Displaying 17 results from an estimated 17 matches for "treatment1".
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2010 Oct 28
1
xyplot and panel.curve
Hi All
I have regression coefficients from an experiment and I want to plot them
in lattice using panel curve but I have run into error messages.
I want an 3 panel conditioned plot of 2 curves of Treatment 2 in each panel
conditioned by Treatment1, the example curve expression is x+value*x^2
A rough toy example to give an idea of what I want is:
Data:
data = expand.grid(Treatment1 = LETTERS[1:3],Treatment2 = letters[1:2])
data$value =seq(1.1,1.6,0.1)
data
Treatment1 Treatment2 value
1 A a 1.1
2 B a...
2002 Sep 11
0
Contrasts with interactions
...I then set up the contrasts;
contrasts(treatment)<-c(1,1,1,1,1,1,-8/3,1,-8/3,1,-8/3)
summary.lm(model);
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.30411 0.22355 5.834 2.11e-08 ***
dryweight 0.05445 0.02556 2.130 0.034358 *
treatment1 -0.01390 0.10283 -0.135 0.892575
treatment2 -0.25015 0.50841 -0.492 0.623230
treatment3 -0.65174 0.50580 -1.289 0.199026
treatment4 0.35105 0.42871 0.819 0.413838
treatment5 -0.15977 0.46738 -0.342 0.732827
treat...
2010 May 18
1
proportion of treatment effect by a surrogate (fitting multivariate survival model)
...989 JASA), and is implemented in Lin's MULCOX2, SAS,
and S-plus.
Is this the way to fit such a model in R?
Suppose I have variables: time, delta, treatment, and surrogate.
Should I repeat the dataset (2x) and stack, creating the variables:
time1 (time repeated 2x), delta1 (delta repeated 2x), treatment1 (same
as treatment, but 0's for the 2nd set), treatment2 (0's in first set,
then same as treatment), and surrogate2 (0's in first set, then same
as treatment), and id (label the subject, so each id should have 2
observations).
Thus, a dataset with n observations will become 2n observat...
2008 Feb 03
1
Effect size of comparison of two levels of a factor in multiple linear regression
...ment==1], outcome[treatment==0])
[1] 3.984774
cohens.d(outcome[treatment==2], outcome[treatment==0])
[1] 6.167798
# Sometimes standardized regression coefficients are recommended
# for determining effect size but that clearly doesn't work here:
coef(lm(scale(outcome) ~ treatment))
(Intercept) treatment1 treatment2
-1.233366 1.453152 2.246946
# The reason it doesn't work is that the difference of outcome
# means is divided by the sd of *all* outcomes:
(mean(outcome[treatment==1])-mean(outcome[treatment==0]))/sd(outcome)
[1] 1.453152
(mean(outcome[treatment==2])-mean(outcome[treatment==...
2008 Oct 09
1
Interpretation in cor()
Hello,
I am performing cor() of some of my data. For example, I'll do 3 corr()
(many variables) operations, one for each of the three treatments.
I then do the following:
i <-lower.tri(treatment1.cor)
cor(cbind(one = treatment1.corr[i], two = treatment2.corr[i], three =
treatment3.corr[i]))
Does this operation above tell me how correlated each of the three
treatments is? Because this how I am interpreting it.
Thanks,
Michael Just
[[alternative HTML version deleted]]
2005 Oct 26
1
Post Hoc Groupings
...create a grouping table. e.g., if I have treatments 1, 2,
and 3, with 1 and 2 being statistically the same and 3 being different
from both
Group Treatment
A 1
A 2
B 3
2) I've been stumbling over the proper syntax for simple effects for a
tukeyHSD test. Is it
TukeyHSD(model.aov, "Treatment1", "Treatment2")
or
TukeyHSD(model, c("Treatment1", "Treatment2"))
or something else, as neither of those seem to really work.
