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Continuation to the above question, if each coin is tossed 10 times (100 tosses are made in total). Will you modify your approach to the test the fairness of the coin or continue with the same?
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Assuming all ten coins are of the same mint, we find P(average of head-counts for the 10 coins is extreme). To do that we need to know the distribution of this average. It can be calculated explicitly with a bit of probability. In fact, the sum of (independent) Binomials with the same probability of success is another Binomial.
And again, if the p-value is too small, we reject the null hypothesis of P(heads)=1/2.
by Diamond (60,546 points)

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