R help - Jun 2008 - logistic regression

Hi everyone,

I'm sorry if this turns out to be more a statistical question than one
specifically about R - but would greatly appreciate your advice anyway.

I've been using a logistic regression model to look at the relationship
between a binary outcome (say, the odds of picking n white balls from a bag
containing m balls in total) and a variety of other binary parameters:

_________________________________________________________________
> a.fit <- glm (data=a, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> summary(a.fit)
glm(formula = cbind(SUCCESS, ALL - SUCCESS) ~ A * B * C * D family binomial(link
= "logit"), data = a)

Deviance Residuals: 
 [1]  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

Coefficients:
	Estimate	Std.	Error	z value	Pr(>|z|)
(Intercept)	-0.69751	0.02697	-25.861	<2.00E-16	***
A	-0.02911	0.05451	-0.534	0.593335	
B	0.39842	0.06871	5.798	6.70E-09	***
C	0.829	0.06745	12.29	<2.00E-16	***
D	0.05928	0.11133	0.532	0.594401	
A:B	-0.44053	0.13807	-3.191	0.001419	**
A:C	-0.49596	0.13664	-3.63	0.000284	***
B:C	-0.62194	0.14164	-4.391	1.13E-05	***
A:D	-0.4031	0.2279	-1.769	0.076938	.
B:D	-0.60238	0.25978	-2.319	0.020407	*
C:D	-0.58467	0.27195	-2.15	0.031558	*
A:B:C	0.5006	0.27364	1.829	0.067335	.
A:B:D	0.51868	0.4683	1.108	0.268049	
A:C:D	0.32882	0.51226	0.642	0.520943	
B:C:D	0.56301	0.49903	1.128	0.259231	
A:B:C:D	-0.32115	0.87969	-0.365	0.715059	

---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 2.2185e+02  on 15  degrees of freedom
Residual deviance: 1.0385e-12  on  0  degrees of freedom
AIC: 124.50

Number of Fisher Scoring iterations: 3

_________________________________________________________________

This seems to produce sensible results given the actual data.
However, there are actually three types of balls in the experiment and I
need to model the relationship between the odds of picking each of the type
and the parameters A,B,C,D. So what I do now is split the initial data table
and just run glm three times:
>all
[fictional data]

TYPE WHITE ALL A B C D 
a	100	400	1	0	0	0
b	200	600	1	0	0	0
c	10	300	1	0	0	0
....
a	30	100	1	1	1	1
b	50	200	1	1	1	1
c	20	120	1	1	1	1
> a<-all[all$type=="a",]
> b<-all[all$type=="b",]
> c<-all[all$type=="c",]
> a.fit <- glm (data=a, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> b.fit <- glm (data=b, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> c.fit <- glm (data=c, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
But it seems to me that I should be able to incorporate TYPE into the model. 

Something like:
>summary(glm(data=example2,family=binomial(link="logit"),formula=cbind(WHITE,ALL-WHITE)~TYPE*A*B*C*D))
[please see the output below]

However, when I do this, it does not seem to give an expected result.
Is this not the right way to do it? 
Or this is actually less powerful than running the three models separately?  

Will greatly appreciate your advice!

