Message-ID: <BANLkTinAZqq5cU4PW74VemBYK-mSVY_xMw@mail.gmail.com>
Date: 2011-04-18T12:37:12Z
From: Dorien Herremans
Subject: Rsquared for anova
In-Reply-To: <BANLkTimV6YmQ4+NR_Ly7gsfxTFr8obn1ug@mail.gmail.com>
Thanks Dieter,
Even if I use lm(), I get the following output:
> summary(fit)
? ? ? ? ? ? ?Df ?Sum Sq Mean Sq F value ? ?Pr(>F)
nh1 ? ? ? ? ? ?1 ? 324.0 ?323.99 ?139.13 < 2.2e-16 ***
nh2 ? ? ? ? ? ?1 ? 723.1 ?723.12 ?310.53 < 2.2e-16 ***
nh3 ? ? ? ? ? ?1 ?1794.2 1794.21 ?770.49 < 2.2e-16 ***
Residuals ? 4604 10721.2 ? ?2.33
---
Signif. codes: ?0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.?
no R squared to be found.
The lm() output gives this:
> summary(fit)
? ? ? ? ? ? ?Df ?Sum Sq Mean Sq F value ? ?Pr(>F)
nh1 ? ? ? ? ? ?1 ? 324.0 ?323.99 ?139.13 < 2.2e-16 ***
nh2 ? ? ? ? ? ?1 ? 723.1 ?723.12 ?310.53 < 2.2e-16 ***
nh3 ? ? ? ? ? ?1 ?1794.2 1794.21 ?770.49 < 2.2e-16 ***
Residuals ? 4604 10721.2 ? ?2.33
---
Signif. codes: ?0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.?
I just want to see how wel the model fits...
Another strange thing. I did the same calculations in SPSS and got
different results, among others, higher p values. lm() and aov() just
do linear regressions (anova) right? I tried with the same factorial
variables (without interaction effects to test).
Thanks!
On 15 April 2011 18:07, Dieter Menne <dieter.menne at menne-biomed.de> wrote:
>
> dorien wrote:
>>
>> I calculate an anova test in the following way:
>>
>> ... aov example
>>
>> I want to check the fit of the model with Rsquared
>>
>
> Try summary(lm(...)) instead.
>
> Dieter
>
>
>
> --
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>
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--
Dorien Herremans
Department of Environment, Technology and Technology Management
Faculty of Applied Economics
University of Antwerp
B.513
Prinsstraat 13
2000 Antwerp
Belgium
+32 3 265 41 25
--
Dorien Herremans
Department of Environment, Technology and Technology Management
Faculty of Applied Economics
University of Antwerp
B.513
Prinsstraat 13
2000 Antwerp
Belgium
+32 3 265 41 25