Dear R-experts,
I am trying to fit a GAM with 2 binary predictors (variables coded 0,1). I
guess I cannot just smooth binary variables. By the way I code them as
0=no,1=yes, then mgcv will think those variables are numeric.
I have tried to change 0 and 1 in no and yes. It does not work.
How to
Dear R-experts,
Here below my R code working with quite a few warnings.
x11 and x12 are dichotomous variable (0=no and 1=yes). I substract 1 to ignore
intercept.
I would like not to ignore intercept. How to modify my R code because if I just
remove -1 it does not work?
y= c(32,45,65,34,23,43,
11 feb. 2023 om 18:35 schreef varin sacha via R-help
:
> Dear R-experts,
>
> I am trying to fit a GAM with 2 binary predictors (variables coded 0,1). I
> guess I cannot just smooth binary variables. By the way I code them as
> 0=no,1=yes, then mgcv will think those variables are
> On 21 Feb 2023, at 22:33 , varin sacha via R-help
> wrote:
>
> Dear R-experts,
>
> Here below my R code working with quite a few warnings.
> x11 and x12 are dichotomous variable (0=no and 1=yes). I substract 1 to
> ignore intercept.
> I would like not to ignore i
Dear R-experts,
Here below my R code working but I would like to plot (get the graph) of the
Causal additive model exactly the same way I get the graph (CPDAG) from the
bnlearn R package. Is it possible? If yes, how?
Many thanks
###
# libraries
library(devtools)
Dear R-experts,
How to solve that problem?
My R version is 4.2.1
Here below trying to install RGBL library found here :
https://bioconductor.org/packages/release/bioc/html/RBGL.html
So, I run this R code :
if (!require("BiocManager", quietly = TRUE))
install.packages("BiocManager")
BiocM
’
The following objects are masked from ‘package:igraph’:
bfs, dfs, transitivity
Le samedi 22 avril 2023 à 18:12:56 UTC+2, Eric Berger a
écrit :
What happens with the command
> library(RBGL)
On Sat, Apr 22, 2023 at 7:08 PM varin sacha via R-help
wrote:
> Dear R-e
> degree, edges, intersection
>
>
> Attaching package: ‘RBGL’
>
> The following objects are masked from ‘package:igraph’:
>
> bfs, dfs, transitivity
>
>
>
>
>
>
>
>
>
>
> Le samedi 22 avril 2023 à 18:12:56 UTC+2, Eric Berger
> a écrit :
Bert,
Thanks ! It works !
Best,
Le samedi 22 avril 2023 à 19:42:18 UTC+2, Bert Gunter
a écrit :
Is lvida.R in your working directory?
Try using the full path name to the file instead in source()
-- Bert
On Sat, Apr 22, 2023 at 9:38 AM varin sacha via R-help
wrote:
>
&g
Me again ! How to solve this?
At the end of this page there is the installation command :
https://gitlab.science.ru.nl/gbucur/RUcausal/-/blob/master/README.Rmd
Working with a MAC, I have tried to install the RUcausal library (copy and
paste the installation command).
It is written that the pack
Dear R-experts,
Here below a toy example to calculate the MSE (mean squared error).
Starting from this equation : MSE = bias^2 + variance + irreducible error
I am trying to get the bias and the variance in addition to the MSE. How to get
them both?
Many thanks for your help.
###
Dear R-experts,
Here below my R code. I get a NaN response for gam with mgcv library. How to
solve that problem?
Many thanks.
#
library(mgcv)
y=c(23,24,34,40,42,43,54,34,52,54,23,32,35,45,46,54,34,36,37,48)
x1=c(0.1,0.3,0.5,0.7,0.8,0.9,0.
Dear Simon,
Thanks ! It works !
Best,
Le lundi 1 mai 2023 à 11:19:26 UTC+2, Simon Wood a écrit
:
try...
sum(residuals(model1)^2)
On 30/04/2023 22:03, varin sacha via R-help wrote:
> Dear R-experts,
>
> Here below my R code. I get a NaN response for gam with mgcv librar
Dear R-experts,
Here below my R code (toy example) working! The only thing missing is in my GAM
plot: I would like to get on the same graph the 2 bands (prediction and
confidence bands) like in my lm model graph!
Is it possible? How to get that?
