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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