Many thanks
Mikhail

-----

	Estimate	Std.	Error	z value	Pr(>|z|)
(Intercept)	-8.90E-01	1.91E-02	-46.553	<2.00E-16	***
TYPE1	1.93E-01	2.47E-02	7.822	5.18E-15	***
TYPE2	1.19E+00	2.42E-02	49.108	<2.00E-16	***
A	1.89E-01	3.34E-02	5.665	1.47E-08	***
B	1.60E-01	4.41E-02	3.627	0.000286	***
C	2.24E-02	4.91E-02	0.455	0.64906	
D	1.96E-01	6.58E-02	2.982	0.002868	**
TYPE1:A	-2.19E-01	4.59E-02	-4.759	1.95E-06	***
TYPE2:A	-9.08E-01	4.50E-02	-20.178	<2.00E-16	***
TYPE1:C	2.39E-01	5.93E-02	4.022	5.77E-05	***
TYPE2:B	-1.82E+00	6.46E-02	-28.178	<2.00E-16	***
A:B	-2.26E-01	8.52E-02	-2.649	0.008066	**
TYPE1:C	8.07E-01	6.27E-02	12.87	<2.00E-16	***
TYPE2:C	-2.51E+00	7.83E-02	-32.039	<2.00E-16	***
A:C	-1.70E-01	9.51E-02	-1.783	0.074512	.
B:C	-3.01E-01	1.12E-01	-2.698	0.006985	**
TYPE1:D	-1.37E-01	9.20E-02	-1.489	0.136548	
TYPE2:D	-1.13E+00	9.19E-02	-12.329	<2.00E-16	***
A:D	-2.11E-01	1.27E-01	-1.655	0.097953	.
B:D	-2.15E-01	1.55E-01	-1.387	0.165472	
C:D	-5.51E-01	2.76E-01	-1.997	0.045829	*
TYPE1:A:B	-2.15E-01	1.17E-01	-1.84	0.065714	.


TYPE2:A:B	7.21E-01	1.28E-01	5.635	1.75E-08	***
TYPE1:A:C	-3.26E-01	1.24E-01	-2.643	0.008221	**
TYPE2:A:C	9.70E-01	1.53E-01	6.36	2.02E-10	***
TYPE1:B:C	-3.21E-01	1.38E-01	-2.321	0.020313	*
TYPE2:B:C	1.35E+00	1.89E-01	7.133	9.85E-13	***
A:B:C	1.80E-01	2.11E-01	0.852	0.394425	
TYPE1:A:D	-1.92E-01	1.83E-01	-1.05	0.293758	
TYPE2:A:D	6.76E-01	1.80E-01	3.75	0.000177	***
TYPE1:B:D	-3.87E-01	2.16E-01	-1.796	0.072443	.
TYPE2:B:D	1.09E+00	2.30E-01	4.709	2.49E-06	***
A:B:D	1.92E-01	2.73E-01	0.702	0.482512	
TYPE1:C:D	-3.33E-02	3.18E-01	-0.105	0.916465	
TYPE2:C:D	1.20E-01	5.05E-01	0.238	0.811914	
A:C:D	-7.37E+00	1.74E+04	-4.23E-04	0.999663	
B:C:D	3.14E-01	4.92E-01	0.638	0.523254	
TYPE1:A:B:C	3.21E-01	2.64E-01	1.218	0.223336	
TYPE2:A:B:C	-8.43E-01	3.59E-01	-2.351	0.018747	*
TYPE1:A:B:D	3.27E-01	3.84E-01	0.85	0.3952	
TYPE2:A:B:D	-6.59E-01	4.08E-01	-1.617	0.105883	
TYPE1:A:C:D	7.69E+00	1.74E+04	4.42E-04	0.999648	
TYPE2:A:C:D	-1.60E+01	3.48E+04	-4.58E-04	0.999634	
TYPE1:B:C:D	2.49E-01	5.70E-01	0.437	0.662288
TYPE2:B:C:D	-7.08E-01	8.97E-01	-0.789	0.430007
A:B:C:D	9.08E-03	2.47E+04	3.67E-07	1
TYPE1:A:B:C:D	-3.30E-01	2.47E+04	-1.34E-05	0.999989
TYPE2:A:B:C:D	1.10E+00	4.94E+04	2.22E-05	0.999982
-- 
View this message in context:
http://www.nabble.com/logistic-regression-tp18102137p18102137.html
Sent from the R help mailing list archive at Nabble.com.

It looks like A*B*C*D is a complete, totally saturated model, (the
residual deviance is effectively zero, and the residual degrees of
freedom is exactly zero - this is a clue).  So when you try to put even
more parameters into the model and even higher way interactions,
something has to give.   

I find 3-factor interactions are about as much as I can think about
without getting a bit giddy.  Do you really need 4- and 5-factor
interactions?  If so, your only option is to get more data.


Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary):  +61 7 3826 7304
Mobile:                         +61 4 8819 4402
Home Phone:                     +61 7 3286 7700
mailto:Bill.Venables at csiro.au
http://www.cmis.csiro.au/bill.venables/ 

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Mikhail Spivakov
Sent: Wednesday, 25 June 2008 9:31 AM
To: r-help at r-project.org
Subject: [R] logistic regression


Hi everyone,

I'm sorry if this turns out to be more a statistical question than one
specifically about R - but would greatly appreciate your advice anyway.

I've been using a logistic regression model to look at the relationship
between a binary outcome (say, the odds of picking n white balls from a
bag
containing m balls in total) and a variety of other binary parameters:

_________________________________________________________________
> a.fit <- glm (data=a, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> summary(a.fit)
glm(formula = cbind(SUCCESS, ALL - SUCCESS) ~ A * B * C * D family binomial(link
= "logit"), data = a)

Deviance Residuals: 
 [1]  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

Coefficients:
	Estimate	Std.	Error	z value	Pr(>|z|)
(Intercept)	-0.69751	0.02697	-25.861	<2.00E-16	***
A	-0.02911	0.05451	-0.534	0.593335	
B	0.39842	0.06871	5.798	6.70E-09	***
C	0.829	0.06745	12.29	<2.00E-16	***
D	0.05928	0.11133	0.532	0.594401	
A:B	-0.44053	0.13807	-3.191	0.001419	**
A:C	-0.49596	0.13664	-3.63	0.000284	***
B:C	-0.62194	0.14164	-4.391	1.13E-05	***
A:D	-0.4031	0.2279	-1.769	0.076938	.
B:D	-0.60238	0.25978	-2.319	0.020407	*
C:D	-0.58467	0.27195	-2.15	0.031558	*
A:B:C	0.5006	0.27364	1.829	0.067335	.
A:B:D	0.51868	0.4683	1.108	0.268049	
A:C:D	0.32882	0.51226	0.642	0.520943	
B:C:D	0.56301	0.49903	1.128	0.259231	
A:B:C:D	-0.32115	0.87969	-0.365	0.715059	

---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05
'.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 2.2185e+02  on 15  degrees of freedom
Residual deviance: 1.0385e-12  on  0  degrees of freedom
AIC: 124.50

Number of Fisher Scoring iterations: 3

_________________________________________________________________

This seems to produce sensible results given the actual data.
However, there are actually three types of balls in the experiment and I
need to model the relationship between the odds of picking each of the
type
and the parameters A,B,C,D. So what I do now is split the initial data
table
and just run glm three times:
>all
[fictional data]

TYPE WHITE ALL A B C D 
a	100	400	1	0	0	0
b	200	600	1	0	0	0
c	10	300	1	0	0	0
....
a	30	100	1	1	1	1
b	50	200	1	1	1	1
c	20	120	1	1	1	1
> a<-all[all$type=="a",]
> b<-all[all$type=="b",]
> c<-all[all$type=="c",]
> a.fit <- glm (data=a, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> b.fit <- glm (data=b, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
> c.fit <- glm (data=c, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
> family=binomial(link="logit"))
But it seems to me that I should be able to incorporate TYPE into the
model. 

Something like:
>summary(glm(data=example2,family=binomial(link="logit"),formula=cbind(W
HITE,ALL-WHITE)~TYPE*A*B*C*D))

[please see the output below]

However, when I do this, it does not seem to give an expected result.
Is this not the right way to do it? 
Or this is actually less powerful than running the three models
separately?  

Will greatly appreciate your advice!

