Q = The sum for i=1 to 10 of ( (N_{i} - np_{i})^{2}/np_{i} )

= The sum for i=1 to 10 of ( (N_{i} - 1000)^{2}/1000 )

= (676+121+16+64+324+64+6241+64+16+1024)/1000 = 8.61

According the the chi-squared table with 9 degrees of freedom a Q statistic of greater than 16.92 shold fail the test. 8.61 is well under 16.92 so it should pass the test. In other words the digits do appear to be random.

At a level of significance such that 10,000 random digits passed 60% of the time the 10,000 digits of e would still pass, but at a 50% level they would not.

Michael Shackleford, A.S.A.