Dear SAS community,
I'm trying to use the empirical option in the proc mixed statement for robust variances in a model because my data doesn*t fulfill homogeneity, even after different transformation (ln, square root). I've also tried it with different groups in the repeated statement, usually i select it via AIC. My model estimates plant numbers of barley plants from a pea-barley intercropping experiment. The experiment design is a randomised incomplete block design with the test factors:
- Behandlung(treatment) (6 levels: pure stand, 90:10 (Pea%:Barley%), 70:30, 80:20, 80:40 and 50:50 )
- NStufe(fertiliser level): 0 (no fertilisation), 1 (100 kg N/ha) (for the treatments 80:20 and 80:40 is just NStufe 0 available)
There are 4 observations for each combination, with one missing value for treatment 80:20.
I've tried to put the subject option to the random statement and with subject=Behandlung*NStufe because this is how my data is grouped, but i've got the following error in the log:
ERROR: Empirical standard errors cannot be computed for this model because there is only one effective subject.
I've tried it with Behandlung, NStufe and Block as subject but for all i get the same error. Then i've put the subject option in the random statement instead and it worked for all subjects that i've already tried with the repeated statement. My random factor is Block so i guess that i should put it as a subject in the random statement but i'm not really know if this is correct. Can someone help me to find the correct way how to implement the subject option in my code and explain it to me how tho decide that?
Here is my code
data a3;
input Datum $ Messtermin Block Parzelle Variante Behandlung NStufe Dichte_Erbse Dichte_Gerste Pflanzenzahl_Gerste;
log_Pflanzenzahl_Gerste=1/(log(Pflanzenzahl_Gerste));
datalines;
06.06.2024 1 1 1 2 1 1 0 100 162.0053691
06.06.2024 1 1 2 10 6 0 80 20 34.6878687
06.06.2024 1 1 3 6 3 0 90 10 26.2
06.06.2024 1 1 4 9 5 0 80 40 29.15329531
06.06.2024 1 1 5 7 4 0 70 30 44.13465347
06.06.2024 1 1 8 5 3 0 90 10 26.71594872
06.06.2024 1 1 9 1 1 0 0 100 159.7818182
06.06.2024 1 1 10 11 7 0 50 50 165.3264368
06.06.2024 1 1 12 8 4 1 70 30 112.1639344
06.06.2024 1 1 14 12 7 1 50 50 76.45019608
06.06.2024 1 2 15 11 7 0 50 50 190.9146341
06.06.2024 1 2 17 5 3 0 90 10 24.98595041
06.06.2024 1 2 18 10 6 0 80 20 .
06.06.2024 1 2 19 7 4 0 70 30 66.45525767
06.06.2024 1 2 20 2 1 1 0 100 262.4124286
06.06.2024 1 2 21 9 5 0 80 40 80.12765957
06.06.2024 1 2 22 6 3 1 90 10 35.16582915
06.06.2024 1 2 25 8 4 1 70 30 111.3658537
06.06.2024 1 2 26 12 7 1 50 50 115.5
06.06.2024 1 2 27 1 1 0 0 100 140.4141414
06.06.2024 1 3 29 8 4 1 70 30 109.1770335
06.06.2024 1 3 30 1 1 0 0 100 160.2692308
06.06.2024 1 3 32 11 7 0 50 50 190.1124051
06.06.2024 1 3 34 10 6 0 80 20 70.88881988
06.06.2024 1 3 35 12 7 1 50 50 57.50456026
06.06.2024 1 3 36 2 1 1 0 100 188.1761006
06.06.2024 1 4 37 9 5 0 80 40 44.62406417
06.06.2024 1 4 38 5 3 0 90 10 23.99574468
06.06.2024 1 4 40 7 4 0 70 30 124.7916667
06.06.2024 1 4 41 6 3 1 90 10 35.54
06.06.2024 1 3 43 1 1 0 0 100 200.3962264
06.06.2024 1 3 44 11 7 0 50 50 92.76923077
06.06.2024 1 3 47 8 4 1 70 30 114.6013072
06.06.2024 1 3 48 9 5 0 80 40 60.27710843
06.06.2024 1 3 49 6 3 1 90 10 29.90909091
06.06.2024 1 3 50 7 4 0 70 30 60.33128834
06.06.2024 1 4 51 5 3 0 90 10 45.22580645
06.06.2024 1 4 52 12 7 1 50 50 75.63299663
06.06.2024 1 4 54 10 6 0 80 20 91.33333333
06.06.2024 1 4 56 2 1 1 0 100 233.2857143
;run;
;
/* Pflanzenzahl Gerste = plant count barley*/
Proc mixed data=a3 empirical;
class Block Behandlung NStufe;
model log_Pflanzenzahl_Gerste = Behandlung*NStufe;
repeated/group=Behandlung*NStufe;
random Block /subject=Block;
lsmeans Behandlung*NStufe/cl;
ods output lsmeans = lsmeans;
run;
;
