NOTE: Most of the tests in DIEHARD return a p-value, which should be uniform on [0,1) if the input file contains truly independent random bits. Those p-values are obtained by p=F(X), where F is the assumed distribution of the sample random variable X---often normal. But that assumed F is just an asymptotic approximation, for which the fit will be worst in the tails. Thus you should not be surprised with occasional p-values near 0 or 1, such as .0012 or .9983. When a bit stream really FAILS BIG, you will get p's of 0 or 1 to six or more places. By all means, do not, as a Statistician might, think that a p < .025 or p> .975 means that the RNG has "failed the test at the .05 level". Such p's happen among the hundreds that DIEHARD produces, even with good RNG's. So keep in mind that " p happens". ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BIRTHDAY SPACINGS TEST :: :: Choose m birthdays in a year of n days. List the spacings :: :: between the birthdays. If j is the number of values that :: :: occur more than once in that list, then j is asymptotically :: :: Poisson distributed with mean m^3/(4n). Experience shows n :: :: must be quite large, say n>=2^18, for comparing the results :: :: to the Poisson distribution with that mean. This test uses :: :: n=2^24 and m=2^9, so that the underlying distribution for j :: :: is taken to be Poisson with lambda=2^27/(2^26)=2. A sample :: :: of 500 j's is taken, and a chi-square goodness of fit test :: :: provides a p value. The first test uses bits 1-24 (counting :: :: from the left) from integers in the specified file. :: :: Then the file is closed and reopened. Next, bits 2-25 are :: :: used to provide birthdays, then 3-26 and so on to bits 9-32. :: :: Each set of bits provides a p-value, and the nine p-values :: :: provide a sample for a KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: BIRTHDAY SPACINGS TEST, M= 512 N=2**24 LAMBDA= 2.0000 Results for o:\tmp\x32s.raw For a sample of size 500: mean o:\tmp\x32s.raw using bits 1 to 24 1.950 duplicate number number spacings observed expected 0 71. 67.668 1 142. 135.335 2 136. 135.335 3 82. 90.224 4 43. 45.112 5 16. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 1.93 p-value= .074167 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 2 to 25 1.996 duplicate number number spacings observed expected 0 63. 67.668 1 136. 135.335 2 138. 135.335 3 98. 90.224 4 40. 45.112 5 18. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 1.83 p-value= .064989 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 3 to 26 2.012 duplicate number number spacings observed expected 0 63. 67.668 1 139. 135.335 2 147. 135.335 3 80. 90.224 4 36. 45.112 5 25. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 7.46 p-value= .719846 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 4 to 27 1.946 duplicate number number spacings observed expected 0 79. 67.668 1 120. 135.335 2 148. 135.335 3 82. 90.224 4 55. 45.112 5 7. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 14.56 p-value= .976032 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 5 to 28 2.010 duplicate number number spacings observed expected 0 60. 67.668 1 145. 135.335 2 125. 135.335 3 98. 90.224 4 51. 45.112 5 15. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 4.93 p-value= .447123 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 6 to 29 2.006 duplicate number number spacings observed expected 0 66. 67.668 1 132. 135.335 2 137. 135.335 3 99. 90.224 4 39. 45.112 5 18. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 1.89 p-value= .070287 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 7 to 30 1.960 duplicate number number spacings observed expected 0 70. 67.668 1 128. 135.335 2 152. 135.335 3 84. 90.224 4 43. 45.112 5 16. