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Chi-square uniformity test, 40 numbers

Short answer

Divide the range 1‑59 into 10 equal classes, expected count Ei=4. Observed frequencies yield χ² = Σ (Oi−Ei)²/Ei = 2.00. With df=9 and α=0.10, critical value χ²0.10,9 = 14.684. Since 2.00 < 14.684, H0 is not rejected; the numbers can be accepted as uniform.

Using Chi-Square test the uniformity at 90% for the given random numbers. Degree of freedom for 6 = 10.645, 7 = 12.017, 8 = 13.362, 9 = 14.684, 10 = 15.987.

2034434214103317611
15164135229463757
5149402759544194155
53293314885628127

Setup. H0: the numbers are uniformly distributed. H1: they are not. The 40 numbers lie between 1 and 59, so the range 1–60 is divided into 10 equal classes of width 6. The number of classes is n = 10, N = 40, so the expected frequency of each class is Ei = N/n = 40/10 = 4.

The test statistic is χ0² = Σ (Oi − Ei)²/Ei.

Counting the observed frequencies

ClassOiEiOi − Ei(Oi − Ei)²/Ei
1–6541.00.250
7–12642.01.000
13–18440.00.000
19–2434-1.00.250
25–3034-1.00.250
31–36440.00.000
37–42440.00.000
43–48440.00.000
49–5434-1.00.250
55–60440.00.000

Σ Oi = 40 (all 40 numbers counted).

χ0² = 2.00

Critical value. Confidence 90% means α = 0.10. Degrees of freedom = n − 1 = 10 − 1 = 9. From the values given, χ²0.10,9 = 14.684.

Conclusion. Since χ0² = 2.0 < 14.684, H0 is not rejected. At the 90% level the numbers can be accepted as uniformly distributed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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