Hi Max,

the design matrix looks ok. It seems that an additional 1 is missing in the permutation blocks. 

1. A global mean should not be included. This is because you are modelling a separate mean for each subject. 
   
2. Whether you code with -1/1 or 0/1 won't have any impact on the p-values. Both options are fine. 
   
3. In terms of correcting for multiuple comparisons, this would depend on the nature of your hypothesis. The strictest option would be to divde by 4. But that may be too conservative. 
   

You may also wanted to test for an interaction between task and drug condition. This would probably be the most interesting contrast. 

Andrew

Originally posted by Max Kathofer :

I have a similar case in which a participant got once a placebo and once a drug and then did two tasks [high and low]. The explanation here was very helpful in creating a design matrix, so thank you very much! My study has 32 participants and since every participant has 4 matrices, I created a 128x34 matrix. The first column codes drug, the second highVSlow and the rest indicate to which participants they belong. So the first 4 files belonging to subject1 would look like this: 

Sub1PH   0 1 1 0 0 ... 
Sub1PL    0 0 1 0 0 ...
Sub1DH   1 1 1 0 0 ...
Sub1DL    1 0 1 0 0 ... 

The permutation blocks would then look like this [1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, ...]

is this correct? (also see attachment) 

However, 3 things are unclear cause they seem to contradict other posts in this forum:

1. In other posts a column with only ones is additionally included to indicate a global mean. Why is this column here omitted? 
   
2. Similar to the post above, I encoded drug: 0 for placebo and 1 for drug. Why do other posts recommend 1 and -1? I guess there is a difference but I could not figure it out...
   
3. When I am interested in the contrast for 
   

main effect drug = [1 0 0 0 ..]  and for the decrease [-1 0 0 0]
main effect task = [0 1 0 0 .. ] (I guess this means high > low right?) 

I would run a t-test, right? And since I explore both drug > placebo [1 0 0 0 ..]  and placebo > drug [-1 0 0 0], I would need to half the significance to 0.025 to compensate for testing twice, right? Or would I actually need to divide alpha by 4 since I run both contrasts for both main effects to safeguard against alpha error accumulation? 

Thank you so much for your help! 

Cheers, 

Max