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 outputCONFOUNDE For a sample of size 500: mean outputCONFOUNDED using bits 1 to 24 1.872 duplicate number number spacings observed expected 0 69. 67.668 1 148. 135.335 2 139. 135.335 3 87. 90.224 4 39. 45.112 5 15. 18.045 6 to INF 3. 8.282 Chisquare with 6 d.o.f. = 6.14 p-value= .591883 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 2 to 25 2.030 duplicate number number spacings observed expected 0 64. 67.668 1 130. 135.335 2 146. 135.335 3 85. 90.224 4 50. 45.112 5 16. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 2.38 p-value= .117877 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 3 to 26 2.036 duplicate number number spacings observed expected 0 55. 67.668 1 138. 135.335 2 143. 135.335 3 95. 90.224 4 44. 45.112 5 19. 18.045 6 to INF 6. 8.282 Chisquare with 6 d.o.f. = 3.82 p-value= .298646 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 4 to 27 1.990 duplicate number number spacings observed expected 0 76. 67.668 1 128. 135.335 2 133. 135.335 3 88. 90.224 4 51. 45.112 5 14. 18.045 6 to INF 10. 8.282 Chisquare with 6 d.o.f. = 3.55 p-value= .262738 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 5 to 28 2.070 duplicate number number spacings observed expected 0 56. 67.668 1 148. 135.335 2 117. 135.335 3 101. 90.224 4 51. 45.112 5 20. 18.045 6 to INF 7. 8.282 Chisquare with 6 d.o.f. = 8.15 p-value= .772473 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 6 to 29 2.018 duplicate number number spacings observed expected 0 69. 67.668 1 131. 135.335 2 142. 135.335 3 82. 90.224 4 49. 45.112 5 16. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 2.70 p-value= .154767 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 7 to 30 1.932 duplicate number number spacings observed expected 0 72. 67.668 1 143. 135.335 2 135. 135.335 3 86. 90.224 4 37. 45.112 5 16. 18.045 6 to INF 11. 8.282 Chisquare with 6 d.o.f. = 3.49 p-value= .255025 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 8 to 31 1.962 duplicate number number spacings observed expected 0 72. 67.668 1 134. 135.335 2 141. 135.335 3 83. 90.224 4 46. 45.112 5 15. 18.045 6 to INF 9. 8.282 Chisquare with 6 d.o.f. = 1.70 p-value= .054840 ::::::::::::::::::::::::::::::::::::::::: For a sample of size 500: mean outputCONFOUNDED using bits 9 to 32 1.862 duplicate number number spacings observed expected 0 91. 67.668 1 132. 135.335 2 131. 135.335 3 81. 90.224 4 40. 45.112 5 17. 18.045 6 to INF 8. 8.282 Chisquare with 6 d.o.f. = 9.86 p-value= .869260 ::::::::::::::::::::::::::::::::::::::::: The 9 p-values were .591883 .117877 .298646 .262738 .772473 .154767 .255025 .054840 .869260 A KSTEST for the 9 p-values yields .698845 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDE For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 82.868; p-value= .121465 OPERM5 test for file outputCONFOUNDE For a sample of 1,000,000 consecutive 5-tuples, chisquare for 99 degrees of freedom= 82.172; p-value= .110567 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDE Rank test for 31x31 binary matrices: rows from leftmost 31 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 28 224 211.4 .748784 .749 29 5152 5134.0 .063037 .812 30 23232 23103.0 .719771 1.532 31 11392 11551.5 2.203003 3.735 chisquare= 3.735 for 3 d. of f.; p-value= .735062 -------------------------------------------------------------- ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDE Rank test for 32x32 binary matrices: rows from leftmost 32 bits of each 