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Re: comparing two aov models

To: "Thompson, David (MNR)" <david.thompson@mnr.gov.on.ca>
Subject: Re: comparing two aov models
From: Prof Brian Ripley <ripley@stats.ox.ac.uk>
Date: Wed, 27 Apr 2005 22:09:31 +0100 (BST)
Cc: s-news <s-news@lists.biostat.wustl.edu>
In-reply-to: <327C2434A3DA2D4CB61554D644BAE48BAB33EF@LRCPSSMBMXMB001.lrc.ad.gov.on.ca>
References: <327C2434A3DA2D4CB61554D644BAE48BAB33EF@LRCPSSMBMXMB001.lrc.ad.gov.on.ca>
You cannot legitimately apply anova() to two models applied to different sets of data. You do have two sets of data, of different sizes.

Cp is Mallow's C_p statistic. It is not a `test', rather a close relative of AIC. Small Cp is good (and Googling will find you more). Once again,
it applies to multiple models and one set of data.

On Wed, 27 Apr 2005, Thompson, David (MNR) wrote:

Hello,

I have run the following ANOVA, both with the same model,
a) first with imputed missing cell values,
b) then with missing cells left in place.

        Call:   aov(formula = ht00 ~ blk + trt * size,
        data = morph, projections = T, qr = T, na.action = na.exclude)

a)                Df Sum of Sq Mean Sq F Value  Pr(F)
              blk  2      35.0   17.52    7.10 0.0032
              trt  4     165.7   41.43   16.80 0.0000
             size  2       9.4    4.70    1.90 0.1677
         trt:size  8      15.0    1.88    0.76 0.6393
        Residuals 28      69.1    2.47
        Residual standard error: 1.571

b)                Df Sum of Sq Mean Sq F Value  Pr(F)
              blk  2     102.5   51.25   18.80 0.0000
              trt  4      86.0   21.49    7.88 0.0006
             size  2       7.8    3.92    1.44 0.2614
         trt:size  8      14.7    1.84    0.67 0.7080
        Residuals 20      54.5    2.73
        Residual standard error: 1.651

I then use anova() to compare the two aov.objects(?):

        > anova(aovht00, aovht00na, test="F")
        Response: ht00
                     Terms Resid. Df   RSS Test Df Sum of Sq F Value
Pr(F)
        1 blk + trt * size        28 69.06

        2 blk + trt * size        20 54.53    =  8     14.53  0.6662
0.7148

        > anova(aovht00, aovht00na, test="Chi")
        Response: ht00
                     Terms Resid. Df   RSS Test Df Sum of Sq Pr(Chi)
        1 blk + trt * size        28 69.06
        2 blk + trt * size        20 54.53    =  8     14.53 0.06892

        > anova(aovht00, aovht00na, test="Cp")
        Response: ht00
                     Terms Resid. Df   RSS Test Df Sum of Sq    Cp
        1 blk + trt * size        28 69.06                   161.8
        2 blk + trt * size        20 54.53    =  8     14.53 190.9

Q1. Do I interpret the lack of significance under F and Chi^2 tests
to indicate that the differences are negligible?

Q2. And can I extend this interpretation to mean that the imbalance
has little, or no, impact on the result?

Q3. What is the "Cp" test? and how do I use it?

Thank you, DaveT.
**********************************************************
Silviculture Data Analyst
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
Sault Ste. Marie, Ontario, Canada
david.thompson@mnr.gov.on.ca
**********************************************************
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--
Brian D. Ripley,                  ripley@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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