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 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.

Rarity	Observed	Test proportion	Expected	Contribution to chi-square
Common	145	0.75	150	0.167
Uncommon	32	0.15	30	0.133
Rare	23	0.10	20	0.450

Chi-square statistic:0.7500

Degrees of freedom:2

If you wish, you may download the data in your preferred format.  

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What is the ?‑value for Nathan's chi-square test for goodness‑of‑fit? Round your answer to three decimal places.

?=

Based on the ?‑value for his test and assuming a significance level of ?=0.05, what should Nathan conclude about the claimed distribution of baseball cards?

a) Because ?<?, there is significant evidence against the claimed distribution of baseball cards.

b) Because ?>?, there is significant evidence against the claimed distribution of baseball cards.

c) Because ?<?, there is insufficient evidence against the claimed distribution of baseball cards.

d) Because ?>?, there is insufficient evidence against the claimed distribution of baseball cards.

Answer 1

p value using excel function CHISQ.DIST(x,df,true)

setting x = chi square statistics = 0.75 and df= degree of freedom = 2

this implies

p value = 1-CHISQ.DIST(x,df,true)

= 1-CHISQ.DIST(0.75,2,true)

= 1 - 0.313

= 0.687

(B) It is clear that p value is greater than alpha level 0.05, failing to reject the null hypothesis.

So, we can conclude that there is no significant difference than the distribution claimed on the card packs

therefore, option D is correct.