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Fluorite | Level 6

## Contrasts for estimate statement for intercropping problem

Dear SAS-Community,

i'm analysing plant dry matter data (Pflanzen_TM_kg_ha) with SAS 9.4 TS Level 1M7 for an intercropping experiment with pea and barley.

I have 7 treatments (Behandlungen) with different mixing ratios (based on seeding density of the pure stands)
1: 100 % barley; 2: 100% pea; 3: 90%pea+10%barley; 4: 70%pea+30%barley; 5: 80%pea+40%barley; 6: 80%pea+20%barley; 7: 50%pea+50%barley

Treatments 1, 2, 3, 4, 7 have 2 fertiliser levels (NStufe): 0 and 1. Treatments 5 and 6 have just 1 fertiliser level (NStufe=0). For every combination of treatment and fertiliser level i have 4 replications.

I calculate the lsmeans for the total mixture plant dry matter with proc mixed and i want to test the differences between the total mixture plant dry matter of the different mixtures with the sum of the corresponding pure stand dry matter, multiplied by the mixing ratio of the respective plant in the mixture (e.g.:
total plant dry matter of the 50:50-mixture with fertiliser level 0 - pea pure stand (fertiliser level 0) dry matter * 0.5 - barley pure stand (f. level 0) * 0.5)

```Proc mixed data=a1;
class Block Behandlung NStufe;
model Pflanzen_TM_kg_ha=Behandlung*NStufe /ddfm=kr outp=r residual s;
lsmeans Behandlung*NStufe;
random Block;
estimate "90:10 N0 vs. RS" Behandlung*NStufe -0.1 0 -0.9 0 1 0 0 0 0 0 0 0;
estimate "90:10 N1 vs. RS" Behandlung*NStufe 0 -0.1 0 -0.9 0 1 0 0 0 0 0 0;
estimate "70:30 N0 vs. RS" Behandlung*NStufe -0.3 0 -0.7 0 0 1 0 0 0 0 0 0;
estimate "70:30 N1 vs. RS" Behandlung*NStufe 0 -0.3 0 -0.7 0 0 0 1 0 0 0 0;
estimate "80:40 N0 vs. RS" Behandlung*NStufe -0.4 0 -0.8 0 0 0 0 0 1.2 0 0 0;
estimate "80:20 N0 vs. RS" Behandlung*NStufe -0.2 0 -0.8 0 0 0 0 0 0 1 0 0;
estimate "50:50 N0 vs. RS" Behandlung*NStufe -0.5 0 -0.5 0 0 0 0 0 0 0 1 0;
estimate "50:50 N1 vs. RS" Behandlung*NStufe 0 -0.5 0 -0.5 0 0 0 0 0 0 0 1;
run;```

I get the following table for the Solution for Fixed Effects:

For all the treatments the contrasts are easy to get, except for treatment 5 (80%pea+40%barley) because it goes above 100% seeding density of the pure stands. I've tried the following estimate statements for treatment 5:

`estimate "80:40 N0 vs. RS" Behandlung*NStufe -0.4 0 -0.8 0 0 0 0 0 1.2 0 0 0;`
`estimate "80:40 N0 vs. RS" Behandlung*NStufe -0.33 0 -0.67 0 0 0 0 0 1.0 0 0 0;`

I think the second one is wrong but i'm not sure if the first estimate statement is correct for the difference which i want to test:
total plant dry matter of the 80%pea+40%barley-mixture with fertiliser level 0 - pea pure stand (fertiliser level 0) dry matter * 0.8 - barley pure stand (f. level 0) * 0.4

The p-value is the same for both estimate statements.

Can you tell me if one of my estimate statements are correct and explain it to me or if not, can you tell me the right estimate statement for my goal?

1 ACCEPTED SOLUTION

Accepted Solutions
Fluorite | Level 6

## Re: Contrasts for estimate statement for intercropping problem

The estimate statement:

`estimate "80:40 N0 vs. RS" Behandlung*NStufe -0.33 0 -0.67 0 0 0 0 0 1.0 0 0 0;`

was the correct statement because the reference unit is the area.

3 REPLIES 3
Super User

## Re: Contrasts for estimate statement for intercropping problem

Is it possible that the "5: 80%pea+40%barley; " is incorrect and should be "5: 60%pea+40%barley;"?

Fluorite | Level 6

## Re: Contrasts for estimate statement for intercropping problem

No, the mixing ratio is correct. The percentages are based on the seeding densities of the pure stands. That means that on the same area in the mixture 40% of the amount of the barley seeds of the barley pure stands were sown and 80% of the amount of seeds in the pea pure stands.

Fluorite | Level 6

## Re: Contrasts for estimate statement for intercropping problem

The estimate statement:

`estimate "80:40 N0 vs. RS" Behandlung*NStufe -0.33 0 -0.67 0 0 0 0 0 1.0 0 0 0;`

was the correct statement because the reference unit is the area.

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