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 ar_16M-1.dat
For a sample of size 500: mean
ar_16M-1.dat using bits 1 to 24 2.018
duplicate number number
spacings observed expected
0 52. 67.668
1 134. 135.335
2 157. 135.335
3 88. 90.224
4 53. 45.112
5 13. 18.045
6 to INF 3. 8.282
Chisquare with 6 d.o.f. = 13.32 p-value= .961801
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 2 to 25 2.022
duplicate number number
spacings observed expected
0 65. 67.668
1 130. 135.335
2 147. 135.335
3 84. 90.224
4 45. 45.112
5 21. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 2.24 p-value= .104060
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 3 to 26 1.950
duplicate number number
spacings observed expected
0 65. 67.668
1 152. 135.335
2 132. 135.335
3 78. 90.224
4 47. 45.112
5 20. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 4.82 p-value= .432255
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 4 to 27 1.980
duplicate number number
spacings observed expected
0 70. 67.668
1 135. 135.335
2 127. 135.335
3 103. 90.224
4 40. 45.112
5 18. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 3.18 p-value= .214265
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 5 to 28 2.024
duplicate number number
spacings observed expected
0 63. 67.668
1 123. 135.335
2 143. 135.335
3 109. 90.224
4 42. 45.112
5 14. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 7.54 p-value= .726052
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 6 to 29 2.090
duplicate number number
spacings observed expected
0 62. 67.668
1 107. 135.335
2 157. 135.335
3 106. 90.224
4 41. 45.112
5 20. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 13.42 p-value= .963157
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 7 to 30 2.094
duplicate number number
spacings observed expected
0 67. 67.668
1 119. 135.335
2 136. 135.335
3 94. 90.224
4 57. 45.112
5 20. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 5.68 p-value= .540354
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 8 to 31 1.958
duplicate number number
spacings observed expected
0 68. 67.668
1 133. 135.335
2 150. 135.335
3 86. 90.224
4 35. 45.112
5 21. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 4.78 p-value= .427366
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
ar_16M-1.dat using bits 9 to 32 2.032
duplicate number number
spacings observed expected
0 64. 67.668
1 152. 135.335
2 122. 135.335
3 78. 90.224
4 50. 45.112
5 26. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 9.27 p-value= .840909
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.961801 .104060 .432255 .214265 .726052
.963157 .540354 .427366 .840909
A KSTEST for the 9 p-values yields .397903
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=126.855; p-value= .968964
OPERM5 test for file ar_16M-1.dat
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=140.892; p-value= .996340
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 235 211.4 2.630382 2.630
29 5074 5134.0 .701446 3.332
30 23061 23103.0 .076524 3.408
31 11630 11551.5 .533126 3.941
chisquare= 3.941 for 3 d. of f.; p-value= .755702
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 227 211.4 1.148427 1.148
30 5241 5134.0 2.229603 3.378
31 23033 23103.0 .212377 3.590
32 11499 11551.5 .238827 3.829
chisquare= 3.829 for 3 d. of f.; p-value= .744679
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 929 944.3 .248 .248
r =5 21828 21743.9 .325 .573
r =6 77243 77311.8 .061 .634
p=1-exp(-SUM/2)= .27183
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 923 944.3 .481 .481
r =5 21845 21743.9 .470 .951
