Here's the results from diehard for yours (seed 123412341234 - assuming order of operations was the same)
While the rate of outlier p values (p < 0.01 or p> 0.99) was high for the number of samples (also there were two p=1.000000 values [See DNA tests for bits 32-30]), its probably sufficient levels of randomness for transcendence as well as being
faster than mine. (I tested; yours is ~3-4 orders of magnitude faster than mine)
Code: Select all
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 pxl_out.rbnf
For a sample of size 500: mean
pxl_out.rbnf using bits 1 to 24 1.998
duplicate number number
spacings observed expected
0 63. 67.668
1 130. 135.335
2 151. 135.335
3 91. 90.224
4 40. 45.112
5 19. 18.045
6 to INF 6. 8.282
Chisquare with 6 d.o.f. = 3.61 p-value= .270802
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 2 to 25 1.920
duplicate number number
spacings observed expected
0 71. 67.668
1 148. 135.335
2 133. 135.335
3 84. 90.224
4 39. 45.112
5 18. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 2.85 p-value= .172013
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 3 to 26 2.036
duplicate number number
spacings observed expected
0 66. 67.668
1 127. 135.335
2 144. 135.335
3 86. 90.224
4 48. 45.112
5 22. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = 2.56 p-value= .137990
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 4 to 27 1.930
duplicate number number
spacings observed expected
0 63. 67.668
1 149. 135.335
2 142. 135.335
3 79. 90.224
4 44. 45.112
5 19. 18.045
6 to INF 4. 8.282
Chisquare with 6 d.o.f. = 5.72 p-value= .544466
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 5 to 28 2.004
duplicate number number
spacings observed expected
0 63. 67.668
1 140. 135.335
2 135. 135.335
3 90. 90.224
4 46. 45.112
5 19. 18.045
6 to INF 7. 8.282
Chisquare with 6 d.o.f. = .75 p-value= .006666
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 6 to 29 2.014
duplicate number number
spacings observed expected
0 75. 67.668
1 134. 135.335
2 120. 135.335
3 99. 90.224
4 44. 45.112
5 17. 18.045
6 to INF 11. 8.282
Chisquare with 6 d.o.f. = 4.38 p-value= .374489
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 7 to 30 1.988
duplicate number number
spacings observed expected
0 65. 67.668
1 138. 135.335
2 142. 135.335
3 86. 90.224
4 40. 45.112
5 21. 18.045
6 to INF 8. 8.282
Chisquare with 6 d.o.f. = 1.76 p-value= .059314
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 8 to 31 1.960
duplicate number number
spacings observed expected
0 77. 67.668
1 137. 135.335
2 125. 135.335
3 76. 90.224
4 65. 45.112
5 15. 18.045
6 to INF 5. 8.282
Chisquare with 6 d.o.f. = 14.92 p-value= .979123
:::::::::::::::::::::::::::::::::::::::::
For a sample of size 500: mean
pxl_out.rbnf using bits 9 to 32 2.056
duplicate number number
spacings observed expected
0 71. 67.668
1 134. 135.335
2 128. 135.335
3 82. 90.224
4 49. 45.112
5 22. 18.045
6 to INF 14. 8.282
Chisquare with 6 d.o.f. = 6.47 p-value= .627831
:::::::::::::::::::::::::::::::::::::::::
The 9 p-values were
.270802 .172013 .137990 .544466 .006666
.374489 .059314 .979123 .627831
A KSTEST for the 9 p-values yields .846869
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom= 81.249; p-value= .097134
OPERM5 test for file pxl_out.rbnf
For a sample of 1,000,000 consecutive 5-tuples,
chisquare for 99 degrees of freedom=102.731; p-value= .621446
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Rank test for 31x31 binary matrices:
rows from leftmost 31 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
28 194 211.4 1.435011 1.435
29 5125 5134.0 .015813 1.451
30 23199 23103.0 .398519 1.849
31 11482 11551.5 .418442 2.268
chisquare= 2.268 for 3 d. of f.; p-value= .547324
--------------------------------------------------------------
