R help - Jul 2010 - Correct statistical inference for linear regression models without intercept in R

Dear R community,

is there a way to get correct t- and p-values and R squared for linear
regression models specified without an intercept?

example model:
summary(lm(y ~ 0 + x))

This gives too low p-values and too high R squared. Is there a way to
correct it? Or should I specify with intercept to get the correct values?

Thank you in advance!

Wojtek Musial
-- 
View this message in context:
http://r.789695.n4.nabble.com/Correct-statistical-inference-for-linear-regression-models-without-intercept-in-R-tp2295193p2295193.html
Sent from the R help mailing list archive at Nabble.com.

What x and y represent? Are they non-stationary, trending? then you would get
very high R2 (~97-99%) and very low p-value. Perhaps you land on the world
of spurious regression.

In this case forcing intercept to zero would not help you. Work with
differenced series instead raw data.

Thanks and regards,
-- 
View this message in context:
http://r.789695.n4.nabble.com/Correct-statistical-inference-for-linear-regression-models-without-intercept-in-R-tp2295193p2295230.html
Sent from the R help mailing list archive at Nabble.com.

Hi:

On Tue, Jul 20, 2010 at 2:41 AM, StatWM <wmusial@gmx.de> wrote:
>
> Dear R community,
>
> is there a way to get correct t- and p-values and R squared for linear
> regression models specified without an intercept?
>
> example model:
> summary(lm(y ~ 0 + x))
>
> This gives too low p-values and too high R squared. Is there a way to
> correct it? Or should I specify with intercept to get the correct values?
>
How do you know that the p-value is too low and R^2 is too high? Too low or
too high compared to what? You've constrained the intercept of the model to
pass through zero, which affects several features of a simple linear
regression model. For example, sum the residuals from your no-intercept
model - I'll bet they don't add to zero. Do you think that might affect 
a
few things? Here's an example:

# Generate some data; notice that the true y-intercept is 2 and the true
slope is 2
dd <- data.frame(x = 1:10, y = 2 + 2 * 1:10 + rnorm(10))
plot(y ~ x, data = dd, xlim = c(0, 10), ylim = c(0, 25))
m1 <- lm(y ~ x, data = dd)
abline(coef(m1))
m2 <- lm(y ~ x + 0, data = dd)
abline(c(0, coef(m2)), lty = 'dotted')

# As you noted, the no-intercept model has a higher R^2,
# even though the 'usual' simple linear regression (SLR)
# model provided a better visual fit. Why?
summary(m1)$r.squared
[1] 0.982328
summary(m2)$r.squared
[1] 0.9946863

# The p-value for the F-test on the slope is higher in the
# no-intercept model is lower than in the SLR model. Why?
anova(m1)

Analysis of Variance Table

Response: y
          Df Sum Sq Mean Sq F value    Pr(>F)
x          1 385.22  385.22  444.69 2.686e-08 ***
Residuals  8   6.93    0.87
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

anova(m2)

Analysis of Variance Table

Response: y
          Df  Sum Sq Mean Sq F value    Pr(>F)
x          1 2164.07 2164.07  1684.7 1.507e-11 ***
Residuals  9   11.56    1.28


Look at the differences in sums of squares between the two models, both in
terms of model SS and error SS. What is responsible for those differences?
Once you understand that, it becomes clear why the apparent anomalies in R^2
and in the F-test occur by applying the definitions. Also try

sum(m1$resid)
sum(m2$resid)

Why is there a difference? Why dies m2$resid not have to sum to zero?

(Hint: The output in each case is correct, so it's not an R problem. You
need to derive the differences among the various quantities in regression
modeling between the intercept and no-intercept models to understand the
paradox.)

HTH,
Dennis

Thank you in advance!
>
> Wojtek Musial
> --
> View this message in context:
>
http://r.789695.n4.nabble.com/Correct-statistical-inference-for-linear-regression-models-without-intercept-in-R-tp2295193p2295193.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

On Jul 20, 2010, at 11:41 AM, StatWM wrote:
> 
> Dear R community,
> 
> is there a way to get correct t- and p-values and R squared for linear
> regression models specified without an intercept?
> 
> example model:
> summary(lm(y ~ 0 + x))
> 
> This gives too low p-values and too high R squared. Is there a way to
> correct it? Or should I specify with intercept to get the correct values?
They are already correct. If you want incorrect ones, please specify their
definition...

> Thank you in advance!
> 
> Wojtek Musial
> -- 
> View this message in context:
http://r.789695.n4.nabble.com/Correct-statistical-inference-for-linear-regression-models-without-intercept-in-R-tp2295193p2295193.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

Thank you very much for your effort!

But is there a measure, which can compare the goodness of fit of regression
models with and without the intercept? Can I only compare them in terms of
sum of squares residual?
-- 
View this message in context:
http://r.789695.n4.nabble.com/Correct-statistical-inference-for-linear-regression-models-without-intercept-in-R-tp2295193p2295960.html
Sent from the R help mailing list archive at Nabble.com.