r =6 77232 77311.8 .082 1.033
p=1-exp(-SUM/2)= .40338
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 969 944.3 .646 .646
r =5 21645 21743.9 .450 1.096
r =6 77386 77311.8 .071 1.167
p=1-exp(-SUM/2)= .44207
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 968 944.3 .595 .595
r =5 21903 21743.9 1.164 1.759
r =6 77129 77311.8 .432 2.191
p=1-exp(-SUM/2)= .66565
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 952 944.3 .063 .063
r =5 21923 21743.9 1.475 1.538
r =6 77125 77311.8 .451 1.989
p=1-exp(-SUM/2)= .63015
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21701 21743.9 .085 .176
r =6 77364 77311.8 .035 .211
p=1-exp(-SUM/2)= .10035
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 976 944.3 1.064 1.064
r =5 21569 21743.9 1.407 2.471
r =6 77455 77311.8 .265 2.736
p=1-exp(-SUM/2)= .74540
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 972 944.3 .812 .812
r =5 21612 21743.9 .800 1.613
r =6 77416 77311.8 .140 1.753
p=1-exp(-SUM/2)= .58377
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 975 944.3 .998 .998
r =5 21525 21743.9 2.204 3.202
r =6 77500 77311.8 .458 3.660
p=1-exp(-SUM/2)= .83957
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 929 944.3 .248 .248
r =5 21712 21743.9 .047 .295
r =6 77359 77311.8 .029 .324
p=1-exp(-SUM/2)= .14937
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 955 944.3 .121 .121
r =5 21793 21743.9 .111 .232
r =6 77252 77311.8 .046 .278
p=1-exp(-SUM/2)= .12992
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 915 944.3 .909 .909
r =5 21578 21743.9 1.266 2.175
r =6 77507 77311.8 .493 2.668
p=1-exp(-SUM/2)= .73655
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 906 944.3 1.554 1.554
r =5 21626 21743.9 .639 2.193
r =6 77468 77311.8 .316 2.508
p=1-exp(-SUM/2)= .71469
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21772 21743.9 .036 .078
r =6 77290 77311.8 .006 .085
p=1-exp(-SUM/2)= .04138
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21956 21743.9 2.069 2.092
r =6 77095 77311.8 .608 2.700
p=1-exp(-SUM/2)= .74080
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21551 21743.9 1.711 1.753
r =6 77511 77311.8 .513 2.267
p=1-exp(-SUM/2)= .67803
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 929 944.3 .248 .248
r =5 21510 21743.9 2.516 2.764
r =6 77561 77311.8 .803 3.567
p=1-exp(-SUM/2)= .83197
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 951 944.3 .048 .048
r =5 21893 21743.9 1.022 1.070
r =6 77156 77311.8 .314 1.384
p=1-exp(-SUM/2)= .49940
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 936 944.3 .073 .073
r =5 22014 21743.9 3.355 3.428
r =6 77050 77311.8 .887 4.315
p=1-exp(-SUM/2)= .88437
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21822 21743.9 .281 .288
r =6 77231 77311.8 .084 .373
p=1-exp(-SUM/2)= .17001
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 932 944.3 .160 .160
r =5 21869 21743.9 .720 .880
r =6 77199 77311.8 .165 1.045
p=1-exp(-SUM/2)= .40684
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 987 944.3 1.931 1.931
r =5 21680 21743.9 .188 2.119
r =6 77333 77311.8 .006 2.124
p=1-exp(-SUM/2)= .65429
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21717 21743.9 .033 .041
r =6 77336 77311.8 .008 .049
p=1-exp(-SUM/2)= .02399
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 913 944.3 1.038 1.038
r =5 21818 21743.9 .253 1.290
r =6 77269 77311.8 .024 1.314
p=1-exp(-SUM/2)= .48154
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG ar_16M-1.dat
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 977 944.3 1.132 1.132
r =5 21749 21743.9 .001 1.133
r =6 77274 77311.8 .018 1.152
p=1-exp(-SUM/2)= .43785