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Rank test for 32x32 binary matrices:
rows from leftmost 32 bits of each 32-bit integer
rank observed expected (o-e)^2/e sum
29 193 211.4 1.604514 1.605
30 5222 5134.0 1.508021 3.113
31 23159 23103.0 .135513 3.248
32 11426 11551.5 1.364009 4.612
chisquare= 4.612 for 3 d. of f.; p-value= .813364
--------------------------------------------------------------
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 1 to 8
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21791 21743.9 .102 .125
r =6 77260 77311.8 .035 .160
p=1-exp(-SUM/2)= .07694
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 2 to 9
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 931 944.3 .187 .187
r =5 21739 21743.9 .001 .188
r =6 77330 77311.8 .004 .193
p=1-exp(-SUM/2)= .09187
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 3 to 10
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 913 944.3 1.038 1.038
r =5 21718 21743.9 .031 1.068
r =6 77369 77311.8 .042 1.111
p=1-exp(-SUM/2)= .42614
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 4 to 11
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 864 944.3 6.829 6.829
r =5 21623 21743.9 .672 7.501
r =6 77513 77311.8 .524 8.024
p=1-exp(-SUM/2)= .98191
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 5 to 12
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 942 944.3 .006 .006
r =5 21643 21743.9 .468 .474
r =6 77415 77311.8 .138 .612
p=1-exp(-SUM/2)= .26346
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 6 to 13
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 911 944.3 1.174 1.174
r =5 21802 21743.9 .155 1.330
r =6 77287 77311.8 .008 1.338
p=1-exp(-SUM/2)= .48767
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 7 to 14
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 954 944.3 .100 .100
r =5 21830 21743.9 .341 .441
r =6 77216 77311.8 .119 .559
p=1-exp(-SUM/2)= .24394
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 8 to 15
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 916 944.3 .848 .848
r =5 21914 21743.9 1.331 2.179
r =6 77170 77311.8 .260 2.439
p=1-exp(-SUM/2)= .70462
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 9 to 16
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 971 944.3 .755 .755
r =5 21644 21743.9 .459 1.214
r =6 77385 77311.8 .069 1.283
p=1-exp(-SUM/2)= .47354
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 10 to 17
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 997 944.3 2.941 2.941
r =5 21458 21743.9 3.759 6.700
r =6 77545 77311.8 .703 7.404
p=1-exp(-SUM/2)= .97532
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 11 to 18
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21684 21743.9 .165 .207
r =6 77378 77311.8 .057 .264
p=1-exp(-SUM/2)= .12354
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 12 to 19
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 992 944.3 2.409 2.409
r =5 21916 21743.9 1.362 3.772
r =6 77092 77311.8 .625 4.396
p=1-exp(-SUM/2)= .88900
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 13 to 20
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 935 944.3 .092 .092
r =5 21967 21743.9 2.289 2.381
r =6 77098 77311.8 .591 2.972
p=1-exp(-SUM/2)= .77372
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 14 to 21
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 957 944.3 .171 .171
r =5 21553 21743.9 1.676 1.847
r =6 77490 77311.8 .411 2.257
p=1-exp(-SUM/2)= .67656
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 15 to 22
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 921 944.3 .575 .575
r =5 21489 21743.9 2.988 3.563
r =6 77590 77311.8 1.001 4.564
p=1-exp(-SUM/2)= .89793
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 16 to 23
