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,
Rui Barradas
Às 11:56 de 27/09/2018, Patrick Barrie escreveu:
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 expression
when the intercept is constrained to be zero. It gives results different
to Excel and other data analysis packages.
As an example (using built-in cars dataframe):
cars.lm=lm(dist ~ 0+speed, data=cars) # linear regression through
origin
summary(cars.lm)$r.squared # report R-squared [1] 0.8962893 >
1-deviance(cars.lm)/sum((cars$dist-mean(cars$dist))^2) # calculates
R-squared directly [1] 0.6018997 > # The latter corresponds to the value
reported by Excel (and other data analysis packages) > > # Note that we
expect R-squared to be smaller for linear regression through the origin
> # than for linear regression without a constraint (which is 0.6511 in
this example)
Does anyone know what R is doing in this case? Is there an option to get
R to return what I termed the "general" expression for R-squared? The
adjusted R-squared value is also affected. [Other parameters all seem
correct.]
Thanks for any help on this issue,
Patrick
P.S. I believe old versions of Excel (before 2003) also had this issue.
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