32-bit integer rank observed expected (o-e)^2/e sum 29 209 211.4 .027655 .028 30 5122 5134.0 .028096 .056 31 23049 23103.0 .126436 .182 32 11620 11551.5 .405912 .588 chisquare= .588 for 3 d. of f.; p-value= .322427 -------------------------------------------------------------- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDE Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 1 to 8 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 919 944.3 .678 .678 r =5 21872 21743.9 .755 1.433 r =6 77209 77311.8 .137 1.569 p=1-exp(-SUM/2)= .54372 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 2 to 9 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 959 944.3 .229 .229 r =5 21635 21743.9 .545 .774 r =6 77406 77311.8 .115 .889 p=1-exp(-SUM/2)= .35885 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 3 to 10 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 885 944.3 3.724 3.724 r =5 21937 21743.9 1.715 5.439 r =6 77178 77311.8 .232 5.670 p=1-exp(-SUM/2)= .94130 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 4 to 11 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 952 944.3 .063 .063 r =5 21951 21743.9 1.973 2.035 r =6 77097 77311.8 .597 2.632 p=1-exp(-SUM/2)= .73181 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 5 to 12 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 929 944.3 .248 .248 r =5 21932 21743.9 1.627 1.875 r =6 77139 77311.8 .386 2.261 p=1-exp(-SUM/2)= .67719 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 6 to 13 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 962 944.3 .332 .332 r =5 21957 21743.9 2.088 2.420 r =6 77081 77311.8 .689 3.109 p=1-exp(-SUM/2)= .78873 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 7 to 14 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 950 944.3 .034 .034 r =5 21816 21743.9 .239 .273 r =6 77234 77311.8 .078 .352 p=1-exp(-SUM/2)= .16128 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 8 to 15 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 967 944.3 .546 .546 r =5 21828 21743.9 .325 .871 r =6 77205 77311.8 .148 1.018 p=1-exp(-SUM/2)= .39904 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 9 to 16 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 919 944.3 .678 .678 r =5 21839 21743.9 .416 1.094 r =6 77242 77311.8 .063 1.157 p=1-exp(-SUM/2)= .43922 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 10 to 17 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 958 944.3 .199 .199 r =5 21612 21743.9 .800 .999 r =6 77430 77311.8 .181 1.180 p=1-exp(-SUM/2)= .44555 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 11 to 18 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 940 944.3 .020 .020 r =5 21909 21743.9 1.254 1.273 r =6 77151 77311.8 .334 1.608 p=1-exp(-SUM/2)= .55239 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 12 to 19 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 949 944.3 .023 .023 r =5 21744 21743.9 .000 .023 r =6 77307 77311.8 .000 .024 p=1-exp(-SUM/2)= .01177 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 13 to 20 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 954 944.3 .100 .100 r =5 21556 21743.9 1.624 1.723 r =6 77490 77311.8 .411 2.134 p=1-exp(-SUM/2)= .65597 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 14 to 21 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 952 944.3 .063 .063 r =5 21423 21743.9 4.736 4.799 r =6 77625 77311.8 1.269 6.067 p=1-exp(-SUM/2)= .95186 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 15 to 22 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 