Good advice!
Cheers,
Ted
From: r-sig-phylo-boun...@r-project.org [r-sig-phylo-boun...@r-project.org] on
behalf of Luke Matthews [lmatth...@activatenetworks.net]
Sent: Tuesday, December 03, 2013 8:29 AM
To: r-sig-phylo@r-project.org
Subject: Re: [R-sig-phylo] best fit vs normality of residuals
Hi Agus,
If I understand your post correctly, you implemented the two models with
exactly the same formula, and phylogenetic tree, and varied only the transform
applied to the variables. In one case the transform was to 'scale and center'
both the dependent and independent variables while in the other case you rank
ordered the all variables. Please clarify if I have this wrong.
In this case you would have to compare the normality of the residuals, as when
you transform the dependent variable in different ways I think the likelihoods
will necessarily shift just because the distribution of the data being
explained is shifted. I don't think you can compare likelihoods, AICs, or BICs
in this way when you alter the dependent, just as you can't compare
likelihood-based values for models that include mostly the same data points but
also some different ones. You can only compare likelihood based values for
models that have the exact same dependent data but differ in the formulation of
the model's independent variables, autocorrelation parameters, etc.
As another point, although ranking seemed to fix your normality problem in this
case, it may be introducing some other issues. Ranked variables are usually
too flat for linear regression analysis, and perhaps the test of residual
normality you are using only detects some deviations like skewness (I'm not
familiar with the Liliefors test). I would recommend that instead of ranking
you plot the distribution of your dependent and independent variables and try
the transforms (log, arcsine, etc.) that a standard stats book would recommend
given the appearance of each distribution. Some variables may not need
transforming at all. It may be just one or two that is causing the problem in
the model residuals, in which case you should transform only those variables.
Remember that since it's the normality of the residuals that matters, it may be
more appropriately fixed by transforming an independent variable than a
dependent one. If the dependent itself is normal, then nonnormal!
ity might be introduced in the residuals because of the distribution of a
particular independent variable.
Best
Luke
Luke J. Matthews | Senior Scientific Director | Activate Networks
------------------------------
Message: 2
Date: Fri, 29 Nov 2013 12:29:06 -0200
From: Agus Camacho <agus.cama...@gmail.com>
To: "r-sig-phylo@r-project.org" <r-sig-phylo@r-project.org>
Subject: [R-sig-phylo] best fit vs normality of residuals
Message-ID:
<calsj7pssmsg5hp7yiiesquejnuqad2zuprqs4ldx6wxagyx...@mail.gmail.com>
Content-Type: text/plain
Dear colleagues,
Im having difficulties to decide whether I choose a phylogenetic GLS model with
a higher fit (lower AIC and BIC), or a model in which normality of the
residuals, after accounting for phylogenetic signal, is compromised. The number
of species is reasonably high (87), but i dont know if that would justify for
allowing a highly significant deviation of normality.
When using scaled and centered data, i get:
AIC BIC logLik
255.2029505 269.5696455 -121.6014753
Correlation Structure: corPagel
Formula: ~1
Parameter estimate(s):
lambda
-0.03313647856
Coefficients:
Value Std.Error t-value p-value
(Intercept) 0.0432999895 0.06632562733 0.6528395024 0.5157
X 0.0425258358 0.03552018760 1.1972300459 0.2347
X2 0.4620358585 0.18471478739 2.5013474287 0.0144
X1:X2 -0.1211398020 0.04969892007 -2.4374735269 0.0170
Liliefors test (thanks Liam for posting on this) gave me:
D = 0.1815, p-value = 2.558e-07
I ranked both, the response variable and the factors. My variables had some
zeros and in some cases negative values, so thought that would be the simplest
and most robust way. But i might be wrong.
When ranking all variables:
AIC BIC logLik
766.7826784 781.1493734 -377.3913392
Correlation Structure: corPagel
Formula: ~1
Parameter estimate(s):
lambda
0.1434096557
Coefficients:
Value Std.Error t-value p-value
(Intercept) 5.615576688 9.195445483 0.610691097 0.5431
X1 0.477054032 0.200882571 2.374790556 0.0199
X2 0.771914482 0.208616720 3.700156356 0.0004
X1:x2 -0.007371999 0.004035148 -1.826946400 0.0714
Lilliefors (Kolmogorov-Smirnov) normality test
data: chol(solve(vcv(tree))) %*% residuals(M2) D = 0.0545, p-value = 0.7709
Would anybody have a hint on this?
Gracias!
Agus
--
Agust?n Camacho Guerrero.
Doutor em Zoologia.
Laborat?rio de Herpetologia, Departamento de Zoologia, Instituto de
Bioci?ncias, USP.
Rua do Mat?o, trav. 14, n? 321, Cidade Universit?ria, S?o Paulo - SP, CEP:
05508-090, Brasil.
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