# st: survey command - interpretation

 From ann e fitzmaurice To statalist@hsphsun2.harvard.edu Subject st: survey command - interpretation Date Thu, 21 Nov 2002 09:46:23 -0500 (EST)

```hi

i have the following output from the survey commands procesure
The psu is a self computed variable based on the number of times
the woman appears in the data set

my question relates to the p value

if i ignore the clustering i get a p value of p<0.001, but with the
clustering i get the output below and a p value of 0.0397, can this
second p value be interpreted as an association between the group
status and education level after adjusting for the the clustering
effect??

comment would be appreciated, i realise that one of the cells does
not have any responses , so a certain amount of amalgamation will
be required

regards

ann

log type:  text
opened on:  21 Nov 2002, 09:21:21

. svyset pweight weight

. svyset psu sib_no1

. svytab sib_grp v106a if level1 ==1, pearson row  count format (%9.2f)

pweight:  weight                                Number of obs      =     12531
Strata:   <one>                                 Number of strata   =         1
PSU:      sib_no1                               Number of PSUs     =         9
Population size    = 12489.897

------------------------------------------------------------
grouping  |
variable  |
for       |           education levels transposed
sibling   |   higher   seconda   primary   no educ     Total
----------+-------------------------------------------------
alive |    12.75    464.89   2471.94   8779.22  11728.80
|     0.00      0.04      0.21      0.75      1.00
|
non-mate |     0.51     15.74     62.05    363.91    442.22
|     0.00      0.04      0.14      0.82      1.00
|
materna |     0.00      5.10     45.77    268.02    318.88
|     0.00      0.02      0.14      0.84      1.00
|
Total |    13.26    485.72   2579.76   9411.15  12489.90
|     0.00      0.04      0.21      0.75      1.00
------------------------------------------------------------
Key:  weighted counts
row proportions

Pearson:
Uncorrected   chi2(6)         =   27.9839
Design-based  F(2.19, 17.49)  =    3.7901     P = 0.0397

. log close
log type:  text
closed on:  21 Nov 2002, 09:22:10
----------------------
ann e fitzmaurice
medical statistician
dugald baird centre
department of obs and gynae
university of aberdeen
a.e.fitzmaurice@abdn.ac.uk
tel 01224 553876 (direct)
fax 01224 404925

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