Dear R-Experts,
In a data simulation, I would like a balanced distribution with a nested
structure for classroom and teacher (not for school). I mean 50 pupils
belonging to C1, 50 other pupils belonging to C2, 50 other pupils belonging to
C3 and so on. Then I want the 50 pupils belonging to C1
","C2","C3","C4","C5","C6"), 50) # make a character
vector, with 50 "C1", 50 "C2", ...
classroom <- tmp[sample(1:300)] # make a random permutation.
Certainly you may also make it into one line:
classroom <- rep(c("C
quot;c", 1:5, sep=""), sep = "."))
}
pupils <- character()
for (class in classes) {
pupils <- c(pupils, paste(class, paste("p", 1:10, sep=""), sep = "."))
}
B.
> On 2019-05-18, at 09:57, varin sacha via R-help wrote:
>
&
, May 18, 2019 at 10:04 PM varin sacha via R-help
wrote:
>
> Dear Boris,
>
> Yes, top-down, no problem. Many thanks, but in your code did you not forget
> "teacher" ? As a reminder teacher has to be nested with classes. I mean the
> 50 pupils belonging to C1 must be
on of
> what exactly you want.
>
> I assume the situation is that you know what a data structure you
> want, but do not know
> how to conveniently create such structure.
> And that is where others can help you.
> So, please, describe the wanted data structure more thoroughly,
>
Dear R-Experts,
Here is a toy example, reproducible example, I get error messages. I have tried
to fix it by myself using "google is my friend", but I did not get it. If
somebody can help me to fix these errors, would be highly appreciated.
##
install.packages("fitdistrp
ist(floor(x), "pois")
plot(f2p)
summary(f2p)
Not so well
##negative binomial distribution
f2n <- fitdist(floor(x), "nbinom")
f2n <- fitdist(ceiling(x), "nbinom")
f2n <- fitdist(round(x), "nbinom")
plot(f2n)
summary(f2n)
Hope this helps,
Rui Ba
Dear Experts,
I have fitted MARS and GAM models on a real dataset. My goal is prediction. I
have run crossvalidation many times to get an idea of the out-of-bag accuracy
value. I use the Mean Squared Error (MSE) as an error evaluation criterion. I
have published my paper and the reviewers ask m
Dear Abby,
Many thanks for your response.
To answer your question. For me better all the x variables (collectively), to
have m% missing values.
When you tell me : "Modify your code so that a single function say sim.test()
computes your simulated statistics, for n sample size and m missing valu
Dear R-helpers,
Doing dput(x) and dput(y_obs), the 2 vectors are not the same length (1800 for
y_obs and 2000 for x)
How can I solve the problem ?
Here is the reproducible R code
# # # # # # # # # #
library(mgcv)
library(earth)
n<-2000
x<-runif(n, 0, 5)
y_model<- 0.1*x^3 -
)^2)
MSE_MARS<-mean((mars_model$fitted.values - y_model)^2)
MSE_GAM
MSE_MARS
Le mardi 17 septembre 2019 à 22:27:54 UTC+2, David Winsemius
a écrit :
On 9/17/19 12:48 PM, varin sacha via R-help wrote:
> Dear R-helpers,
>
> Doing dput(x) and dput(y_obs), the 2 vectors are not
obs <- c( rnorm(n*0.9, y_model, 0.1), rnorm(n*0.1, y_model, 0.5) )
then y_obs:
> length(y_obs)
[1] 2000
De: varin sacha via R-help
Enviado: martes, 17 de septiembre de 2019 21:49
Para: R-help Mailing List
Asunto: [R] Not the same length
Dear R-helpers,
Doing dput(x) and dput(y_ob
s <- y_model + c(rnorm(0.9 * n, 0, 0.1), rnorm(0.1 * n, 0, 0.5))
or
y_obs <- rnorm(n, y_model, rep(c(0.1, 0.5), c(.9 * n, .1 * n)))
-pd
> On 17 Sep 2019, at 22:27 , David Winsemius wrote:
>
>
> On 9/17/19 12:48 PM, varin sacha via R-help wrote:
>> Dear R-helpers,
>>
Ar ! What a pity, I have not seen it ! Many thanks !
Le mercredi 18 septembre 2019 à 19:07:42 UTC+2, peter dalgaard
a écrit :
Redefining n is probably not a good idea...
[...snip...]
> m <-runif(n, 0, 5)
> n <-rnorm(n, 2, 3)
Oops! n is now a vector of length 2000.
[...snip...]