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 3.49 p-value= .254466 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 8 to 31 1.966 duplicate number number spacings observed expected 0 68. 67.668 1 138. 135.335 2 135. 135.335 3 97. 90.224 4 34. 45.112 5 21. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 3.98 p-value= .321064 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\x32s.raw using bits 9 to 32 2.060 duplicate number number spacings observed expected 0 56. 67.668 1 149. 135.335 2 125. 135.335 3 94. 90.224 4 47. 45.112 5 17. 18.045 6 to INF 12. 8.282 Chisquare with 6 d.o.f. = 6.15 p-value= .593151 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .074167 .064989 .719846 .976032 .447123 .070287 .254466 .321064 .593151 A KSTEST for the 9 p-values yields .678304 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE OVERLAPPING 5-PERMUTATION TEST :: :: This is the OPERM5 test. It looks at a sequence of one mill- :: :: ion 32-bit random integers. Each set of five consecutive :: :: integers can be in one of 120 states, for the 5! possible or- :: :: derings of five numbers. Thus the 5th, 6th, 7th,...numbers :: :: each provide a state. As many thousands of state transitions :: :: are observed, cumulative counts are made of the number of :: :: occurences of each state. Then the quadratic form in the :: :: weak inverse of the 120x120 covariance matrix yields a test :: :: equivalent to the likelihood ratio test that the 120 cell :: :: counts came from the specified (asymptotically) normal dis- :: :: tribution with the specified 120x120 covariance matrix (with :: :: rank 99). This version uses 1,000,000 integers, twice. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPERM5 test for file o:\tmp\x32s.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 92.641; p-value= .339216 OPERM5 test for file o:\tmp\x32s.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 86.653; p-value= .192345 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 31x31 matrices. The leftmost :: :: 31 bits of 31 random integers from the test sequence are used :: :: to form a 31x31 binary matrix over the field {0,1}. The rank :: :: is determined. That rank can be from 0 to 31, but ranks< 28 :: :: are rare, and their counts are pooled with those for rank 28. :: :: Ranks are found for 40,000 such random matrices and a chisqua-:: :: re test is performed on counts for ranks 31,30,29 and <=28. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for o:\tmp\x32s.raw Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 215 211.4 .060688 .061 29 5076 5134.0 .655470 .716 30 23052 23103.0 .112790 .829 31 11657 11551.5 .963085 1.792 chisquare= 1.792 for 3 d. of f.; p-value= .473670 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 32x32 matrices. A random 32x :: :: 32 binary matrix is formed, each row a 32-bit random integer. :: :: The rank is determined. That rank can be from 0 to 32, ranks :: :: less than 29 are rare, and their counts are pooled with those :: :: for rank 29. Ranks are found for 40,000 such random matrices :: :: and a chisquare test is performed on counts for ranks 32,31, :: :: 30 and <=29. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary rank test for o:\tmp\x32s.raw Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 224 211.4 .748784 .749 30 5163 5134.0 .163694 .912 31 23116 23103.0 .007262 .920 32 11497 11551.5 .257361 1.177 chisquare= 1.177 for 3 d. of f.; p-value= .380472 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the BINARY RANK TEST for 6x8 matrices. From each of :: :: six random 32-bit integers from the generator under test, a :: :: specified byte is chosen, and the resulting six bytes form a :: :: 6x8 binary matrix whose rank is determined. That rank can be :: :: from 0 to 6, but ranks 0,1,2,3 are rare; their counts are :: :: pooled with those for rank 4. Ranks are found for 100,000 :: :: random matrices, and a chi-square test is performed on :: :: counts for ranks 6,5 and <=4. