@David

R1, R2 are results collected in two measurement runs so r=2. Other words two responses that I obtained in the experiment. So yes only two replicates.

@ Fausto

Do you mean that both R1 and R2 are so different that we cannot assume that they are from the same distribution and we have to calculate  SS of replications. Other words we have to include influence of replication in ANOVA table? If yes please have a look right now at attached file. Is it correct?  I had only table of Fisher coefficient to 0.001. So I used it.

I answered faster than I thought about Fausto's comment. Hence I modified my last post please have a look at it I made new computations.

You need to refine your model; the residuals of the model with main effect and all interactions are not normally distributed, consequently, your p-values and significance may be called into question. See attached file.

@Fausto

1) How high have to be Fisher value to make an assumption that we do not have to bother with "normal distribution"?

2) Is there any rule for a selection of a alpha for Fisher test if we take account an influence of replications? I am asking because I found the critical F value distribution for F(1,7,0.00005)=77.1324. According to this significance level the only significant factors are replications and factor B. I would like to use alpha=0.05 or 0.001 then I would be able to show significance of many more factors. Will I make a mistake to use such values?

3) I am puzzled with my new computations results. I made separate analysis for response R1 and R2. Therefore I divided by 2 number of tests and I found that after this only factor B is significant for both R1 and R2 (I am attaching you files). I wanted to show the fact that you mentioned. That more factors re significant for r1 than for R2 and in the same way to show the difference between these two distributions. But I failed. Where I did an mistake?

Thank you Fausto, I will try to do this as soon as possible and I will upload the results once more.

@Fausto,

I am attaching next version. Please let me know if you were talking about such analysis. If yes I have in my computations doubts about DOF of SS A@R1, SS A @R2 etc. According to info that I have all of them should be 1. I am also not sure if I should include SS of replications in this analysis but I did it. According to results that I obtained still only factors A, B,C and replications are significant all interactions are insignificant for both R1 and R2.

@Dominik Jurków: Your explanation "R1, R2 are results collected in two measurement runs so r=2. Other words two responses that I obtained in the experiment. So yes only two replicates." is confusing to me ... Having an experiment with two responses is very different from having an experiment with two replicates ... Which one is it?

@Noel, thank you for your contribution to the topic. You are right it is confusing. I meant replications. I have an experiment with two replications.

@Fausto,

1) If I understood correctly following document

http://www.ndsu.nodak.edu/ndsu/horsley/Polycnst.pdf

Each of my SS A@R1, SS A@R2 etc. and as well interactions will have 1 DOF (when I will be using contrasts). Is it correct?

2) yes I tried. The problem is that I do not feel it. Previously I was examined trends using comparisons (trends comparisons), moreover I had 2 DOF per SS so I could examined two comparisons let's say quadratic and linear. Here I have to conduct class comparison and moreover I have only 1 DOF per SS A etc. So when I am trying to do two comparisons per SS A I got wrong result. I tried to use formulas given below but as you previously spotted number of DOF is unequal and SS of separated comparisons do not sum to e.g. SS A. Other words the formulas in my opinion are wrong. Please help me with it.

SS A@R1=4*[(A-/R-)-(A+/R-)]^2/2

SS A@R2=4*[(A+/R+)-(A-/R+)]^2/2

SS AB@R1=4*[(AB-/R-)-(AB+/R-)]^2/2

3) The matter that I am also do not feel is the DOF number of a factor replications. Each of factors A,B, C has two replications in case of factor replications I have 8 replications, have not I? I do not now how to include it in computations.

@Fausto,

Thank you I did not notice this value earlier, sorry. I will try once more.

@ Fausto,

1) I computed SS A, B, C and R using both usual formulas and contrasts and both results are the same. Thanks

2) When I am computing SS AR

SS AR= (v*y)^2/Iv|

where:

v - vector (1,-1,-1,1)

A/R            - 1                                  +1

-1            SUM (green)   SUM(blue)

+1          SUM(RED)       SUM(yellow)

the meaning of colours is given in attached file.

So I am achieving:

SS AR=(SUM(green)+SUM(yellow)-SUM(blue)-SUM(RED))^2/(4*4)=272.25

@Fausto

my SS AR and yours is the same,  I made so stupid calculation mistake, so the formula that I gave before has to be okay. I modified my last post. Thanks for support.

@Fausto

I computed SS AR, SS BR, SS CR and I included these values in ANOVA. I tried also computed SS ABR ACR BCR but these values were so small that I decided not to include them. Can you have a look right now in this document.

Yes you are right  I found it as well yesterday. I used wrong data in the table and then I lost the whole day to find what was wrong :). It was this stupid mistake to which I confessed today.  Hence, I do not have to bother with my last post because this time I used correct data.

The results in the actual version of the attached document are correct. Can you have a look at them and say if it is the thing that you were recommending?

@Fausto,

I am very glad to hear it. Thank you once more.