan wrote:
* Dear List, I have tried a
I can hopefully save bandwidth here by suggesting that this belongs on
the R-sig-mixed-models list.
-- Bert
As an aside, shouldn't you be figuring this out yourself or seeking
local consulting expertise?
On Fri, Feb 25, 2011 at 9:08 AM, Ben Ward benjamin.w...@bathspa.org wrote:
Hi, As part of
See comments inline.
On 2011-02-06 03:17, Jinsong Zhao wrote:
Hi there,
I have a data frame as listed below:
Ca.P.Biomass.A
P Biomass
1 334.5567 0.287
2 737.5400 0.571
3 894.5300 0.639
4 782.3800 0.5836667
5 857.5900 0.600
6 829.2700 0.588
I have fit the
I'm not sure what a real ANOVA diagram is supposed to look like, nor
do I know what your data look like.
But this might get you started:
fakedata - runif(100)
fakegroups - sample(rep(letters[1:5], each=20))
boxplot(fakedata ~ fakegroups)
If that isn't what you're after, a clearer explanation
I recently started using R and I have a simple question. I am running R
(v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I
know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI. However,
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