2008 Oct 10
1
Correlation among correlation matrices cor() - Interpretation
...lt;- c(11, 222, 333, 4444, 55555, 1.1, 2.2, 3.3, 4.4, .55)
> var3 <-c(22.2, 66.7, 99.9, 1000008, 123123, .1, .2, .3, .4, .5)
> var4<- c(.000001,.00001,.0001, .001, .1, .12345, .56789, .67890, .78901,
.89012)
> dat <- cbind(var1,var2,var3,var4)
> dat.d <- data.frame(dat)
> treatment1 <- dat.d[1:5,]
> treatment2 <-dat.d[6:10,]
> t1.d.cor <- cor(treatment1)
> t2.d.cor <- cor(treatment2)
> I <-lower.tri(t1.d.cor)
> t1.t2 <- cor(cbind(T1 = t1.d.cor[I], T2 = t2.d.cor[I]))
> t1.t2
T1 T2
T1 1.0000000 0.2802750
T2 0.2802750 1.0000000...
2008 Jun 05
1
quite complicated case(the repeated data arranage~)
...y<-c(20,40,30,11,23,24,56,65,60)
id<-c(1,1,3,4,4,6,8,9,9)
table1<-cbind(treatment,id,age,y)
*the actual data are way more than this*,the id is from 1~500,and not in
regular ,some number missing~
all I want to do is
put the cases to variable according the id
for example when id =1
we have
treatment1 age1 y
low 50 20
high 60 40
this will generate a new matrix
for this example I will have 6 new matrix,according to id.
it is reasonable to do this in loop for,but the I met some problem:
1:how to automatically generate the new title such as treatment1 and age1
un...
2006 Sep 05
1
help: advice on the structuring of ReML models for analysing growth curves
...ume that this should be a
random factor? Growth is not linear exactly (more quadratic), so I thought
rather than put time in the fixed model I want to control for the effects
of time as a random factor....
The resulting model is this
where id=chick identity and brood=nest box
model1<-lmer(weight~treatment1*treatment2*brood
size*sex+(id|brood)+(1|brood)+(1|age), data=H)
Is this the "right" approach or am I barking up the wrong tree?
Any suggestions much appreciated,
Simon
Simon Pickett
PhD student
Centre For Ecology and Conservation
Tremough Campus
University of Exeter in Cornwall
TR109EZ...
2005 May 23
0
using lme in csimtest
...onse ~ treatment * (site + time), whichf =
"treatment", type = "Tukey")
#
# Tukey contrasts for factor treatment, covariables: site +time
+treatment:site + treatment:time
#
#Coefficients:
# Estimate t value Std.Err. p raw p Bonf p adj
#treatment3-treatment1 -0.655 -2.004 0.327 0.050 0.149 0.120
#treatment3-treatment2 -0.581 -1.777 0.327 0.081 0.162 0.143
#treatment2-treatment1 -0.074 -0.227 0.327 0.821 0.821 0.821
___
drs. René Eschen
CABI Bioscience Switzerland Centre
1 Rue des Grillons
CH-2800 Delémont
Switzerland
+41 32 421 48...
2013 Feb 13
2
NA/NaN/Inf in foreign function call (arg 6) error from coxph function
...2] 0.8925+ 1.8836+ 2.1191+ 5.3744+
1.6099+ 5.2567 0.2081+ 0.2108+ 0.2683+ 0.4873+ ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:2] "time" "status"
..- attr(*, "type")= chr "right"
$ therapy : Factor w/ 2 levels "treatment1","treatment2": 1 1 1 1 1
1 1 2 2 1 ...
$ ReceptorA: Factor w/ 2 levels "0","1": 1 2 2 2 1 2 2 2 2 1 ...
$ ReceptorB: Factor w/ 2 levels "0","1": 1 2 1 1 2 1 1 1 1 1 ...
But when I tried to fit a multivariate Cox proportional model, I got
the...