Many thanks
Mikhail

-----

	Estimate	Std.	Error	z value	Pr(>|z|)
(Intercept)	-8.90E-01	1.91E-02	-46.553	<2.00E-16
***
TYPE1	1.93E-01	2.47E-02	7.822	5.18E-15	***
TYPE2	1.19E+00	2.42E-02	49.108	<2.00E-16	***
A	1.89E-01	3.34E-02	5.665	1.47E-08	***
B	1.60E-01	4.41E-02	3.627	0.000286	***
C	2.24E-02	4.91E-02	0.455	0.64906	
D	1.96E-01	6.58E-02	2.982	0.002868	**
TYPE1:A	-2.19E-01	4.59E-02	-4.759	1.95E-06	***
TYPE2:A	-9.08E-01	4.50E-02	-20.178	<2.00E-16	***
TYPE1:C	2.39E-01	5.93E-02	4.022	5.77E-05	***
TYPE2:B	-1.82E+00	6.46E-02	-28.178	<2.00E-16	***
A:B	-2.26E-01	8.52E-02	-2.649	0.008066	**
TYPE1:C	8.07E-01	6.27E-02	12.87	<2.00E-16	***
TYPE2:C	-2.51E+00	7.83E-02	-32.039	<2.00E-16	***
A:C	-1.70E-01	9.51E-02	-1.783	0.074512	.
B:C	-3.01E-01	1.12E-01	-2.698	0.006985	**
TYPE1:D	-1.37E-01	9.20E-02	-1.489	0.136548	
TYPE2:D	-1.13E+00	9.19E-02	-12.329	<2.00E-16	***
A:D	-2.11E-01	1.27E-01	-1.655	0.097953	.
B:D	-2.15E-01	1.55E-01	-1.387	0.165472	
C:D	-5.51E-01	2.76E-01	-1.997	0.045829	*
TYPE1:A:B	-2.15E-01	1.17E-01	-1.84	0.065714
.


TYPE2:A:B	7.21E-01	1.28E-01	5.635	1.75E-08
***
TYPE1:A:C	-3.26E-01	1.24E-01	-2.643	0.008221
**
TYPE2:A:C	9.70E-01	1.53E-01	6.36	2.02E-10
***
TYPE1:B:C	-3.21E-01	1.38E-01	-2.321	0.020313
*
TYPE2:B:C	1.35E+00	1.89E-01	7.133	9.85E-13
***
A:B:C	1.80E-01	2.11E-01	0.852	0.394425	
TYPE1:A:D	-1.92E-01	1.83E-01	-1.05	0.293758	
TYPE2:A:D	6.76E-01	1.80E-01	3.75	0.000177
***
TYPE1:B:D	-3.87E-01	2.16E-01	-1.796	0.072443
.
TYPE2:B:D	1.09E+00	2.30E-01	4.709	2.49E-06
***
A:B:D	1.92E-01	2.73E-01	0.702	0.482512	
TYPE1:C:D	-3.33E-02	3.18E-01	-0.105	0.916465	
TYPE2:C:D	1.20E-01	5.05E-01	0.238	0.811914	
A:C:D	-7.37E+00	1.74E+04	-4.23E-04	0.999663	
B:C:D	3.14E-01	4.92E-01	0.638	0.523254	
TYPE1:A:B:C	3.21E-01	2.64E-01	1.218	0.223336	
TYPE2:A:B:C	-8.43E-01	3.59E-01	-2.351	0.018747
*
TYPE1:A:B:D	3.27E-01	3.84E-01	0.85	0.3952	
TYPE2:A:B:D	-6.59E-01	4.08E-01	-1.617	0.105883	
TYPE1:A:C:D	7.69E+00	1.74E+04	4.42E-04	0.999648

TYPE2:A:C:D	-1.60E+01	3.48E+04	-4.58E-04	0.999634

TYPE1:B:C:D	2.49E-01	5.70E-01	0.437	0.662288
TYPE2:B:C:D	-7.08E-01	8.97E-01	-0.789	0.430007
A:B:C:D	9.08E-03	2.47E+04	3.67E-07	1
TYPE1:A:B:C:D	-3.30E-01	2.47E+04	-1.34E-05	0.999989
TYPE2:A:B:C:D	1.10E+00	4.94E+04	2.22E-05	0.999982
-- 
View this message in context:
http://www.nabble.com/logistic-regression-tp18102137p18102137.html
Sent from the R help mailing list archive at Nabble.com.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

oh, I'm sorry...
here's my affiliation:

