p>[z] in nlcom results

[Arnold Levinson]

I have a large survey data set. After estimating means (proportions), I use nlcom to estimate rate ratios and use post to store results. The first output includes p>[z], which is almost always <0.000, while the confidence intervals cover 1. The test command produces a more expected p value, usually >>0.05. What is p>[z] reporting?
Thanks,
Arnold

[Andrew Musau]

It would be helpful if you showed your output to better illustrate your question. For nlcom and most commands, the null hypothesis of the test is whether the estimated coefficient (in your case, the ratio) is equal to zero. This may be uninteresting if you are instead interested in whether the ratio is equal to one. In that case, simply use the test command following nlcom.

sysuse auto, clear
regress mpg weight displacement
nlcom _b[weight]/ _b[displacement], post
test _b[_nl_1]=0
test _b[_nl_1]=1

Res.:

. nlcom _b[weight]/ _b[displacement], post

       _nl_1: _b[weight]/ _b[displacement]

------------------------------------------------------------------------------
         mpg | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       _nl_1 |  -1.243587   2.128868    -0.58   0.559    -5.416092    2.928917
------------------------------------------------------------------------------

. 
. test _b[_nl_1]=0

 ( 1)  _nl_1 = 0

           chi2(  1) =    0.34
         Prob > chi2 =    0.5591

. 
. test _b[_nl_1]=1

 ( 1)  _nl_1 = 1

           chi2(  1) =    1.11
         Prob > chi2 =    0.2919

[Arnold Levinson]

Andrew,
Yes this is what I was asking, and your example is like my situation. I see that the reported P>|z| is for coeffiicient==0, and for a rate ratio the test is against coefficient==1. Thanks for the helpful feedback.

[Andrew Musau]

I see that the reported P>|z| is for coeffiicient==0, and for a rate ratio the test is against coefficient==1

Subtracting 1 from the ratio would also serve as an appropriate reparameterization to test the hypothesis that the ratio is equal to 1.

sysuse auto, clear
regress mpg weight displacement
nlcom _b[weight]/ _b[displacement], post
test _b[_nl_1]=1

qui regress mpg weight displacement
*REPARAMETRIZE
nlcom _b[weight]/ _b[displacement] -1, post

Res.:

. nlcom _b[weight]/ _b[displacement], post

       _nl_1: _b[weight]/ _b[displacement]

------------------------------------------------------------------------------
         mpg | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       _nl_1 |  -1.243587   2.128868    -0.58   0.559    -5.416092    2.928917
------------------------------------------------------------------------------

. 
. test _b[_nl_1]=1

 ( 1)  _nl_1 = 1

           chi2(  1) =    1.11
         Prob > chi2 =    0.2919

. 
. 
. 
. qui regress mpg weight displacement

. 
. *REPARAMETRIZE

. 
. nlcom _b[weight]/ _b[displacement] -1, post

       _nl_1: _b[weight]/ _b[displacement] -1

------------------------------------------------------------------------------
         mpg | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       _nl_1 |  -2.243587   2.128868    -1.05   0.292    -6.416092    1.928917
------------------------------------------------------------------------------

.

[Arnold Levinson]

Yes, I see. Thank you again. Now I have a new question, please advise whether I should create a new thread.
I have calculated weight ratios from survey data, and ignoring z and P>|z| because they're irrelevant, I want to determine whether the rate-ratio increasing trend is significant. Test won't accept a series of inequalities, or even a single inequality. I tried creating a data set of the rate ratios and standard errors, then regressing with pweights = 1/se^2. But there are only five observations that way, which seems fishy.

Here's the code and output


svyset hh_id_allyr [pw=phonewt], strata(geowt)
svy, subpop(if single!=1): mean quitterb, over(year inequity)

nlcom (Rate_ratio_2008: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2012: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2015: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2018: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2022: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]), post


Survey: Mean estimation

Number of strata = 113 Number of obs = 4,883
Number of PSUs = 4,842 Population size = 1,950,480
Subpop. no. obs = 4,883
Subpop. size = 1,950,480
Design df = 4,729

--------------------------------------------------------------------------
| Linearized
| Mean Std. Err. [95% Conf. Interval]
-------------------------+------------------------------------------------
c.quitterb@year#inequity |
2008 0 | .1676293 .0299719 .1088705 .2263882
2008 1 | .074061 .0110659 .0523667 .0957552
2012 0 | .1251159 .0364989 .0535611 .1966707
2012 1 | .0878328 .0146444 .0591229 .1165427
2015 0 | .1406765 .0416307 .0590609 .2222921
2015 1 | .0981557 .0164493 .0659075 .1304039
2018 0 | .2122044 .0416318 .1305866 .2938222
2018 1 | .1602063 .0181833 .1245585 .195854
2022 0 | .1902246 .0596587 .0732657 .3071835
2022 1 | .1772962 .0249485 .1283854 .226207
--------------------------------------------------------------------------
Note: 2 strata omitted because they contain no subpopulation members.

.
.
.
. nlcom (Rate_ratio_2008: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2012: _b[[email protected]#1.inequity] / _b[quitterb@20
> 12.year#0.inequity]) (Rate_ratio_2015: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]) (Rate_ratio_2018: _b[[email protected]#1.inequity] /
> _b[[email protected]#0.inequity]) (Rate_ratio_2022: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]), post

Rate_ra~2008: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]
Rate_ra~2012: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]
Rate_ra~2015: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]
Rate_ra~2018: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]
Rate_ra~2022: _b[[email protected]#1.inequity] / _b[[email protected]#0.inequity]

---------------------------------------------------------------------------------
Mean | Coef. Std. Err. z P>|z| [95% Conf. Interval]
----------------+----------------------------------------------------------------
Rate_ratio_2008 | .4418138 .1019554 4.33 0.000 .2419849 .6416426
Rate_ratio_2012 | .7020116 .2357582 2.98 0.003 .239934 1.164089
Rate_ratio_2015 | .6977409 .2377938 2.93 0.003 .2316736 1.163808
Rate_ratio_2018 | .754962 .1711076 4.41 0.000 .4195974 1.090327
Rate_ratio_2022 | .932036 .3217958 2.90 0.004 .3013277 1.562744
---------------------------------------------------------------------------------

[Arnold Levinson]

Correction: Rate ratios (actually prevalence ratios), not "weight ratios".

[Andrew Musau]

I want to determine whether the rate-ratio increasing trend is significant.

It depends on how you define the slope—for example, whether you take the difference between the first and last estimates and divide by the number of survey waves, or instead compute an average slope based on all successive differences across waves. For the latter:

\[
\text{slope} \;=\; \frac{1}{4} \Big[
\big( \hat{\theta}_{2012} - \hat{\theta}_{2008} \big) \;+\;
\big( \hat{\theta}_{2015} - \hat{\theta}_{2012} \big) \;+\;
\big( \hat{\theta}_{2018} - \hat{\theta}_{2015} \big) \;+\;
\big( \hat{\theta}_{2022} - \hat{\theta}_{2018} \big)
\Big].
\]

To test that it is significantly > 0, you can use nlcom as follows:

nlcom (slope: ( (_b[Rate_ratio_2012] - _b[Rate_ratio_2008]) ///
              + (_b[Rate_ratio_2015] - _b[Rate_ratio_2012]) ///
              + (_b[Rate_ratio_2018] - _b[Rate_ratio_2015]) ///
              + (_b[Rate_ratio_2022] - _b[Rate_ratio_2018]) ) / 4 )

[Arnold Levinson]

Thanks! The second option is what I wanted.