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.271832 .403381 .442073 .665649 .630154
.100349 .745402 .583767 .839573 .149368
.129923 .736553 .714691 .041375 .740796
.678029 .831971 .499400 .884367 .170011
.406838 .654292 .023990 .481538 .437845
brank test summary for ar_16M-1.dat
The KS test for those 25 supposed UNI's yields
KS p-value= .322356
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 141371 missing words, -1.26 sigmas from mean, p-value= .10424
tst no 2: 141803 missing words, -.25 sigmas from mean, p-value= .40190
tst no 3: 141577 missing words, -.78 sigmas from mean, p-value= .21874
tst no 4: 142218 missing words, .72 sigmas from mean, p-value= .76461
tst no 5: 141937 missing words, .06 sigmas from mean, p-value= .52578
tst no 6: 142359 missing words, 1.05 sigmas from mean, p-value= .85329
tst no 7: 142819 missing words, 2.13 sigmas from mean, p-value= .98322
tst no 8: 141701 missing words, -.49 sigmas from mean, p-value= .31322
tst no 9: 142047 missing words, .32 sigmas from mean, p-value= .62615
tst no 10: 142055 missing words, .34 sigmas from mean, p-value= .63321
tst no 11: 141907 missing words, -.01 sigmas from mean, p-value= .49783
tst no 12: 142028 missing words, .28 sigmas from mean, p-value= .60921
tst no 13: 142625 missing words, 1.67 sigmas from mean, p-value= .95275
tst no 14: 142060 missing words, .35 sigmas from mean, p-value= .63759
tst no 15: 141418 missing words, -1.15 sigmas from mean, p-value= .12549
tst no 16: 142010 missing words, .24 sigmas from mean, p-value= .59298
tst no 17: 141816 missing words, -.22 sigmas from mean, p-value= .41369
tst no 18: 141971 missing words, .14 sigmas from mean, p-value= .55729
tst no 19: 142350 missing words, 1.03 sigmas from mean, p-value= .84840
tst no 20: 142092 missing words, .43 sigmas from mean, p-value= .66524
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for ar_16M-1.dat using bits 23 to 32 141883 -.091 .4638
OPSO for ar_16M-1.dat using bits 22 to 31 142027 .406 .6575
OPSO for ar_16M-1.dat using bits 21 to 30 141761 -.511 .3045
OPSO for ar_16M-1.dat using bits 20 to 29 141571 -1.167 .1217
OPSO for ar_16M-1.dat using bits 19 to 28 141432 -1.646 .0499
OPSO for ar_16M-1.dat using bits 18 to 27 141712 -.680 .2481
OPSO for ar_16M-1.dat using bits 17 to 26 142225 1.089 .8618
OPSO for ar_16M-1.dat using bits 16 to 25 141518 -1.349 .0886
OPSO for ar_16M-1.dat using bits 15 to 24 141564 -1.191 .1169
OPSO for ar_16M-1.dat using bits 14 to 23 142132 .768 .7787
OPSO for ar_16M-1.dat using bits 13 to 22 142385 1.640 .9495
OPSO for ar_16M-1.dat using bits 12 to 21 141982 .251 .5989
OPSO for ar_16M-1.dat using bits 11 to 20 142518 2.099 .9821
OPSO for ar_16M-1.dat using bits 10 to 19 141777 -.456 .3241
OPSO for ar_16M-1.dat using bits 9 to 18 141786 -.425 .3353
OPSO for ar_16M-1.dat using bits 8 to 17 141939 .102 .5407
OPSO for ar_16M-1.dat using bits 7 to 16 142356 1.540 .9383
OPSO for ar_16M-1.dat using bits 6 to 15 142138 .789 .7848
OPSO for ar_16M-1.dat using bits 5 to 14 141816 -.322 .3738
OPSO for ar_16M-1.dat using bits 4 to 13 141471 -1.511 .0653
OPSO for ar_16M-1.dat using bits 3 to 12 141484 -1.467 .0712
OPSO for ar_16M-1.dat using bits 2 to 11 142080 .589 .7219
OPSO for ar_16M-1.dat using bits 1 to 10 142028 .409 .6588
OQSO test for generator ar_16M-1.dat
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for ar_16M-1.dat using bits 28 to 32 141502 -1.381 .0837
OQSO for ar_16M-1.dat using bits 27 to 31 141587 -1.093 .1373
OQSO for ar_16M-1.dat using bits 26 to 30 141977 .229 .5907
OQSO for ar_16M-1.dat using bits 25 to 29 141732 -.601 .2739