>
Dear R-Experts,
Here is my reproducible R code to get the Mean squared error of GAM and MARS
after I = 50 iterations/replications.
If I want to get the 95% bootstrap CIs around the MSE of GAM and around the MSE
of MARS, how can I complete/modify my R code ?
Many thanks for your precious help.
ootResults <-boot(data=data,statistic=mse,R=1000)
##
Le lundi 23 septembre 2019 à 21:42:56 UTC+2, varin sacha via R-help
a écrit :
Dear R-Experts,
Here is my reproducible R code to get the Mean squared error of GAM and MARS
after I =
type)
mse <- function(data,i) {
boot.earth <- earth((y_obs~x+z+a),data=data[i,])
mean(boot.earth$residuals^2)
}
bootResults <- boot(data=data, statistic=mse, R=1000)
boot.ci(bootResults, type = boot.ci.type)
Hope this helps,
Rui Barradas
Às 13:43 de 25/09/19, varin sacha via R-help es
Dear R-Experts,
Here below my reproducible example working but not entirely (working). What I
understand is that there is a problem of libraries library(hbrfit) and ... ?
How can I make it work entirely, many thanks for your precious help.
SIMULATION STUDY 3 variables with 10% outliers
1] hbrfit_0.02 Rfit_0.23.0 RobStatTM_1.0.1 fit.models_0.5-14
[5] RobPer_1.2.2 rgenoud_5.8-3.0 BB_2019.10-1 quantreg_5.51
[9] SparseM_1.77 MASS_7.3-51.4 robustbase_0.93-5
```
There is no error or warning, except that MSE_fastTau is an NaN. What problem
are you l
Dear David, Dear Jiefei,
Many thanks for your comments. I got it now. It works.
Best,
Sacha
Le lundi 21 octobre 2019 à 22:00:39 UTC+2, David Winsemius
a écrit :
On 10/21/19 9:40 AM, varin sacha via R-help wrote:
> Dear R-Experts,
>
> Here below my reproducible example wo
Dear R-experts,
My reproducible example here below is not working because of an error message :
Erreur : vecteurs de mémoire épuisés (limite atteinte ?)
My code perfectly works when n=3000 or n=5000 but as soon as n=1 my code
does not work anymore. By the way, my code takes a very long time
Dear R-experts,
Here below a toy example with some error messages, especially at the end of the
code (Tuning the parameters). Your help to correct my R code would be highly
appreciated.
###
#libraries
library(lattice)
library(ggplot2)
library(caret)
library(
, May 9, 2023 at 6:40 AM Eric Berger wrote:
> Hi,
> One problem you have is with the command:
> regr<-randomForest(y~x1+x2, data=X_train, proximity=TRUE)
>
> What you need is something like this:
>
> X2 <- cbind(X,y)
> regr<-randomForest(y~x1+x2, data=X2, proxi
R-experts,
I am trying to find the exact 95% confidence intervals of the mean (and the
median) for the weibull distribution. Here below I have the exact 95% CIs of
the meanlog and the median for a lognormal using EnvStats library (ready-to-run
function in library) but I don't find how to get th
Dear R-experts,
Here below my R code with an error message. Can somebody help me to fix this
error?
Really appreciate your help.
Best,
# MSE CROSSVALIDATION Lasso regression
library(glmnet)
x1=c(34,35,12,13,15,37,65,45,47,67,87,
o that it
>is
> >> > not necessary to rerun your code. This might enable someone to see
>the
> >> > problem without running the code (e.g. downloading packages, etc.)
> >>
> >> And it's not necessarily true
Dear Rui,
I really thank you a lot for your response and your R code.
Best,
Sacha
Le mardi 24 octobre 2023 à 16:37:56 UTC+2, Rui Barradas
a écrit :
Às 20:12 de 23/10/2023, varin sacha via R-help escreveu:
> Dear R-experts,
>
> I really thank you all a lot for your responses.
Dear R-Experts,
Here below my R code working but I don't know how to complete/finish my R code
to get the final plot with the extrapolation for the10 more years.
Indeed, I try to extrapolate my data with a linear fit over the next 10 years.
So I create a date sequence for the next 10 years and
Dear Rui,
I really thank you a lot for your precious R help. It is exactly what I was
trying to do! Once more, many thanks!
Best,
Sacha
Le vendredi 27 octobre 2023 à 09:36:18 UTC+2, Rui Barradas
a écrit :
Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
> Dear R-Expe
R-Experts,
Here below my R code working without error message but I don't get the results
I am expecting.
