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Ross Culloch

2009-Aug-13 20:45 UTC

[R] lm coefficients output confusing

Hi all,

I have an issue with the lm() function regarding the listing of the
coefficients. My data are below, showing a list of hours (HR) relating to
the time spent resting (R) by an individual animal. Simply i want to run a
lm() to run in an anova() to see if there is a significant difference in
resting between hours. 

   HR         R
1   2 0.6666667
2   2 0.4666667
3   2 0.8000000
4   2 0.6333333
5   2 0.7333333
6   2 0.8000000
7   2 0.8666667
8   2 0.7857143
9   2 0.7826087
10  2 0.6666667
11  2 0.9166667
12  2 0.6666667
13  3 0.5294118
14  3 0.8541667
15  3 0.4583333
16  3 0.5882353
17  3 0.9347826
18  3 0.7878788
19  3 0.7857143
20  3 0.6944444
21  3 0.8333333
22  3 0.7450980
23  3 0.9230769
24  3 0.7222222
25  4 0.6571429
26  4 0.7241379
27  4 0.7391304
28  4 0.6571429
29  4 0.8000000
30  4 0.9130435
31  4 0.7187500
32  4 0.8437500
33  4 0.9230769
34  4 0.8571429
35  4 0.8695652
36  4 0.8888889
37  5 0.3333333
38  5 0.5365854
39  5 0.6774194
40  5 0.7142857
41  5 0.6904762
42  5 0.5483871
43  5 0.5952381
44  5 0.4166667
45  5 0.5666667
46  5 0.5952381
47  5 0.7894737
48  5 0.7500000
49  6 0.6268657
50  6 0.7187500
51  6 0.5500000
52  6 0.7164179
53  6 0.7656250
54  6 0.5869565
55  6 0.7164179
56  6 0.7031250
57  6 0.7230769
58  6 0.7462687
59  6 0.9200000
60  6 0.8536585
61  7 0.6379310
62  7 0.5357143
63  7 0.5227273
64  7 0.8000000
65  7 0.6724138
66  7 0.7083333
67  7 0.7241379
68  7 0.6938776
69  7 0.6545455
70  7 0.7931034
71  7 0.7560976
72  7 0.8684211
73  8 0.6727273
74  8 0.6000000
75  8 0.8333333
76  8 0.8181818
77  8 0.7818182
78  8 0.7647059
79  8 0.5818182
80  8 0.5918367
81  8 0.7450980
82  8 0.7818182
83  8 0.8048780
84  8 0.8684211


The script i'm using and output is as follows:
> anova(rdayml <- lm(R ~ HR, data=rdata2, na.action=na.exclude)) Analysis of Variance Table

Response: R
          Df  Sum Sq Mean Sq F value  Pr(>F)   
HR         6 0.25992 0.04332  3.1762 0.00774 **
Residuals 77 1.05021 0.01364                   
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
> 
> summary(rdayml <- lm(R ~ HR,data=rdata2))
Call:
lm(formula = R ~ HR, data = rdata2)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.279725 -0.065416  0.005593  0.077486  0.201070 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.732082   0.033713  21.715   <2e-16 ***
HR3          0.005976   0.047678   0.125   0.9006    
HR4          0.067232   0.047678   1.410   0.1625    
HR5         -0.130935   0.047678  -2.746   0.0075 ** 
HR6         -0.013152   0.047678  -0.276   0.7834    
HR7         -0.034807   0.047678  -0.730   0.4676    
HR8          0.004971   0.047678   0.104   0.9172    
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 

Residual standard error: 0.1168 on 77 degrees of freedom
Multiple R-squared: 0.1984,     Adjusted R-squared: 0.1359 
F-statistic: 3.176 on 6 and 77 DF,  p-value: 0.00774 


What i really don't understand is why the lm summary lists the hour numbers
in the coefficient of the lm, as apposed to just reading HR? On top of that
if R does display the data like this then i don't understand why it omits
hour 2? If i can get this to work correctly can I use the p value to
determine which of the hours is significantly different to the others - so
in this example hour 5 is significantly different? Or is it just a case of
using the p value from the anova to determine that there is a significant
difference between hours (in this case) and use a plot to determine which
hour(s) are likely to be the cause?