OQSO for ar_16M-1.dat using bits 24 to 28 141931 .073 .5293
OQSO for ar_16M-1.dat using bits 23 to 27 142049 .473 .6821
OQSO for ar_16M-1.dat using bits 22 to 26 141888 -.072 .4712
OQSO for ar_16M-1.dat using bits 21 to 25 142158 .843 .8004
OQSO for ar_16M-1.dat using bits 20 to 24 141679 -.781 .2175
OQSO for ar_16M-1.dat using bits 19 to 23 141527 -1.296 .0975
OQSO for ar_16M-1.dat using bits 18 to 22 141822 -.296 .3836
OQSO for ar_16M-1.dat using bits 17 to 21 141879 -.103 .4591
OQSO for ar_16M-1.dat using bits 16 to 20 141315 -2.015 .0220
OQSO for ar_16M-1.dat using bits 15 to 19 141862 -.160 .4363
OQSO for ar_16M-1.dat using bits 14 to 18 141832 -.262 .3966
OQSO for ar_16M-1.dat using bits 13 to 17 141901 -.028 .4887
OQSO for ar_16M-1.dat using bits 12 to 16 142120 .714 .7624
OQSO for ar_16M-1.dat using bits 11 to 15 141832 -.262 .3966
OQSO for ar_16M-1.dat using bits 10 to 14 142127 .738 .7697
OQSO for ar_16M-1.dat using bits 9 to 13 141534 -1.272 .1016
OQSO for ar_16M-1.dat using bits 8 to 12 141852 -.194 .4230
OQSO for ar_16M-1.dat using bits 7 to 11 141457 -1.533 .0626
OQSO for ar_16M-1.dat using bits 6 to 10 141685 -.760 .2235
OQSO for ar_16M-1.dat using bits 5 to 9 142093 .623 .7332
OQSO for ar_16M-1.dat using bits 4 to 8 142417 1.721 .9574
OQSO for ar_16M-1.dat using bits 3 to 7 141995 .290 .6142
OQSO for ar_16M-1.dat using bits 2 to 6 141976 .226 .5894
OQSO for ar_16M-1.dat using bits 1 to 5 142175 .901 .8161
DNA test for generator ar_16M-1.dat
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for ar_16M-1.dat using bits 31 to 32 141918 .026 .5102
DNA for ar_16M-1.dat using bits 30 to 31 142210 .887 .8124
DNA for ar_16M-1.dat using bits 29 to 30 142269 1.061 .8557
DNA for ar_16M-1.dat using bits 28 to 29 141766 -.423 .3362
DNA for ar_16M-1.dat using bits 27 to 28 142245 .990 .8390
DNA for ar_16M-1.dat using bits 26 to 27 141772 -.405 .3427
DNA for ar_16M-1.dat using bits 25 to 26 141650 -.765 .2221
DNA for ar_16M-1.dat using bits 24 to 25 141859 -.148 .4410
DNA for ar_16M-1.dat using bits 23 to 24 141322 -1.733 .0416
DNA for ar_16M-1.dat using bits 22 to 23 142411 1.480 .9305
DNA for ar_16M-1.dat using bits 21 to 22 141998 .262 .6032
DNA for ar_16M-1.dat using bits 20 to 21 142363 1.338 .9096
DNA for ar_16M-1.dat using bits 19 to 20 142527 1.822 .9658
DNA for ar_16M-1.dat using bits 18 to 19 142183 .807 .7903
DNA for ar_16M-1.dat using bits 17 to 18 142312 1.188 .8825
DNA for ar_16M-1.dat using bits 16 to 17 142365 1.344 .9106
DNA for ar_16M-1.dat using bits 15 to 16 141494 -1.225 .1103
DNA for ar_16M-1.dat using bits 14 to 15 142454 1.607 .9459
DNA for ar_16M-1.dat using bits 13 to 14 141921 .034 .5137
DNA for ar_16M-1.dat using bits 12 to 13 141695 -.632 .2636
DNA for ar_16M-1.dat using bits 11 to 12 142703 2.341 .9904
DNA for ar_16M-1.dat using bits 10 to 11 142532 1.837 .9669
DNA for ar_16M-1.dat using bits 9 to 10 141561 -1.028 .1521
DNA for ar_16M-1.dat using bits 8 to 9 142346 1.288 .9011
DNA for ar_16M-1.dat using bits 7 to 8 142313 1.191 .8831
DNA for ar_16M-1.dat using bits 6 to 7 141902 -.022 .4914
DNA for ar_16M-1.dat using bits 5 to 6 141665 -.721 .2355
DNA for ar_16M-1.dat using bits 4 to 5 141161 -2.207 .0136
DNA for ar_16M-1.dat using bits 3 to 4 141583 -.963 .1679
DNA for ar_16M-1.dat using bits 2 to 3 141909 -.001 .4996
DNA for ar_16M-1.dat using bits 1 to 2 142376 1.377 .9157
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the COUNT-THE-1's TEST on a stream of bytes. ::
:: Consider the file under test as a stream of bytes (four per ::
:: 32 bit integer). Each byte can contain from 0 to 8 1's, ::