925 944.3 .395 .395 r =5 21779 21743.9 .057 .451 r =6 77296 77311.8 .003 .454 p=1-exp(-SUM/2)= .20324 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 16 to 23 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 945 944.3 .001 .001 r =5 21812 21743.9 .213 .214 r =6 77243 77311.8 .061 .275 p=1-exp(-SUM/2)= .12848 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 17 to 24 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 937 944.3 .056 .056 r =5 22105 21743.9 5.997 6.053 r =6 76958 77311.8 1.619 7.672 p=1-exp(-SUM/2)= .97842 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 18 to 25 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 904 944.3 1.720 1.720 r =5 21941 21743.9 1.787 3.507 r =6 77155 77311.8 .318 3.825 p=1-exp(-SUM/2)= .85226 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 19 to 26 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 916 944.3 .848 .848 r =5 21877 21743.9 .815 1.663 r =6 77207 77311.8 .142 1.805 p=1-exp(-SUM/2)= .59445 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 20 to 27 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 920 944.3 .625 .625 r =5 21735 21743.9 .004 .629 r =6 77345 77311.8 .014 .643 p=1-exp(-SUM/2)= .27504 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 21 to 28 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 938 944.3 .042 .042 r =5 21909 21743.9 1.254 1.296 r =6 77153 77311.8 .326 1.622 p=1-exp(-SUM/2)= .55555 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 22 to 29 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 975 944.3 .998 .998 r =5 21820 21743.9 .266 1.264 r =6 77205 77311.8 .148 1.412 p=1-exp(-SUM/2)= .50636 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 23 to 30 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 952 944.3 .063 .063 r =5 22072 21743.9 4.951 5.014 r =6 76976 77311.8 1.459 6.472 p=1-exp(-SUM/2)= .96068 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 24 to 31 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 975 944.3 .998 .998 r =5 21880 21743.9 .852 1.850 r =6 77145 77311.8 .360 2.210 p=1-exp(-SUM/2)= .66875 Rank of a 6x8 binary matrix, rows formed from eight bits of the RNG outputCONFOUNDE b-rank test for bits 25 to 32 OBSERVED EXPECTED (O-E)^2/E SUM r<=4 934 944.3 .112 .112 r =5 21677 21743.9 .206 .318 r =6 77389 77311.8 .077 .395 p=1-exp(-SUM/2)= .17934 TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices These should be 25 uniform [0,1] random variables: .543718 .358846 .941296 .731808 .677189 .788729 .161283 .399038 .439225 .445546 .552385 .011770 .655975 .951864 .203240 .128479 .978424 .852264 .594449 .275041 .555550 .506356 .960682 .668750 .179339 brank test summary for outputCONFOUNDE The KS test for those 25 supposed UNI's yields KS p-value= .284756 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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, 2^21 words. This test samples the bitstream 20 times. No. missing words should average 141909. with sigma=428. ----------------------------------- --------------- tst no 1: 141676 missing words, -.55 sigmas from mean, p-value= .29282 tst no 2: 142504 missing words, 1.39 sigmas from mean, p-value= .91765 tst no 3: 142196 missing words, .67 sigmas from mean, p-value= .74851 tst no 4: 142311 missing words, .94 sigmas from mean, p-value= .82600 tst no 5: 142361 missing words, 1.06 sigmas from mean, p-value= .85436 tst no 6: 142017 missing words, .25 sigmas from mean, p-value= .59931 tst no 7: 142159 missing words, .58 sigmas