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 949 944.3 .023 .023
r =5 21596 21743.9 1.006 1.029
r =6 77455 77311.8 .265 1.295
p=1-exp(-SUM/2)= .47655
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 17 to 24
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 938 944.3 .042 .042
r =5 21531 21743.9 2.085 2.127
r =6 77531 77311.8 .621 2.748
p=1-exp(-SUM/2)= .74692
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 18 to 25
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 884 944.3 3.851 3.851
r =5 21744 21743.9 .000 3.851
r =6 77372 77311.8 .047 3.898
p=1-exp(-SUM/2)= .85755
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 19 to 26
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 912 944.3 1.105 1.105
r =5 21678 21743.9 .200 1.305
r =6 77410 77311.8 .125 1.429
p=1-exp(-SUM/2)= .51065
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 20 to 27
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 930 944.3 .217 .217
r =5 21719 21743.9 .029 .245
r =6 77351 77311.8 .020 .265
p=1-exp(-SUM/2)= .12409
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 21 to 28
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 926 944.3 .355 .355
r =5 21768 21743.9 .027 .381
r =6 77306 77311.8 .000 .382
p=1-exp(-SUM/2)= .17380
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 22 to 29
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 947 944.3 .008 .008
r =5 21793 21743.9 .111 .119
r =6 77260 77311.8 .035 .153
p=1-exp(-SUM/2)= .07378
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 23 to 30
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 927 944.3 .317 .317
r =5 21900 21743.9 1.121 1.438
r =6 77173 77311.8 .249 1.687
p=1-exp(-SUM/2)= .56976
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 24 to 31
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 911 944.3 1.174 1.174
r =5 21782 21743.9 .067 1.241
r =6 77307 77311.8 .000 1.241
p=1-exp(-SUM/2)= .46244
Rank of a 6x8 binary matrix,
rows formed from eight bits of the RNG pxl_out.rbnf
b-rank test for bits 25 to 32
OBSERVED EXPECTED (O-E)^2/E SUM
r<=4 959 944.3 .229 .229
r =5 21822 21743.9 .281 .509
r =6 77219 77311.8 .111 .621
p=1-exp(-SUM/2)= .26682
TEST SUMMARY, 25 tests on 100,000 random 6x8 matrices
These should be 25 uniform [0,1] random variables:
.076937 .091875 .426136 .981907 .263455
.487674 .243938 .704618 .473537 .975320
.123545 .888999 .773720 .676562 .897929
.476546 .746917 .857555 .510652 .124087
.173801 .073785 .569763 .462444 .266816
brank test summary for pxl_out.rbnf
The KS test for those 25 supposed UNI's yields
KS p-value= .057610
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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: 142375 missing words, 1.09 sigmas from mean, p-value= .86171
tst no 2: 141977 missing words, .16 sigmas from mean, p-value= .56282
tst no 3: 142098 missing words, .44 sigmas from mean, p-value= .67033
tst no 4: 142733 missing words, 1.92 sigmas from mean, p-value= .97285
tst no 5: 142001 missing words, .21 sigmas from mean, p-value= .58480
tst no 6: 142032 missing words, .29 sigmas from mean, p-value= .61280
tst no 7: 142444 missing words, 1.25 sigmas from mean, p-value= .89421
tst no 8: 141918 missing words, .02 sigmas from mean, p-value= .50808
tst no 9: 141131 missing words, -1.82 sigmas from mean, p-value= .03449
tst no 10: 141558 missing words, -.82 sigmas from mean, p-value= .20586
tst no 11: 141643 missing words, -.62 sigmas from mean, p-value= .26688
tst no 12: 141339 missing words, -1.33 sigmas from mean, p-value= .09134
tst no 13: 142243 missing words, .78 sigmas from mean, p-value= .78219
tst no 14: 141973 missing words, .15 sigmas from mean, p-value= .55913
tst no 15: 142963 missing words, 2.46 sigmas from mean, p-value= .99309
tst no 16: 142042 missing words, .31 sigmas from mean, p-value= .62171
