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 [tex]\alpha = 0.01[/tex] 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, there is convincing evidence, and the conditions for inference are met.
B. Yes, there is convincing evidence, but the conditions for inference are not met.
C. No, there is not convincing evidence, and the conditions for inference are met.
D. No, there is not convincing evidence, and the conditions for inference are not met.

To determine if the proportion of rolls that will land 6 spots up is different from one-sixth, we can conduct a hypothesis test using a chi-square goodness-of-fit test. The null hypothesis is that the proportion is equal to one-sixth, and the alternative hypothesis is that the proportion is different from one-sixth.

Explanation:

To determine if the proportion of rolls that will land 6 spots up is different from one-sixth, we can conduct a hypothesis test. The null hypothesis (H0) is that the proportion is equal to one-sixth, and the alternative hypothesis (Ha) is that the proportion is different from one-sixth.

Calculate the expected number of times a 6 will land up if the die is fair. The expected number is (total rolls) * (probability of landing 6). In this case, the expected number is 150 * (1/6) = 25.
Next, we perform a chi-square goodness-of-fit test. We calculate the test statistic, which is χ² = Σ((observed frequency - expected frequency)² / expected frequency). The observed frequency of 6 landing up is 35, and the expected frequency is 25. Calculate the test statistic using these values and the other observed frequencies and expected frequencies.
Finally, we compare the test statistic to the critical value of the chi-square distribution with 5 degrees of freedom (6 categories - 1). If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is convincing evidence that the proportion of rolls that will land 6 spots up is different from one-sixth.

In this case, we would compare the test statistic to the critical value at the 0.01 significance level. If the test statistic is greater than the critical value, we would reject the null hypothesis and conclude that there is convincing evidence that the proportion is different from one-sixth.