from mean, p-value= .72017 tst no 8: 141469 missing words, -1.03 sigmas from mean, p-value= .15179 tst no 9: 142195 missing words, .67 sigmas from mean, p-value= .74776 tst no 10: 141807 missing words, -.24 sigmas from mean, p-value= .40552 tst no 11: 140924 missing words, -2.30 sigmas from mean, p-value= .01066 tst no 12: 141965 missing words, .13 sigmas from mean, p-value= .55175 tst no 13: 140984 missing words, -2.16 sigmas from mean, p-value= .01531 tst no 14: 141975 missing words, .15 sigmas from mean, p-value= .56097 tst no 15: 141997 missing words, .20 sigmas from mean, p-value= .58115 tst no 16: 142033 missing words, .29 sigmas from mean, p-value= .61369 tst no 17: 142142 missing words, .54 sigmas from mean, p-value= .70665 tst no 18: 141946 missing words, .09 sigmas from mean, p-value= .53414 tst no 19: 142611 missing words, 1.64 sigmas from mean, p-value= .94944 tst no 20: 142588 missing words, 1.59 sigmas from mean, p-value= .94359 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDEDv4 Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OPSO for outputCONFOUNDE using bits 23 to 32 142437 1.820 .9656 OPSO for outputCONFOUNDE using bits 22 to 31 141233 -2.332 .0098 OPSO for outputCONFOUNDE using bits 21 to 30 142015 .364 .6422 OPSO for outputCONFOUNDE using bits 20 to 29 141943 .116 .5462 OPSO for outputCONFOUNDE using bits 19 to 28 141750 -.549 .2914 OPSO for outputCONFOUNDE using bits 18 to 27 141710 -.687 .2459 OPSO for outputCONFOUNDE using bits 17 to 26 141565 -1.187 .1175 OPSO for outputCONFOUNDE using bits 16 to 25 141922 .044 .5174 OPSO for outputCONFOUNDE using bits 15 to 24 141289 -2.139 .0162 OPSO for outputCONFOUNDE using bits 14 to 23 141733 -.608 .2716 OPSO for outputCONFOUNDE using bits 13 to 22 141589 -1.105 .1347 OPSO for outputCONFOUNDE using bits 12 to 21 142076 .575 .7173 OPSO for outputCONFOUNDE using bits 11 to 20 142080 .589 .7219 OPSO for outputCONFOUNDE using bits 10 to 19 142114 .706 .7598 OPSO for outputCONFOUNDE using bits 9 to 18 142563 2.254 .9879 OPSO for outputCONFOUNDE using bits 8 to 17 141797 -.387 .3493 OPSO for outputCONFOUNDE using bits 7 to 16 142279 1.275 .8988 OPSO for outputCONFOUNDE using bits 6 to 15 141683 -.780 .2176 OPSO for outputCONFOUNDE using bits 5 to 14 141888 -.074 .4707 OPSO for outputCONFOUNDE using bits 4 to 13 141485 -1.463 .0717 OPSO for outputCONFOUNDE using bits 3 to 12 141664 -.846 .1988 OPSO for outputCONFOUNDE using bits 2 to 11 141894 -.053 .4789 OPSO for outputCONFOUNDE using bits 1 to 10 141992 .285 .6122 OQSO test for generator outputCONFOUNDEDv4 Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p OQSO for outputCONFOUNDE using bits 28 to 32 142184 .931 .8241 OQSO for outputCONFOUNDE using bits 27 to 31 142198 .979 .8361 OQSO for outputCONFOUNDE using bits 26 to 30 142246 1.141 .8731 OQSO for outputCONFOUNDE using bits 25 to 29 141448 -1.564 .0589 OQSO for outputCONFOUNDE using bits 24 to 28 141767 -.482 .3147 OQSO for outputCONFOUNDE using bits 23 to 27 142276 1.243 .8931 OQSO for outputCONFOUNDE using bits 22 to 26 141234 -2.289 .0110 OQSO for outputCONFOUNDE using bits 21 to 25 142389 1.626 .9480 OQSO for outputCONFOUNDE using bits 20 to 24 142167 .873 .8088 OQSO for outputCONFOUNDE using bits 19 to 23 142150 .816 .7927 OQSO for outputCONFOUNDE using bits 18 to 22 142081 .582 .7197 OQSO for outputCONFOUNDE using bits 17 to 21 141961 .175 .5695 OQSO for outputCONFOUNDE using bits 16 to 20 142036 .429 .6662 OQSO for outputCONFOUNDE using bits 15 to 19 141940 .104 .5414 