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\f48h.raw For a sample of size 500: mean o:\tmp\f48h.raw using bits 1 to 24 1.852 duplicate number number spacings observed expected 0 82. 67.668 1 141. 135.335 2 129. 135.335 3 93. 90.224 4 38. 45.112 5 11. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 8.15 p-value= .773026 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 2 to 25 1.884 duplicate number number spacings observed expected 0 73. 67.668 1 150. 135.335 2 128. 135.335 3 96. 90.224 4 32. 45.112 5 11. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 9.69 p-value= .861872 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 3 to 26 1.942 duplicate number number spacings observed expected 0 62. 67.668 1 140. 135.335 2 153. 135.335 3 83. 90.224 4 40. 45.112 5 17. 18.045 6 to INF 5. 8.282 Chisquare with 6 d.o.f. = 5.46 p-value= .513660 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 4 to 27 1.922 duplicate number number spacings observed expected 0 68. 67.668 1 158. 135.335 2 122. 135.335 3 80. 90.224 4 43. 45.112 5 27. 18.045 6 to INF 2. 8.282 Chisquare with 6 d.o.f. = 15.58 p-value= .983791 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 5 to 28 1.838 duplicate number number spacings observed expected 0 80. 67.668 1 143. 135.335 2 133. 135.335 3 89. 90.224 4 36. 45.112 5 16. 18.045 6 to INF 3. 8.282 Chisquare with 6 d.o.f. = 8.18 p-value= .774732 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 6 to 29 1.920 duplicate number number spacings observed expected 0 73. 67.668 1 142. 135.335 2 131. 135.335 3 92. 90.224 4 37. 45.112 5 22. 18.045 6 to INF 3. 8.282 Chisquare with 6 d.o.f. = 6.62 p-value= .642216 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 7 to 30 1.936 duplicate number number spacings observed expected 0 63. 67.668 1 140. 135.335 2 156. 135.335 3 82. 90.224 4 34. 45.112 5 17. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 7.19 p-value= .696786 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 8 to 31 1.962 duplicate number number spacings observed expected 0 69. 67.668 1 148. 135.335 2 131. 135.335 3 79. 90.224 4 45. 45.112 5 19. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 2.86 p-value= .173746 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean o:\tmp\f48h.raw using bits 9 to 32 1.996 duplicate number number spacings observed expected 0 89. 67.668 1 121. 135.335 2 114. 135.335 3 98. 90.224 4 46. 45.112 5 23. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 13.72 p-value= .967048 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .773026 .861872 .513660 .983791 .774732 .642216 .696786 .173746 .967048 A KSTEST for the 9 p-values yields .958966 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom=117.497; p-value= .901061 OPERM5 test for file o:\tmp\f48h.raw For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 90.960; p-value= .294535 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.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 28 208 211.4 .055259 .055 29 5105 5134.0 .163925 .219 30 23144 23103.0 .072595 .292 31 11543 11551.5 .006291 .298 chisquare= .298 for 3 d. of f.; p-value= .332157 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.