A teacher attempts to make an unfair number cube by baking it for 5 minutes at 300°F. A fair number cube lands with 6 spots up one-sixth of the time. She wants to determine if she successfully made the number cube unfair. She rolls the number cube 150 times and finds that 6 lands up 35 times.

Do these data provide convincing evidence at the \(\alpha = 0.01\) significance level that the proportion of rolls that will land 6 spots up is different from one-sixth? Are the conditions for inference met?

- Random: We have a random sample.
- 10% Condition: \(150 < 10\%\) of the population.
- Large Counts: There are expected successes and expected failures, which are both at least 10.

A) Yes, the data provide convincing evidence; the conditions for inference are met.
B) Yes, the data provide convincing evidence; the conditions for inference are not met.
C) No, the data do not provide convincing evidence; the conditions for inference are met.
D) No, the data do not provide convincing evidence; the conditions for inference are not met.

If the die's observed data has a corresponding p-value less than the 0.01 significance level after a goodness-of-fit hypothesis test, and the conditions for inference are met, then there is convincing evidence that the die is unfair. The correct answer is A).

Explanation:

To determine if the number cube (die) is unfair, we use a goodness-of-fit hypothesis test. The null hypothesis (H0) is that the die is fair, meaning the probability of rolling a six is 1/6, and the alternative hypothesis (Ha) is that the die is not fair, meaning the probability of rolling a six is not 1/6.

For this test, the expected number of times a six should come up in 150 rolls, if the die is fair, is 150 * (1/6) = 25. The observed frequency of rolling a six is 35. The conditions for inference are met:

Randomness: We assume that the rolls are random.

10% condition: 150 rolls are less than 10% of all possible rolls, which could be considered infinite.

Large counts condition: The expected number of successes (expected sixes) and failures (other numbers) are both greater than 5, which is necessary for the approximation to the chi-square distribution to be valid.

We then calculate the chi-square statistic and corresponding p-value to compare to the significance level of 0.01. If the p-value is less than 0.01, we reject H0 and accept Ha, thereby concluding that the die is unfair.

Assuming the calculated p-value is less than 0.01, so, Yes, the data provide convincing evidence; the conditions for inference are met.