OQSO for outputCONFOUNDE using bits 14 to 18 141904 -.018 .4928 OQSO for outputCONFOUNDE using bits 13 to 17 141496 -1.401 .0806 OQSO for outputCONFOUNDE using bits 12 to 16 142350 1.494 .9324 OQSO for outputCONFOUNDE using bits 11 to 15 142253 1.165 .8780 OQSO for outputCONFOUNDE using bits 10 to 14 141751 -.537 .2957 OQSO for outputCONFOUNDE using bits 9 to 13 142069 .541 .7058 OQSO for outputCONFOUNDE using bits 8 to 12 142139 .779 .7819 OQSO for outputCONFOUNDE using bits 7 to 11 141606 -1.028 .1519 OQSO for outputCONFOUNDE using bits 6 to 10 141979 .236 .5934 OQSO for outputCONFOUNDE using bits 5 to 9 142488 1.962 .9751 OQSO for outputCONFOUNDE using bits 4 to 8 141861 -.164 .4349 OQSO for outputCONFOUNDE using bits 3 to 7 142658 2.538 .9944 OQSO for outputCONFOUNDE using bits 2 to 6 142635 2.460 .9931 OQSO for outputCONFOUNDE using bits 1 to 5 142074 .558 .7116 DNA test for generator outputCONFOUNDEDv4 Output: No. missing words (mw), equiv normal variate (z), p-value (p) mw z p DNA for outputCONFOUNDE using bits 31 to 32 141445 -1.370 .0854 DNA for outputCONFOUNDE using bits 30 to 31 141906 -.010 .4961 DNA for outputCONFOUNDE using bits 29 to 30 142064 .456 .6759 DNA for outputCONFOUNDE using bits 28 to 29 142034 .368 .6435 DNA for outputCONFOUNDE using bits 27 to 28 142148 .704 .7593 DNA for outputCONFOUNDE using bits 26 to 27 141701 -.615 .2694 DNA for outputCONFOUNDE using bits 25 to 26 141524 -1.137 .1278 DNA for outputCONFOUNDE using bits 24 to 25 141828 -.240 .4052 DNA for outputCONFOUNDE using bits 23 to 24 142462 1.630 .9485 DNA for outputCONFOUNDE using bits 22 to 23 142168 .763 .7773 DNA for outputCONFOUNDE using bits 21 to 22 141966 .167 .5664 DNA for outputCONFOUNDE using bits 20 to 21 141447 -1.364 .0863 DNA for outputCONFOUNDE using bits 19 to 20 141963 .158 .5629 DNA for outputCONFOUNDE using bits 18 to 19 141294 -1.815 .0348 DNA for outputCONFOUNDE using bits 17 to 18 141894 -.045 .4820 DNA for outputCONFOUNDE using bits 16 to 17 141565 -1.016 .1549 DNA for outputCONFOUNDE using bits 15 to 16 141845 -.190 .4247 DNA for outputCONFOUNDE using bits 14 to 15 141710 -.588 .2783 DNA for outputCONFOUNDE using bits 13 to 14 141943 .099 .5396 DNA for outputCONFOUNDE using bits 12 to 13 141822 -.258 .3984 DNA for outputCONFOUNDE using bits 11 to 12 141504 -1.196 .1159 DNA for outputCONFOUNDE using bits 10 to 11 141875 -.101 .4597 DNA for outputCONFOUNDE using bits 9 to 10 141954 .132 .5524 DNA for outputCONFOUNDE using bits 8 to 9 141757 -.449 .3266 DNA for outputCONFOUNDE using bits 7 to 8 142350 1.300 .9032 DNA for outputCONFOUNDE using bits 6 to 7 141369 -1.594 .0555 DNA for outputCONFOUNDE using bits 5 to 6 141789 -.355 .3613 DNA for outputCONFOUNDE using bits 4 to 5 142038 .380 .6479 DNA for outputCONFOUNDE using bits 3 to 4 141824 -.252 .4006 DNA for outputCONFOUNDE using bits 2 to 3 141872 -.110 .4562 DNA for outputCONFOUNDE using bits 1 to 2 141492 -1.231 .1092 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the COUNT-THE-1's TEST on a stream of bytes. :: :: Consider the file under test as a stream of bytes (four per :: :: 32 bit integer). Each byte can contain from 0 to 8 1's, :: :: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let :: :: the stream of bytes provide a string of overlapping 5-letter :: :: words, each "letter" taking values A,B,C,D,E. The letters are :: :: determined by the number of 1's in a byte:: 0,1,or 2 yield A,:: :: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus :: :: we have a monkey at a typewriter