Question

Suppose Nathan, an avid baseball card collector, is interested in studying the proportions of common, uncommon, and rare baseball cards found in newly purchased card packs. Each card pack contains exactly 10 baseball cards. The fine print on each I pack of cards says that, on average, 75% of the cards in each pack are common, 15% are uncommon, and 10% are rare. Nathan wishes to test the validity of this claimed distribution, so he randomly selects 20 packs of baseball cards and looks at the rarity of each of the 200 total cards. To determine if the distribution of rarity levels in his sample is significantly different than the distribution claimed on the card packs, Nathan decides to perform a chi-square test for goodness-of-fit. His results are shown in the table.

Answer 1

The p-value for Nathan's chi-square test for goodness-of-fit is approximately 0.493, indicating no significant difference between observed and expected rarity distributions in the baseball card packs.

To find the p-value for Nathan's chi-square test for goodness-of-fit, you can use a chi-square distribution table or a statistical software package. The formula for the p-value is:

`\[ p \text{-value} = P(\chi^2 \geq \text{chi-square statistic}) \]`

Given that Nathan's chi-square statistic is 1.4143 with 2 degrees of freedom, you would look up this value in a chi-square distribution table or use statistical software to find the corresponding p-value.

Using a chi-square distribution table, the p-value for a chi-square statistic of 1.4143 with 2 degrees of freedom is approximately 0.493.

Therefore, the p-value for Nathan's chi-square test for goodness-of-fit is approximately 0.493 (rounded to three decimal places). This p-value suggests that there is no significant difference between the observed and expected distributions of rarity levels in the baseball card packs.

The complete question is:
(attached)