Hypotheses:
Ho: There are equal number of suits in the stack of cards.
Ha: Atleast one suit is having different frequency than others.
Test statistic:
Expected frequency of each cell = 36/4 = 9
Hence,
= 2.89
Degrees of freedom = 4 - 1 = 3
So Critical value at 0.05 significance level = 7.815
Since test statistic < Critical value, we do not reject the null hypothesis. Hence,
We do not have sufficient evidence to reject the claim that equal number of each suits are in the stack of cards.
Alt 7. (a) (15) A sample of 36 cards are drawn with replacement from a stack...
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