topic Re: LSD in Proc Mixed in SAS Procedures

LSD in Proc Mixed

cken — Thu, 07 Feb 2013 22:39:13 GMT

Can someone help me update this code for SAS 9.2 or suggest better. I found it on the web in Google Groups from 1998.

*****************

Proc mixed doesn't compute lsd directly, but it can be done using info. generated by the make statement. Look at this example:

proc mixed;
...
make "diffs" out=diffs;
quit;

data calc_lsd;
set;
lsd=_se_*tinv(1-.05/2,_df_);

**************

Thanks

Re: LSD in Proc Mixed

SteveDenham — Mon, 11 Feb 2013 14:26:01 GMT

Replace the old make statement with an ODS OUTPUT statement:

ods output diffs=diffs;

And then use:

data calc_lsd;

set diffs;

lsd=stderr*tinv(1 -0.5/2,df);

run;

Steve Denham

Re: LSD in Proc Mixed

cken — Mon, 11 Feb 2013 17:43:41 GMT

Thank you very much!

The program ran the code, but produced no output.

Can you provide code to generate output?

Re: LSD in Proc Mixed

SteveDenham — Tue, 12 Feb 2013 12:56:32 GMT

Add a proc print.

proc print data=calc_lsd;

var lsd estimate stderr df;

run;

Steve Denham

Re: LSD in Proc Mixed

gaston — Wed, 15 May 2013 13:20:01 GMT

Many thanks for this.

Different values of LSD have been printed. which one to consider. The min of all the LSD?

Thanks

Re: LSD in Proc Mixed

SteveDenham — Thu, 16 May 2013 14:00:16 GMT

You probably have different values for the standard error of the differences, as a result of unequal observations in the effects you are comparing. Consequently, the LSD will differ from case to case. Sharing some more code, especially your PROC MIXED code, and some info on the design could help clarify this.

Steve Denham

Re: LSD in Proc Mixed

gaston — Fri, 17 May 2013 13:52:41 GMT

Many thanks Steve.

below is the dode:

Proc mixed;

class block variety;

model yield=variety;

random block;

lsmeans variety/pdiff;

ods output Diffs=d;

run;

This is an augmented design where 200 varieties and 5 checks have been tested in an Augmented randomized complete block design. The 200 varieties have been unreplicated in 20 blocks of 10 plots each. Each block has been completed by the 5 checks. Hence a total of 300 plots (200+5*20). Thanks

Gaston

Re: LSD in Proc Mixed

SteveDenham — Fri, 17 May 2013 14:44:50 GMT

So in the lsmeans table, are the standard errors of the varieties all equal? If not, then there is probably some missing data (variety within a block), so that the standard error of the difference is not constant. Also, are the 5 checks all identical check varieties? This also leads to unequal replication, unequal standard errors of the differences, and consequently, different LSDs.

Also, in a design of this type, block by variety is often included, since not all varieties are seen in each block, but the check varieties are, and this may represent a better source of residual error. Emphasis on the word may, there.

Steve Denham

Re: LSD in Proc Mixed

gaston — Sat, 18 May 2013 21:57:30 GMT

Thanks and well noted. The 5 checks are all identical, but some missing data. Then in this case, it makes sense to have different LSDs. many thanks again. Gaston

Re: LSD in Proc Mixed

ALSU — Tue, 24 Nov 2015 14:39:37 GMT

hi is the above code used for LSD at 0.05 or 0.1 ? I would like to obtain the LSD value at 0.1 thanks

Re: LSD in Proc Mixed

Rick_SAS — Mon, 30 Nov 2015 18:00:18 GMT

In the equation

lsd=_se_*tinv(1-.05/2,_df_);

the value 0.05 represents the significance level, often denote by "alpha." So the answer to your question is that the LSD is for alpha=0.05. Use alpha=0.1 to get the LSD value that you are asking for.

Re: LSD in Proc Mixed

Olanike — Wed, 17 Feb 2016 17:27:04 GMT

Hi Steve, Please is the 0.05 in the lsd equation for 5% significant level?, can it be changed to lsd=stderr*tinv(1-0.1/2,df). Thanks

Re: LSD in Proc Mixed

SteveDenham — Thu, 18 Feb 2016 14:47:11 GMT

See @Rick_SAS's post right above. The answers to your questions are yes and yes.

Steve Denham

Re: LSD in Proc Mixed

Olanike — Thu, 18 Feb 2016 16:56:17 GMT

Thanks!!!

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Re: LSD in Proc Mixed

alimon — Mon, 13 Sep 2021 14:16:43 GMT

I want to estimate LSD, can you tell me what is wrong ?

ods output diffs=ppp;
data calc_lsd;
set diffs;
lsd=stderr*tinv(1 -0.05/2,df);
proc print data=calc_lsd;
var lsd stderr df;
run;

For some reason that I ignore SAS indicates that the file diffs can not be created

Tkanks !!!

Re: LSD in Proc Mixed

Rick_SAS — Mon, 13 Sep 2021 15:23:07 GMT

1. In the future, please don't reopen a thread from 5 years ago. Open your own thread if you have a question.

2. Always include the SAS log so that we know what you mean by "SAS indicates that the file diffs can not be created".

Presumably, you are getting a warning such as
WARNING: Output 'diffs' was not created.
This warning appears because you are not displaying a table called 'diffs'. Consequently, your ODS OUTPUT statement cannot find a table from which to create a data set.

You are directly creating a SAS data set 'calc_lsd' that contains the information you need. I think you can simply delete the ODS OUTPUT statement.

Re: LSD in Proc Mixed

alimon — Mon, 13 Sep 2021 18:22:27 GMT

Thanks a lot !