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Binary Rank Test for o:\tmp\x32s.raw Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 886 944.3 3.600 3.600 r =5 21708 21743.9 .059 3.659 r =6 77406 77311.8 .115 3.774 p=1-exp(-SUM/2)= .84844 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 906 944.3 1.554 1.554 r =5 21861 21743.9 .631 2.184 r =6 77233 77311.8 .080 2.264 p=1-exp(-SUM/2)= .67769 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 941 944.3 .012 .012 r =5 21831 21743.9 .349 .360 r =6 77228 77311.8 .091 .451 p=1-exp(-SUM/2)= .20199 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 950 944.3 .034 .034 r =5 21701 21743.9 .085 .119 r =6 77349 77311.8 .018 .137 p=1-exp(-SUM/2)= .06617 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 971 944.3 .755 .755 r =5 21729 21743.9 .010 .765 r =6 77300 77311.8 .002 .767 p=1-exp(-SUM/2)= .31849 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 950 944.3 .034 .034 r =5 21740 21743.9 .001 .035 r =6 77310 77311.8 .000 .035 p=1-exp(-SUM/2)= .01741 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 936 944.3 .073 .073 r =5 21555 21743.9 1.641 1.714 r =6 77509 77311.8 .503 2.217 p=1-exp(-SUM/2)= .66995 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 944 944.3 .000 .000 r =5 21540 21743.9 1.912 1.912 r =6 77516 77311.8 .539 2.451 p=1-exp(-SUM/2)= .70646 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 946 944.3 .003 .003 r =5 21691 21743.9 .129 .132 r =6 77363 77311.8 .034 .166 p=1-exp(-SUM/2)= .07949 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 964 944.3 .411 .411 r =5 21664 21743.9 .294 .705 r =6 77372 77311.8 .047 .751 p=1-exp(-SUM/2)= .31319 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 949 944.3 .023 .023 r =5 21838 21743.9 .407 .431 r =6 77213 77311.8 .126 .557 p=1-exp(-SUM/2)= .24304 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21539 21743.9 1.931 2.212 r =6 77533 77311.8 .633 2.845 p=1-exp(-SUM/2)= .75890 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 946 944.3 .003 .003 r =5 21707 21743.9 .063 .066 r =6 77347 77311.8 .016 .082 p=1-exp(-SUM/2)= .04003 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 938 944.3 .042 .042 r =5 21705 21743.9 .070 .112 r =6 77357 77311.8 .026 .138 p=1-exp(-SUM/2)= .06670 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 912 944.3 1.105 1.105 r =5 21819 21743.9 .259 1.364 r =6 77269 77311.8 .024 1.388 p=1-exp(-SUM/2)= .50042 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 965 944.3 .454 .454 r =5 21789 21743.9 .094 .547 r =6 77246 77311.8 .056 .603 p=1-exp(-SUM/2)= .26039 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 950 944.3 .034 .034 r =5 21754 21743.9 .005 .039 r =6 77296 77311.8 .003 .042 p=1-exp(-SUM/2)= .02093 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 907 944.3 1.473 1.473 r =5 21785 21743.9 .078 1.551 r =6 77308 77311.8 .000 1.551 p=1-exp(-SUM/2)= .53960 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 962 944.3 .332 .332 r =5 21991 21743.9 2.808 3.140 r =6 77047 77311.8 .907 4.047 p=1-exp(-SUM/2)= .86779 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 905 944.3 1.636 1.636 r =5 21865 21743.9 .674 2.310 r =6 77230 77311.8 .087 2.397 p=1-exp(-SUM/2)= .69831 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 902 944.3 1.895 1.895 r =5 21783 21743.9 .070 1.965 r =6 77315 77311.8 .000 1.965 p=1-exp(-SUM/2)= .62570 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 966 944.3 .499 .499 r =5 21691 21743.9 .129 .627 r =6 77343 77311.8 .013 .640 p=1-exp(-SUM/2)= .27381 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21586 21743.9 1.147 1.428 r =6 77486 77311.8 .392 1.821 p=1-exp(-SUM/2)= .59758 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 