You playa game with your friend everyday that involves flipping a coin 10 times.

Question

You playa game with your friend everyday that involves flipping a coin 10 times. Your friend always provides the coin and, you begin to suspect that they are cheating by using weighted coins. In order to test this, you decide to record the number of heads in these 10 flips each day. Suppose you steal one of your friend's coins. Devise a specific experiment that you could use to test your hypothesis that the coin is unfairly weighted. Specifically, pick some number of times to flip the coin, n, and some criteria for analyzing the results such that there is less than a 0.1% chance of your test leading you to incorrectly accuse your friend of using a weighted coin. i) explain your test ii) If the coin were in fact weighted such that the expected number of heads per flip were 0.8(i.e. on average 8/10 flips were heads), what would be the probability that your test would lead you to correctly conclude that your friend is using a weighted coin?

Explanation / Answer

We will be doing a hypothesis testing , with null hypothesis that the coin is fair at 0.1% significance level. We will have to perform the test by approximating the distribution to a normal distribution and find the Z score at 0.1% significance level and check of the observed output ( no of heads in this case ) were more than the critical z score. Z score at P = 0.001.Since the coin ca be weighted either ways, it will be a two tailed test. with significance level at both ends

Hence the p limits are 0.0005 and 0.9995.

To approximate the distribution which is a binomial distribution, to normal. np and nq should be at least 5. n = no of fips and p and q are probability of heads and tails respectively which will be 0.5 for a fair coin

for np or nq to be at least 5. n has to be at least 10. Lets pick n as 50 to maximize the range. ( I tried with 10 and the range was 0-10 heads in 10 flips, which is not useful)

Mean, mu = np = 50*0.5 = 25

standard deviation , sd = sqrt(npq) =50*0.5*0.5 = sqrt (12.5 ) = 3.536

Hence z score at p = 0.0005 and 0.9995 from normal distribution is = +/- 3.291

Hence no of heads = z*sd + mean = 3.291*12.5 + 25 = 36.64~ 37 heads

Hence if we flip the coin for 50 times and we get more than 37 heads or tails , we can state with 99.9 % confidence that the coin is weighted.

ii) if the coin gives 8 heads in 10 flips : We calculate type II error ( incorrectly accepting the null hypothesis)

and

1 - type II error will be the probability of correctly concuding that a weighted coin is used.

mean for the biased coin for 50 flips = np = 50*0.8 = 40

sd = 50*0.8*0.2 = 8

z score of 37 heads ( critical z) in weighted coin = (37-40)/8 = -3/8 = -0.375

probaility of z of -0.375 = 0.354

Hence type II error 35.4%

Hence the probability that we correctly conclude the coin is weighted in the above experiment = 1-0.354 = 0.646

The probability that we correctly conclude the coin is weighted is 64.6%

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