given dataset below, i want to compute the pooled sigma by Group. Already tried Ttest but it is limited to two class variables
Group | COUNT | MEAN | SIGMA | MEAS_DATE |
a | 501 | 101.1664 | 12.62792 | 14-Apr-15 |
a | 123 | 103.33 | 10.545 | 8-Apr-15 |
b | 164 | 99.21071 | 16.14705 | 8-Apr-15 |
b | 697 | 86.24266 | 9.794989 | 9-Apr-15 |
c | 223 | 82.22 | 7.422 | 7-Apr-15 |
c | 780 | 84.10039 | 8.913545 | 8-Apr-15 |
Thanks Dr. Rick - Yes, we need one extra set of parentheses to formulate this correctly.
proc sql;
select *,sqrt(sum((count-1)*sigma**2) / (sum(count)-count(*))) as pool_sigma from have
group by group;
quit;
Assuming sigma is standard deviation and therefore variance I would look into PROC GLM.
PS. Stats questions are best posted under Statistical Procedure forum.
data have;
input Group $ COUNT MEAN SIGMA MEAS_DATE :anydtdte.;
format meas_date ddmmyy8.;
datalines;
a 501 101.1664 12.62792 14-Apr-15
a 123 103.33 10.545 8-Apr-15
b 164 99.21071 16.14705 8-Apr-15
b 697 86.24266 9.794989 9-Apr-15
c 223 82.22 7.422 7-Apr-15
c 780 84.10039 8.913545 8-Apr-15
;
proc sql;
select *,sqrt(sum((count-1)*sigma**2)/sum(count)-count(*)) as pool_sigma from have
group by group;
quit;
Are you missing a set of parentheses? Seem like it should be
sqrt(sum((count-1)*sigma**2) / (sum(count)-count(*)))
Thanks Dr. Rick - Yes, we need one extra set of parentheses to formulate this correctly.
proc sql;
select *,sqrt(sum((count-1)*sigma**2) / (sum(count)-count(*))) as pool_sigma from have
group by group;
quit;
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