factorial split-plot combined experiment

[lvm]

Too bad you don't have a newer version of SAS, because the LINES option and SLICE statement in GLIMMIX do exactly what you want! If you are in education or noncommercial activity, you can always install SAS University Edition (it is free),. and it is always the newest version. But that would take a while for you to learn about. With only 9.1, the table of mean differences you showed would have to be translated by you, by hand, into the letter (lines) display. You sort the means from high to low (or low to high), and then figure out which ones are different (different letters) and which ones are the same using the difference table. This can be tedious and take a while.

You have an alternative that I think (?) should work in 9.1 (not absolutely sure). You can use the macro written by Hans-Peter Piepho. See the article here.

You would need to download the macro and run it first (every time you launch SAS for the mean separation analysis). Then you follow the directions in the article to get the letter display. Note that you have to use the ODS output statement to save LSMEANS and mean differences from a MIXED run. With the interaction, you do slices of one factor at fixed values of the other.  Good luck.(By the way, the macro can also do things not possible with the LINES option in current versions of SAS, but that is not relevant for your application).

http://agrobiol.sggw.waw.pl/~cbcs/articles/CBCS_7_1_2.pdf

You do need PROC IML for the macro to work. Message was edited by: Larry Madden

[Herwibawa]

A very big thank you @Larry Madden, I have installed SAS University Edition.

anyway I'm facing a new problem, in which several plants had died, even of plants in one of the treatment combination. So, what should I do ?

Ok, I'll explain my data;

A1     B1     1      died

A1     B1     2      20

A1     B1     3      died

A1     B1     4      died

A1     B1     5      died

A1     B2     1      died

A1     B2     2      13

A1     B2     3      died

A1     B2     4      12

A1     B2     5      10

A1     B3     1      died

A1     B3     2      died

A1     B3     3      died

A1     B3     4      15

A1     B3     5      17

A2     B1     1      died

A2     B1     2      died

A2     B1     3      died

A2     B1     4      died

A2     B1     5      died

A2     B2     1      died

A2     B2     2      28

A2     B2     3      died

A2     B2     4      died

A2     B2     5      30

A2     B3     1      17

A2     B3     2      15

A2     B3     3      16

A2     B3     4      died

A2     B3     5      18

A3     B1     1      5

A3     B1     2      died

A3     B1     3      10

A3     B1     4      died

A3     B1     5      died

A3     B2     1      died

A3     B2     2      10

A3     B2     3      8

A3     B2     4      9

A3     B2     5      died

A3     B3     1      died

A3     B3     2      5

A3     B3     3      died

A3     B3     4      10

A3     B3     5      6

It would be appreciated if you could help me

[SteveDenham]

All that I have to offer comes from digging through the documentation.  This looks like a mixture of distributions: one is a binomial as to whether a plant lived or died, the other is a normal that describes the response.  It may be possible to use something like PROC FMM to do this, but it does not handle random effects, so all of the old tricks used in GLM would have to be applied to get reasonable estimates.

The easiest answer is to assume that plants died at random, so that the data are what is called missing at random.  In that case, recode 'died' as missing values ( . ), and analyze as previously outlined.

Steve Denham

[Herwibawa]

My experiment arranged in Split Plot design. How to make a Charts/ Boxplots for Proportions of a Binary Variable ?

data Survival;

  input A $ B $ Block Y;  if Y=1 then 1='Survive'; else 0='Died';

datalines;

A1     B1     1      0

A1     B1     2      1

A1     B1     3      0

A1     B1     4      0

A1     B1     5      0

A1     B2     1      1

A1     B2     2      1

A1     B2     3      1

A1     B2     4      1

A1     B2     5      0

A1     B3     1      1

A1     B3     2      0

A1     B3     3      0

A1     B3     4      1

A1     B3     5      1

A2     B1     1      0

A2     B1     2      0

A2     B1     3      0

A2     B1     4      0

A2     B1     5      0

A2     B2     1      1

A2     B2     2      0

A2     B2     3      0

A2     B2     4      1

A2     B2     5      0

A2     B3     1      1

A2     B3     2      1

A2     B3     3      1

A2     B3     4      0

A2     B3     5      1

A3     B1     1      1

A3     B1     2      0

A3     B1     3      1

A3     B1     4      0

A3     B1     5      0

A3     B2     1      0

A3     B2     2      1

A3     B2     3      1

A3     B2     4      1

A3     B2     5      0

A3     B3     1      0

A3     B3     2      1

A3     B3     3      0

A3     B3     4      1

A3     B3     5      1

;

Is it true when it displayed in as shown below ? how to write the codes ?

Please help me it is so important for me...

[Thanhlong]

I have the same problem as Mr/Mrs. Hervibawa because I use SAS 9.1. But I don't know how to install the macro for SAS?

[SteveDenham]

I am going to argue against the use of the Dunnett adjustment in the face of a model that has an interaction, as it is designed for comparison to a single control (see for example Chapters 3, 4 and 9 in Multiple Comparisons and Multiple Tests Using SAS by Westfall et al.).  They make a very strong case for the use of the adjust=simulate option for higher order ANOVAs and mixed models.

Steve Denham