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]]
>
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David Winsemius, MD
West Hartford, CT