Hello @Visiting,
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!