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 29 220 211.4 .348364 .348 30 5032 5134.0 2.026893 2.375 31 23202 23103.0 .423828 2.799 32 11546 11551.5 .002642 2.802 chisquare= 2.802 for 3 d. of f.; p-value= .623866 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.raw Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1010 944.3 4.571 4.571 r =5 21678 21743.9 .200 4.771 r =6 77312 77311.8 .000 4.771 p=1-exp(-SUM/2)= .90794 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 993 944.3 2.511 2.511 r =5 21725 21743.9 .016 2.528 r =6 77282 77311.8 .011 2.539 p=1-exp(-SUM/2)= .71908 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 969 944.3 .646 .646 r =5 21968 21743.9 2.310 2.956 r =6 77063 77311.8 .801 3.756 p=1-exp(-SUM/2)= .84713 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 957 944.3 .171 .171 r =5 21869 21743.9 .720 .891 r =6 77174 77311.8 .246 1.136 p=1-exp(-SUM/2)= .43338 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 875 944.3 5.086 5.086 r =5 21783 21743.9 .070 5.156 r =6 77342 77311.8 .012 5.168 p=1-exp(-SUM/2)= .92453 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 922 944.3 .527 .527 r =5 21795 21743.9 .120 .647 r =6 77283 77311.8 .011 .658 p=1-exp(-SUM/2)= .28018 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 917 944.3 .789 .789 r =5 21960 21743.9 2.148 2.937 r =6 77123 77311.8 .461 3.398 p=1-exp(-SUM/2)= .81714 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 1001 944.3 3.404 3.404 r =5 21811 21743.9 .207 3.611 r =6 77188 77311.8 .198 3.810 p=1-exp(-SUM/2)= .85115 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 969 944.3 .646 .646 r =5 21807 21743.9 .183 .829 r =6 77224 77311.8 .100 .929 p=1-exp(-SUM/2)= .37150 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 977 944.3 1.132 1.132 r =5 21823 21743.9 .288 1.420 r =6 77200 77311.8 .162 1.582 p=1-exp(-SUM/2)= .54654 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 959 944.3 .229 .229 r =5 21867 21743.9 .697 .926 r =6 77174 77311.8 .246 1.171 p=1-exp(-SUM/2)= .44327 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 965 944.3 .454 .454 r =5 21684 21743.9 .165 .619 r =6 77351 77311.8 .020 .639 p=1-exp(-SUM/2)= .27334 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 966 944.3 .499 .499 r =5 21796 21743.9 .125 .623 r =6 77238 77311.8 .070 .694 p=1-exp(-SUM/2)= .29316 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 925 944.3 .395 .395 r =5 22126 21743.9 6.715 7.109 r =6 76949 77311.8 1.703 8.812 p=1-exp(-SUM/2)= .98779 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 979 944.3 1.275 1.275 r =5 21871 21743.9 .743 2.018 r =6 77150 77311.8 .339 2.357 p=1-exp(-SUM/2)= .69220 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 928 944.3 .281 .281 r =5 21695 21743.9 .110 .391 r =6 77377 77311.8 .055 .446 p=1-exp(-SUM/2)= .20003 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 938 944.3 .042 .042 r =5 21649 21743.9 .414 .456 r =6 77413 77311.8 .132 .589 p=1-exp(-SUM/2)= .25498 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 952 944.3 .063 .063 r =5 21777 21743.9 .050 .113 r =6 77271 77311.8 .022 .135 p=1-exp(-SUM/2)= .06513 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 946 944.3 .003 .003 r =5 21813 21743.9 .220 .223 r =6 77241 77311.8 .065 .287 p=1-exp(-SUM/2)= .13389 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 890 944.3 3.123 3.123 r =5 21782 21743.9 .067 3.189 r =6 77328 77311.8 .003 3.193 p=1-exp(-SUM/2)= .79737 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 931 944.3 .187 .187 r =5 21548 21743.9 1.765 1.952 r =6 77521 77311.8 .566 2.518 p=1-exp(-SUM/2)= .71611 