Hello,
As for R^2 in Excel for models without an intercept, maybe the following
are relevant.
https://support.microsoft.com/en-us/help/829249/you-will-receive-an-incorrect-r-squared-value-in-the-chart-tool-in-exc
https://stat.ethz.ch/pipermail/r-help/2012-July/318347.html
Hope this helps,
This is an old discussion. The thing that R is doing is to compare the model to
the model without any regressors, which in the no-intercept case is the
constant zero. Otherwise, you would be comparing non-nested models and the R^2
would not satisfy the property of being between 0 and 1.
A
See also this thread in stats.stackexchange
https://stats.stackexchange.com/questions/26176/removal-of-statistically-significant-intercept-term-increases-r2-in-linear-mo
On Thu, Sep 27, 2018 at 3:43 PM, J C Nash wrote:
> This issue that traces back to the very unfortunate use
> of R-squared
This issue that traces back to the very unfortunate use
of R-squared as the name of a tool to simply compare a model to the model that
is a single number (the mean). The mean can be shown to be the optimal choice
for a model that is a single number, so it makes sense to try to do better.
The OP
I have a query on the R-squared correlation coefficient for linear
regression through the origin.
The general expression for R-squared in regression (whether linear or
non-linear) is
R-squared = 1 - sum(y-ypredicted)^2 / sum(y-ybar)^2
However, the lm function within R does not seem to use this
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