Dear all
I have the following data (a shortened extract
shown; some replictates of time deleted) to which I
fitted the linear model given below:
time group mass
11 control 0.019
11 control 0.014
14 control 0.0306
14 control 0.0289
14 control 0.0236
17 control 0.0469
17 control 0.0709
11 five 0.0077
11 five 0.0101
14 five 0.0092
14 five 0.0125
17 five 0.0134
17 five 0.0143
11 ten 0.0088
11 ten 0.0092
14 ten 0.0114
14 ten 0.0113
17 ten 0.0131
17 ten 0.0154
11 twenty 0.0076
11 twenty 0.0081
14 twenty 0.0093
14 twenty 0.0091
17 twenty 0.0108
17 twenty 0.0097
> options(contrasts = c("contr.treatment",
+ "contr.poly"))> mass.lm <- lm(mass ~ group * time, data = growth)
The question I am asking is whether the slopes
differ, which they do. This is the output:
> round(summary(mass.lm)$coef, 3)
Estimate SE t value Pr(>|t|)
(Intercept) -0.024 0.010 -2.530 0.014
groupfive 0.027 0.013 1.998 0.051
groupten 0.025 0.013 1.840 0.071
grouptwenty 0.027 0.013 1.988 0.052
time 0.004 0.001 5.995 0.000
groupfive:time -0.003 0.001 -3.551 0.001
groupten:time -0.003 0.001 -3.407 0.001
grouptwenty:time -0.004 0.001 -3.719 0.000
This essentially compares the intercepts and slopes
of groups 'five', 'ten' and 'twenty' to that of the
'control' group. But what is a simple way to
compare the slopes of groups 'five', 'ten' and
'twenty' to each other without having to do a
series of pairwise comparisons? Is there some
method involving the TukeyHSD procedure or some
other means that can be used instead?
Any help or comments appreciated.
Regards,
Albertus
Dr. Albertus J. Smit
Department of Botany
University of Cape Town
Private Bag Rondebosch
7700
South Africa
Tel. +27 21 689 3032
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