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 939 944.3 .030 .030 r =5 21606 21743.9 .875 .904 r =6 77455 77311.8 .265 1.170 p=1-exp(-SUM/2)= .44277 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 948 944.3 .014 .014 r =5 21676 21743.9 .212 .227 r =6 77376 77311.8 .053 .280 p=1-exp(-SUM/2)= .13057 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 942 944.3 .006 .006 r =5 21682 21743.9 .176 .182 r =6 77376 77311.8 .053 .235 p=1-exp(-SUM/2)= .11092 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG o:\tmp\f48h.raw b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 953 944.3 .080 .080 r =5 21629 21743.9 .607 .687 r =6 77418 77311.8 .146 .833 p=1-exp(-SUM/2)= .34070 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .907941 .719080 .847131 .433381 .924530 .280178 .817142 .851154 .371501 .546542 .443266 .273340 .293158 .987794 .692197 .200028 .254982 .065127 .133892 .797365 .716114 .442769 .130566 .110917 .340704 brank test summary for o:\tmp\f48h.raw The KS test for those 25 supposed UNI's yields KS p-value= .063467 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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: 142152 missing words, .57 sigmas from mean, p-value= .71464 tst no 2: 142078 missing words, .39 sigmas from mean, p-value= .65324 tst no 3: 142068 missing words, .37 sigmas from mean, p-value= .64458 tst no 4: 142230 missing words, .75 sigmas from mean, p-value= .77314 tst no 5: 142294 missing words, .90 sigmas from mean, p-value= .81561 tst no 6: 142024 missing words, .27 sigmas from mean, p-value= .60562 tst no 7: 141865 missing words, -.10 sigmas from mean, p-value= .45876 tst no 8: 141755 missing words, -.36 sigmas from mean, p-value= .35921 tst no 9: 141383 missing words, -1.23 sigmas from mean, p-value= .10940 tst no 10: 141553 missing words, -.83 sigmas from mean, p-value= .20255 tst no 11: 141938 missing words, .07 sigmas from mean, p-value= .52671 tst no 12: 141978 missing words, .16 sigmas from mean, p-value= .56374 tst no 13: 142231 missing words, .75 sigmas from mean, p-value= .77385 tst no 14: 141766 missing words, -.33 sigmas from mean, p-value= .36886 tst no 15: 141688 missing words, -.52 sigmas from mean, p-value= .30254 tst no 16: 141249 missing words, -1.54 sigmas from mean, p-value= .06144 tst no 17: 142053 missing words, .34 sigmas from mean, p-value= .63144 tst no 18: 141728 missing words, -.42 sigmas from mean, p-value= .33591 tst no 19: 141774 missing words, -.32 sigmas from mean, p-value= .37593 tst no 20: 142193 missing words, .66 sigmas from mean, p-value= .74627 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for o:\tmp\f48h.raw using bits 23 to 32 141851 -.201 .4203 OPSO for o:\tmp\f48h.raw using bits 22 to 31 141772 -.474 .3179 OPSO for o:\tmp\f48h.raw using bits 21 to 30 141808 -.349 .3634 OPSO for o:\tmp\f48h.raw using bits 20 to 29 141554 -1.225 .1102 OPSO for o:\tmp\f48h.raw using bits 19 to 28 141893 -.056 .4775 OPSO for o:\tmp\f48h.raw using bits 18 to 27 141757 -.525 .2997 OPSO for o:\tmp\f48h.raw using bits 17 to 26 141791 -.408 .3416 OPSO for o:\tmp\f48h.raw using bits 16 to 25 141979 .240 .5949 OPSO for o:\tmp\f48h.raw using bits 15 to 24 141754 -.536 .2961 OPSO for o:\tmp\f48h.raw using bits 14 to 23 141916 .023 .5092 OPSO for o:\tmp\f48h.raw using bits 13 to 22 141466 -1.529 .0632 OPSO for o:\tmp\f48h.raw using bits 12 to 21 141697 -.732 .2320 OPSO for o:\tmp\f48h.raw using