tst no 17: 141591 missing words, -.74 sigmas from mean, p-value= .22851
tst no 18: 142126 missing words, .51 sigmas from mean, p-value= .69366
tst no 19: 141795 missing words, -.27 sigmas from mean, p-value= .39469
tst no 20: 141858 missing words, -.12 sigmas from mean, p-value= .45227
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OPSO for pxl_out.rbnf using bits 23 to 32 142170 .899 .8156
OPSO for pxl_out.rbnf using bits 22 to 31 141836 -.253 .4002
OPSO for pxl_out.rbnf using bits 21 to 30 141805 -.360 .3595
OPSO for pxl_out.rbnf using bits 20 to 29 142079 .585 .7208
OPSO for pxl_out.rbnf using bits 19 to 28 142195 .985 .8377
OPSO for pxl_out.rbnf using bits 18 to 27 141534 -1.294 .0978
OPSO for pxl_out.rbnf using bits 17 to 26 142044 .464 .6788
OPSO for pxl_out.rbnf using bits 16 to 25 142158 .857 .8044
OPSO for pxl_out.rbnf using bits 15 to 24 141648 -.901 .1838
OPSO for pxl_out.rbnf using bits 14 to 23 141405 -1.739 .0410
OPSO for pxl_out.rbnf using bits 13 to 22 141582 -1.129 .1295
OPSO for pxl_out.rbnf using bits 12 to 21 142001 .316 .6240
OPSO for pxl_out.rbnf using bits 11 to 20 141936 .092 .5366
OPSO for pxl_out.rbnf using bits 10 to 19 142088 .616 .7311
OPSO for pxl_out.rbnf using bits 9 to 18 141839 -.243 .4042
OPSO for pxl_out.rbnf using bits 8 to 17 141888 -.074 .4707
OPSO for pxl_out.rbnf using bits 7 to 16 142163 .875 .8091
OPSO for pxl_out.rbnf using bits 6 to 15 141597 -1.077 .1407
OPSO for pxl_out.rbnf using bits 5 to 14 141966 .195 .5775
OPSO for pxl_out.rbnf using bits 4 to 13 142254 1.189 .8827
OPSO for pxl_out.rbnf using bits 3 to 12 142377 1.613 .9466
OPSO for pxl_out.rbnf using bits 2 to 11 141916 .023 .5092
OPSO for pxl_out.rbnf using bits 1 to 10 141579 -1.139 .1273
OQSO test for generator pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
OQSO for pxl_out.rbnf using bits 28 to 32 142194 .965 .8327
OQSO for pxl_out.rbnf using bits 27 to 31 142041 .446 .6723
OQSO for pxl_out.rbnf using bits 26 to 30 141826 -.282 .3888
OQSO for pxl_out.rbnf using bits 25 to 29 141467 -1.499 .0669
OQSO for pxl_out.rbnf using bits 24 to 28 141725 -.625 .2660
OQSO for pxl_out.rbnf using bits 23 to 27 142053 .487 .6869
OQSO for pxl_out.rbnf using bits 22 to 26 141978 .233 .5920
OQSO for pxl_out.rbnf using bits 21 to 25 142068 .538 .7047
OQSO for pxl_out.rbnf using bits 20 to 24 142175 .901 .8161
OQSO for pxl_out.rbnf using bits 19 to 23 141648 -.886 .1878
OQSO for pxl_out.rbnf using bits 18 to 22 141993 .284 .6117
OQSO for pxl_out.rbnf using bits 17 to 21 142200 .985 .8378
OQSO for pxl_out.rbnf using bits 16 to 20 142035 .426 .6649
OQSO for pxl_out.rbnf using bits 15 to 19 141482 -1.449 .0737
OQSO for pxl_out.rbnf using bits 14 to 18 141736 -.588 .2784
OQSO for pxl_out.rbnf using bits 13 to 17 142196 .972 .8344
OQSO for pxl_out.rbnf using bits 12 to 16 141593 -1.072 .1418
OQSO for pxl_out.rbnf using bits 11 to 15 141317 -2.008 .0223
OQSO for pxl_out.rbnf using bits 10 to 14 141444 -1.577 .0574
OQSO for pxl_out.rbnf using bits 9 to 13 142662 2.551 .9946
OQSO for pxl_out.rbnf using bits 8 to 12 142596 2.328 .9900
OQSO for pxl_out.rbnf using bits 7 to 11 142165 .867 .8069
OQSO for pxl_out.rbnf using bits 6 to 10 142339 1.457 .9274
OQSO for pxl_out.rbnf using bits 5 to 9 141519 -1.323 .0929
OQSO for pxl_out.rbnf using bits 4 to 8 141804 -.357 .3605
OQSO for pxl_out.rbnf using bits 3 to 7 141734 -.594 .2761
OQSO for pxl_out.rbnf using bits 2 to 6 141803 -.360 .3593
OQSO for pxl_out.rbnf using bits 1 to 5 141243 -2.259 .0119
DNA test for generator pxl_out.rbnf
Output: No. missing words (mw), equiv normal variate (z), p-value (p)
mw z p
DNA for pxl_out.rbnf using bits 31 to 32 145144 9.542 1.0000
DNA for pxl_out.rbnf using bits 30 to 31 143875 5.798 1.0000