923 944.3 .481 .481 r =5 21853 21743.9 .547 1.028 r =6 77224 77311.8 .100 1.128 p=1-exp(-SUM/2)= .43097 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\x32s.raw b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 919 944.3 .678 .678 r =5 21842 21743.9 .443 1.120 r =6 77239 77311.8 .069 1.189 p=1-exp(-SUM/2)= .44818 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .848441 .677688 .201994 .066173 .318487 .017413 .669951 .706457 .079491 .313191 .243037 .758903 .040027 .066702 .500425 .260389 .020934 .539602 .867793 .698309 .625698 .273813 .597584 .430967 .448178 brank test summary for o:\tmp\x32s.raw The KS test for those 25 supposed UNI's yields KS p-value= .849580 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE BITSTREAM TEST :: :: The file under test is viewed as a stream of bits. Call them :: :: b1,b2,... . Consider an alphabet with two "letters", 0 and 1 :: :: and think of the stream of bits as a succession of 20-letter :: :: "words", overlapping. Thus the first word is b1b2...b20, the :: :: second is b2b3...b21, and so on. The bitstream test counts :: :: the number of missing 20-letter (20-bit) words in a string of :: :: 2^21 overlapping 20-letter words. There are 2^20 possible 20 :: :: letter words. For a truly random string of 2^21+19 bits, the :: :: number of missing words j should be (very close to) normally :: :: distributed with mean 141,909 and sigma 428. Thus :: :: (j-141909)/428 should be a standard normal variate (z score) :: :: that leads to a uniform [0,1) p value. The test is repeated :: :: twenty times. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: THE OVERLAPPING 20-tuples BITSTREAM TEST, 20 BITS PER WORD, N words This test uses N=2^21 and samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. --------------------------------------------------------- tst no 1: 141028 missing words, -2.06 sigmas from mean, p-value= .01974 tst no 2: 141983 missing words, .17 sigmas from mean, p-value= .56833 tst no 3: 142156 missing words, .58 sigmas from mean, p-value= .71781 tst no 4: 141855 missing words, -.13 sigmas from mean, p-value= .44950 tst no 5: 141673 missing words, -.55 sigmas from mean, p-value= .29042 tst no 6: 141388 missing words, -1.22 sigmas from mean, p-value= .11160 tst no 7: 141969 missing words, .14 sigmas from mean, p-value= .55544 tst no 8: 141052 missing words, -2.00 sigmas from mean, p-value= .02258 tst no 9: 142160 missing words, .59 sigmas from mean, p-value= .72096 tst no 10: 141925 missing words, .04 sigmas from mean, p-value= .51460 tst no 11: 141991 missing words, .19 sigmas from mean, p-value= .57567 tst no 12: 140927 missing words, -2.30 sigmas from mean, p-value= .01086 tst no 13: 141877 missing words, -.08 sigmas from mean, p-value= .46990 tst no 14: 141700 missing words, -.49 sigmas from mean, p-value= .31239 tst no 15: 142271 missing words, .85 sigmas from mean, p-value= .80095 tst no 16: 142605 missing words, 1.63 sigmas from mean, p-value= .94796 tst no 17: 142162 missing words, .59 sigmas from mean, p-value= .72252 tst no 18: 141406 missing words, -1.18 sigmas from mean, p-value= .11980 tst no 19: 141438 missing words, -1.10 sigmas from mean, p-value= .13540 tst no 20: 141555 missing words, -.83 sigmas from mean, p-value= .20387 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The tests OPSO, OQSO and DNA :: :: OPSO means Overlapping-Pairs-Sparse-Occupancy :: :: The OPSO test considers 2-letter words from an alphabet of :: :: 1024 letters. Each letter is determined by a specified ten :: :: bits from a 32-bit integer in the sequence to be tested. OPSO :: :: generates 2^21 (overlapping) 2-letter words (from 2^21+1 :: :: "keystrokes") and counts the number of missing words---that :: :: is 2-letter words which do not appear in the entire sequence. :: :: That