Any help or advice would be most useful!

Best wishes,

Ross


-- 
View this message in context:
http://www.nabble.com/lm-coefficients-output-confusing-tp24958398p24958398.html
Sent from the R help mailing list archive at Nabble.com.

Charles C. Berry

2009-Aug-13 21:48 UTC

head link

[R] lm coefficients output confusing

On Thu, 13 Aug 2009, Ross Culloch wrote:
> 
> Hi all,
> 
> I have an issue with the lm() function regarding the listing of the
> coefficients. My data are below, showing a list of hours (HR) relating to
> the time spent resting (R) by an individual animal. Simply i want to run a
> lm() to run in an anova() to see if there is a significant difference in
> resting between hours.
The problem is not with lm(), but with your data.

The listing below does not precisely represent your data; when I copy it 
to the clipboard and then use

 	dat <- read.table("clipboard")
 	lm(R~HR,dat)

I get a different result.

OTOH, when I use

 	summary(lm(R~as.factor(HR),rdata2))

I recap your results (up to labelling).

I suggest you try

 	str( rdata2 )

to get more insight into your data. I suspect that one or more of the 
values in HR was quoted in your data file or that you used colClasses to 
determine the class of data in each column and turned HR into a factor.

HTH,

Chuck
>
>    HR         R
> 1   2 0.6666667
> 2   2 0.4666667
> 3   2 0.8000000
> 4   2 0.6333333
> 5   2 0.7333333
> 6   2 0.8000000
> 7   2 0.8666667
> 8   2 0.7857143
> 9   2 0.7826087
> 10  2 0.6666667
> 11  2 0.9166667
> 12  2 0.6666667
> 13  3 0.5294118
> 14  3 0.8541667
> 15  3 0.4583333
> 16  3 0.5882353
> 17  3 0.9347826
> 18  3 0.7878788
> 19  3 0.7857143
> 20  3 0.6944444
> 21  3 0.8333333
> 22  3 0.7450980
> 23  3 0.9230769
> 24  3 0.7222222
> 25  4 0.6571429
> 26  4 0.7241379
> 27  4 0.7391304
> 28  4 0.6571429
> 29  4 0.8000000
> 30  4 0.9130435
> 31  4 0.7187500
> 32  4 0.8437500
> 33  4 0.9230769
> 34  4 0.8571429
> 35  4 0.8695652
> 36  4 0.8888889
> 37  5 0.3333333
> 38  5 0.5365854
> 39  5 0.6774194
> 40  5 0.7142857
> 41  5 0.6904762
> 42  5 0.5483871
> 43  5 0.5952381
> 44  5 0.4166667
> 45  5 0.5666667
> 46  5 0.5952381
> 47  5 0.7894737
> 48  5 0.7500000
> 49  6 0.6268657
> 50  6 0.7187500
> 51  6 0.5500000
> 52  6 0.7164179
> 53  6 0.7656250
> 54  6 0.5869565
> 55  6 0.7164179
> 56  6 0.7031250
> 57  6 0.7230769
> 58  6 0.7462687
> 59  6 0.9200000
> 60  6 0.8536585
> 61  7 0.6379310
> 62  7 0.5357143
> 63  7 0.5227273
> 64  7 0.8000000
> 65  7 0.6724138
> 66  7 0.7083333
> 67  7 0.7241379
> 68  7 0.6938776
> 69  7 0.6545455
