Looking for sample data for example

[Jerrynetwork]

Found a paper about logistic regression for small sample size, here is the link for the paper:

Rare Events or Non-Convergence with a Binary Outcome?

https://www.sas.com/content/dam/SAS/support/en/sas-global-forum-proceedings/2020/4654-2020.pdf 

Does anyone know where can find the datasets SPARSE and SPREAD that were used in the exampleS? Thank you!

[Ksharp]

Write e-mail to author ?

[Ksharp]

Or you could re-produce it by the result of PROC FREQ in paper .

[FreelanceReinh]

---

@Ksharp wrote:
Or you could re-produce it by the result of PROC FREQ in paper .

---

Exactly. But the challenging (and interesting!) part is the AGE distribution in dataset SPARSE. 🙂 I'm working on that and I see chances to find solutions for the subgroup with vvalue(Complication)='Yes' (N=9).

[Jerrynetwork]

Thank you!

[FreelanceReinh · Accepted solution]

Hello @Jerrynetwork,

As @Ksharp has already suggested, it's easy to reproduce the (formatted) categorical variables of the two datasets by using the PROC FREQ outputs shown in the paper.

proc format;
value yesno
1='Yes'
2='No';
run;

data sparse;
do complication=1, 2;
  do procedure='New', 'Old';
    input _n_ @@;
    do _n_=1 to _n_;
      output;
    end;
  end;
end;
format complication yesno.;
cards;
0 9 30 191
;

data spread;
do event=1, 2;
  do group=1, 2;
    input _n_ @@;
    do _n_=1 to _n_;
      output;
    end;
  end;
end;
format event yesno.;
cards;
4 15 11 195
;

With these datasets you can reproduce tables 1 - 7, 10 and 11 of the paper. For table 4 add the option order=formatted to the PROC FREQ statement.

So, only tables 8 and 9 (involving variable AGE) remain. I'm sure other people have worked on this type of problem before -- creating data from given summary statistics -- so there must be more advanced techniques for this than I'm aware of.

From table 8 we get N=9, Mean=43.44 and Std=5.81 for AGE in the subgroup with vvalue(complication)='Yes'. Mostly, age values in clinical studies are integers. This together with the combination of N=9 and the decimals .44 of the mean suggest that the sum of the nine age values is 9*43.4444444...=391. The formula Var(X)=E(X²)-E(X)² applied to the discrete uniform distribution on the nine age values x1, ..., x9 yields (after multiplying with N²=81):

9*uss(of x1-x9) = sum(of x1-x9)**2 + 8*9*std(of x1-x9)**2

Given the inequality 5.805<=std(of x1-x9)<5.815 from the rounded Std value of 5.81, we conclude that

155308 <= 9*uss(of x1-x9) <= 155315

since 9*uss(of x1-x9) is an integer. But uss(of x1-x9) itself is an integer, too, and only one of the integers 155308, ..., 155315 is divisible by 9, namely 155313, hence:

uss(of x1-x9)=17257  (and std(of x1-x9)=sqrt(304)/3=5.811865...)

Number theorists could certainly tell us more about the ways 17257 can be written as a sum of 9 squares ... and even with the constraint sum(of x1-x9)=391 there will be a number of solutions.

Arranging and shifting integers with sum 391, centered around the rounded mean value 43 I found this particular solution for x1, ..., x9 even without letting the computer search through large numbers of combinations:

36 38 39 41 43 46 46 47 55

If you (unlike me) have SAS/OR, I think you can find all possible solutions for the above nine age values (assuming a reasonable age range, say, 18 - 90), tackle the second subgroup (N=221) in a similar way and ideally take table 9 of the paper into account in the optimization. Good luck and thanks for asking this inspiring question!