High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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.

Answer :

Final answer:

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.