> On Mar 29, 2025, at 9:59 AM, Kevin Zembower via R-help
> wrote:
>
> Hi, Rui and Michael, thank you both for replying.
>
> Yeah, I'm not supposed to know about Chi-squared yet. So far, all of
> our work with hypothesis tests has involved creating the sample data,
> then resampling it to cre
As always, I would like to thank all who responded for their insights and
suggestions. I have learned from them.
Thus far, my own aesthetic preference -- and therefore not to be considered
in any sense as a "best" approach -- is to use Duncan's suggestion to
produce the call directly with call() r
I am confused. Richard's answer that Bert did not like did not use parse
explicitly. Richard pasted together a string that a function like lm() will
have to parse to run the analysis. However, the answers so far do not use
parse(). In the reply to Richard, Bert indicated we cannot use strings. E
my take of the assignment was to avoid 'parse' specifically.
we start with a character vector, so avoiding characters is not possible. i was
dealing with the fortune "if parse is the answer, you have the wrong question"
Sent from my iPhone
On Mar 29, 2025, at 15:39, Bert Gunter wrote:
Thanks
Your approach seems reasonable to me, though I wouldn't do it that way.
The way I work this sort of thing out is to get R to do a simple
example, then see what it did, and duplicate that.
For example,
f <- ~ (Heigh + Ho + Silver + Away)^2
Then look at as.list(f), as.list(f[[2]]), as.list(
The general formula is y ~ a + b + c + ...
There is this approach:
formula <- reformulate(independent_vars, response = "y")
model <- lm(formula, data = mydata)
summary(model)
It does not generate a string object, but the formula is still a string even if
it is of class formula. Also, in this app
> somenames <- c("Heigh", "Ho", "Silver", "Away")
> as.formula(paste("~(",paste(somenames, collapse="+"),")^2"))
~(Heigh + Ho + Silver + Away)^2
>
> On Mar 29, 2025, at 14:30, Bert Gunter wrote:
>
> somenames <- c("Heigh", "Ho", "Silver", "Away")
__
The computer intensive approaches (randomization, permutation, bootstrap,
jackknife) are awesome when you have enough data. In this age we are all about
huge data sets. Yet basic agricultural research often does not come close. I
have three to ten replicates per treatment.
-Original Message
Às 19:19 de 29/03/2025, Ebert,Timothy Aaron escreveu:
How about calculating a 95% confidence interval about the estimated proportion
in favor. The PooledInfRate package will do this for you. If confidence
intervals overlap then there is no significant difference.
-Original Message-
Fro
Note: I am almost certain that this has been asked and answered here
before, so my apologies for the redundant query.
I also know that there are several packages that will do this, but I wish
to do it using base R functions only (see below).
The query: Suppose I have a character vector of names l
>
> I've been setting up problems like this with code similar to:
> ===
> df <- data.frame(
> survey = c(rep("1980", 1000), rep("2010", 1000)),
> DP = c(rep("Y", 0.66*1000), rep("N", 1000 - (0.66*1000)),
>rep("Y", 0.64*1000), rep("N", 1000 - (0.64*1000)))
How about calculating a 95% confidence interval about the estimated proportion
in favor. The PooledInfRate package will do this for you. If confidence
intervals overlap then there is no significant difference.
-Original Message-
From: R-help On Behalf Of Kevin Zembower via
R-help
Sent:
Às 16:09 de 29/03/2025, Kevin Zembower via R-help escreveu:
Hello, all,
We're now starting to cover hypothesis tests in my Stats 101 course. As
usual in courses using the Lock5 textbook, 3rd ed., the homework
answers are calculated using their StatKey application. In addition
(and for no extra c
Hello, all,
We're now starting to cover hypothesis tests in my Stats 101 course. As
usual in courses using the Lock5 textbook, 3rd ed., the homework
answers are calculated using their StatKey application. In addition
(and for no extra credit), I'm trying to solve the problems using R. In
the case
Dear Kevin
Unless it is a course requirement that you do it this way it would be
easier to use the chisq.test function. You can then just use the
frequencies which you have (660, 340, 640, 360). I will not give you
example code since your learning would be enhanced by having to do it
yourself
This is great, thank you so much for your help!
I found the problem, the data contains 3 and 4 numbers. I solved it.
Thank you
On Sat, Mar 29, 2025 at 2:21 AM CALUM POLWART wrote:
> So either:
>
> XX contains some non-finite numbers
>
> Or
>
> Something produced from it does.
>
> Non finite mea
So either:
XX contains some non-finite numbers
Or
Something produced from it does.
Non finite means things like 1÷0. R might show Inf or -Inf. But if there is
maths happening in the function it might be passing something to it the is
a 0 and so creating a Inf.
But if there is a test of is.fini
How are you dear Helper?
I am from Africa, Ethiopia , I have faced this type of problem
Error in optim(seq(-3, 3, length.out = k), wle.fit, method = "BFGS",
control = list(maxit = max.it)) :
initial value in 'vmmin' is not finite
when I want to run RM.w
> wt = beha$wt> rr = RM.w(XX, wt)
libra
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