Question

stion 2 of 10> Suppose Nathan, an avid baseball card collector, is interested in studying the proportions of common, uncommon, and rar baseball cards found in newly purchased card packs. Each card pack contains exactly 10 baseball cards. The fine print on each pack of cards says that, on average, 75% of the cards in each pack are common, 20% are uncommon, and 5% 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. Rarity Observed Test proportion Expected Contribution to chi-square Common Uncommon 148 0.75 0.20 0.05 150 40 10 0.027 0.025 0.100 Rare Chi-square statistic:O.151 Degrees of freedom: 2 If you wish, you may download the data in your preferred format Crunchilt CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc What is the p-value for Nathan's chi-square test for goodness-of-fit? Round your answer to three decimal places. Based on the pvalue for his test and assuming a significance level of α-005, what should Nathan conclude about the claimed distribution of bascball cards? 0.75 0.20 0.05 150 40 10 0.027 0.025 0.100 Uncommon 4 Rare Chi-square statistic:0.1517 Degrees of freedom: 2 If you wish, you may download the data in your preferred format. Crunchlt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc What is the p-value for Nathan's chi-square test for goodness-of-fit? Round your answer to three decimal places. Based on the p-value for his test and assuming a significance level of α-0.05, what should Nathan conclude about the claimed distribution of baseball cards? O Because p> a, there is significant evidence against the claimed distribution of baseball cards. O Because p

Answer 1

Given,

to test, for the chi-square goodness of fit of the data,
Chi-Square statistic = 0.1517
Degrees of freedom = 2

Under the null hypothesis, the test statistic follows a $\chi ^2_2$ distribution,

Now, p-value is the upper quantile of the test statistic value, based on the given data, under the null hypothesis, and is obtained as,

Pa(X > 0.1517) = 0.927

Now since the p-value is greater than , there is insufficient evidence against the claimed distribution of baseball cards.

a-0.05