"Nasser M. Abbasi" schrieb:   
   >   
   > Lets say there are 14,000 integrals which are passed.   
   >   
   > Then for each one of these integrals, I divide its leaf size by the   
   > leaf size of the optimal result for it. The reasume is new   
   > list of 14,000 numbers (each is ratio).   
   >   
   > Then find the mean of this long list of numbers using the Mean   
   > command.   
   >   
   > So I am doing the second method you mentioned above. Is this the   
   > right way? I am not a statistic person and only managed to get   
   > B in my probability and statics course with lots of struggle.   
   >   
   > The code is this one line:   
   >   
   >    meanNormalized = Mean[  allPassed[[All,3]] / allPassed[[All,4]]  ];   
   >   
   > The 3rd field is the leaf size of the result from CAS, and the 4th   
   > field is the leaf size of the optimal from your input file.   
   >   
   > allPassed is the array which contains all the integrals which passed   
   > from all the 200 test files.   
   >   
   > I can change it if it not the correct way.   
   >   
      
   I think this is the correct way, but the column heading in your table   
   should then read "mean normalized leaf size" and not "normalized mean   
   leaf size" as quoted in your preceding post.   
      
   Compare my message of 11th June 2015 in the thread "fyi, rebuild CAS   
   integration tests, Rubi 4.8, Maple 2015, Mathematica 10.1".   
      
   How about computing the "median normalized leaf size" too for   
   comparison?   
      
   Martin.   
      
   PS: By the way, that thread "fyi, rebuild CAS integration tests, Rubi   
   4.8, Maple 2015, Mathematica 10.1" of summer 2015 is a real gem towards   
   the end.   
      
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