bits 11 to 20 142096 .644 .7401 OPSO for o:\tmp\f48h.raw using bits 10 to 19 141687 -.767 .2216 OPSO for o:\tmp\f48h.raw using bits 9 to 18 141979 .240 .5949 OPSO for o:\tmp\f48h.raw using bits 8 to 17 141909 -.001 .4995 OPSO for o:\tmp\f48h.raw using bits 7 to 16 142169 .895 .8147 OPSO for o:\tmp\f48h.raw using bits 6 to 15 141901 -.029 .4885 OPSO for o:\tmp\f48h.raw using bits 5 to 14 141884 -.087 .4652 OPSO for o:\tmp\f48h.raw using bits 4 to 13 142141 .799 .7878 OPSO for o:\tmp\f48h.raw using bits 3 to 12 142060 .520 .6983 OPSO for o:\tmp\f48h.raw using bits 2 to 11 141493 -1.436 .0756 OPSO for o:\tmp\f48h.raw using bits 1 to 10 141401 -1.753 .0398 OQSO test for generator o:\tmp\f48h.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for o:\tmp\f48h.raw using bits 28 to 32 142523 2.080 .9812 OQSO for o:\tmp\f48h.raw using bits 27 to 31 141789 -.408 .3417 OQSO for o:\tmp\f48h.raw using bits 26 to 30 142124 .728 .7666 OQSO for o:\tmp\f48h.raw using bits 25 to 29 142061 .514 .6964 OQSO for o:\tmp\f48h.raw using bits 24 to 28 142619 2.406 .9919 OQSO for o:\tmp\f48h.raw using bits 23 to 27 142115 .697 .7572 OQSO for o:\tmp\f48h.raw using bits 22 to 26 141844 -.221 .4124 OQSO for o:\tmp\f48h.raw using bits 21 to 25 141914 .016 .5063 OQSO for o:\tmp\f48h.raw using bits 20 to 24 141784 -.425 .3355 OQSO for o:\tmp\f48h.raw using bits 19 to 23 142527 2.094 .9819 OQSO for o:\tmp\f48h.raw using bits 18 to 22 141592 -1.076 .1410 OQSO for o:\tmp\f48h.raw using bits 17 to 21 141563 -1.174 .1202 OQSO for o:\tmp\f48h.raw using bits 16 to 20 141780 -.438 .3305 OQSO for o:\tmp\f48h.raw using bits 15 to 19 141731 -.605 .2728 OQSO for o:\tmp\f48h.raw using bits 14 to 18 142132 .755 .7748 OQSO for o:\tmp\f48h.raw using bits 13 to 17 142037 .433 .6674 OQSO for o:\tmp\f48h.raw using bits 12 to 16 141766 -.486 .3135 OQSO for o:\tmp\f48h.raw using bits 11 to 15 141570 -1.150 .1250 OQSO for o:\tmp\f48h.raw using bits 10 to 14 142192 .958 .8310 OQSO for o:\tmp\f48h.raw using bits 9 to 13 141595 -1.066 .1433 OQSO for o:\tmp\f48h.raw using bits 8 to 12 141265 -2.184 .0145 OQSO for o:\tmp\f48h.raw using bits 7 to 11 141868 -.140 .4443 OQSO for o:\tmp\f48h.raw using bits 6 to 10 142272 1.229 .8905 OQSO for o:\tmp\f48h.raw using bits 5 to 9 141814 -.323 .3733 OQSO for o:\tmp\f48h.raw using bits 4 to 8 141993 .284 .6117 OQSO for o:\tmp\f48h.raw using bits 3 to 7 141821 -.299 .3823 OQSO for o:\tmp\f48h.raw using bits 2 to 6 142286 1.277 .8992 OQSO for o:\tmp\f48h.raw using bits 1 to 5 142031 .412 .6600 DNA test for generator o:\tmp\f48h.raw Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for o:\tmp\f48h.raw using bits 31 to 32 141678 -.682 .2475 DNA for o:\tmp\f48h.raw using bits 30 to 31 141986 .226 .5895 DNA for o:\tmp\f48h.raw using bits 29 to 30 141564 -1.019 .1542 DNA for o:\tmp\f48h.raw using bits 28 to 29 141875 -.101 .4597 DNA for o:\tmp\f48h.raw using bits 27 to 28 141841 -.202 .4201 DNA for o:\tmp\f48h.raw using bits 26 to 27 141745 -.485 .3139 DNA for o:\tmp\f48h.raw using bits 25 to 26 142327 1.232 .8910 DNA for o:\tmp\f48h.raw using bits 24 to 25 141777 -.390 .3481 DNA for o:\tmp\f48h.raw using bits 23 to 24 141735 -.514 .3035 DNA for o:\tmp\f48h.raw using bits 22 to 23 141960 .149 .5594 DNA for o:\tmp\f48h.raw using bits 21 to 22 142532 1.837 .9669 DNA for o:\tmp\f48h.raw using bits 20 to 21 142231 .949 .8287 DNA for o:\tmp\f48h.raw using bits 19 to 20 142279 