Dear Peter;
This is an exact duplicate of a question posted on SO. Cross-posting
is deprecated on Rhelp.
--
David.
On Jul 6, 2012, at 11:06 AM, mails wrote:
the study design of the data I have to analyse is simple. There is 1
control group (CTRL) and 2 different treatment groups (TREAT_1 and
TREAT_2).
The data also includes 2 covariates COV1 and COV2. I have been asked
to check if there is a linear or quadratic treatment effect in the
data.
I created a dummy data set to explain my situation:
df1 <- data.frame(
Observation = c(rep("CTRL",15), rep("TREAT_1",13), rep("TREAT_2",
12)),
COV1 = c(rep("A1", 30), rep("A2", 10)),
COV2 = c(rep("B1", 5), rep("B2", 5), rep("B3", 10), rep("B1", 5),
rep("B2", 5), rep("B3", 10)),
Variable = c(3944133, 3632461, 3351754, 3655975, 3487722, 3644783,
3491138, 3328894,
3654507, 3465627, 3511446, 3507249, 3373233,
3432867, 3640888,
3677593, 3585096, 3441775, 3608574, 3669114,
4000812, 3503511, 3423968,
3647391, 3584604, 3548256, 3505411, 3665138,
4049955, 3425512, 3834061, 3639699, 3522208,
3711928, 3576597, 3786781,
3591042, 3995802, 3493091, 3674475)
)
plot(Variable ~ Observation, data = df1)
As you can see from the plot there is a linear relationship between
the control and the treatment groups. To check if this linear effect
is statistical
significant I change the contrasts using the contr.poly() function
and fit a linear model like this:
contrasts(df1$Observation) <- contr.poly(levels(df1$Observation))
lm1 <- lm(log(Variable) ~ Observation, data = df1)
summary.lm(lm1)
From the summary we can see that the linear effect is statistically
significant:
Observation.L 0.029141 0.012377 2.355 0.024 *
Observation.Q 0.002233 0.012482 0.179 0.859
However, this first model does not include any of the two
covariates. Including them results in a non-significant p-value for
the linear relationship:
lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1)
summary.lm(lm2)
Observation.L 0.04116 0.02624 1.568 0.126
Observation.Q 0.01003 0.01894 0.530 0.600
COV1A2 -0.01203 0.04202 -0.286 0.776
COV2B2 -0.02071 0.02202 -0.941 0.354
COV2B3 -0.02083 0.02066 -1.008 0.320
So far so good. However, I have been told to conduct a Type II Anova
rather than a Type I. To conduct a Type II Anova I used the Anova()
function
from the car package.
Anova(lm2, type="II")
Anova Table (Type II tests)
Response: log(Variable)
Sum Sq Df F value Pr(>F)
Observation 0.006253 2 1.4651 0.2453
COV1 0.000175 1 0.0820 0.7763
COV2 0.002768 2 0.6485 0.5292
Residuals 0.072555 34
The problem here with using Type II is that you do not get a p-value
for the linear and quadratic effect.
So I do not know if the treatment effect is statistically linear and
or quadratic.
I found out that the following code produces the same p-value for
Observation as the Anova() function. However, the result also does
not include
any p-values for the linear or quadratic effect:
lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1)
lm3 <- lm(log(Variable) ~ COV1 + COV2, data = df1)
anova(lm2, lm3)
Does anybody know how to conduct a Type II anova and the contrasts
function to obtain the p-values for the linear and quadratic effects?
Help would be very much appreciated.
Best Peter
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David Winsemius, MD
West Hartford, CT
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