Mikhail Spivakov, PhD     spivakov@ebi.ac.uk
Interdisciplinary Postdoctoral Fellow    http://www.ebi.ac.uk/~spivakov/
European Bioinformatics Institute   Tel: +44 1223 492660 (office)
Wellcome Trust Genome Campus        +44 7985 09 6675 (mob)
Cambridge CB10 1SD, UK

On Thu, Jun 26, 2008 at 7:50 AM, Prof Brian Ripley <ripley@stats.ox.ac.uk>
wrote:
> On Wed, 25 Jun 2008, Mikhail Spivakov wrote:
>
>  Sorry for flooding this forum, but I think I've realised that I need
to do
>> multinomial logistic regression for my problem...
>> Would be interested in your opinion as to whether this is actually any
>> better than running three binomial logistic regressions separately..
>>
>
> Sometimes, depending on the problem.  We haven't seen a real-world
problem
> in this thread, and (see the posting guide) this is not a statistical
> consultancy forum, so please ask your statsitical advisor for help.
>
> You won't get consultancy help here unless you give your affiliation in
a
> proper signature block (as the posting guide says).
>
>
>
>> Thanks
>> M
>>
>> On Wed, Jun 25, 2008 at 1:17 AM, <Bill.Venables@csiro.au> wrote:
>>
>>  It looks like A*B*C*D is a complete, totally saturated model, (the
>>> residual deviance is effectively zero, and the residual degrees of
>>> freedom is exactly zero - this is a clue).  So when you try to put
even
>>> more parameters into the model and even higher way interactions,
>>> something has to give.
>>>
>>> I find 3-factor interactions are about as much as I can think about
>>> without getting a bit giddy.  Do you really need 4- and 5-factor
>>> interactions?  If so, your only option is to get more data.
>>>
>>>
>>> Bill Venables
>>> CSIRO Laboratories
>>> PO Box 120, Cleveland, 4163
>>> AUSTRALIA
>>> Office Phone (email preferred): +61 7 3826 7251
>>> Fax (if absolutely necessary):  +61 7 3826 7304
>>> Mobile:                         +61 4 8819 4402
>>> Home Phone:                     +61 7 3286 7700
>>> mailto:Bill.Venables@csiro.au
>>> http://www.cmis.csiro.au/bill.venables/
>>>
>>> -----Original Message-----
>>> From: r-help-bounces@r-project.org
[mailto:r-help-bounces@r-project.org]
>>> On Behalf Of Mikhail Spivakov
>>> Sent: Wednesday, 25 June 2008 9:31 AM
>>> To: r-help@r-project.org
>>> Subject: [R] logistic regression
>>>
>>>
>>> Hi everyone,
>>>
>>> I'm sorry if this turns out to be more a statistical question
than one
>>> specifically about R - but would greatly appreciate your advice
anyway.
>>>
>>> I've been using a logistic regression model to look at the
relationship
>>> between a binary outcome (say, the odds of picking n white balls
from a
>>> bag
>>> containing m balls in total) and a variety of other binary
parameters:
>>>
>>> _________________________________________________________________
>>>
>>>  a.fit <- glm (data=a, formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
>>>> family=binomial(link="logit"))
>>>> summary(a.fit)
>>>>
>>>
>>> glm(formula = cbind(SUCCESS, ALL - SUCCESS) ~ A * B * C * D family
>>> binomial(link = "logit"), data = a)
>>>
>>> Deviance Residuals:
>>>  [1]  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
>>>
>>> Coefficients:
>>>       Estimate        Std.    Error   z value Pr(>|z|)
>>> (Intercept)     -0.69751        0.02697 -25.861 <2.00E-16      
***
>>> A       -0.02911        0.05451 -0.534  0.593335
