[Limdep Nlogit List] interpreting Poe test result

[Kolady, Deepthi]

Hello All,

I have a discrete choice experiment model where I estimate marginal willingness to pay (mWTP) using WTP space model.  I have included an attribute on whether the livestock is raised using conventional production method ot not. I have a control and information treatment for the survey. My objective is to test whether the mWTP for conventional control sample is statistically different from mWTP for conventional treatment. I ran the following code for Poe Test where vector 1=CONV (conventional control) and Vector 2=CONVT(conventional treatment). I am new to NLOGIT and has not used Poe test before. I have used the code below after reading through many papers. Can somebody confirm whether the code is correct? I got the result for Poe test using the code, but I am not sure how to interpret the results. I do not see any p Values to see whether we accept or reject the null hypothesis.

|-> IMPORT;FILE="C:\Users\mariam.ishaq\OneDrive - South Dakota State University - SDSU\Desktop\Mariam Ishaq\Consumer behavior for sustaianable beef\Pork data\Porkdata for Poe test in N logitCSV.csv"$
Last observation read from data file was     328
|-> crea   ; x=CONV ; y = CONVT $
|-> calc   ; m = 328    ; l = 328 ; ixy=0 ; nd0 = 0 $
|-> matrix ; ones = init(m,1,1)$
|-> matr   ; my = y $
|-> crea   ; vd = 0 $
|-> sample ; 1 - m $
|-> proc $
|-> calc   ; ixy=ixy+1 ; xi = x(ixy) $
|-> matrix ; md = xi*ones - my $
|-> crea   ; vd = md $
|-> crea   ; vd = vd < 0 $
|-> calc   ; nd0 = nd0 + sum(vd) $
|-> endproc
|-> exec;n=m$
|-> calc; list ; h0= nd0 / (m*l)$

Maximum repetitions of PROC
*************************************************************************************
The output I got from the code above was
|-> calc; list ; h0= nd0 / (m*l)$
[CALC] H0      =         .3534262
|-> calc; list ; Log_prob = 1 - h0 $
[CALC] LOG_PROB=         .6465738
|->                           |-----------> ?Kr_Prob is essentially the p-value


Thanks,

Deepthi

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