> 70  7 0.7931034
> 71  7 0.7560976
> 72  7 0.8684211
> 73  8 0.6727273
> 74  8 0.6000000
> 75  8 0.8333333
> 76  8 0.8181818
> 77  8 0.7818182
> 78  8 0.7647059
> 79  8 0.5818182
> 80  8 0.5918367
> 81  8 0.7450980
> 82  8 0.7818182
> 83  8 0.8048780
> 84  8 0.8684211
> 
> 
> The script i'm using and output is as follows:
> 
> > anova(rdayml <- lm(R ~ HR, data=rdata2, na.action=na.exclude)) 
> Analysis of Variance Table
> 
> Response: R
>           Df  Sum Sq Mean Sq F value  Pr(>F) 
> HR         6 0.25992 0.04332  3.1762 0.00774 **
> Residuals 77 1.05021 0.01364 
> ---
> Signif. codes:  0 ???***??? 0.001 ???**??? 0.01 ???*??? 0.05 ???.??? 0.1
??? ??? 1
> > 
> > summary(rdayml <- lm(R ~ HR,data=rdata2))
> 
> Call:
> lm(formula = R ~ HR, data = rdata2)
> 
> Residuals:
>       Min        1Q    Median        3Q       Max 
> -0.279725 -0.065416  0.005593  0.077486  0.201070 
> 
> Coefficients:
>              Estimate Std. Error t value Pr(>|t|) 
> (Intercept)  0.732082   0.033713  21.715   <2e-16 ***
> HR3          0.005976   0.047678   0.125   0.9006 
> HR4          0.067232   0.047678   1.410   0.1625 
> HR5         -0.130935   0.047678  -2.746   0.0075 ** 
> HR6         -0.013152   0.047678  -0.276   0.7834 
> HR7         -0.034807   0.047678  -0.730   0.4676 
> HR8          0.004971   0.047678   0.104   0.9172 
> ---
> Signif. codes:  0 ???***??? 0.001 ???**??? 0.01 ???*??? 0.05 ???.??? 0.1
??? ??? 1
> 
> Residual standard error: 0.1168 on 77 degrees of freedom
> Multiple R-squared: 0.1984,     Adjusted R-squared: 0.1359 
> F-statistic: 3.176 on 6 and 77 DF,  p-value: 0.00774 
> 
> 
> What i really don't understand is why the lm summary lists the hour
numbers
> in the coefficient of the lm, as apposed to just reading HR? On top of that
> if R does display the data like this then i don't understand why it
omits
> hour 2? If i can get this to work correctly can I use the p value to
> determine which of the hours is significantly different to the others - so
> in this example hour 5 is significantly different? Or is it just a case of
> using the p value from the anova to determine that there is a significant
> difference between hours (in this case) and use a plot to determine which
> hour(s) are likely to be the cause?
> 
> Any help or advice would be most useful!
> 
> Best wishes,
> 
> Ross
> 
> 
> -- 
> View this message in context:
http://www.nabble.com/lm-coefficients-output-confusing-tp24958398p24958398.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.
>
Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