>>> B       0.39842 0.06871 5.798   6.70E-09        ***
>>> C       0.829   0.06745 12.29   <2.00E-16       ***
>>> D       0.05928 0.11133 0.532   0.594401
>>> A:B     -0.44053        0.13807 -3.191  0.001419        **
>>> A:C     -0.49596        0.13664 -3.63   0.000284        ***
>>> B:C     -0.62194        0.14164 -4.391  1.13E-05        ***
>>> A:D     -0.4031 0.2279  -1.769  0.076938        .
>>> B:D     -0.60238        0.25978 -2.319  0.020407        *
>>> C:D     -0.58467        0.27195 -2.15   0.031558        *
>>> A:B:C   0.5006  0.27364 1.829   0.067335        .
>>> A:B:D   0.51868 0.4683  1.108   0.268049
>>> A:C:D   0.32882 0.51226 0.642   0.520943
>>> B:C:D   0.56301 0.49903 1.128   0.259231
>>> A:B:C:D -0.32115        0.87969 -0.365  0.715059
>>>
>>> ---
>>> Signif. codes:  0 '***' 0.001 '**' 0.01 '*'
0.05 '.' 0.1 ' ' 1
>>>
>>> (Dispersion parameter for binomial family taken to be 1)
>>>
>>>   Null deviance: 2.2185e+02  on 15  degrees of freedom
>>> Residual deviance: 1.0385e-12  on  0  degrees of freedom
>>> AIC: 124.50
>>>
>>> Number of Fisher Scoring iterations: 3
>>>
>>> _________________________________________________________________
>>>
>>> This seems to produce sensible results given the actual data.
>>> However, there are actually three types of balls in the experiment
and I
>>> need to model the relationship between the odds of picking each of
the
>>> type
>>> and the parameters A,B,C,D. So what I do now is split the initial
data
>>> table
>>> and just run glm three times:
>>>
>>>  all
>>>>
>>>
>>> [fictional data]
>>>
>>> TYPE WHITE ALL A B C D
>>> a       100     400     1       0       0       0
>>> b       200     600     1       0       0       0
>>> c       10      300     1       0       0       0
>>> ....
>>> a       30      100     1       1       1       1
>>> b       50      200     1       1       1       1
>>> c       20      120     1       1       1       1
>>>
>>>  a<-all[all$type=="a",]
>>>> b<-all[all$type=="b",]
>>>> c<-all[all$type=="c",]
>>>> a.fit <- glm (data=a,
formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
>>>> family=binomial(link="logit"))
>>>> b.fit <- glm (data=b,
formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
>>>> family=binomial(link="logit"))
>>>> c.fit <- glm (data=c,
formula=cbind(WHITE,ALL-WHITE)~A*B*C*D,
>>>> family=binomial(link="logit"))
>>>>
>>>
>>> But it seems to me that I should be able to incorporate TYPE into
the
>>> model.
>>>
>>> Something like:
>>>
>>> 
summary(glm(data=example2,family=binomial(link="logit"),formula=cbind(W
>>>>
>>> HITE,ALL-WHITE)~TYPE*A*B*C*D))
>>>
>>> [please see the output below]
>>>
>>> However, when I do this, it does not seem to give an expected
result.
>>> Is this not the right way to do it?
>>> Or this is actually less powerful than running the three models
>>> separately?
>>>
>>> Will greatly appreciate your advice!
>>>
>>> Many thanks
>>> Mikhail
>>>
>>> -----
>>>
>>>       Estimate        Std.    Error   z value Pr(>|z|)
>>> (Intercept)     -8.90E-01       1.91E-02        -46.553
<2.00E-16
>>> ***
>>> TYPE1   1.93E-01        2.47E-02        7.822   5.18E-15        ***
>>> TYPE2   1.19E+00        2.42E-02        49.108  <2.00E-16      
***
>>> A       1.89E-01        3.34E-02        5.665   1.47E-08        ***
>>> B       1.60E-01        4.41E-02        3.627   0.000286        ***
