proc survey mean, standard deviation of square root

[Maria01]

I am somewhat new to survey analysis. I am running a code for survey sample and want to report the standard deviation, I am getting this big numbers in comparison to an unweighted code, I am pretty sure is something easy I am not looking at,can anyone help?

proc means data=ms19.ms19 MEAN STD STDERR ALPHA=0.05 CLM;;
VAR AGE;
RUN ;

SAS Output

Analysis Variable : AGE Age Mean Std Dev Std Error Lower 95%
CL for Mean Upper 95%
CL for Mean

45.8475175

13.9318244

0.1087396

45.6343762

46.0606588

 proc surveymeans data=ms19.ms19 MEAN RMSSTD STDERR ALPHA=0.05 CLM;
strata strat; cluster PSU_ID; weight WEIGHT_PA_IPW_OVERALL;
var AGE;
run;

SAS Output

SAS Output

Data Summary Number of Strata Number of Clusters Number of Observations Sum of Weights

20

664

16415

23413.2548

Statistics Variable Label Mean Std Error of Mean 95% CL for Mean Std Dev AGE

Age

39.867382

0.369889

39.1410467

40.5937167

40229

[art297]

My initial guess is that you have one or more AGEs that is impossibly high and also has a very high value in the variable you are using to weight the analysis.

Art, CEO, AnalystFinder.com

[Maria01]

Hi Art297, 

thank you for your help, I have read up on it and it seems I have a code the is not appropriate. The age range I have is from 45 to 75 years old and have no outliers, and my weight is to adjust for age as well. 

it seems the code I need is someting similar to this:

proc means data=xxx n MIN MAX MEAN STD STDERR ALPHA=0.05 CLM;
weight xxx;
BY xxx;
var AGE;
run;

SAS Output

Analysis Variable : AGE Age N Minimum Maximum Mean Std Dev Std Error Lower 95%
CL for Mean Upper 95%
CL for Mean

7675

45

75

56.50

6.63

0.09

56.31

56.68

see the results seem something more realistic.

Thank you. 

[lopezr]

PROC SURVEYMEANS does not directly compute the standard deviation for the mean, the reported st. dev. is for the SUM. 

You might find this SAS Note useful: http://support.sas.com/kb/45/701.html.