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
[[alternative HTML version deleted]]
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 > [[alternative HTML version deleted]] > > ______________________________________________ > 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.David Winsemius, MD West Hartford, CT
Dear Peter, Because your model is additive, "type-II" and "type-III" tests are identical, and the t-tests for the linear and quadratic coefficients are interpretable. I hope this helps, John ------------------------------------------------ John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 6 Jul 2012 16:06:26 +0100 mails <mails00000 at gmail.com> 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 > [[alternative HTML version deleted]] > > ______________________________________________ > 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.