:: with probabilities 1,8,28,56,70,56,28,8,1 over 256. Now let ::
:: the stream of bytes provide a string of overlapping 5-letter ::
:: words, each "letter" taking values A,B,C,D,E. The letters are ::
:: determined by the number of 1's in a byte:: 0,1,or 2 yield A,::
:: 3 yields B, 4 yields C, 5 yields D and 6,7 or 8 yield E. Thus ::
:: we have a monkey at a typewriter hitting five keys with vari- ::
:: ous probabilities (37,56,70,56,37 over 256). There are 5^5 ::
:: possible 5-letter words, and from a string of 256,000 (over- ::
:: lapping) 5-letter words, counts are made on the frequencies ::
:: for each word. The quadratic form in the weak inverse of ::
:: the covariance matrix of the cell counts provides a chisquare ::
:: test:: Q5-Q4, the difference of the naive Pearson sums of ::
:: (OBS-EXP)^2/EXP on counts for 5- and 4-letter cell counts. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test results for ar_16M-1.dat
Chi-square with 5^5-5^4=2500 d.of f. for sample size:2560000
chisquare equiv normal p-value
Results fo COUNT-THE-1's in successive bytes:
byte stream for ar_16M-1.dat 2529.60 .419 .662274
byte stream for ar_16M-1.dat 2573.44 1.039 .850497
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2458.62 -.585 .279219
bits 2 to 9 2631.82 1.864 .968859
bits 3 to 10 2598.38 1.391 .917943
bits 4 to 11 2542.63 .603 .726686
bits 5 to 12 2488.24 -.166 .433976
bits 6 to 13 2630.06 1.839 .967065
bits 7 to 14 2555.76 .789 .784834
bits 8 to 15 2422.90 -1.090 .137777
bits 9 to 16 2589.89 1.271 .898189
bits 10 to 17 2522.24 .314 .623420
bits 11 to 18 2553.39 .755 .774902
bits 12 to 19 2507.88 .111 .544339
bits 13 to 20 2477.52 -.318 .375286
bits 14 to 21 2592.53 1.309 .904668
bits 15 to 22 2545.51 .644 .740090
bits 16 to 23 2410.22 -1.270 .102090
bits 17 to 24 2751.02 3.550 .999807
bits 18 to 25 2510.42 .147 .558591
bits 19 to 26 2450.58 -.699 .242319
bits 20 to 27 2562.51 .884 .811654
bits 21 to 28 2449.01 -.721 .235441
bits 22 to 29 2340.87 -2.250 .012211
bits 23 to 30 2460.39 -.560 .287692
bits 24 to 31 2472.82 -.384 .350331
bits 25 to 32 2459.71 -.570 .284415
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 ar_16M-1.dat
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3529 z-score: .274 p-value: .607947
Successes: 3519 z-score: -.183 p-value: .427537
Successes: 3558 z-score: 1.598 p-value: .944998
Successes: 3543 z-score: .913 p-value: .819442
Successes: 3499 z-score: -1.096 p-value: .136563
Successes: 3534 z-score: .502 p-value: .692266
Successes: 3490 z-score: -1.507 p-value: .065925
Successes: 3525 z-score: .091 p-value: .536382
Successes: 3506 z-score: -.776 p-value: .218799
Successes: 3524 z-score: .046 p-value: .518210
square size avg. no. parked sample sigma
100. 3522.700 19.368
KSTEST for the above 10: p= .002653
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE MINIMUM DISTANCE TEST ::
:: It does this 100 times:: choose n=8000 random points in a ::
:: square of side 10000. Find d, the minimum distance between ::
:: the (n^2-n)/2 pairs of points. If the points are truly inde- ::
:: pendent uniform, then d^2, the square of the minimum distance ::
:: should be (very close to) exponentially distributed with mean ::
:: .995 . Thus 1-exp(-d^2/.995) should be uniform on [0,1) and ::
:: a KSTEST on the resulting 100 values serves as a test of uni- ::
:: formity for random points in the square. Test numbers=0 mod 5 ::
:: are printed but the KSTEST is based on the full set of 100 ::
:: random choices of 8000 points in the 10000x10000 square. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
