[R] lm without intercept

[Mike Marchywka]

----------------------------------------
Date: Fri, 18 Feb 2011 12:25:36 +0100
From: Achim.Zeileis at uibk.ac.at
To: jrheinlaender at gmx.de
CC: r-help at r-project.org
Subject: Re: [R] lm without intercept


On Fri, 18 Feb 2011, Jan wrote:

> Hi,
>
> I am not a statistics expert, so I have this question. A linear model
> gives me the following summary:

You should be able to find the equations presumably
used for the computer output you question, as suggested by the earlier response, 
but then I would add that you should make up data with known
attributes, and verify that the output makes sense to you. Paper, pencil,
and books still work and they don't run out of memory :) 

In any case, eventually you will not to modify your real data slightly
to explore sensitivity so what you did above will not just be a learning
effort but produce useful stuff. 

Also, it doesn't appear that you have even plotted your data.

As someone once told me just as a general philosophical point
rather than an implied insult, " do things that make sense." ( not
asking " are you an idiot" just along the lines of "buy low, sell high").
All of this computer stuff comes from something, you just need to
dig into that. 



>
> Call:
> lm(formula = N ~ N_alt)
>
> Residuals:
>    Min      1Q  Median      3Q     Max
> -110.30  -35.80  -22.77   38.07  122.76
>
> Coefficients:
>            Estimate Std. Error t value Pr(>|t|)
> (Intercept)  13.5177   229.0764   0.059   0.9535
> N_alt         0.2832     0.1501   1.886   0.0739 .
> ---
> Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
>
> Residual standard error: 56.77 on 20 degrees of freedom
>  (16 observations deleted due to missingness)
> Multiple R-squared: 0.151, Adjusted R-squared: 0.1086
> F-statistic: 3.558 on 1 and 20 DF,  p-value: 0.07386
>
> The regression is not very good (high p-value, low R-squared).

Yes.

> The Pr value for the intercept seems to indicate that it is zero with a
> very high probability (95.35%).

Not quite. Consult your statistics textbook for the correct interpretation
of p-values. Under the null hypothesis of a true intercept of zero, it is
very likely to observe an intercept as large as 13.52 or larger.

> So I repeat the regression forcing the intercept to zero:

Do you have a good interpretation for that?

> Call:
> lm(formula = N ~ N_alt - 1)
>
> Residuals:
>    Min      1Q  Median      3Q     Max
> -110.11  -36.35  -22.13   38.59  123.23
>
> Coefficients:
>      Estimate Std. Error t value Pr(>|t|)
> N_alt 0.292046   0.007742   37.72   <2e-16 ***
> ---
> Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
>
> Residual standard error: 55.41 on 21 degrees of freedom
>  (16 observations deleted due to missingness)
> Multiple R-squared: 0.9855, Adjusted R-squared: 0.9848
> F-statistic:  1423 on 1 and 21 DF,  p-value: < 2.2e-16
>
> 1. Is my interpretation correct?
> 2. Is it possible that just by forcing the intercept to become zero, a
> bad regression becomes an extremely good one?
> 3. Why doesn't lm suggest a value of zero (or near zero) by itself if
> the regression is so much better with it?

The model without intercept needs to be interpreted differently. The
p-value pertains to a regression with intercept zero and slope 0.292
against a model with both intercept zero and slope zero. If I had to
guess, I would say this is not a very meaningful comparison for your data.
The same is true for the R-squared (see also ?summary.lm for its
definition in the case without intercept).

hth,
Z

> Please excuse my ignorance.
>
> Jan Rheinländer
>
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