>>> C       2.24E-02        4.91E-02        0.455   0.64906
>>> D       1.96E-01        6.58E-02        2.982   0.002868        **
>>> TYPE1:A -2.19E-01       4.59E-02        -4.759  1.95E-06        ***
>>> TYPE2:A -9.08E-01       4.50E-02        -20.178 <2.00E-16      
***
>>> TYPE1:C 2.39E-01        5.93E-02        4.022   5.77E-05        ***
>>> TYPE2:B -1.82E+00       6.46E-02        -28.178 <2.00E-16      
***
>>> A:B     -2.26E-01       8.52E-02        -2.649  0.008066        **
>>> TYPE1:C 8.07E-01        6.27E-02        12.87   <2.00E-16      
***
>>> TYPE2:C -2.51E+00       7.83E-02        -32.039 <2.00E-16      
***
>>> A:C     -1.70E-01       9.51E-02        -1.783  0.074512        .
>>> B:C     -3.01E-01       1.12E-01        -2.698  0.006985        **
>>> TYPE1:D -1.37E-01       9.20E-02        -1.489  0.136548
>>> TYPE2:D -1.13E+00       9.19E-02        -12.329 <2.00E-16      
***
>>> A:D     -2.11E-01       1.27E-01        -1.655  0.097953        .
>>> B:D     -2.15E-01       1.55E-01        -1.387  0.165472
>>> C:D     -5.51E-01       2.76E-01        -1.997  0.045829        *
>>> TYPE1:A:B       -2.15E-01       1.17E-01        -1.84   0.065714
>>> .
>>>
>>>
>>> TYPE2:A:B       7.21E-01        1.28E-01        5.635   1.75E-08
>>> ***
>>> TYPE1:A:C       -3.26E-01       1.24E-01        -2.643  0.008221
>>> **
>>> TYPE2:A:C       9.70E-01        1.53E-01        6.36    2.02E-10
>>> ***
>>> TYPE1:B:C       -3.21E-01       1.38E-01        -2.321  0.020313
>>> *
>>> TYPE2:B:C       1.35E+00        1.89E-01        7.133   9.85E-13
>>> ***
>>> A:B:C   1.80E-01        2.11E-01        0.852   0.394425
>>> TYPE1:A:D       -1.92E-01       1.83E-01        -1.05   0.293758
>>> TYPE2:A:D       6.76E-01        1.80E-01        3.75    0.000177
>>> ***
>>> TYPE1:B:D       -3.87E-01       2.16E-01        -1.796  0.072443
>>> .
>>> TYPE2:B:D       1.09E+00        2.30E-01        4.709   2.49E-06
>>> ***
>>> A:B:D   1.92E-01        2.73E-01        0.702   0.482512
>>> TYPE1:C:D       -3.33E-02       3.18E-01        -0.105  0.916465
>>> TYPE2:C:D       1.20E-01        5.05E-01        0.238   0.811914
>>> A:C:D   -7.37E+00       1.74E+04        -4.23E-04       0.999663
>>> B:C:D   3.14E-01        4.92E-01        0.638   0.523254
>>> TYPE1:A:B:C     3.21E-01        2.64E-01        1.218   0.223336
>>> TYPE2:A:B:C     -8.43E-01       3.59E-01        -2.351  0.018747
>>> *
>>> TYPE1:A:B:D     3.27E-01        3.84E-01        0.85    0.3952
>>> TYPE2:A:B:D     -6.59E-01       4.08E-01        -1.617  0.105883
>>> TYPE1:A:C:D     7.69E+00        1.74E+04        4.42E-04       
0.999648
>>>
>>> TYPE2:A:C:D     -1.60E+01       3.48E+04        -4.58E-04      
0.999634
>>>
>>> TYPE1:B:C:D     2.49E-01        5.70E-01        0.437   0.662288
>>> TYPE2:B:C:D     -7.08E-01       8.97E-01        -0.789  0.430007
>>> A:B:C:D 9.08E-03        2.47E+04        3.67E-07        1
>>> TYPE1:A:B:C:D   -3.30E-01       2.47E+04        -1.34E-05      
0.999989
>>> TYPE2:A:B:C:D   1.10E+00        4.94E+04        2.22E-05       
0.999982
>>>
>>
>
> --
> Brian D. Ripley,                  ripley@stats.ox.ac.uk
> Professor of Applied Statistics, 
http://www.stats.ox.ac.uk/~ripley/<http://www.stats.ox.ac.uk/%7Eripley/>
> 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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