Displaying 20 results from an estimated 200 matches similar to: "Extracting standard errors for adjusted fixed effect sizes in lmer"
2006 May 15
1
anova statistics in lmer
Dear list members,
I am new to R and to the R-help list. I am trying to perform a
mixed-model analysis using the lmer() function. I have a problem with
the output anova table when using the anova() function on the lmer
output object: I only get the numerator d.f., the sum of squares and the
mean squares, but not the denominator d.f., F statistics and P values.
Below is a sample output, following
2006 Oct 05
2
treatment effect at specific time point within mixedeffects model
Hi David:
In looking at your original post it is a bit difficult to ascertain
exactly what your null hypothesis was. That is, you want to assess
whether there is a treatment effect at time 3, but compared to what. I
think your second post clears this up. You should refer to pages 224-
225 of Pinhiero and Bates for your answer. This shows how to specify
contrasts.
> -----Original Message-----
2018 Mar 05
0
data analysis for partial two-by-two factorial design
> On Mar 5, 2018, at 2:27 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>
> David:
>
> I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected:
>> three groups, no drugA/no drugB, yes drugA/no drugB,
2018 Mar 05
2
data analysis for partial two-by-two factorial design
But of course the whole point of additivity is to decompose the combined
effect as the sum of individual effects.
"Mislead" is a subjective judgment, so no comment. The explanation I
provided is standard. I used it for decades when I taught in industry.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into
2018 Mar 05
0
data analysis for partial two-by-two factorial design
> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>
> But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects.
Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a
2018 Mar 05
0
data analysis for partial two-by-two factorial design
> On Mar 5, 2018, at 8:52 AM, Ding, Yuan Chun <ycding at coh.org> wrote:
>
> Hi Bert,
>
> I am very sorry to bother you again.
>
> For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply.
>
> I really hope that you can do me a great favor to share your points about how to explain the
2018 Mar 05
2
data analysis for partial two-by-two factorial design
Hi Bert,
I am very sorry to bother you again.
For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply.
I really hope that you can do me a great favor to share your points about how to explain the coefficients for drug A and drug B if run anova model (response variable = drug A + drug B). is it different from running three
2018 Mar 05
0
data analysis for partial two-by-two factorial design
Hi Bert and David,
Thank you so much for willingness to spend some time on my problem!!! I have some statistical knowledge (going to get a master in applied statisitics), but do not have a chance to purse a phD for statistics, so I am always be careful before starting to do analysis and hope to gather supportive information from real statisticians.
Sorry that I did not tell more info about
2018 Mar 05
5
data analysis for partial two-by-two factorial design
David:
I believe your response on SO is incorrect. This is a standard OFAT (one
factor at a time) design, so that assuming additivity (no interactions),
the effects of drugA and drugB can be determined via the model you rejected:
For example, if baseline control (no drugs) has a response of 0, drugA has
an effect of 1, drugB has an effect of 2, and the effects are additive,
with no noise we
2018 Mar 02
3
data analysis for partial two-by-two factorial design
Dear R users,
I need to analyze data generated from a partial two-by-two factorial design: two levels for drug A (yes, no), two levels for drug B (yes, no); however, data points are available only for three groups, no drugA/no drugB, yes drugA/no drugB, yes drugA/yes drug B, omitting the fourth group of no drugA/yes drugB. I think we can not investigate interaction between drug A and drug B,
2018 Mar 02
0
data analysis for partial two-by-two factorial design
This list provides help on R programming (see the posting guide linked
below for details on what is/is not considered on topic), and generally
avoids discussion of purely statistical issues, which is what your query
appears to be. The simple answer is yes, you can fit the model as
described, but you clearly need the off topic discussion as to what it
does or does not mean. For that, you might try
2007 Jul 06
1
maintaining specified factor contrasts when subsetting in lmer
All,
I'm using lmer for some repeated measures data and have specified
the contrasts for a time factor such that say time 3 is the base. This
works fine. However, when
I next use the subset argument to remove the last two time values, the
output indicates that
the specified contrast is not maintained (see below). I can solve this
by creating a new dataframe
for the subset of interest
2011 Oct 17
3
Extracting results from a function output
Hello,
I am having hard time obtaining a value from a function. "fit" is a survival
function that produces some results, such as "median", "confidence
intervals" etc. But str() function does not list these values. How can I
extract these to be able use them? For example, I need "median" value for
the group DrugA which is 48. "Print" function does
2006 Sep 26
2
treatment effect at specific time point within mixed effects model
All,
The code below is for a pseudo dataset of repeated measures on patients
where there is also a treatment factor called "drug". Time is treated
as categorical.
What code is necessary to test for a treatment effect at a single time
point,
e.g., time = 3? Does the answer matter if the design is a crossover
design,
i.e, each patient received drug and placebo?
Finally, what would
2009 Oct 10
2
easy way to find all extractor functions and the datatypes of what they return
Am I asking for too much:
for any object that a stat proc returns ( y <- lm( y~x) , etc ) ) , is there
a super convenient function like give_all_extractors( y ) that lists all
extractor functions , the datatype returned , and a text descriptor
field ("pairwisepval" "lsmean" etc)
That would just be so convenient.
What are my options for querying an object so that I can
2011 Oct 20
3
Survival analysis
Hello,
I need some results from the survival analysis of my data
that I do not know whether exist in Survival Package or how to obtain if
they do:
1. The Mean survival time
2. The standard error of the mean
3. Point and 95% Lower & Upper Confidence Intervals estimates
Any help will be greatly appreciated.
Cem
[[alternative HTML version
2009 Oct 27
0
syntax for estimable(gmodels package) and glht(multcomp package)
Hello,
I have a question as to how the syntax for glht(package multcomp) and
estimable (gmodels) works, since I'm not getting everything from the
documents I've googled so far, especially with models with 2nd order
terms.
A modestly complex model:
2-way anova with one continuous covariate, no random effects(and no
repeated measures) to keep it modestly complex:
Y = treatmentgroup + sex
2003 May 02
2
stepAIC/lme (1.6.2)
Based on the stepAIC help, I have assumed that it only was for lm, aov, and
glm models. I gather from the following correspondence that it also works
with lme models.
Thomas Lumley 07:40 a.m. 28/04/03 -0700 4 Re: [R] stepAIC/lme problem
(1.7.0 only)
Prof Brian Ripley 04:19 p.m. 28/04/03 +0100 6 Re: [R] stepAIC/lme problem
(1.7.0 only)
Prof Brian Ripley 06:09 p.m. 29/04/03 +0100 6 Re: [R]
2007 Mar 29
1
ccf time units
Hi,
I am using ccf but I could not figure out how to calculate the actual lag in
number of periods from the returned results. The documentation for ccf
says:"The lag is returned and plotted in units of time". What does "units of
time" mean? For example:
> x=ldeaths
> x1=lag(ldeaths,1)
> results=ccf(x,x1)
> results
Autocorrelations of series 'X', by lag
2010 Jul 06
1
acf
Hi list,
I have the following code to compute the acf of a time series
acfresid <- acf(residfit), where residfit is the series
when I type acfresid at the prompt the follwoing is displayed
Autocorrelations of series ?residfit?, by lag
0.0000 0.0833 0.1667 0.2500 0.3333 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333
1.000 -0.015 0.010 0.099 0.048 -0.014 -0.039 -0.019 0.040 0.018