(Ted Harding)

2009-Aug-13 21:53 UTC

head link

[R] lm coefficients output confusing

On 13-Aug-09 20:45:31, Ross Culloch wrote:> 
> Hi all,
> 
> I have an issue with the lm() function regarding the listing of
> the coefficients. My data are below, showing a list of hours (HR)
> relating to the time spent resting (R) by an individual animal.
> Simply i want to run a lm() to run in an anova() to see if there
> is a significant difference in resting between hours. 
> 
>    HR         R
> 1   2 0.6666667
> 2   2 0.4666667
> 3   2 0.8000000
> 4   2 0.6333333
> 5   2 0.7333333
> 6   2 0.8000000
> 7   2 0.8666667
> 8   2 0.7857143
> 9   2 0.7826087
> 10  2 0.6666667
> 11  2 0.9166667
> 12  2 0.6666667
> 13  3 0.5294118
> 14  3 0.8541667
> 15  3 0.4583333
> 16  3 0.5882353
> 17  3 0.9347826
> 18  3 0.7878788
> 19  3 0.7857143
> 20  3 0.6944444
> 21  3 0.8333333
> 22  3 0.7450980
> 23  3 0.9230769
> 24  3 0.7222222
> 25  4 0.6571429
> 26  4 0.7241379
> 27  4 0.7391304
> 28  4 0.6571429
> 29  4 0.8000000
> 30  4 0.9130435
> 31  4 0.7187500
> 32  4 0.8437500
> 33  4 0.9230769
> 34  4 0.8571429
> 35  4 0.8695652
> 36  4 0.8888889
> 37  5 0.3333333
> 38  5 0.5365854
> 39  5 0.6774194
> 40  5 0.7142857
> 41  5 0.6904762
> 42  5 0.5483871
> 43  5 0.5952381
> 44  5 0.4166667
> 45  5 0.5666667
> 46  5 0.5952381
> 47  5 0.7894737
> 48  5 0.7500000
> 49  6 0.6268657
> 50  6 0.7187500
> 51  6 0.5500000
> 52  6 0.7164179
> 53  6 0.7656250
> 54  6 0.5869565
> 55  6 0.7164179
> 56  6 0.7031250
> 57  6 0.7230769
> 58  6 0.7462687
> 59  6 0.9200000
> 60  6 0.8536585
> 61  7 0.6379310
> 62  7 0.5357143
> 63  7 0.5227273
> 64  7 0.8000000
> 65  7 0.6724138
> 66  7 0.7083333
> 67  7 0.7241379
> 68  7 0.6938776
> 69  7 0.6545455
> 70  7 0.7931034
> 71  7 0.7560976
> 72  7 0.8684211
> 73  8 0.6727273
> 74  8 0.6000000
> 75  8 0.8333333
> 76  8 0.8181818
> 77  8 0.7818182
> 78  8 0.7647059
> 79  8 0.5818182
> 80  8 0.5918367
> 81  8 0.7450980
> 82  8 0.7818182
> 83  8 0.8048780
> 84  8 0.8684211
> 
> 
> The script i'm using and output is as follows:
> 
>> anova(rdayml <- lm(R ~ HR, data=rdata2, na.action=na.exclude)) 
> Analysis of Variance Table
> 
> Response: R
>           Df  Sum Sq Mean Sq F value  Pr(>F)   
> HR         6 0.25992 0.04332  3.1762 0.00774 **
> Residuals 77 1.05021 0.01364                   
> ---
> Signif. codes:  0 ???***??? 0.001 ???**??? 0.01 ???*??? 0.05 ???.???
> 0.1 ??? ??? 1 
>> 
>> summary(rdayml <- lm(R ~ HR,data=rdata2))
> 
> Call:
> lm(formula = R ~ HR, data = rdata2)
> 
> Residuals:
>       Min     >    1Q    Median        3Q       Max 
> -0.279725 -0.065416  0.005593  0.077486  0.201070 
> 
> Coefficients:
>              Estimate Std. Error t value Pr(>|t|)    
> (Intercept)  0.732082   0.033713  21.715   <2e-16 ***
> HR3          0.005976   0.047678   0.125   0.9006    
> HR4          0.067232   0.047678   1.410   0.1625    
> HR5         -0.130935   0.047678  -2.746   0.0075 ** 
> HR6         -0.013152   0.047678  -0.276   0.7834    
> HR7         -0.034807   0.047678  -0.730   0.4676    
> HR8          0.004971   0.047678   0.104   0.9172    
> ---> Signif. codes:  0 ???***??? 0.001 ???**??? .01 ???*??? 0.05
???.???> 0.1 ??? ??? 1 
> 
> Residual standard error: 0.1168 on 77 degrees of freedom
> Multiple R-squared: 0.1984,     Adjusted R-squared: 0.1359 
> F-statistic: 3.176 on 6 and 77 DF,  p-value: 0.00774 
> 
> 
> What i really don't understand is why the lm summary lists the hour
> numbers in the coefficient of the lm, as apposed to just reading HR?
What has apparently happened here is that HR has been read in as
a factor, nor a numerical covariate (which is probably what you wanted
anyway, since you seem to want to compare hours).

Hence HR is a factor with 7 levels:
"2","3","4","5","6","7","8"
(I have wirtten them between "" to show that they will not be
interpreted numerically). When you get a coefficient from lm(),
e.g. for HR6, this is the coefficient for level "6" of HR, and
that is the standard way of writing a factor with its level in
the output of ln().
> On top of that if R does display the data like this then i don't
> understand why it omits hour 2?
This is because your have the Intercept in the model, so along
with the 7 levels of HR you now have 8 variables present for a
model with 7 coefficients. So one has to go (being redundant).
Conventionally this is the first level of the factor (though you
can arrange it differently if you want to). Hence HR2 is, in effect,
absorbed into the Intercept. In this case, the Intercept is equal
to the mean of the values for Hour 2.