This is the MINIMUM DISTANCE test
for random integers in the file ar_16M-1.dat
Sample no. d^2 avg equiv uni
5 .6119 .7368 .459348
10 .1540 .6757 .143373
15 .9849 1.0203 .628376
20 .2902 1.0998 .252980
25 1.9284 1.1880 .856023
30 2.4383 1.1669 .913758
35 1.1180 1.0983 .674891
40 .2217 1.0697 .199708
45 .6159 1.1123 .461535
50 .6674 1.1050 .488684
55 1.6371 1.1096 .807060
60 1.8427 1.0957 .843065
65 1.0388 1.1394 .647961
70 .6352 1.1179 .471837
75 .0051 1.0846 .005071
80 1.1848 1.0590 .695999
85 3.5088 1.0745 .970591
90 1.0914 1.0776 .666094
95 .5709 1.0578 .436616
100 2.3403 1.0814 .904823
MINIMUM DISTANCE TEST for ar_16M-1.dat
Result of KS test on 20 transformed mindist^2's:
p-value= .905873
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: THE 3DSPHERES TEST ::
:: Choose 4000 random points in a cube of edge 1000. At each ::
:: point, center a sphere large enough to reach the next closest ::
:: point. Then the volume of the smallest such sphere is (very ::
:: close to) exponentially distributed with mean 120pi/3. Thus ::
:: the radius cubed is exponential with mean 30. (The mean is ::
:: obtained by extensive simulation). The 3DSPHERES test gener- ::
:: ates 4000 such spheres 20 times. Each min radius cubed leads ::
:: to a uniform variable by means of 1-exp(-r^3/30.), then a ::
:: KSTEST is done on the 20 p-values. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The 3DSPHERES test for file ar_16M-1.dat
sample no: 1 r^3= 17.953 p-value= .45034
sample no: 2 r^3= 9.727 p-value= .27691
sample no: 3 r^3= 18.916 p-value= .46769
sample no: 4 r^3= 35.754 p-value= .69633
sample no: 5 r^3= 44.738 p-value= .77491
sample no: 6 r^3= 28.068 p-value= .60765
sample no: 7 r^3= 24.225 p-value= .55403
sample no: 8 r^3= 15.282 p-value= .39914
sample no: 9 r^3= 50.553 p-value= .81458
sample no: 10 r^3= 41.715 p-value= .75105
sample no: 11 r^3= 6.813 p-value= .20317
sample no: 12 r^3= 27.827 p-value= .60449
sample no: 13 r^3= 30.114 p-value= .63351
sample no: 14 r^3= 38.841 p-value= .72602
sample no: 15 r^3= 25.924 p-value= .57858
sample no: 16 r^3= 14.378 p-value= .38076
sample no: 17 r^3= 5.093 p-value= .15614
sample no: 18 r^3= 43.901 p-value= .76855
sample no: 19 r^3= 160.006 p-value= .99517
sample no: 20 r^3= 13.431 p-value= .36090
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file ar_16M-1.dat p-value= .772462
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the SQEEZE test ::
:: Random integers are floated to get uniforms on [0,1). Start- ::
:: ing with k=2^31=2147483647, the test finds j, the number of ::
:: iterations necessary to reduce k to 1, using the reduction ::
:: k=ceiling(k*U), with U provided by floating integers from ::
:: the file being tested. Such j's are found 100,000 times, ::
:: then counts for the number of times j was <=6,7,...,47,>=48 ::
:: are used to provide a chi-square test for cell frequencies. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESULTS OF SQUEEZE TEST FOR ar_16M-1.dat
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
-.8 -1.6 -.4 -.4 -.9 -.6
-.3 1.4 2.5 .0 -.8 -1.0
.7 1.1 -1.5 -.6 1.3 -.6
-.9 .6 1.6 .4 -.3 -1.2
-.2 -1.0 1.1 1.0 -.6 -.2
-.3 -.3 -.5 -1.6 -.6 -.7
.0 .2 -1.6 -1.8 .1 -1.0
.8
Chi-square with 42 degrees of freedom: 40.870
z-score= -.123 p-value= .479403
______________________________________________________________
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: The OVERLAPPING SUMS test ::
:: Integers are floated to get a sequence U(1),U(2),... of uni- ::
:: form [0,1) variables. Then overlapping sums, ::
