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comparing two aov models

To: "s-news" <s-news@lists.biostat.wustl.edu>
Subject: comparing two aov models
From: "Thompson, David (MNR)" <david.thompson@mnr.gov.on.ca>
Date: Wed, 27 Apr 2005 16:21:05 -0400
Thread-index: AcVLZprDYd9/X92LRnKV9vJvNVz5ew==
Thread-topic: comparing two aov models
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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