1.090 .8622 DNA for o:\tmp\f48h.raw using bits 18 to 19 142533 1.840 .9671 DNA for o:\tmp\f48h.raw using bits 17 to 18 142202 .863 .8060 DNA for o:\tmp\f48h.raw using bits 16 to 17 142170 .769 .7790 DNA for o:\tmp\f48h.raw using bits 15 to 16 141259 -1.918 .0275 DNA for o:\tmp\f48h.raw using bits 14 to 15 141478 -1.272 .1016 DNA for o:\tmp\f48h.raw using bits 13 to 14 141845 -.190 .4247 DNA for o:\tmp\f48h.raw using bits 12 to 13 142121 .624 .7338 DNA for o:\tmp\f48h.raw using bits 11 to 12 141618 -.859 .1951 DNA for o:\tmp\f48h.raw using bits 10 to 11 141694 -.635 .2627 DNA for o:\tmp\f48h.raw using bits 9 to 10 141888 -.063 .4749 DNA for o:\tmp\f48h.raw using bits 8 to 9 141277 -1.865 .0311 DNA for o:\tmp\f48h.raw using bits 7 to 8 141390 -1.532 .0628 DNA for o:\tmp\f48h.raw using bits 6 to 7 142406 1.465 .9286 DNA for o:\tmp\f48h.raw using bits 5 to 6 141318 -1.744 .0406 DNA for o:\tmp\f48h.raw using bits 4 to 5 142133 .660 .7453 DNA for o:\tmp\f48h.raw using bits 3 to 4 142074 .486 .6864 DNA for o:\tmp\f48h.raw using bits 2 to 3 141400 -1.502 .0665 DNA for o:\tmp\f48h.raw using bits 1 to 2 142601 2.040 .9793 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for o:\tmp\f48h.raw Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for o:\tmp\f48h.raw 2440.49 -.842 .199996 byte stream for o:\tmp\f48h.raw 2467.67 -.457 .323771 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2349.40 -2.130 .016592 bits 2 to 9 2519.99 .283 .611274 bits 3 to 10 2461.07 -.551 .290964 bits 4 to 11 2491.73 -.117 .453437 bits 5 to 12 2434.85 -.921 .178417 bits 6 to 13 2533.97 .480 .684556 bits 7 to 14 2711.63 2.993 .998618 bits 8 to 15 2536.46 .516 .696939 bits 9 to 16 2539.10 .553 .709848 bits 10 to 17 2515.00 .212 .584022 bits 11 to 18 2305.88 -2.745 .003023 bits 12 to 19 2337.21 -2.302 .010661 bits 13 to 20 2600.38 1.420 .922131 bits 14 to 21 2305.92 -2.745 .003028 bits 15 to 22 2508.26 .117 .546472 bits 16 to 23 2565.73 .930 .823692 bits 17 to 24 2520.75 .293 .615412 bits 18 to 25 2576.68 1.084 .860915 bits 19 to 26 2483.74 -.230 .409066 bits 20 to 27 2458.99 -.580 .280968 bits 21 to 28 2418.79 -1.148 .125394 bits 22 to 29 2551.79 .732 .768039 bits 23 to 30 2477.93 -.312 .377483 bits 24 to 31 2590.89 1.285 .900668 bits 25 to 32 2574.21 1.050 .853030 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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\f48h.raw Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3508 z-score: -.685 p-value: .246694 Successes: 3517 z-score: -.274 p-value: .392053 Successes: 3558 z-score: 1.598 p-value: .944998 Successes: 3545 z-score: 1.005 p-value: .842447 Successes: 3474 z-score: -2.237 p-value: .012629 Successes: 3524 z-score: .046 p-value: .518210 Successes: 3529 z-score: .274 p-value: .607947 Successes: 3533 z-score: .457 p-value: .676028 Successes: 3561 z-score: 1.735 p-value: .958644 Successes: 3521 z-score: -.091 p-value: .463618 square size avg. no. parked sample sigma 100. 