In: Statistics and Probability
Answer: A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 28, 27, 40, 38, 26, 41. Use a 0.01 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?
Solution:
The hypothesis test:
Ho: Outcomes are equally likely.
Ha: Outcomes aren't equally likely.
Test statistic χ2:
χ2 = (Observed value - Expected value)^2/Expected value
Outcomes | Observed (O) | Expected (E) | (O-E)^2/E |
1 | 28 | 200/6 = 33.3333 | 0.8533 |
2 | 27 | 33.3333 | 1.2033 |
3 | 40 | 33.3333 | 1.3333 |
4 | 38 | 33.3333 | 0.6533 |
5 | 26 | 33.3333 | 1.6133 |
6 | 41 | 33.3333 | 1.7634 |
Total | 200 | 200 | 7.4200 |
Therefore,
Test statistic χ2 = 7.420
Critical value of χ2:
df = k-1 = 6-1 = 5
Critical value of χ2(df, α) = 15.086
Since, test statistic χ2 (7.420) < critical value (15.086)
We fail to reject null hypothesis Ho.
Conclusion:
Do not reject Ho . There is not sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes appears to be equally likely, so the loaded die does not appear to behave differently from a fair die.
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