[R] Anova Type II and Contrasts

[David Winsemius]

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]]
>
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David Winsemius, MD
West Hartford, CT