Here is the result I get:
[1] "All values of t are equal to 0.28611928397257 \n Cannot calculate
confidence intervals"
NULL
If someone knows how to solve my problem, really appreciate.
Bes
y's are not looked for in data (i.e. NSE) but in the environment
where the function was defined, which is standard evaluation. Change the above
to:
cor1 <- with(d, cor(x1, y1, method="spearman"))
cor2 <- with(d, cor(x2, y2, method="spearman"))
and all should be fi
Dear R-experts,
Here below my R code, as my X-axis is "year", I must be missing one or more
steps! I am trying to get the regression line with the 95% confidence bands
around the regression line. Any help would be appreciated.
Best,
S.
#
library(g
; On 12/10/2023 2:50 PM, Rui Barradas wrote:
>> Às 22:35 de 10/12/2023, varin sacha via R-help escreveu:
>>>
>>> Dear R-experts,
>>>
>>> Here below my R code, as my X-axis is "year", I must be missing one
>>> or more steps! I am trying to
Dear R-experts,
Here below, my R code working BUT I get a strange result I was not expecting!
Indeed, the 95% percentile bootstrap CIs is (-54.81, -54.81 ). Is anything
going wrong?
Best,
##
Score=c(345,564,467,675,432,346,476,512,567,543,234,435,654,411
Time)
library(boot)
func= function(data, idx) {
coef(lm(Score~ Time + factor(Country),data=data[idx,]))
}
B= boot(e, func, R=1000)
boot.ci(B, index=2, type="perc")
#
Le samedi 13 janvier 2024 à 21:56:58 UTC+1, Ivan Kr
Dear R-experts,
I really thank you all for your responses.
Best,
Le dimanche 14 janvier 2024 à 10:22:12 UTC+1, Duncan Murdoch
a écrit :
On 13/01/2024 8:58 p.m., Rolf Turner wrote:
> On Sat, 13 Jan 2024 17:59:16 -0500
> Duncan Murdoch wrote:
>
>
>
>> My guess is that one of the boot
Dear R-experts,
I write to you to know if somebody is aware of a R package (or function) able
to plot graphs for extrapolation.
I need to be clear on what extrapolation really is to me. It is when we use the
model for X variables outside the range of X variables that were used to
construct the
Good afternoon,
Here below my reproducible R code. I don't get any results. I am looking for
MSE_fastMM value and the bootstrap CIs around MSE_fastMM value. How can I
finish/correct my R code to get the results ?
Many thanks for your help.
install.packages( "robustbase",dep
t.MM <- lmrob(y_obs~b+z+a,data=data[i,])
mean(boot.MM$residuals^2)
}
bootResults_MM <-boot(data=data, statistic=MSE_fastMM, R=100)
boot.ci(bootResults_MM, type = boot.ci.type)
}
Le mercredi 18 mars 2020 à 09:44:19 UTC+1, peter dalgaard a
écrit :
The double
Dear R-experts,
Here below my R code giving an error message that I don't understand. If
somebody can help me to fix it, it would be highly appreciated.
# # # # # # # # # # # # # # # # # # # # # # # #
install.packages( "robustbase",dependencies=TRUE )
install.packages( "boot",dependencies=TRUE )
AM, varin sacha via R-help wrote:
> Dear R-experts,
>
> Here below my R code giving an error message that I don't understand. If
> somebody can help me to fix it, it would be highly appreciated.
>
> # # # # # # # # # # # # # # # # # # # # # # # #
> install.pack
Hi David,
Perfect I got it now. One last precision, if I want to get the MSE value how
can I get it ?
Le samedi 21 mars 2020 à 20:26:37 UTC+1, David Winsemius
a écrit :
On 3/21/20 12:18 PM, varin sacha wrote:
> mean((d[["y_obs "]] - ypred)^2)
__
David,
Great... exactly what I was looking for... Many thanks.
Le samedi 21 mars 2020 à 21:12:31 UTC+1, David Winsemius
a écrit :
On 3/21/20 12:56 PM, varin sacha wrote:
> Hi David,
>
> Perfect I got it now. One last precision, if I want to get the MSE value how
> can I get it ?
If
Dear R-helpers,
Using the HBR (high breakdown rank-based) robust estimator and the hbrfit
function, I get an error saying Error in UseMethod("predict") for hbrfit. How
can I solve the problem ? Many thanks for your help.