If you were to add the coefficint for Intercept to the coefficient
for any level of HR, e.g.

   Intercept           HR5
  (0.732082) + (-0.130935)  =  0.601147

you should find that it is the same as the mean of the values in
Hour 5:

  (0.3333333+0.5365854+0.6774194+0.7142857+0.6904762+0.5483871+
   0.5952381+0.4166667+0.5666667+0.5952381+0.7894737+0.7500000)/12

  = 0.6011475

as indeed it is! The coefficient of a given level of HR is here
the difference between the mean for that hour and the mean for
Hour 2. This depends on the system of contrasts ... by default,
lm() uses "treatment contrasts" for factors, which express the
difference between each "treatment" and the "base level"
treatment.
Alternativve systems of cintrasts give different representations.
> If i can get this to work correctly
> can I use the p value to determine which of the hours is significantly
> different to the others - so in this example hour 5 is significantly
> different? Or is it just a case of using the p value from the
anova> to determine that there is a significant difference between hours
> (in this case) and use a plot to determine which hour(s) are likely
> to be the cause?
Don't trust the individual P-values in the first instance, since with
several levels somce are bound to have more extreme P-values than
others.

However, your anova() test is an overall test for whether one or more
of the means jointly depart significantly from the Null Hypothesis
that they are all equal. Here, your anova() has given a significant
result, so the hour means are not all equal. *Now* you can look at
the individual P-values to see where this may be coming from.
In the list of coefficients, there is only one candidate for
"being different from Intercept", and this is Hour 5. Indeed the
P-value for HR5 and the P-value for the anova() are nearly equal
(the latter being slightly larger, allowing for the fact that it
is 1 out of 7).

Hoping this helps!
Ted.

--------------------------------------------------------------------
E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 13-Aug-09                                       Time: 22:53:22
------------------------------ XFMail ------------------------------

Daniel Malter

2009-Aug-13 21:56 UTC

head link

[R] lm coefficients output confusing

I have answered a similar question just hours ago. Your question either
indicates an unfamiliarity with R or, more generally, an unfamiliarity with
regression analysis, or both.

Your anova indicates that the overall model is significant. That is, your
model is an improvement over the null model with just an intercept.  Thus,
the anova output tells you that there are significant differences between
the HR groups. This is also indicated by the F-test and p-value for the
F-test in the second model output (the summary.lm).

The question why R does not show HR2 or not just only HR indicates the lack
of understanding of R or regression. The fact that there are seven
coefficient estimates for HR3 through HR8 indicates that the HR variable is
coded as a factor. Consequently, R estimates the model using dummy variables
for each level of HR, except for the baseline, which is coded zero (this is
basically an analysis of variance, not of covariance). By standard, the
baseline is the smallest value (or lowest in alphabetical order) of a
factor-coded variable, which is HR=2 in your case. Since it is the baseline,
there is no estimate for a coefficient for HR2. All other estimated
coefficients compare the effect of HR=3 through HR8 relative to the group in
which HR=2.

If you want to include HR as a numeric variable in the regression (analysis of
covariance), you have to code it as.numeric. Note, however, that
your results for the factor-coded HR variable indicate that the trend (if
any) is not linear.

For a comparison of all groups against one another from an analysis of
variance, I think there other methods, like the Bonferroni-Dunn test (as a
post-hoc test).

Best,
Daniel

-------------------------
cuncta stricte discussurus
-------------------------

-----Urspr?ngliche Nachricht-----
Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im
Auftrag von Ross Culloch
Gesendet: Thursday, August 13, 2009 4:46 PM
An: r-help at r-project.org
Betreff: [R] lm coefficients output confusing


Hi all,

I have an issue with the lm() function regarding the listing of the
coefficients. My data are below, showing a list of hours (HR) relating to
the time spent resting (R) by an individual animal. Simply i want to run a
lm() to run in an anova() to see if there is a significant difference in
resting between hours. 