2006 May 09
2
post hoc comparison in repeated measure
Hi, I have a simple dataset with repeated measures.
one factor is treatment with 3 levels (treatment1,
treatment2 and control), the other factor is time (15
time points). Each treatment group has 10 subjects
with each followed up at each time points, the
response variable is numeric, serum protein amount. So
the between subject factor is treatment, and the
within subject factor is time. I ran a 2-w...
2008 Sep 17
1
ANOVA contrast matrix vs. TukeyHSD?
...not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 328.73096 0.26303 1249.770 < 2e-16 ***
genders -37.39069 0.19661 -190.179 < 2e-16 ***
condu -37.47740 0.19693 -190.308 < 2e-16 ***
treatment1 0.51026 0.40084 1.273 0.203079
treatment2 -0.17333 0.23175 -0.748 0.454541
treatment3 0.07761 0.22535 0.344 0.730566
treatment4 -1.96020 0.38524 -5.088 3.73e-07 ***
treatment5 NA NA NA NA...
2010 Jul 07
0
interaction post hoc/ lme repeated measures
...iously, since the subjects only have two legs, they can’t receive each combination of treatment. The groups are unbalanced so I can’t use aov. As I understand it, lme should work but I am having a tough time figuring out what to use as a random term. Is this correct?
x.lme <- lme(effect~bone*treatment1*treatment2, random= ~1|subject, data=x)
anova(x.lme)
I’ve tried a number of different random terms and get very similar results (and even get similar results running aov) but I’d like to know what the proper way of doing it is.
My next question is how to do post hoc tests on this. One website r...
2013 Feb 12
0
NA/NaN/Inf in foreign function call (arg 6) error from coxph
...2] 0.8925+ 1.8836+ 2.1191+ 5.3744+
1.6099+ 5.2567 0.2081+ 0.2108+ 0.2683+ 0.4873+ ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:2] "time" "status"
..- attr(*, "type")= chr "right"
$ therapy : Factor w/ 2 levels "treatment1","treatment2": 1 1 1 1 1
1 1 2 2 1 ...
$ ReceptorA: Factor w/ 2 levels "0","1": 1 2 2 2 1 2 2 2 2 1 ...
$ ReceptorB: Factor w/ 2 levels "0","1": 1 2 1 1 2 1 1 1 1 1 ...
But when I tried to fit a multivariate Cox proportional model, I got
the...
2009 May 01
0
Confusion going from Stata -> R
...ets base level to "ME"
attributes(d$i.st)
# Difference-in-difference estimate
reg1 <- lm(hwover ~ i.treatment*i.year + i.st, data=d, weights=weight1)
summary(reg1)
# Output (correctly) omit state "14" ("ME")
# Estimates of average treatment effect (coefficient on i.treatment1:i.year1985) same as Stata's
# So is the coefficient on i.year1985.
# All other params are off.
For comparison, I am pasting Stata's output below.
Any ideas would be much appreciated!
xi: reg hwover i.treatment*i.year i.st if st1976==1 [pw=weight1]
i.treatment _Itreatment_0-1 (...
2007 Aug 03
4
FW: Selecting undefined column of a data frame (was [BioC] read.phenoData vs read.AnnotatedDataFrame)
...filenames=pData(pd)$FileName,phenoData=pd)
> ##normalisation
> eset.rma<-rma(Data)
> ##analysis
> targs<-factor(pData(pd)$Target)
> design<-model.matrix(~0+targs)
> colnames(design)<-levels(targs)
> fit<-lmFit(eset.rma,design)
> cont.wt<-makeContrasts("treatment1-control","treatment2-control",level
> s=
> design)
> fit2<-contrasts.fit(fit,cont.wt)
> fit2.eb<-eBayes(fit2)
> testconts<-classifyTestsF(fit2.eb,p.value=0.01)
> topTable(fit2.eb,coef=2,n=300)
> topTable(fit2.eb,coef=1,n=300)
>
>
> [[alternat...