# # # # # # # # # # # # # # # # # # # # # # # #
install.packages( "robustb
Dear R-helpers,
Another problem with FastTau function from the RobPer packages. Any solution to
solve my problem would be highly appreciated.
# # # # # # # # # # # # # # # # # # # # # # # #
install.packages( "boot",dependencies=TRUE )
install.packages( "RobPer",dependencies=TRUE )
library(boo
e results reproducible
set.seed(1234)
# bootstrapping with 60 replications
results <- boot(data = df, statistic = MSE,
R = 60, formula = y_obs ~ b+z+a)
str(results)
boot.ci(results, type="bca" )
# # # # # # # # # # # # # # # # # # # # # # # #
Le dimanche 22 m
= MSE,
> R = 10, formula = ~b+z+a)
>
> type <- c("norm","basic", "stud", "perc", "bca")
> boot.ci(results, type = type[-5])
>
>
> Hope this helps,
>
> Rui Barradas
>
> Às 23:14 de 21/
Dear R-experts,
The rlm command in the MASS package command implements several versions of
robust regression, for example the Huber and the Tukey (bisquare weighting
function) estimators.
In my R code here below I try to get the Tukey (bisquare weighting function)
estimation, R gives me an erro
n fantastic when it was introduced (into S /
S-plus, before R existed [in a publicly visible way]) but it had
been based of what was available back then, end of the 80's, beginning 90's.
Martin
> On 23/03/2020 12:39, varin sacha via R-help wrote:
>> Dear R-expert
Dear R-experts,
Here below my "toy" reproducible example showing many warnings and an error
message. What I am interested in is the error message.
Of course I can access and view the results doing : summary(results$t) to give
me an idea of what is going on. But I don't know how to correct/solve
s)
boot.ci(results, type="norm" )
str(Dataset)
str(newdata)
summary(results$t)
# # # # # # # # # # # # # # # # # # # # # # # # #
Le lundi 30 mars 2020 à 00:40:33 UTC+2, David Winsemius
a écrit :
On 3/28/20 12:25 PM, varin sacha via R-help wrote:
> Dear R-experts,
>
> Here below my
R = 100, formula = crp ~ bmi+glucose+age+sex,method="MM")
>
> str(results)
> boot.ci(results, type="norm" )
>
> str(Dataset)
> str(newdata)
> summary(results$t)
> # # # # # # # # # # # # # # # # # # # # # # # # #
>
>
>
>
>
>
Hi,
A google search does not give me any hint :=(
Maybe somebody can help me to fix the error message I get : Error in
robdist.hbrfit(x) : x is probably collinear
# # # # # # # # # # # # # # # # # # # ## # # # # # # # # # # # # # #
bmi=c(23,43,21,23,45,65,45,11,12,13,23,34,NA,NA,34,35,45,65,43
coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Mon, Mar 30, 2020 at 12:10 PM varin sacha via R-help
wrote:
>
> Hi,
>
> A google search does not give me any hint :=(
> Maybe somebody can help me to fix
Dear R-experts,
Here below my reproducible example. I would like to fit/add the Gauss normal
curve to this data.
I don't get it. There is no error message but I don't get what I am looking
for.
Many thanks for your help.
mydates <-
gaard a
écrit :
Two obvious problems:
1. mean(nc) is a count, not a date, sd likewise
2. the scale of dnorm() is density, not count
So (slightly inefficient, but who cares...):
y <- rep(mydates, nc)
n <- sum(nc)
curve(n*dnorm(x, mean(y), sd(y)), add=TRUE, col="
Dear R-experts,
I am trying to find the best span for my loess regression. Here below a
reproducible example. I don't get the result. Am I missing something ?
Many thanks for your help.
a<-c(2,3,4,3,2,6,5,7,4,5,12,13,21,6,4,5,6,7)
b<-
Dear Ivan,
Many thanks I got it now.
Best,
Le mercredi 15 avril 2020 à 09:49:26 UTC+2, Ivan Krylov
a écrit :
On Tue, 14 Apr 2020 21:00:34 + (UTC)
varin sacha via R-help wrote:
> Here below a reproducible example. I don't get the result.
Thanks for providing a conci
Dear R-experts,
Here below the reproducible example. I can not add the "green" gam curve on the
plot. The last line of my R code gives me problem. There is a message error
that I don't understand. Many thanks for your precious help.