DNA for pxl_out.rbnf using bits 29 to 30 142390 1.418 .9219
DNA for pxl_out.rbnf using bits 28 to 29 142277 1.085 .8609
DNA for pxl_out.rbnf using bits 27 to 28 142144 .692 .7556
DNA for pxl_out.rbnf using bits 26 to 27 142371 1.362 .9134
DNA for pxl_out.rbnf using bits 25 to 26 141998 .262 .6032
DNA for pxl_out.rbnf using bits 24 to 25 141948 .114 .5454
DNA for pxl_out.rbnf using bits 23 to 24 141560 -1.030 .1514
DNA for pxl_out.rbnf using bits 22 to 23 141927 .052 .5208
DNA for pxl_out.rbnf using bits 21 to 22 142071 .477 .6833
DNA for pxl_out.rbnf using bits 20 to 21 141631 -.821 .2058
DNA for pxl_out.rbnf using bits 19 to 20 142239 .972 .8346
DNA for pxl_out.rbnf using bits 18 to 19 142113 .601 .7260
DNA for pxl_out.rbnf using bits 17 to 18 142182 .804 .7894
DNA for pxl_out.rbnf using bits 16 to 17 142296 1.141 .8730
DNA for pxl_out.rbnf using bits 15 to 16 141620 -.853 .1967
DNA for pxl_out.rbnf using bits 14 to 15 141755 -.455 .3245
DNA for pxl_out.rbnf using bits 13 to 14 142224 .928 .8234
DNA for pxl_out.rbnf using bits 12 to 13 141150 -2.240 .0125
DNA for pxl_out.rbnf using bits 11 to 12 142648 2.179 .9853
DNA for pxl_out.rbnf using bits 10 to 11 142009 .294 .6156
DNA for pxl_out.rbnf using bits 9 to 10 142114 .604 .7270
DNA for pxl_out.rbnf using bits 8 to 9 142300 1.152 .8754
DNA for pxl_out.rbnf using bits 7 to 8 142361 1.332 .9086
DNA for pxl_out.rbnf using bits 6 to 7 142144 .692 .7556
DNA for pxl_out.rbnf using bits 5 to 6 142193 .837 .7986
DNA for pxl_out.rbnf using bits 4 to 5 141593 -.933 .1754
DNA for pxl_out.rbnf using bits 3 to 4 141863 -.137 .4456
DNA for pxl_out.rbnf using bits 2 to 3 141599 -.915 .1800
DNA for pxl_out.rbnf using bits 1 to 2 141740 -.499 .3087
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
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 pxl_out.rbnf 2503.60 .051 .520298
byte stream for pxl_out.rbnf 2499.89 -.002 .499391
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 2649.09 2.108 .982505
bits 2 to 9 2464.47 -.502 .307691
bits 3 to 10 2508.01 .113 .545073
bits 4 to 11 2459.82 -.568 .284947
bits 5 to 12 2531.40 .444 .671514
bits 6 to 13 2533.45 .473 .681897
bits 7 to 14 2427.66 -1.023 .153138
bits 8 to 15 2495.70 -.061 .475736
bits 9 to 16 2525.30 .358 .639753
bits 10 to 17 2483.34 -.236 .406858
bits 11 to 18 2493.98 -.085 .466052
bits 12 to 19 2525.69 .363 .641815
bits 13 to 20 2491.37 -.122 .451442
bits 14 to 21 2684.34 2.607 .995433
bits 15 to 22 2664.91 2.332 .990153
bits 16 to 23 2526.33 .372 .645175
bits 17 to 24 2549.44 .699 .757781
bits 18 to 25 2493.49 -.092 .463333
bits 19 to 26 2633.46 1.887 .970447
bits 20 to 27 2556.29 .796 .787002
bits 21 to 28 2423.79 -1.078 .140572
bits 22 to 29 2559.75 .845 .800959
bits 23 to 30 2579.56 1.125 .869725
bits 24 to 31 2425.69 -1.051 .146651
bits 25 to 32 2431.76 -.965 .167257
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Of 12,000 tries, the average no. of successes
should be 3523 with sigma=21.9
Successes: 3558 z-score: 1.598 p-value: .944998
Successes: 3522 z-score: -.046 p-value: .481790
Successes: 3533 z-score: .457 p-value: .676028
Successes: 3560 z-score: 1.689 p-value: .954438
Successes: 3548 z-score: 1.142 p-value: .873180
Successes: 3477 z-score: -2.100 p-value: .017844
Successes: 3525 z-score: .091 p-value: .536382
Successes: 3544 z-score: .959 p-value: .831196
Successes: 3504 z-score: -.868 p-value: .192812
Successes: 3528 z-score: .228 p-value: .590298
square size avg. no. parked sample sigma
100. 3529.900 24.058
KSTEST for the above 10: p= .676456
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Sample no. d^2 avg equiv uni
5 .4158 .5669 .341561
10 .6495 .7930 .479413
15 1.4765 .9208 .773248
20 .1684 .8054 .155709
25 .8864 .7717 .589707
30 .1966 .7591 .179272
35 .5151 .7458 .404086
40 .2314 .6891 .207473
