SOLUTION: A playing-card drawing machine is supposed to draw playing cards at random, as if from an infinite deck of cards. In a draw, you counted 3200 cards, and observed the following: S

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Question 1202767: A playing-card drawing machine is supposed to draw playing cards at random, as if from an infinite deck of cards. In a draw, you counted 3200 cards, and observed the following:
Spades 808
Hearts 840
Diamonds 800
Clubs 752
Could it be that the draws are equally likely? Or are these discrepancies too much to be random? Test the hypothesis at  = 0.05.
Answer by jekoishun(6) About Me  (Show Source):
You can put this solution on YOUR website!
In this case, a chi-square goodness-of-fit test is an appropriate method to determine if the observed frequencies significantly deviate from the expected frequencies.
The null hypothesis (H0) is that the draw of each card suit (Spades, Hearts, Diamonds, Clubs) is equally likely. The alternative hypothesis (H1) is that the draw is not equally likely for each suit.
If the drawing is perfectly random, we would expect each suit to appear 1/4 of the time, since there are 4 suits.
Expected Frequencies Calculation: With a total of 3200 cards, we would expect each suit to occur 3200 / 4 = 800 times if the draws are indeed equally likely.
Chi-square statistic Calculation: For each suit, calculate the value: (Observed - Expected)^2 / Expected.
Let's do the calculation:
For Spades: (808 - 800)^2 / 800 = 0.08
For Hearts: (840 - 800)^2 / 800 = 2.0
For Diamonds: (800 - 800)^2 / 800 = 0
For Clubs: (752 - 800)^2 / 800 = 2.88
Summing these values together gives the Chi-square statistic: χ^2 = 0.08 + 2.0 + 0 + 2.88 = 4.96
Degrees of Freedom and P-Value Calculation: The degrees of freedom for this test are number of categories - 1, so here it would be 4 - 1 = 3.
Look up the p-value corresponding to χ^2 = 4.96 with df = 3 in a Chi-square distribution table or use statistical software to calculate it. As of my knowledge cutoff in September 2021, this value is roughly between 0.1 and 0.2.
Since this p-value is larger than our significance level (α = 0.05), we fail to reject the null hypothesis. This means that the observed discrepancies in the draw frequencies for each suit are not statistically significant at the 0.05 level, and could be due to random chance. Thus, it is plausible that the draws are equally likely for each suit.