###
a <- a
David,
Rui,
Many thanks for your response. Thanks Rui it perfectly works.
Best,
Le jeudi 23 avril 2020 à 06:18:46 UTC+2, Rui Barradas a
écrit :
Hello,
Inline.
Às 23:29 de 22/04/20, varin sacha via R-help escreveu:
> Dear R-experts,
>
> Here below the reproducible e
Dear R-experts,
I am trying to fit a gaussian density curve. More precisely, I would like to
obtain the fitted "Gaussian curve".
"m" is the gaussian mean, "sd" is the standard deviation and "k" is an
arbitrary scaling parameter (since the gaussian density is constrained to
integrate to 1, wher
till could be a function representing a probability
distribution...
On Sun, Apr 26, 2020 at 7:09 AM varin sacha via R-help
wrote:
>
> Dear R-experts,
>
> I am trying to fit a gaussian density curve. More precisely, I would like to
> obtain the fitted "Gaussian curve".
Dear R-experts,
My goal is to get only 1 value : the average/ the mean of the 100 MSE values.
How can I finish my R code ?
###
my.experiment <- function() {
n<-500
x<-runif(n, 0, 5)
z <- rnorm(n, 2, 3)
a <- runif(n, 0, 5)
y_mode
Dear R-experts,
Here is my R code, I get a result but I also get an error message so I doubt I
can trust the result I get.
What is going wrong ? Many thanks.
a<-c(2,4,3,4,6,5,3,1,2,3,4,3,4,5,65)
b<-c(23,45,32,12,23,43,56,44,33,11,12,54,23,34,54)
d<-c(9,4
0
On Tue, May 19, 2020 at 3:51 PM varin sacha via R-help
wrote:
> Dear R-experts,
>
> Here is my R code, I get a result but I also get an error message so I doubt
> I can trust the result I get.
> What is going wrong ? Many thanks.
>
>
C+2, Rui Barradas a
écrit :
Hello,
Inline.
Às 21:38 de 19/05/20, varin sacha via R-help escreveu:
>
> Hi Richard,
>
> Thanks for your response.
> However, how can I correct my R code knowing that I want, as a result, only
> one value : the mean of the 500 MSE_OLS values ?
ent <- function() {
>
> OLS <- lm( a ~ b+d )
>
> MSE_OLS<-mean(OLS$residuals^2)
>
> return( c(MSE_OLS) )
> }
>
> my.data = replicate( 500, my.experiment() )
> colnames(my.data) <- c("MSE_OLS")
> mean(my.data)
> ###########
Dear R-experts,
I am trying to do a permutation test for the Ramsey RESET test. More precisely,
I am interested in the "exact" p-value of the test.
I have checked the coin package and all the functions (oneway_test; ...). There
are plenty of functions but no one is helping me for RESET test. I
Dear R-Experts,
Is there an all-ready function to calculate the Double MAD (Median absolute
deviation) as there is an easy function to calculate the MAD "mad function". Or
I have to write my own function for Double MAD ?
To calculate the double MAD, the idea is the following : for the obtained
))
}
out
}
double.mad(x)
# lo hi
#0.81543 0.44478
double.mad(c(x, 1))
# lo hi
#2.29803 0.44478
double.mad(c(x, 1), include.right = TRUE)
# lo hi
#1.03782 1.63086
Hope this helps,
Rui Barradas
Às 15:22 de 03/08/2020, varin sacha via R-help escreveu:
> Dear R-Exper
Dear R-experts,
Using the bootES package I can easily calculate the bootstrap confidence
intervals of the means like in the toy example here below. Now, I am looking
for the confidence intervals for the difference between group means. In my
case, the point estimate of the mean difference is 64.
Dear Matthias,
Many thanks for your response.
Best,
SV
Le mardi 4 août 2020 à 16:22:41 UTC+2, Prof. Dr. Matthias Kohl
a écrit :
you could try:
library(MKinfer)
meanDiffCI(a, b, boot = TRUE)
Best
Matthias
Am 04.08.20 um 16:08 schrieb varin sacha via R-help:
> Dear R-expe
Dear R-experts,
I have fitted an orthogonal regression and have some difficulties to get the
adjusted R^2 and R^2, the AICc, the coefficients and R^2 bootstrap confidence
intervals. Here below my R codes.
Many thanks for your precious help.
y=c(231,212,112,143,154,165,1
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