45 1.5354 .7787 .786285
50 .3581 .7369 .302228
55 2.1589 .7722 .885796
60 .7537 .7463 .531158
65 .5487 .7249 .423875
70 .3815 .7405 .318455
75 .4195 .7450 .343980
80 .2645 .7724 .233401
85 1.5017 .8094 .778917
90 .3021 .8095 .261850
95 .5573 .7981 .428828
100 .4472 .8611 .361995
MINIMUM DISTANCE TEST for pxl_out.rbnf
Result of KS test on 20 transformed mindist^2's:
p-value= .777127
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
sample no: 1 r^3= 62.595 p-value= .87588
sample no: 2 r^3= 9.175 p-value= .26349
sample no: 3 r^3= .822 p-value= .02704
sample no: 4 r^3= 1.004 p-value= .03291
sample no: 5 r^3= 6.329 p-value= .19021
sample no: 6 r^3= 26.698 p-value= .58931
sample no: 7 r^3= 31.373 p-value= .64857
sample no: 8 r^3= 59.190 p-value= .86096
sample no: 9 r^3= 6.742 p-value= .20128
sample no: 10 r^3= 15.087 p-value= .39522
sample no: 11 r^3= 4.257 p-value= .13229
sample no: 12 r^3= 37.958 p-value= .71783
sample no: 13 r^3= 18.202 p-value= .45487
sample no: 14 r^3= 44.742 p-value= .77494
sample no: 15 r^3= 66.820 p-value= .89218
sample no: 16 r^3= 99.728 p-value= .96400
sample no: 17 r^3= 17.679 p-value= .44528
sample no: 18 r^3= 3.214 p-value= .10161
sample no: 19 r^3= 58.317 p-value= .85686
sample no: 20 r^3= 6.864 p-value= .20450
A KS test is applied to those 20 p-values.
---------------------------------------------------------
3DSPHERES test for file pxl_out.rbnf p-value= .161536
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Table of standardized frequency counts
( (obs-exp)/sqrt(exp) )^2
for j taking values <=6,7,8,...,47,>=48:
-.1 -.3 .1 .2 -.3 .7
.1 .2 -1.7 .8 -.3 .6
-.3 .7 -.8 -.4 2.5 .8
.7 -.7 -3.0 1.0 .2 .6
-.2 .0 -1.5 -.5 -1.5 .2
.1 -1.4 -.1 .4 1.6 -.5
-.7 1.5 -.8 1.5 -1.3 -1.0
-1.1
Chi-square with 42 degrees of freedom: 43.397
z-score= .152 p-value= .588427
______________________________________________________________
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 .471938
Test no. 2 p-value .661696
Test no. 3 p-value .127312
Test no. 4 p-value .138076
Test no. 5 p-value .252853
Test no. 6 p-value .966210
Test no. 7 p-value .981441
Test no. 8 p-value .961647
Test no. 9 p-value .877290
Test no. 10 p-value .974672
Results of the OSUM test for pxl_out.rbnf
KSTEST on the above 10 p-values: .974109
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
Up and down runs in a sample of 10000
_________________________________________________
Run test for pxl_out.rbnf :
runs up; ks test for 10 p's: .137403
runs down; ks test for 10 p's: .458294
Run test for pxl_out.rbnf :
runs up; ks test for 10 p's: .896885
runs down; ks test for 10 p's: .347245
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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:: 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 pxl_out.rbnf
No. of wins: Observed Expected
98448 98585.86
98448= No. of wins, z-score= -.617 pvalue= .26875
Analysis of Throws-per-Game:
Chisq= 25.52 for 20 degrees of freedom, p= .81772
Throws Observed Expected Chisq Sum
1 66332 66666.7 1.680 1.680
2 37909 37654.3 1.723 3.403
3 26940 26954.7 .008 3.411
4 19418 19313.5 .566 3.976
5 13744 13851.4 .833 4.810
6 9937 9943.5 .004 4.814
7 7026 7145.0 1.983 6.797
8 5303 5139.1 5.229 12.026
9 3782 3699.9 1.823 13.849
10 2708 2666.3 .652 14.501
11 1929 1923.3 .017 14.518
12 1352 1388.7 .972 15.490
13 1015 1003.7 .127 15.617
14 708 726.1 .453 16.070
15 529 525.8 .019 16.089
16 373 381.2 .174 16.263
17 244 276.5 3.829 20.092
18 179 200.8 2.373 22.465
19 157 146.0 .831 23.296
20 121 106.2 2.058 25.354
21 294 287.1 .165 25.519
SUMMARY FOR pxl_out.rbnf
p-value for no. of wins: .268753
p-value for throws/game: .817722
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Results of DIEHARD battery of tests sent to file pxl_out.txt