count should be very close to normally distributed with :: :: mean 141,909, sigma 290. Thus (missingwrds-141909)/290 should :: :: be a standard normal variable. The OPSO test takes 32 bits at :: :: a time from the test file and uses a designated set of ten :: :: consecutive bits. It then restarts the file for the next de- :: :: signated 10 bits, and so on. :: :: :: :: OQSO means Overlapping-Quadruples-Sparse-Occupancy :: :: The test OQSO is similar, except that it considers 4-letter :: :: words from an alphabet of 32 letters, each letter determined :: :: by a designated string of 5 consecutive bits from the test :: :: file, elements of which are assumed 32-bit random integers. :: :: The mean number of missing words in a sequence of 2^21 four- :: :: letter words, (2^21+3 "keystrokes"), is again 141909, with :: :: sigma = 295. The mean is based on theory; sigma comes from :: :: extensive simulation. :: :: :: :: The DNA test considers an alphabet of 4 letters:: C,G,A,T,:: :: determined by two designated bits in the sequence of random :: :: integers being tested. It considers 10-letter words, so that :: :: as in OPSO and OQSO, there are 2^20 possible words, and the :: :: mean number of missing words from a string of 2^21 (over- :: :: lapping) 10-letter words (2^21+9 "keystrokes") is 141909. :: :: The standard deviation sigma=339 was determined as for OQSO :: :: by simulation. (Sigma for OPSO, 290, is the true value (to :: :: three places), not determined by simulation. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: OPSO test for generator o:\tmp\x32s.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for o:\tmp\x32s.raw using bits 23 to 32 142411 1.730 .9582 OPSO for o:\tmp\x32s.raw using bits 22 to 31 141518 -1.349 .0886 OPSO for o:\tmp\x32s.raw using bits 21 to 30 141817 -.318 .3751 OPSO for o:\tmp\x32s.raw using bits 20 to 29 141724 -.639 .2614 OPSO for o:\tmp\x32s.raw using bits 19 to 28 141431 -1.649 .0495 OPSO for o:\tmp\x32s.raw using bits 18 to 27 141825 -.291 .3856 OPSO for o:\tmp\x32s.raw using bits 17 to 26 142084 .602 .7265 OPSO for o:\tmp\x32s.raw using bits 16 to 25 141909 -.001 .4995 OPSO for o:\tmp\x32s.raw using bits 15 to 24 141438 -1.625 .0521 OPSO for o:\tmp\x32s.raw using bits 14 to 23 142121 .730 .7673 OPSO for o:\tmp\x32s.raw using bits 13 to 22 141678 -.798 .2125 OPSO for o:\tmp\x32s.raw using bits 12 to 21 141789 -.415 .3391 OPSO for o:\tmp\x32s.raw using bits 11 to 20 141758 -.522 .3009 OPSO for o:\tmp\x32s.raw using bits 10 to 19 141396 -1.770 .0384 OPSO for o:\tmp\x32s.raw using bits 9 to 18 141854 -.191 .4243 OPSO for o:\tmp\x32s.raw using bits 8 to 17 141842 -.232 .4082 OPSO for o:\tmp\x32s.raw using bits 7 to 16 141764 -.501 .3081 OPSO for o:\tmp\x32s.raw using bits 6 to 15 142166 .885 .8119 OPSO for o:\tmp\x32s.raw using bits 5 to 14 141578 -1.143 .1266 OPSO for o:\tmp\x32s.raw using bits 4 to 13 142086 .609 .7288 OPSO for o:\tmp\x32s.raw using bits 3 to 12 141943 .116 .5462 OPSO for o:\tmp\x32s.raw using bits 2 to 11 142252 1.182 .8813 OPSO for o:\tmp\x32s.raw using bits 1 to 10 141729 -.622 .2670 OQSO test for generator o:\tmp\x32s.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for o:\tmp\x32s.raw using bits 28 to 32 142015 .358 .6399 OQSO for o:\tmp\x32s.raw using bits 27 to 31 141855 -.184 .4269 OQSO for o:\tmp\x32s.raw using bits 26 to 30 141805 -.354 .3618 OQSO for o:\tmp\x32s.raw using bits 25 to 29 142202 .992 .8394 OQSO for o:\tmp\x32s.raw using bits 24 to 28 141876 -.113 .4550 OQSO for o:\tmp\x32s.raw using bits 23 to 27 142121 .718 .7635 OQSO for o:\tmp\x32s.raw using bits 22 to 26 141417 -1.669 .0476 OQSO for o:\tmp\x32s.raw using bits 21 to 25 142172 .890 .8134 OQSO for o:\tmp\x32s.raw using bits 20 to 24 141013 -3.038 .0012 OQSO for o:\tmp\x32s.raw using bits 19 to 23 141542 -1.245 .1065 OQSO for o:\tmp\x32s.raw using bits 18 to 22 142036 .429 .6662 OQSO for o:\tmp\x32s.raw using bits 17 to 21 142287 1.280 .8998 OQSO for o:\tmp\x32s.raw using bits 16 to 20 141335 -1.947 .0258 OQSO for o:\tmp\x32s.raw using bits 15 to 19 141943 .114 .5454 OQSO for o:\tmp\x32s.raw using bits 14 to 18 141474 -1.476 .0700 OQSO for o:\tmp\x32s.raw using bits 13 to 17 141578 -1.123 .1307 OQSO for o:\tmp\x32s.raw using bits 12 to 16 142133 .758 .7758 OQSO for o:\tmp\x32s.raw using bits 11 to 15 142273 1.233 .8912 OQSO for o:\tmp\x32s.raw using bits 10 to 14 141832 -.262 .3966 OQSO for o:\tmp\x32s.raw using bits 9 to 13 141930 .070 .5279 OQSO for o:\tmp\x32s.raw using bits 8 to 12 142082 .585 .7208 OQSO for o:\tmp\x32s.raw using bits 7 to 11 142364 1.541 .9384 OQSO for o:\tmp\x32s.raw using bits 6 to 10 142108 .673 .7497 OQSO for o:\tmp\x32s.raw using bits 5 to 9 141811 -.333 .3694 OQSO for o:\tmp\x32s.raw using bits 4 to 8 141781 -.435 .3318 OQSO for o:\tmp\x32s.raw using bits 3 to 7 142198 .979 .8361 OQSO for o:\tmp\x32s.raw using bits 2 to 6 142348 1.487 .9315 OQSO for o:\tmp\x32s.raw using bits 1 to 5 141990 .273 .6078 DNA test for generator o:\tmp\x32s.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for o:\tmp\x32s.raw using bits 31 to 32 142317 1.203 .8854 DNA for o:\tmp\x32s.raw using bits 30 to 31 142363 1.338 .9096 DNA for o:\tmp\x32s.raw using bits 29 to 30 141461 -1.323 .0930 DNA for o:\tmp\x32s.raw using bits 28 to 29 141928 .055 .5220 DNA for o:\tmp\x32s.raw using bits 27 to 28 141610 -.883 .1886 DNA for o:\tmp\x32s.raw using bits 26 to 27 141736 -.511 .3046 DNA for o:\tmp\x32s.raw using bits 25 to 26 141762 -.435 .3319 DNA for o:\tmp\x32s.raw using bits 24 to 25 141463 -1.317 .0940 DNA for o:\tmp\x32s.raw using bits 23 to 24 141735 -.514 .3035 DNA for o:\tmp\x32s.raw using bits 22 to 23 142259 1.031 .8488 DNA for o:\tmp\x32s.raw using bits 21 to 22 142246 .993 .8397 DNA for o:\tmp\x32s.raw using bits 20 to 21 142416 1.495 .9325 DNA for o:\tmp\x32s.raw using bits 19 to 20 141766 -.423 .3362 DNA for o:\tmp\x32s.raw using bits 18 to 19 142230 .946 .8279 DNA for o:\tmp\x32s.raw using bits 17 to 18 141787 -.361 .3591 DNA for o:\tmp\x32s.raw using bits 16 to 17 142050 .415 .6609 DNA for o:\tmp\x32s.raw using bits 15 to 16 141908 -.004 .4984 DNA for o:\tmp\x32s.raw using bits 14 to 15 142040 .385 .6501 DNA for o:\tmp\x32s.raw using bits 13 to 14 141580 -.971 .1657 DNA for o:\tmp\x32s.raw using bits 12 to 13 142007 .288 .6134 DNA for o:\tmp\x32s.raw using bits 11 to 12 142052 .421 .6631 DNA for o:\tmp\x32s.raw using bits 10 to 11 141399 -1.505 .0661 DNA for o:\tmp\x32s.raw using bits 9 to 10 142064 .456 .6759 DNA for o:\tmp\x32s.raw using bits 8 to 9 141976 .197 .5780 DNA for o:\tmp\x32s.raw using bits 7 to 8 141827 -.243 .4041 DNA for o:\tmp\x32s.raw using bits 6 to 7 141554 -1.048 .1473 DNA for o:\tmp\x32s.raw using bits 5 to 6 142050 .415 .6609 DNA for o:\tmp\x32s.raw using bits 4 to 5 142151 .713 .7620 DNA for o:\tmp\x32s.raw using bits 3 to 4 142179 .795 .7868 DNA for o:\tmp\x32s.raw using bits 2 to 3 142025 .341 .6335 DNA for o:\tmp\x32s.raw using bits 1 to 2 141575 -.986 .1620 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for o:\tmp\x32s.raw Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for o:\tmp\x32s.raw 2500.57 .008 .503226 byte stream for o:\tmp\x32s.raw 2581.26 1.149 .874773 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST for specific bytes. :: :: Consider the file under test as a stream of 32-bit integers. :: :: From each integer, a specific byte is chosen , say the left- :: :: most:: bits 1 to 8. Each byte can contain from 0 to 8 1's, :: :: with probabilitie 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the specified bytes from successive integers provide a string :: :: of (overlapping) 5-letter