hitting five keys with vari- :: :: ous probabilities (37,56,70,56,37 over 256). There are 5^5 :: :: possible 5-letter words, and from a string of 256,000 (over- :: :: lapping) 5-letter words, counts are made on the frequencies :: :: for each word. The quadratic form in the weak inverse of :: :: the covariance matrix of the cell counts provides a chisquare :: :: test:: Q5-Q4, the difference of the naive Pearson sums of :: :: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test results for outputCONFOUNDE Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000 chisquare equiv normal p-value Results fo COUNT-THE-1's in successive bytes: byte stream for outputCONFOUNDE 2471.15 -.408 .341612 byte stream for outputCONFOUNDE 2473.10 -.380 .351805 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 2400.31 -1.410 .079293 bits 2 to 9 2521.90 .310 .621627 bits 3 to 10 2497.22 -.039 .484335 bits 4 to 11 2550.85 .719 .763959 bits 5 to 12 2448.52 -.728 .233303 bits 6 to 13 2650.82 2.133 .983536 bits 7 to 14 2378.05 -1.725 .042301 bits 8 to 15 2490.65 -.132 .447416 bits 9 to 16 2694.66 2.753 .997046 bits 10 to 17 2541.71 .590 .722350 bits 11 to 18 2532.74 .463 .678329 bits 12 to 19 2478.51 -.304 .380614 bits 13 to 20 2455.71 -.626 .265527 bits 14 to 21 2443.56 -.798 .212389 bits 15 to 22 2391.70 -1.532 .062816 bits 16 to 23 2531.71 .448 .673096 bits 17 to 24 2679.72 2.542 .994483 bits 18 to 25 2431.84 -.964 .167544 bits 19 to 26 2462.94 -.524 .300097 bits 20 to 27 2497.12 -.041 .483737 bits 21 to 28 2449.67 -.712 .238291 bits 22 to 29 2506.72 .095 .537848 bits 23 to 30 2625.99 1.782 .962605 bits 24 to 31 2624.93 1.767 .961362 bits 25 to 32 2534.30 .485 .686183 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 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 outputCONFOUNDE Of 12,000 tries, the average no. of successes should be 3523 with sigma=21.9 Successes: 3489 z-score: -1.553 p-value: .060270 Successes: 3556 z-score: 1.507 p-value: .934075 Successes: 3553 z-score: 1.370 p-value: .914635 Successes: 3510 z-score: -.594 p-value: .276387 Successes: 3522 z-score: -.046 p-value: .481790 Successes: 3531 z-score: .365 p-value: .642555 Successes: 3537 z-score: .639 p-value: .738676 Successes: 3551 z-score: 1.279 p-value: .899470 Successes: 3550 z-score: 1.233 p-value: .891189 Successes: 3506 z-score: -.776 p-value: .218799 square size avg. no. parked sample sigma 100. 3530.500 21.909 KSTEST for the above 10: p= .708924 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE MINIMUM DISTANCE TEST :: :: It does this 100 times:: choose n=8000 random points in a :: :: square of side 10000. Find d, the minimum distance between :: :: the (n^2-n)/2 pairs of points. If the points are truly inde- :: :: pendent uniform, then d^2, the square of the minimum distance :: :: should be (very close to) exponentially distributed with mean :: :: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and :: :: a KSTEST on the resulting 100 values serves as a test of uni- :: :: formity for random points in the square. Test numbers=0 mod 5 :: :: are printed but the KSTEST is based on the full set of 100 :: :: random choices of 8000 points in the 10000x10000 square. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: This is the MINIMUM DISTANCE test for random integers in the file outputCONFOUNDE Sample no. d^2 avg equiv uni 5 .9356 1.6351 .609507 10 .8843 1.2633 .588825 15 .4431 1.7168 .359356 20 .4697 1.3829 .376299 25 .5573 1.4805 .428822 30 .4024 1.5266 .332658 35 1.2626 1.3683 .718874 40 .4711 1.3292 .377159 45 .8497 1.2946 .574297 50 1.8482 1.2368 .843933 55 4.1484 1.2845 .984535 60 .7912 1.5556 .548507 65 3.6789 1.6693 .975211 70 .6508 1.6457 .480055 75 .2109 1.7419 .190999 80 1.6782 1.7135 .814866 85 .2574 1.6371 .227944 90 1.8629 1.6721 .846219 95 1.4555 1.6440 .768409 100 1.2197 1.7017 .706472 MINIMUM DISTANCE TEST for outputCONFOUNDE Result of KS test on 20 transformed mindist^2's: p-value= .997262 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: THE 3DSPHERES TEST :: :: Choose 4000 random points in a cube of edge 1000. At each :: :: point, center a sphere large enough to reach the next closest :: :: point. Then the volume of the smallest such sphere is (very :: :: close to) exponentially distributed with mean 120pi/3. Thus :: :: the radius cubed is exponential with mean 30. (The mean is :: :: obtained by extensive simulation). The 3DSPHERES test gener- :: :: ates 4000 such spheres 20 times. Each min radius cubed leads :: :: to a uniform variable by means of 1-exp(-r^3/30.), then a :: :: KSTEST is done on the 20 p-values. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The 3DSPHERES test for file outputCONFOUNDE sample no: 1 r^3= 58.547 p-value= .85795 sample no: 2 r^3= 3.847 p-value= .12035 sample no: 3 r^3= 36.138 p-value= .70019 sample no: 4 r^3= 118.296 p-value= .98061 sample no: 5 r^3= 6.392 p-value= .19190 sample no: 6 r^3= 45.241 p-value= .77865 sample no: 7 r^3= 8.865 p-value= .25584 sample no: 8 r^3= 7.216 p-value= .21378 sample no: 9 r^3= 15.531 p-value= .40411 sample no: 10 r^3= 21.265 p-value= .50778 sample no: 11 r^3= 30.383 p-value= .63679 sample no: 12 r^3= 43.053 p-value= .76191 sample no: 13 r^3= 11.844 p-value= .32619 sample no: 14 r^3= 77.251 p-value= .92385 sample no: 15 r^3= 27.482 p-value= .59991 sample no: 16 r^3= 10.159 p-value= .28727 sample no: 17 r^3= 17.772 p-value= .44700 sample no: 18 r^3= 88.415 p-value= .94751 sample no: 19 r^3= 48.687 p-value= .80267 sample no: 20 r^3= 59.139 p-value= .86073 A KS test is applied to those 20 p-values. --------------------------------------------------------- 3DSPHERES test for file outputCONFOUNDE p-value= .611191 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the SQEEZE test :: :: Random integers are floated to get uniforms on [0,1). Start- :: :: ing with k=2^31=2147483647, the test finds j, the number of :: :: iterations necessary to reduce k to 1, using the reduction :: :: k=ceiling(k*U), with U provided by floating integers from :: :: the file being tested. Such j's are found 100,000 times, :: :: then counts for the number of times j was <=6,7,...,47,>=48 :: :: are used to provide a chi-square test for cell frequencies. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: RESULTS OF SQUEEZE TEST FOR outputCONFOUNDE Table of standardized frequency counts ( (obs-exp)/sqrt(exp) )^2 for j taking values <=6,7,8,...,47,>=48: -.1 .1 .1 -1.7 -1.9 2.1 .1 .8 -.2 .6 1.8 -.4 -2.1 1.0 .4 1.1 -.8 1.3 -.5 -.2 -1.1 -.1 .7 -.2 .2 -1.7 -.1 -.2 -2.1 .5 -.3 .9 1.1 -.1 -.4 -.1 .0 -1.3 1.7 -1.3 .1 .0 -1.1 Chi-square with 42 degrees of freedom: 44.059 z-score= .225 p-value= .615552 ______________________________________________________________ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: The OVERLAPPING SUMS test :: :: Integers are floated to get a sequence U(1),U(2),... of uni- :: :: form [0,1) variables. Then overlapping sums, :: :: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. :: :: The S's are virtually normal with a certain covariance mat- :: :: rix. A linear transformation of the S's converts them to a :: :: sequence of independent standard normals, which are converted :: :: to uniform variables for a KSTEST. The p-values from ten :: :: KSTESTs are given still another KSTEST. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Test no. 1 p-value .373834 Test no. 2 p-value .943529 Test no. 3 p-value .879475 Test no. 4 p-value .516920 Test no. 5 p-value .460795 Test no. 6 p-value .897212 Test no. 7 p-value .016615 Test no. 8 p-value .799306 Test no. 9 p-value .968487 Test no. 10 p-value .782876 Results of the OSUM test for outputCONFOUNDE KSTEST on the above 10 p-values: .925061 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the RUNS test. It counts runs up, and runs down, :: :: in a sequence of uniform [0,1) variables, obtained by float- :: :: ing the 32-bit integers in the specified file. This example :: :: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95:: :: contains an up-run of length 3, a down-run of length 2 and an :: :: up-run of (at least) 2, depending on the next values. The :: :: covariance matrices for the runs-up and runs-down are well :: :: known, leading to chisquare tests for quadratic forms in the :: :: weak inverses of the covariance matrices. Runs are counted :: :: for sequences of length 10,000. This is done ten times. Then :: :: repeated. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: The RUNS test for file outputCONFOUNDE Up and down runs in a sample of 10000 _________________________________________________ Run test for outputCONFOUNDE: runs up; ks test for 10 p's: .678268 runs down; ks test for 10 p's: .395328 Run test for outputCONFOUNDE: runs up; ks test for 10 p's: .186333 runs down; ks test for 10 p's: .311758 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: This is the CRAPS TEST. It plays 200,000 games of craps, finds:: :: the number of wins and the number of throws necessary to end :: :: each game. The number of wins should be (very close to) a :: :: normal with mean 200000p and variance 200000p(1-p), with :: :: p=244/495. Throws necessary to complete the game can vary :: :: from 1 to infinity, but counts for all>21 are lumped with 21. :: :: A chi-square test is made on the no.-of-throws cell counts. :: :: Each 32-bit integer from the test file provides the value for :: :: the throw of a die, by floating to [0,1), multiplying by 6 :: :: and taking 1 plus the integer part of the result. :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Results of craps test for outputCONFOUNDE No. of wins: Observed Expected 98282 98585.86 98282= No. of wins, z-score=-1.359 pvalue= .08707 Analysis of Throws-per-Game: Chisq= 21.53 for 20 degrees of freedom, p= .63323 Throws Observed Expected Chisq Sum 1 66490 66666.7 .468 .468 2 37686 37654.3 .027 .495 3 26826 26954.7 .615 1.110 4 19305 19313.5 .004 1.113 5 13875 13851.4 .040 1.153 6 10083 9943.5 1.956 3.109 7 7262 7145.0 1.915 5.024 8 5063 5139.1 1.126 6.150 9 3732 3699.9 .279 6.430 10 2604 2666.3 1.456 7.885 11 2056 1923.3 9.152 17.037 12 1383 1388.7 .024 17.061 13 998 1003.7 .033 17.093 14 726 726.1 .000 17.093 15 521 525.8 .044 17.138 16 397 381.2 .659 17.797 17 269 276.5 .206 18.002 18 203 200.8 .023 18.026 19 129 146.0 1.976 20.002 20 95 106.2 1.184 21.186 21 297 287.1 .340 21.526 SUMMARY FOR outputCONFOUNDE p-value for no. of wins: .087068 p-value for throws/game: .633234 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Results of DIEHARD battery of tests sent to file confounded.out