[R] Correct statistical inference for linear regression modelswithout intercept in R

Setlhare Lekgatlhamang SetlhareL at bob.bw
Fri Jul 23 17:34:01 CEST 2010


In addition, there are 'theoretical' reasons for excluding intercept
from the model that must be considered. The reasons related to the
regressor(s) and depend on the phenomenon being modelled. For example,
whereas the intercept can be excluded in a bivariate model on the
expenditure of an individual as determined by income, the same would be
senseless when applying the same model to the population.

Hope this helps a little

Lexi

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Dennis Murphy
Sent: Tuesday, July 20, 2010 12:34 PM
To: StatWM
Cc: r-help at r-project.org
Subject: Re: [R] Correct statistical inference for linear regression
modelswithout intercept in R

Hi:

On Tue, Jul 20, 2010 at 2:41 AM, StatWM <wmusial at 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.
>
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