[Carlo Lazzaro]

Thinking about this thread during the last few days, perhaps -testnl- can also contribute to decide about the evidence of a -poisson- mechanism.
Starting off from the first lines of Nick Cox 's code:

. set obs 1000
number of observations (_N) was 0, now 1,000

. set seed 2803

. gen y = rpoisson(3)

. sum y

    Variable |        Obs        Mean    Std. Dev.       Min        Max
-------------+---------------------------------------------------------
           y |      1,000       3.029    1.746198          0         10

. testnl (_se[y]^2*1000)/_b[y]=1

  (1)  (_se[y]^2*1000)/_b[y] = 1

               chi2(1) =        0.13
           Prob > chi2 =        0.7162

In this example, the outcome of -testnl- does not reject the null of equidispersion between variance and mean (ie, the main feature of a -poisson- distribution).

[Gaurav Dhamija]

Dear Nick,
I was able to do the exercise with the help of your code. However, I am getting all the expected values of less than 5 for one variable.

In this scenario, should I rely on the chi-square test statistic?
If not, then what should I do in this case to test the fit?

[Gaurav Dhamija]

Dear Carlo,

Thank you for this alternative solution in #16.
Why do you multiply 1000(=N) with variance in (1)?
Since the index of dispersion is variance divided by mean then why do we need to multiply by 1000 in the numerator?

[Nick Cox]

You have a sample size of 10. You won't get clearcut signals from a sample so small. In essence you don't have enough information to reject Poissonness.


Here is what your analysis implies:

* Example generated by -dataex-. To install: ssc install dataex
clear
input float(x frequency)
 0 0
 1 0
 2 1
 3 1
 4 2
 5 2
 6 0
 7 3
 8 0
 9 0
10 0
11 0
12 1
end

su x [w=freq]

gen expected = 10 * poissonp(5.6, x)

set obs 14 
replace x = 13 in L
label def x 13 "13 up"
label val x x 

su expected
replace expected = 10 - r(sum) in L

set scheme s1color
twoway bar freq x, xla(0/13, valuelabel) barw(0.8) bfcolor(eltgreen*0.3) ///
|| connected exp x, ytitle(frequency) legend(order(2) pos(1) ring(0)) yla(, ang(h))

However, back in #1 you said

I have a variable - number of people who are malnourished in a PSU. I have approximately 2000 independently surveyed PSUs. Data would like as follows:
X
20
30
50
70
.
.
.
.
80

so this analysis seems to bear absolutely no relation to your original question.

[Carlo Lazzaro]

Gaurav:

. . sum y

    Variable |        Obs        Mean    Std. Dev.       Min        Max
-------------+---------------------------------------------------------
           y |      1,000       3.029    1.746198          0         10

.
*Variance of y calculated from standard deviation*
.
. di r(sd)^2
3.0492082

. mean y

Mean estimation                   Number of obs   =      1,000

--------------------------------------------------------------
             |       Mean   Std. Err.     [95% Conf. Interval]
-------------+------------------------------------------------
           y |      3.029   .0552196       2.92064     3.13736
--------------------------------------------------------------

*Variance of y calculated from standard error of the mean with N=1000*

. di (.0552196^2*1000)
3.0492042

.

[Gaurav Dhamija]

Dear Nick,

Thank you for further explanation.
Yes, you are right. In the first case (#1) I was trying to fit the Poisson distribution for the PSUs at the national level which are approximately 2000 in number. However, in the last query (#17) I did it at the district level and hence the number of PSUs reduce to 10 in the latter case.

[Gaurav Dhamija]

Dear Carlo,

Thank you for making it more clear.

[Nick Cox]

A full analysis would surely feature population at risk too.