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Briank988
Calcite | Level 5

Mean seperation procedures will not be carried out due to zero degrees of freedom for the specified error source. Using 9.4. Below is my setup in excel, attached is the sas output

DietSpan 
13.03.42.62.92.4 
22.83.33.63.53.4 
33.63.94.13.84.0 
       

Proc GLM;

class diet;

model span=diet;

means diet/lsd;

run;


SAS rat diet.PNG
1 ACCEPTED SOLUTION

Accepted Solutions
Reeza
Super User

Do you only have a single record for each diet?

Or are all the values in the row, values for span? If you haven't transposed the data then SAS will not be using all the span values, only the first one.

View solution in original post

1 REPLY 1
Reeza
Super User

Do you only have a single record for each diet?

Or are all the values in the row, values for span? If you haven't transposed the data then SAS will not be using all the span values, only the first one.

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