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Pseudo-random; Kolmogorov-Smirnov numerical

Short answer

Pseudo-random numbers are numbers generated by a deterministic algorithm that mimic true random numbers, reproducible from a seed, uniformly distributed on [0,1) and passing statistical tests. For the K-S test, D = 0.17, less than critical value 0.410 (α=0.05, N=10), so the uniformity hypothesis cannot be rejected.

Define pseudo random numbers. The following numbers have been generated 0.44, 0.19, 0.88, 0.27, 0.55, 0.13, 0.63, 0.74, 0.11 and 0.33. Use the Kolmogorov-Smirnov test with α = 0.05 to determine, if the hypothesis that the numbers are uniformly distributed on the interval [0, 1] can be rejected. (Note that the critical value of D for α 0.05 and N = 10 is 0.410.)

Pseudo-random numbers

Pseudo-random numbers are numbers generated by a deterministic algorithm (such as the linear congruential method Xn+1 = (aXn + c) mod m) that imitate true random numbers. They are designed to be uniformly distributed on [0, 1) and to pass statistical tests of uniformity and independence, but the sequence can be reproduced from the seed and eventually cycles.

K-S test

H0: the numbers are uniformly distributed on [0, 1]. H1: they are not. N = 10, α = 0.05.

Rank the numbers and compute the differences:

iR(i)i/Ni/N − R(i)R(i) − (i−1)/N
10.110.1-0.010.11
20.130.20.070.03
30.190.30.11-0.01
40.270.40.13-0.03
50.330.50.17-0.07
60.440.60.16-0.06
70.550.70.15-0.05
80.630.80.17-0.07
90.740.90.16-0.06
100.881.00.12-0.02

D⁺ = max(i/N − R(i)) = 0.17 (at i = 5 and i = 8) D⁻ = max(R(i) − (i−1)/N) = 0.11 (at i = 1)

D = max(D⁺, D⁻) = 0.17

Critical value D0.05 = 0.410 (N = 10).

Conclusion. Since D = 0.17 < 0.410, the hypothesis of uniformity cannot be rejected. The numbers can be treated as uniformly distributed on [0, 1].

✓ Answer checked by Claude Opus 5.5 (high effort)

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