   HR         R
1   2 0.6666667
2   2 0.4666667
3   2 0.8000000
4   2 0.6333333
5   2 0.7333333
6   2 0.8000000
7   2 0.8666667
8   2 0.7857143
9   2 0.7826087
10  2 0.6666667
11  2 0.9166667
12  2 0.6666667
13  3 0.5294118
14  3 0.8541667
15  3 0.4583333
16  3 0.5882353
17  3 0.9347826
18  3 0.7878788
19  3 0.7857143
20  3 0.6944444
21  3 0.8333333
22  3 0.7450980
23  3 0.9230769
24  3 0.7222222
25  4 0.6571429
26  4 0.7241379
27  4 0.7391304
28  4 0.6571429
29  4 0.8000000
30  4 0.9130435
31  4 0.7187500
32  4 0.8437500
33  4 0.9230769
34  4 0.8571429
35  4 0.8695652
36  4 0.8888889
37  5 0.3333333
38  5 0.5365854
39  5 0.6774194
40  5 0.7142857
41  5 0.6904762
42  5 0.5483871
43  5 0.5952381
44  5 0.4166667
45  5 0.5666667
46  5 0.5952381
47  5 0.7894737
48  5 0.7500000
49  6 0.6268657
50  6 0.7187500
51  6 0.5500000
52  6 0.7164179
53  6 0.7656250
54  6 0.5869565
55  6 0.7164179
56  6 0.7031250
57  6 0.7230769
58  6 0.7462687
59  6 0.9200000
60  6 0.8536585
61  7 0.6379310
62  7 0.5357143
63  7 0.5227273
64  7 0.8000000
65  7 0.6724138
66  7 0.7083333
67  7 0.7241379
68  7 0.6938776
69  7 0.6545455
70  7 0.7931034
71  7 0.7560976
72  7 0.8684211
73  8 0.6727273
74  8 0.6000000
75  8 0.8333333
76  8 0.8181818
77  8 0.7818182
78  8 0.7647059
79  8 0.5818182
80  8 0.5918367
81  8 0.7450980
82  8 0.7818182
83  8 0.8048780
84  8 0.8684211


The script i'm using and output is as follows:
> anova(rdayml <- lm(R ~ HR, data=rdata2, na.action=na.exclude))Analysis of Variance Table

Response: R
          Df  Sum Sq Mean Sq F value  Pr(>F)   
HR         6 0.25992 0.04332  3.1762 0.00774 **
Residuals 77 1.05021 0.01364                   
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
> 
> summary(rdayml <- lm(R ~ HR,data=rdata2))
Call:
lm(formula = R ~ HR, data = rdata2)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.279725 -0.065416  0.005593  0.077486  0.201070 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.732082   0.033713  21.715   <2e-16 ***
HR3          0.005976   0.047678   0.125   0.9006    
HR4          0.067232   0.047678   1.410   0.1625    
HR5         -0.130935   0.047678  -2.746   0.0075 ** 
HR6         -0.013152   0.047678  -0.276   0.7834    
HR7         -0.034807   0.047678  -0.730   0.4676    
HR8          0.004971   0.047678   0.104   0.9172    
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 

Residual standard error: 0.1168 on 77 degrees of freedom
Multiple R-squared: 0.1984,     Adjusted R-squared: 0.1359 
F-statistic: 3.176 on 6 and 77 DF,  p-value: 0.00774 


What i really don't understand is why the lm summary lists the hour numbers
in the coefficient of the lm, as apposed to just reading HR? On top of that
if R does display the data like this then i don't understand why it omits
hour 2? If i can get this to work correctly can I use the p value to
determine which of the hours is significantly different to the others - so
in this example hour 5 is significantly different? Or is it just a case of
using the p value from the anova to determine that there is a significant
difference between hours (in this case) and use a plot to determine which
hour(s) are likely to be the cause?

Any help or advice would be most useful!

Best wishes,

Ross


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and provide commented, minimal, self-contained, reproducible code.

Ross Culloch

2009-Aug-26 21:04 UTC

head link

[R] lm coefficients output confusing

Hi Ted,

Firstly, so sorry for the late reply - thank you very much for your detailed
and patient answer. It was a big help for me and gave me a big step forward
in understanding the lm function! 

I really appreciate the time you took (and everyone on here takes!),

Best wishes,

Ross

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