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Lars Bergemann

2009-Jul-08 14:34 UTC

[R] Two-way ANOVA gives different results using anova(lm()) than doing it by hand

Hey!

 

Could you please take a quick look at what I have done? Somehow I get wrong
results using the anova(lm()) combination compared to doing a two way ANOVA by
hand.

 

Running:

 

Data<-read.table("Data.txt");
g<-lm(ExM~S1*S2,Data);
anova(g);

 

Gives:

 

Analysis of Variance Table

Response: ExM
           Df Sum Sq Mean Sq F value    Pr(>F)    
S1          1 4.3679  4.3679 167.045 < 2.2e-16 ***
S2          1 0.9427  0.9427  36.053 8.236e-09 ***
S1:S2       1 0.3231  0.3231  12.357 0.0005371 ***
Residuals 212 5.5434  0.0261                      


I compared it to the work done by hand, ie calculated all the different square
sums using sum() and tapply().

So I know that anova(lm()) gets the degrees of freedom equal two 1, 1, 1 and 212
when it should be 5, 5, 25 and 180. Also, the square sums are quite different
... I get 4.xx, 4.xx, 1.xx, 0.xx ... as you see, what anova(lm()) gets is
different.

 

The data: S1 has 6 levels, so has S2. On average, each cell has 6 values, most
cells have actually 6 values, and there are two of each: 5, 7, 4, 8 - so average
6.

 

Could you please help me, why it does not work with anova(lm())? I tried quite a
few thinks found with Google, but it all gave me the same result as anova(lm())
...

 

Thanks a lot!

 

Lars

_________________________________________________________________



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Greg Snow

2009-Jul-08 17:11 UTC

head link

[R] Two-way ANOVA gives different results using anova(lm()) than doing it by hand

Well, since we don't have Data.txt it is kind of hard for us to replicate
what you have done.

Here goes a guess as to what the problem may be.

Have you told R anywhere that S1 and S2 are factors with 6 levels rather than
numeric vectors? Or are you just hoping that the computer can read your mind to
find out this information?

(reading minds is one of the things that R and computers in general are not very
good at yet.  I have made a note to my future self to use the TimeTravel package
to send a copy of the ESP package back to my past self, but I have not received
it yet).


-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Lars Bergemann
> Sent: Wednesday, July 08, 2009 8:35 AM
> To: r-help at r-project.org
> Subject: [R] Two-way ANOVA gives different results using anova(lm())
> than doing it by hand
> 
> 
> Hey!
> 
> 
> 
> Could you please take a quick look at what I have done? Somehow I get
> wrong results using the anova(lm()) combination compared to doing a two
> way ANOVA by hand.
> 
> 
> 
> Running:
> 
> 
> 
> Data<-read.table("Data.txt");
> g<-lm(ExM~S1*S2,Data);
> anova(g);
> 
> 
> 
> Gives:
> 
> 
> 
> Analysis of Variance Table
> 
> Response: ExM
>            Df Sum Sq Mean Sq F value    Pr(>F)
> S1          1 4.3679  4.3679 167.045 < 2.2e-16 ***
> S2          1 0.9427  0.9427  36.053 8.236e-09 ***
> S1:S2       1 0.3231  0.3231  12.357 0.0005371 ***
> Residuals 212 5.5434  0.0261
> 
> 
> I compared it to the work done by hand, ie calculated all the different
> square sums using sum() and tapply().
> 
> So I know that anova(lm()) gets the degrees of freedom equal two 1, 1,
> 1 and 212 when it should be 5, 5, 25 and 180. Also, the square sums are
> quite different ... I get 4.xx, 4.xx, 1.xx, 0.xx ... as you see, what
> anova(lm()) gets is different.
> 
> 
> 
> The data: S1 has 6 levels, so has S2. On average, each cell has 6
> values, most cells have actually 6 values, and there are two of each:
> 5, 7, 4, 8 - so average 6.
> 
> 
> 
> Could you please help me, why it does not work with anova(lm())? I
> tried quite a few thinks found with Google, but it all gave me the same
> result as anova(lm()) ...
> 
> 
> 
> Thanks a lot!
> 
> 
> 
> Lars
> 
> _________________________________________________________________
> 
>

Zhiliang Ma

2009-Jul-08 21:25 UTC

head link

[R] Two-way ANOVA gives different results using anova(lm()) than doing it by hand

the following works. i don't exactly what happens here. I guess
"lm"
might treat S1 and S2 as quantitative variables, not qualitative
variables.

cheers,
Zhiliang

S1 <- as.character(Data[,1])
S1 <- as.factor(S1)
S2 <- as.character(Data[,2])
S2 <- as.factor(S2)
data <- data.frame(S1=S1, S2=S2, ExM=Data[,4])

g <- lm(ExM ~ S1*S2, data)
 anova(g)
Analysis of Variance Table

Response: ExM
           Df Sum Sq Mean Sq F value    Pr(>F)
S1          5 4.7454  0.9491  961.66 < 2.2e-16 ***
S2          5 4.9548  0.9910 1004.10 < 2.2e-16 ***
S1:S2      25 1.2993  0.0520   52.66 < 2.2e-16 ***
Residuals 180 0.1776  0.0010
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1



On Wed, Jul 8, 2009 at 10:34 AM, Lars
Bergemann<lars.bergemann at hotmail.com> wrote:>
> Hey!
>
>
>
> Could you please take a quick look at what I have done? Somehow I get wrong
results using the anova(lm()) combination compared to doing a two way ANOVA by
hand.
>
>
>
> Running:
>
>
>
> Data<-read.table("Data.txt");
> g<-lm(ExM~S1*S2,Data);
> anova(g);
>
>
>
> Gives:
>
>
>
> Analysis of Variance Table
>
> Response: ExM
> ? ? ? ? ? Df Sum Sq Mean Sq F value ? ?Pr(>F)
> S1 ? ? ? ? ?1 4.3679 ?4.3679 167.045 < 2.2e-16 ***
> S2 ? ? ? ? ?1 0.9427 ?0.9427 ?36.053 8.236e-09 ***
> S1:S2 ? ? ? 1 0.3231 ?0.3231 ?12.357 0.0005371 ***
> Residuals 212 5.5434 ?0.0261
>
>
> I compared it to the work done by hand, ie calculated all the different
square sums using sum() and tapply().
>
> So I know that anova(lm()) gets the degrees of freedom equal two 1, 1, 1
and 212 when it should be 5, 5, 25 and 180. Also, the square sums are quite
different ... I get 4.xx, 4.xx, 1.xx, 0.xx ... as you see, what anova(lm()) gets
is different.
>
>
>
> The data: S1 has 6 levels, so has S2. On average, each cell has 6 values,
most cells have actually 6 values, and there are two of each: 5, 7, 4, 8 - so
average 6.
>
>
>
> Could you please help me, why it does not work with anova(lm())? I tried
quite a few thinks found with Google, but it all gave me the same result as
anova(lm()) ...
>
>
>
> Thanks a lot!
>
>
>
> Lars
>
> _________________________________________________________________
>
>
>
>
> ______________________________________________
> 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.
>
>

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