words, each "letter" taking values :: :: A,B,C,D,E. The letters are determined by the number of 1's, :: :: in that byte:: 0,1,or 2 ---> A, 3 ---> B, 4 ---> C, 5 ---> D,:: :: and 6,7 or 8 ---> E. Thus we have a monkey at a typewriter :: :: hitting five keys with with various probabilities:: 37,56,70,:: :: 56,37 over 256. There are 5^5 possible 5-letter words, and :: :: from a string of 256,000 (overlapping) 5-letter words, counts :: :: are made on the frequencies for each word. The quadratic form :: :: in the weak inverse of the covariance matrix of the cell :: :: counts provides a chisquare test:: Q5-Q4, the difference of :: :: the naive Pearson sums of (OBS-EXP)^2/EXP on counts for 5- :: :: and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Chi-square with 5^5-5^4=2500 d.of f. for sample size: 256000 chisquare equiv normal p value Results for COUNT-THE-1's in specified bytes: bits 1 to 8 2481.82 -.257 .398559 bits 2 to 9 2519.84 .281 .610495 bits 3 to 10 2524.85 .351 .637362 bits 4 to 11 2559.54 .842 .800100 bits 5 to 12 2637.72 1.948 .974276 bits 6 to 13 2425.02 -1.060 .144495 bits 7 to 14 2506.11 .086 .534457 bits 8 to 15 2418.80 -1.148 .125421 bits 9 to 16 2548.74 .689 .754677 bits 10 to 17 2520.03 .283 .611490 bits 11 to 18 2575.03 1.061 .855669 bits 12 to 19 2496.46 -.050 .480016 bits 13 to 20 2463.05 -.523 .300638 bits 14 to 21 2467.04 -.466 .320559 bits 15 to 22 2404.67 -1.348 .088812 bits 16 to 23 2472.91 -.383 .350811 bits 17 to 24 2520.01 .283 .611430 bits 18 to 25 2488.87 -.157 .437485 bits 19 to 26 2430.95 -.977 .164399 bits 20 to 27 2546.01 .651 .742372 bits 21 to 28 2456.15 -.620 .267592 bits 22 to 29 2440.46 -.842 .199890 bits 23 to 30 2597.57 1.380 .916179 bits 24 to 31 2528.70 .406 .657587 bits 25 to 32 2480.24 -.279 .389958 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THIS IS A PARKING LOT TEST :: :: In a square of side 100, randomly "park" a car---a circle of :: :: radius 1. Then try to park a 2nd, a 3rd, and so on, each :: :: time parking "by ear". That is, if an attempt to park a car :: :: causes a crash with one already parked, try again at a new :: :: random location. (To avoid path problems, consider parking :: :: helicopters rather than cars.) Each attempt leads to either :: :: a crash or a success, the latter followed by an increment to :: :: the list of cars already parked. If we plot n: the number of :: :: attempts, versus k:: the number successfully parked, we get a:: :: curve that should be similar to those provided by a perfect :: :: random number generator. Theory for the behavior of such a :: :: random curve seems beyond reach, and as graphics displays are :: :: not available for this battery of tests, a simple characteriz :: :: ation of the random experiment is used: k, the number of cars :: :: successfully parked after n=12,000 attempts. Simulation shows :: :: that k should average 3523 with sigma 21.9 and is very close :: :: to normally distributed. Thus (k-3523)/21.9 should be a st- :: :: andard normal variable, which, converted to a uniform varia- :: :: ble, provides input to a KSTEST based on a sample of 10. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: CDPARK: result of ten tests on file o:\tmp\x32s.raw Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3518 z-score: -.228 p-value: .409702 Successes: 3525 z-score: .091 p-value: .536382 Successes: 3509 z-score: -.639 p-value: .261324 Successes: 3526 z-score: .137 p-value: .554479 Successes: 3509 z-score: -.639 p-value: .261324 Successes: 3501 z-score: -1.005 p-value: .157553 Successes: 3528 z-score: .228 p-value: .590298 Successes: 3528 z-score: .228 p-value: .590298 Successes: 3512 z-score: -.502 p-value: .307734 Successes: 3567 z-score: 2.009 p-value: .977738 square size avg. no. parked sample sigma 100. 