Asked by zoie
Isabella is studying the fairness of a six-sided numbered cube with numbers and .
The numbered cube is rolled times, and the numbers on the top side are recorded in the table below.
1,3,2,5,3,1 4,3,5,6,4,2 1,5,4,1,3,5 3,6,6,5,6,2 6,1,3,2,4,5 2,6,2,6,1,3
Based on this data, what conclusion would you make about the fairness of the numbered cube?
Justify your answer.
All Answers 1
Answered by
GPT-5 mini
AI
First count how often each face (1–6) appeared in the 36 rolls:
- 1: 6
- 2: 6
- 3: 7
- 4: 4
- 5: 6
- 6: 7
If the cube is fair we expect 36/6 = 6 occurrences of each face. Use a chi-square goodness-of-fit test:
chi-square = sum[(observed − expected)^2 / expected]
= (6−6)^2/6 + (6−6)^2/6 + (7−6)^2/6 + (4−6)^2/6 + (6−6)^2/6 + (7−6)^2/6
= 0 + 0 + 1/6 + 4/6 + 0 + 1/6 = 1.0.
For df = 5 the 5% critical chi-square ≈ 11.07, and 1.0 ≪ 11.07, so we do not reject the hypothesis of equal probabilities. The observed deviations from 6 are very small and easily attributable to chance.
Conclusion: There is no convincing evidence from these 36 rolls that the cube is unfair.
- 1: 6
- 2: 6
- 3: 7
- 4: 4
- 5: 6
- 6: 7
If the cube is fair we expect 36/6 = 6 occurrences of each face. Use a chi-square goodness-of-fit test:
chi-square = sum[(observed − expected)^2 / expected]
= (6−6)^2/6 + (6−6)^2/6 + (7−6)^2/6 + (4−6)^2/6 + (6−6)^2/6 + (7−6)^2/6
= 0 + 0 + 1/6 + 4/6 + 0 + 1/6 = 1.0.
For df = 5 the 5% critical chi-square ≈ 11.07, and 1.0 ≪ 11.07, so we do not reject the hypothesis of equal probabilities. The observed deviations from 6 are very small and easily attributable to chance.
Conclusion: There is no convincing evidence from these 36 rolls that the cube is unfair.
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