3527.000 24.062 KSTEST for the above 10: p= .308149 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file o:\tmp\f48h.raw Sample no. d^2 avg equiv uni 5 2.3942 .8272 .909842 10 .5886 .6785 .446537 15 .2552 .8042 .226195 20 .8087 .9199 .556350 25 3.8903 .9550 .979957 30 .0737 .9112 .071401 35 .2731 .9445 .240006 40 1.5733 .9208 .794264 45 1.8236 .9586 .840024 50 .4269 .8998 .348873 55 1.4503 .8723 .767198 60 .6944 .9464 .502348 65 .6878 .9074 .499052 70 .0859 .9477 .082700 75 .0718 .9171 .069634 80 .2252 .8847 .202563 85 .4040 .9027 .333698 90 .2980 .8793 .258815 95 1.0396 .8837 .648229 100 .7104 .9163 .510313 MINIMUM DISTANCE TEST for o:\tmp\f48h.raw Result of KS test on 20 transformed mindist^2's: p-value= .660907 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file o:\tmp\f48h.raw sample no: 1 r^3= 8.949 p-value= .25792 sample no: 2 r^3= 7.534 p-value= .22209 sample no: 3 r^3= 7.717 p-value= .22682 sample no: 4 r^3= 13.363 p-value= .35946 sample no: 5 r^3= 62.379 p-value= .87498 sample no: 6 r^3= 8.938 p-value= .25764 sample no: 7 r^3= 7.113 p-value= .21107 sample no: 8 r^3= 6.290 p-value= .18914 sample no: 9 r^3= 7.392 p-value= .21838 sample no: 10 r^3= 46.192 p-value= .78556 sample no: 11 r^3= 100.703 p-value= .96515 sample no: 12 r^3= 33.231 p-value= .66968 sample no: 13 r^3= 60.523 p-value= .86700 sample no: 14 r^3= 4.215 p-value= .13108 sample no: 15 r^3= 70.448 p-value= .90447 sample no: 16 r^3= 24.422 p-value= .55695 sample no: 17 r^3= 6.519 p-value= .19531 sample no: 18 r^3= 3.215 p-value= .10161 sample no: 19 r^3= 34.994 p-value= .68854 sample no: 20 r^3= 13.655 p-value= .36565 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file o:\tmp\f48h.raw p-value= .679786 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR o:\tmp\f48h.raw Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: 2.0 -.7 -.4 -1.1 -1.0 .5 .2 -1.8 -.7 -.2 -1.7 .1 -.3 .6 .2 1.5 .7 .2 -.1 -1.2 -.1 -.2 .1 -.8 .1 .4 1.5 -.9 -.2 .6 .5 1.2 -.2 1.7 .1 .2 -.7 .8 -.4 -1.8 -.6 1.0 -1.1 Chi-square with 42 degrees of freedom: 35.018 z-score= -.762 p-value= .230956 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .134985 Test no. 2 p-value .505769 Test no. 3 p-value .102591 Test no. 4 p-value .075882 Test no. 5 p-value .238573 Test no. 6 p-value .787957 Test no. 7 p-value .353805 Test no. 8 p-value .143879 Test no. 9 p-value .018865 Test no. 10 p-value .501902 Results of the OSUM test for o:\tmp\f48h.raw KSTEST on the above 10 p-values: .972208 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file o:\tmp\f48h.raw Up and down runs in a sample of 10000 _________________________________________________ Run test for o:\tmp\f48h.raw: runs up; ks test for 10 p's: .988098 runs down; ks test for 10 p's: .693896 Run test for o:\tmp\f48h.raw: runs up; ks test for 10 p's: .075595 runs down; ks test for 10 p's: .941348 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for o:\tmp\f48h.raw No. of wins: Observed Expected 98499 98585.86 98499= No. of wins, z-score= -.388 pvalue= .34883 Analysis of Throws-per-Game: Chisq= 30.01 for 20 degrees of freedom, p= .93025 Throws Observed Expected Chisq Sum 1 66199 66666.7 3.281 3.281 2 37868 37654.3 1.213 4.493 3 27056 26954.7 .380 4.874 4 19460 19313.5 1.112 5.986 5 13630 13851.4 3.539 9.525 6 10054 9943.5 1.227 10.752 7 7093 7145.0 .379 11.131 8 5313 5139.1 5.887 17.017 9 3713 3699.9 .047 17.064 10 2643 2666.3 .204 17.268 11 1967 1923.3 .992 18.259 12 1381 1388.7 .043 18.302 13 1037 1003.7 1.104 19.406 14 719 726.1 .070 19.476 15 493 525.8 2.050 21.527 16 336 381.2 5.348 26.875 17 296 276.5 1.369 28.245 18 199 200.8 .017 28.261 19 159 146.0 1.160 29.422 20 109 106.2 .073 29.495 21 275 287.1 .511 30.006 SUMMARY FOR o:\tmp\f48h.raw p-value for no. of wins: .348828 p-value for throws/game: .930245 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file o:\tmp\f48h.txt