3522.300 17.353 KSTEST for the above 10: p= .499148 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file o:\tmp\x32s.raw Sample no. d^2 avg equiv uni 5 1.5606 .8068 .791637 10 .6498 1.0268 .479558 15 .1627 .9543 .150825 20 1.1880 1.1113 .696985 25 3.9640 1.1371 .981388 30 .3731 1.1798 .312704 35 1.9361 1.1572 .857130 40 .2100 1.1245 .190302 45 .1330 1.0719 .125116 50 .7546 1.0373 .531561 55 1.0964 .9997 .667778 60 .5868 .9797 .445530 65 .7519 .9333 .530294 70 1.2706 .9172 .721122 75 .2840 .8853 .248278 80 .8320 .8777 .566637 85 .7171 .8691 .513575 90 .0316 .8401 .031278 95 .9295 .8778 .607076 100 .0401 .8718 .039533 MINIMUM DISTANCE TEST for o:\tmp\x32s.raw Result of KS test on 20 transformed mindist^2's: p-value= .492926 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file o:\tmp\x32s.raw sample no: 1 r^3= 20.774 p-value= .49966 sample no: 2 r^3= 15.700 p-value= .40746 sample no: 3 r^3= 10.239 p-value= .28915 sample no: 4 r^3= 78.658 p-value= .92734 sample no: 5 r^3= 25.428 p-value= .57156 sample no: 6 r^3= 38.805 p-value= .72569 sample no: 7 r^3= 24.863 p-value= .56341 sample no: 8 r^3= 3.054 p-value= .09680 sample no: 9 r^3= 4.812 p-value= .14818 sample no: 10 r^3= 17.048 p-value= .43349 sample no: 11 r^3= 44.909 p-value= .77619 sample no: 12 r^3= 9.714 p-value= .27660 sample no: 13 r^3= 59.252 p-value= .86125 sample no: 14 r^3= 50.737 p-value= .81571 sample no: 15 r^3= 31.489 p-value= .64993 sample no: 16 r^3= 78.093 p-value= .92596 sample no: 17 r^3= 13.792 p-value= .36854 sample no: 18 r^3= 34.403 p-value= .68234 sample no: 19 r^3= 9.231 p-value= .26488 sample no: 20 r^3= 17.372 p-value= .43958 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file o:\tmp\x32s.raw p-value= .265256 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR o:\tmp\x32s.raw Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -.1 -2.0 -1.6 .0 -.2 -.3 -.5 -.5 -2.8 .2 -1.3 -.1 .6 2.9 -.2 -.3 -.4 -.3 .8 -.5 -1.9 1.8 .1 .4 -.5 -.4 2.7 -1.4 .3 -.2 .7 -1.3 -.4 -.3 -.1 .4 -.5 -1.7 -.4 .4 2.4 -1.0 -1.1 Chi-square with 42 degrees of freedom: 56.578 z-score= 1.591 p-value= .934254 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .575020 Test no. 2 p-value .343907 Test no. 3 p-value .581041 Test no. 4 p-value .306547 Test no. 5 p-value .373924 Test no. 6 p-value .659522 Test no. 7 p-value .131555 Test no. 8 p-value .563936 Test no. 9 p-value .921395 Test no. 10 p-value .911699 Results of the OSUM test for o:\tmp\x32s.raw KSTEST on the above 10 p-values: .244460 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file o:\tmp\x32s.raw Up and down runs in a sample of 10000 _________________________________________________ Run test for o:\tmp\x32s.raw: runs up; ks test for 10 p's: .036923 runs down; ks test for 10 p's: .670716 Run test for o:\tmp\x32s.raw: runs up; ks test for 10 p's: .179201 runs down; ks test for 10 p's: .773075 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for o:\tmp\x32s.raw No. of wins: Observed Expected 98618 98585.86 98618= No. of wins, z-score= .144 pvalue= .55715 Analysis of Throws-per-Game: Chisq= 17.56 for 20 degrees of freedom, p= .38333 Throws Observed Expected Chisq Sum 1 66699 66666.7 .016 .016 2 37421 37654.3 1.446 1.461 3 26891 26954.7 .151 1.612 4 19377 19313.5 .209 1.821 5 13909 13851.4 .239 2.061 6 9943 9943.5 .000 2.061 7 7283 7145.0 2.664 4.725 8 5138 5139.1 .000 4.725 9 3734 3699.9 .315 5.040 10 2708 2666.3 .652 5.692 11 1951 1923.3 .398 6.091 12 1374 1388.7 .156 6.247 13 1029 1003.7 .637 6.884 14 666 726.1 4.981 11.865 15 522 525.8 .028 11.893 16 376 381.2 .070 11.962 17 259 276.5 1.112 13.075 18 196 200.8 .116 13.191 19 150 146.0 .110 13.302 20 117 106.2 1.095 14.397 21 257 287.1 3.159 17.555 SUMMARY FOR o:\tmp\x32s.raw p-value for no. of wins: .557152 p-value for throws/game: .383330 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests