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lm(y ~ x1) vs. lm(y ~ x0 + x1 - 1) with x0 <- rep(1, length(y))

4 messages · David Winsemius, Bert Gunter, Jochen Laubrock

Jochen Laubrock
# 
Dear list,

the following came up in an introductory class. Please help me understand the -1 (or 0+) syntax in formulae: Why do the enumerator dfs, F-statisics etc. differ between the models lm(y ~ x1) and lm(y ~ x0 + x1 - 1), if x0 is a vector containing simply ones?

Example:

N  <- 40
x0 <- rep(1,N)
x1 <- 1:N
vare <- N/8
set.seed(4)
e <- rnorm(N, 0, vare^2)

X <- cbind(x0, x1)
beta <- c(.4, 1)
y <- X %*% beta + e

summary(lm(y ~ x1))
# [...]
# Residual standard error: 20.92 on 38 degrees of freedom
# Multiple R-squared: 0.1151,	Adjusted R-squared: 0.09182 
# F-statistic: 4.943 on 1 and 38 DF,  p-value: 0.03222 

summary(lm(y ~ x0 + x1 - 1))        # or summary(lm(y ~ 0 + x0 + x1))
# [...]
# Residual standard error: 20.92 on 38 degrees of freedom
# Multiple R-squared: 0.6888,	Adjusted R-squared: 0.6724
# F-statistic: 42.05 on 2 and 38 DF,  p-value: 2.338e-10 


Thanks in advance,
Jochen


----
Jochen Laubrock, Dept. of Psychology, University of Potsdam,
Karl-Liebknecht-Strasse 24-25, 14476 Potsdam, Germany
phone: +49-331-977-2346, fax: +49-331-977-2793
David Winsemius
#
On Jan 21, 2011, at 9:03 PM, jochen laubrock wrote:

            

      
      
      
    

        
You are testing something different. In the first case you are testing  
the difference between the baseline and the second level of x1 (so  
there is only one d.f.), while in the second case you are testing for  
both of the coefficients being zero (so the numerator has 2 d.f.). It  
would be easier to see if you did print() on the fit object. The first  
model would give you an estimate for an "Intercept", which is really  
an estimate for the first level of x1.  Having been taught to think of  
anova as just a special case of regression is helpful here. Look at  
the model first  and only then look at the anova table.

            

      
      
      
    

        

      
      
    

        
David Winsemius, MD
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      Bert Gunter
    

    

      
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Well ... as x1 is continuous(numeric), it has no levels. So ...??

Note that the fits are identical for both models. The issue is only
what is the Null that you are testing in the two cases. In the first
case, it is just y = constant, so you are testing the 1 df for x1. In
the second, it is y = 0 (which rarely makes any sense) and you are
testing the 2 df for the two terms (x0 and x1). Etc. etc.

-- Bert

        
On Fri, Jan 21, 2011 at 7:48 PM, David Winsemius <dwinsemius at comcast.net> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Jochen Laubrock
    

    

      
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Thank you all (including Dennis), this was elucidating. 

I would have (maybe naively) anticipated that in this somewhat pathological case of fitting without an intercept and re-introducing it via constant x1, R might check whether the design matrix includes a column of ones, and adjust the degrees of freedom accordingly. But now I can see that by explicitly requesting via the formula interface not to fit a constant, I am implicitly stating my hypothesis that y==0, even if I re-introduce my suspicion that y==mu via x1 <- 1. If I understood correctly, x1 is treated as a variable in the latter case, right?

        
On Jan 22, 2011, at 5:18 , Bert Gunter wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        
----
Jochen Laubrock, Dept. of Psychology, University of Potsdam,
Karl-Liebknecht-Strasse 24-25, 14476 Potsdam, Germany
phone: +49-331-977-2346, fax: +49-331-977-2793