Wald test on STATA to show that distribution among two groups is statistically different as a result of a parameter.

[Tia Denek]

How do I perform a wald test on STATA to achieve the following? I have a table of that look like the one below. I am trying to write down the distribution of people with 0,1,2,3,4,5 cardiovascular disease risk factors in my sample. Under "All" I will write number of people with these risk factors and under "smoking" and "non-smoking" I will write how many of these people smoke and how many don't. Now, I want to show that the distribution of smokers and non-smokers is significantly different in each row to show that that smoking plays a role in the number of risk factors and wanted to do so with a Wald type test. I wanted help on how I could perform this.
No. of Risk Factors **** All **** Smoking ***** Not smoking
0
1
2
3
4
5

[Carlo Lazzaro]

Tia:
why not -reshape-ing your data in -long- format and going -poisson- here?

poissson <CVD_risk_factor> i.smoker

[Tia Denek]

That makes sense, I will do that. Thank you. However, if I were asked to do a Wald test, how do you suggest I go about it? Asking because I was asked to do a wald test to figure this out although I feel there may be other alternatives.

[Carlo Lazzaro]

Tia:
if you mean a Wald test to check whether coefficients of -poisson-regression show evidence of a statistically significant difference, plese see the following toy-example:

use "C:\Program Files\Stata16\ado\base\a\auto.dta"
. poisson rep78 i.foreign

Iteration 0:   log likelihood = -112.88724 
Iteration 1:   log likelihood = -112.88724 

Poisson regression                              Number of obs     =         69
                                                LR chi2(1)        =       6.58
                                                Prob > chi2       =     0.0103
Log likelihood = -112.88724                     Pseudo R2         =     0.0283

------------------------------------------------------------------------------
       rep78 |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
     foreign |
    Foreign  |   .3497545   .1341926     2.61   0.009     .0867418    .6127672
       _cons |   1.105533   .0830455    13.31   0.000     .9427666    1.268299
------------------------------------------------------------------------------

. test 0.foreign=1.foreign

 ( 1)  [rep78]0b.foreign - [rep78]1.foreign = 0

           chi2(  1) =    6.79
         Prob > chi2 =    0.0092

.

Otherwise, if you want to stop at counts and related proportions of the CVD risk factor in smokers and non-smokers, elaborating on the previous toy-example you may want to consider somethng along the following lines:

use "C:\Program Files\Stata16\ado\base\a\auto.dta"
. tabulate foreign rep78, chi2 expected

+--------------------+
| Key                |
|--------------------|
|     frequency      |
| expected frequency |
+--------------------+

           |                   Repair Record 1978
  Car type |         1          2          3          4          5 |     Total
-----------+-------------------------------------------------------+----------
  Domestic |         2          8         27          9          2 |        48
           |       1.4        5.6       20.9       12.5        7.7 |      48.0
-----------+-------------------------------------------------------+----------
   Foreign |         0          0          3          9          9 |        21
           |       0.6        2.4        9.1        5.5        3.3 |      21.0
-----------+-------------------------------------------------------+----------
     Total |         2          8         30         18         11 |        69
           |       2.0        8.0       30.0       18.0       11.0 |      69.0

          Pearson chi2(4) =  27.2640   Pr = 0.000

.

[Tia Denek]

This clears things up a lot...thank you so much for the swift and very helpful response. I will follow up if I have further questions.

[Tia Denek]

Hello, so I could perform a poisson regression and use the wald p-value to determine if two coefficients show a statistically significant difference. Is there a way to do a prevalence ratio and use wald pvalue to determine the statistical difference between smokers and non smokers for each risk factor level?

[Carlo Lazzaro]

Tia:
1) if you only have CVD risk-factor and -i.smoking- (and the latter is your only predictor), you can go:

bysort <CVD_risk>: poisson <CVD_risk> i.smoking

2) if your regressand is a categorical variable such as -disease_yes_no-, you can go -logit- or -logistic- and interact CVD risk-factor with smoking addiction:

logit <disease_yes_no> i.CVD_risk##i.smoking