You can run that as it is. The term to search for on Google is 'dummy
coding'.
Jeremy
On 28 December 2012 07:45, Lorenzo Isella wrote:
>
> where x3 is a dichotomous variable assuming only 0 and 1 values (x1 and x2
> are continuous variables).
> Is there any particular caveat I should be aware o
Dear All,
A semi-trivial question: suppose you want to carry out a linear regression
of the kind
y~x1+x2+x3
where x3 is a dichotomous variable assuming only 0 and 1 values (x1 and x2
are continuous variables).
Is there any particular caveat I should be aware of? Can I code this as a
simpl
-requ...@r-project.org; r-help@r-project.org
> Subject: [R] Question about linear regression in R
>
> Hi all,
> I wrote a r program as below:
>
> x <- 1:10
> y <- c(3,3,3,3,3,3,3,3,3,3)
>
> fit <- lm(log(y) ~ x)
> summary(fit)
>
> And I expect to ge
On Nov 14, 2011, at 10:49 PM, Miles Yang wrote:
Hi all,
I wrote a r program as below:
x <- 1:10
y <- c(3,3,3,3,3,3,3,3,3,3)
fit <- lm(log(y) ~ x)
summary(fit)
And I expect to get some error message from R, because "y" is
constant.
But, I got the message as below:
You are asking R to tel
What exactly is it that's worrying you? It's a problematic regression
for a few reasons, but ultimately it seems pretty ok, though I'd be
ever so slightly worried about the R^2 value being misinterpreted.
Michael
On Mon, Nov 14, 2011 at 10:49 PM, Miles Yang wrote:
> Hi all,
> I wrote a r program
Hi all,
I wrote a r program as below:
x <- 1:10
y <- c(3,3,3,3,3,3,3,3,3,3)
fit <- lm(log(y) ~ x)
summary(fit)
And I expect to get some error message from R, because "y" is constant.
But, I got the message as below:
> summary(fit)
Call:
lm(formula = log(y) ~ x)
Residuals:
Min 1
You really really need to consult with a local statistician for help.
You are making a valiant effort, but it is clear that you have
insufficient background and experience. Get help from an expert if you
can. It is no dishonor, you will learn a lot, and you will avoid
incorrect conclusions.
Cheers
Dear David,
Thanks for your answer. Yes now that you mentioned these points are in
the beginning of a variable range. From the plot of the residuals seems
to have non constant variance which is solved by a transformation. I
checked also for interactions by using the symbol : between two
variables
On Jun 21, 2011, at 3:49 AM, George Markomanolis wrote:
Dear all,
I am new to this field and I have a question about a linear
regression.
I have a dataset of around to 31000 points and I want to apply a
linear
regression. The R-squared is 0.9 however when I check the diagnostic
plots I ca
Dear all,
I am new to this field and I have a question about a linear regression.
I have a dataset of around to 31000 points and I want to apply a linear
regression. The R-squared is 0.9 however when I check the diagnostic
plots I can see that there are around to 250 points with big leverage
value
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