Hi, All,
I have one question about "Smep-Montreal08-rPruzek.pdf".
For the 2nd graph "2 x 2 factorial design".
Is the disign like
A | |||
Yes | No | ||
B | Yes | 1 | 3 |
NO | 2 | 4 |
So we can have the contrast as
$contrasts (standardized)
A B A*B
[1,] 1/2 1/2 1/2
[2,] 1/2 -1/2 -1/2
[3,] -1/2 1/2 -1/2
[4,] -1/2 -1/2 1/2
If I am right, can any one explain the result (also calculation) as
$means.pos.neg.coeff
neg pos diff stEftSze
A 10.58 10.0 -0.54 -0.18
B 10.38 10.2 -0.14 -0.05
A*B 8.88 11.7 2.86 0.94
ANSWER:
Please go to the pdfs I gave you 2 weeks ago on your flash drives. You will see there
is a document I wrote w/ the approximate title CSDAtalk08... In it you will find more information
about the same data (possibly with one value changed, so about the same). Use the contrast coefficient weights above to find the averages of the FOUR MEANS in the 2 x 2 layout, as per the particular contrasts. You should find that the neg and pos columns shown here contain the means of the two negative coefficient means, same for positive, for each column (as rows A, B and A*B). Use the granoova function itself with your own (small?) data set (say n =3 in each group, w/ four groups, inputing as con the three columns here. Then compute the means, as per my suggestion here, and then see what you get. You should be able to reproduce the means as printed when you run the function.
I might generate the initial vector using, say, sample(2:7,12, repl=T) to keep it simple. Let me know what you get. BP
ps. The talk was very well received, and many people said the really liked the graphics.
See third panel above
Thank you. (Yi Sun)