BookmarkSubscribeRSS Feed
🔒 This topic is solved and locked. Need further help from the community? Please sign in and ask a new question.
Miracle
Barite | Level 11

Dear all, wish you all well.


Can anyone please enlighten me on my understanding of the partial genmod output below?


The result shows that there is a decreasing trend in y over the 2-year period and the decrease does not vary from responses to A1.

I tried to insert the interaction plot here to illustrate my question but it doesn't work and I don't know why given the small image size.

According to the interaction plot, all the lines joining the LSM mean values from 2010 and 2011 by A1 seem to be parallel and show a decreasing trend. However, 2=Disagree a little line intersects with 3=Neither agree nor disagree line. How do I find out/quantify which of the deceases from A1 responses is different given the output?


Your insight is greatly appreciated. Thank you very much.


proc genmod data=temp order=internal plots=all;

  class id time var1;

  MODEL y= var1 time var1*time / type3 dist=gamma link=log;

  lsmeans time*var1 / om diff ilink cl plots=all;

  slice time*var1 / sliceby=var1 ilink diff nof;

  repeated subject=id / type=cs;

run;



Score Statistics For Type 3 GEE Analysis
SourceDFChi-SquarePr > ChiSq
time128.17<.0001p-value<0.05 indicates that there is a change in y over the 2-year period.
A1413.720.0082p-value<0.05 indicates A1 responses do change in y.
time*A143.560.4685p-value>0.05 indicates that A1 responses do not change in y in different ways over the 2-year period.

time*A1 Least Squares Means
YearA1MarginsEstimateStandard Errorz ValuePr > |z|AlphaLowerUpperMeanStandard Error of MeanLower MeanUpper Mean
20101=Disagree a lotWORK.LONG3.7790.0427688.38<.00010.053.69523.862843.77391.871840.254847.6007
20102=Disagree a littleWORK.LONG3.80230.03689103.06<.00010.053.733.874644.80551.653141.679948.1654
20103=Neither agree nor disagreeWORK.LONG3.86490.0667757.88<.00010.053.7343.995747.69623.184641.845754.3647
20104=Agree a littleWORK.LONG3.85830.0504276.53<.00010.053.75953.957147.3832.388942.924952.3043
20105=Agree a lotWORK.LONG4.14240.0772853.6<.00010.053.99094.293862.95284.864854.104973.2476
20111=Disagree a lotWORK.LONG3.62290.03579101.23<.00010.053.55283.69337.44591.340134.909440.1667
20112=Disagree a littleWORK.LONG3.73040.0463280.53<.00010.053.63973.821241.69771.931638.078645.6608
20113=Neither agree nor disagreeWORK.LONG3.65910.0702852.07<.00010.053.52133.796838.82562.728533.829844.5592
20114=Agree a littleWORK.LONG3.72490.053369.89<.00010.053.62053.829441.46872.210337.355246.0353
20115=Agree a lotWORK.LONG3.97320.0875245.4<.00010.053.80174.144753.15464.652244.775763.1015

Simple Differences of time*A1 Least Squares Means
Slicetime_timeEstimateStandard Errorz ValuePr > |z|
A1 1=Disagree a lot201020110.15610.044323.520.0004
A1 2=Disagree a little201020110.071880.040151.790.0734
A1 3=Neither agree nor disagree201020110.20580.08172.520.0118
A1 4=Agree a little201020110.13330.051852.570.0101
A1 5=Agree a lot201020110.16920.064912.610.0092
1 ACCEPTED SOLUTION

Accepted Solutions
SteveDenham
Jade | Level 19

The simple difference tests follow from the overall test--the two years differ, and it is roughly the same difference for all categories of A1.  The estimate values all indicate a change with a range of 0.07 to 0.21, where all changes have 'effect sizes' (Z scores) of at least 1.79.  (I'm weaseling around the p values here, because I don't know what your cutoff for significance is).  There really does not look like there is any strong evidence for an interaction.

Am I making sense, or did I miss the point of your question?

Steve Denham

View solution in original post

3 REPLIES 3
SteveDenham
Jade | Level 19

The simple difference tests follow from the overall test--the two years differ, and it is roughly the same difference for all categories of A1.  The estimate values all indicate a change with a range of 0.07 to 0.21, where all changes have 'effect sizes' (Z scores) of at least 1.79.  (I'm weaseling around the p values here, because I don't know what your cutoff for significance is).  There really does not look like there is any strong evidence for an interaction.

Am I making sense, or did I miss the point of your question?

Steve Denham

Miracle
Barite | Level 11

Hi Steve. I really appreciate you taking the time answering my question.


My analysis is based on 5% level of significance FYI.

If the p-value of the interaction term is > 0.05 indicates the change for all levels of var1 is no difference.

Does each of the significant simple difference tests by var1 reflect from the significant main time effect?  

Also,  is it still sensible to quantify the simple differences and 95% C.I.  on the original scale as you previously posted here ()?


Thank you very much.

SteveDenham
Jade | Level 19

Hi KC,

Some answers:

If the p-value of the interaction term is > 0.05 indicates the change for all levels of var1 is no difference.

A non-significant interaction, but significant main effects, implies that the differences between the levels of var1 is the same at all times, and that the differences between levels of time is the same for all levels of var1.  Therefore, the best estimates of effects are the marginal means (main effect means).

Does each of the significant simple difference tests by var1 reflect from the significant main time effect?  

Yes, that is what is being shown.

Also,  is it still sensible to quantify the simple differences and 95% C.I.  on the original scale as you previously posted here

While you can report these, the marginal differences are probably more meaningful.  Try adding the following to your code:

lsmeans time var1/diff cl ilink;

to get the marginal means and differences.

Steve Denham

sas-innovate-2024.png

Join us for SAS Innovate April 16-19 at the Aria in Las Vegas. Bring the team and save big with our group pricing for a limited time only.

Pre-conference courses and tutorials are filling up fast and are always a sellout. Register today to reserve your seat.

 

Register now!

What is ANOVA?

ANOVA, or Analysis Of Variance, is used to compare the averages or means of two or more populations to better understand how they differ. Watch this tutorial for more.

Find more tutorials on the SAS Users YouTube channel.

Discussion stats
  • 3 replies
  • 1546 views
  • 3 likes
  • 2 in conversation