:: S(1)=U(1)+...+U(100), S2=U(2)+...+U(101),... are formed. ::
:: The S's are virtually normal with a certain covariance mat- ::
:: rix. A linear transformation of the S's converts them to a ::
:: sequence of independent standard normals, which are converted ::
:: to uniform variables for a KSTEST. The p-values from ten ::
:: KSTESTs are given still another KSTEST. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Test no. 1 p-value .579860
Test no. 2 p-value .998262
Test no. 3 p-value .238195
Test no. 4 p-value .060052
Test no. 5 p-value .658996
Test no. 6 p-value .360096
Test no. 7 p-value .136065
Test no. 8 p-value .200156
Test no. 9 p-value .376184
Test no. 10 p-value .536828
Results of the OSUM test for ar_16M-1.dat
KSTEST on the above 10 p-values: .599633
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the RUNS test. It counts runs up, and runs down, ::
:: in a sequence of uniform [0,1) variables, obtained by float- ::
:: ing the 32-bit integers in the specified file. This example ::
:: shows how runs are counted: .123,.357,.789,.425,.224,.416,.95::
:: contains an up-run of length 3, a down-run of length 2 and an ::
:: up-run of (at least) 2, depending on the next values. The ::
:: covariance matrices for the runs-up and runs-down are well ::
:: known, leading to chisquare tests for quadratic forms in the ::
:: weak inverses of the covariance matrices. Runs are counted ::
:: for sequences of length 10,000. This is done ten times. Then ::
:: repeated. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
The RUNS test for file ar_16M-1.dat
Up and down runs in a sample of 10000
_________________________________________________
Run test for ar_16M-1.dat :
runs up; ks test for 10 p's: .891401
runs down; ks test for 10 p's: .863398
Run test for ar_16M-1.dat :
runs up; ks test for 10 p's: .875685
runs down; ks test for 10 p's: .294747
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: This is the CRAPS TEST. It plays 200,000 games of craps, finds::
:: the number of wins and the number of throws necessary to end ::
:: each game. The number of wins should be (very close to) a ::
:: normal with mean 200000p and variance 200000p(1-p), with ::
:: p=244/495. Throws necessary to complete the game can vary ::
:: from 1 to infinity, but counts for all>21 are lumped with 21. ::
:: A chi-square test is made on the no.-of-throws cell counts. ::
:: Each 32-bit integer from the test file provides the value for ::
:: the throw of a die, by floating to [0,1), multiplying by 6 ::
:: and taking 1 plus the integer part of the result. ::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Results of craps test for ar_16M-1.dat
No. of wins: Observed Expected
98161 98585.86
98161= No. of wins, z-score=-1.900 pvalue= .02870
Analysis of Throws-per-Game:
Chisq= 16.68 for 20 degrees of freedom, p= .32634
Throws Observed Expected Chisq Sum
1 66693 66666.7 .010 .010
2 37770 37654.3 .355 .366
3 26943 26954.7 .005 .371
4 19056 19313.5 3.432 3.803
5 13984 13851.4 1.269 5.072
6 9859 9943.5 .719 5.791
7 7235 7145.0 1.133 6.924
8 5174 5139.1 .237 7.161
9 3723 3699.9 .145 7.306
10 2689 2666.3 .193 7.499
11 1896 1923.3 .388 7.888
12 1355 1388.7 .820 8.707
13 1006 1003.7 .005 8.712
14 749 726.1 .720 9.432
15 541 525.8 .437 9.869
16 345 381.2 3.429 13.298
17 273 276.5 .045 13.343
18 178 200.8 2.595 15.939
19 147 146.0 .007 15.946
20 110 106.2 .135 16.081
21 274 287.1 .599 16.680
SUMMARY FOR ar_16M-1.dat
p-value for no. of wins: .028702
p-value for throws/game: .326338
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
Results of DIEHARD battery of tests sent to file results