Suppose two coins $C_1$ and $C_2$ are tossed and suppose further that $C_1$ was tossed $40$ times and $C_2$ only $20$ times. The outcome for $C_1$ is $8$ times heads and $32$ times tails. For $C_2$ we have once head and $19$ times tails. My question is, what test do I use in order to try to show that the probabilities for having heads are significantly different for $C_1$ and $C_2$?

In addition I'd be interested if anyone could explain the following confusion to me. If I estimate the probability for heads of $C_1$ by $p_1=1/5$ then the probability of the outcome of $C_2$ under the hypothesis that the probability is $1/5$ as well equals $$ \binom{20}{1}\frac{1}{5}\left(\frac{4}{5}\right)^{19}\approx 5.7\% $$ which is not significant (I would like to have a significance level of $<5\%$). On the other hand If I reverse the situation I obtain for the probability of the outcome of $C_1$ assuming a probability for heads of $p_2=1/20$ $$ \binom{40}{8}\frac{1}{20^8}\left(\frac{19}{20}\right)^{32}\approx 0.06\% $$ which is by far smaller than $5\%$. So one test tells me the probabilities are significantly different, while the other doesn't (this is the source of confusion) but in fact I would just like to know how I can tell that $p_1\ne p_2$ significantly.


marked as duplicate by whuber Apr 2 '18 at 14:38

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  • $\begingroup$ Re the second part of the question: neither of your tests is standard or even any good. You might want to review the concepts of hypothesis tests and p-values to see what the problems are. stats.stackexchange.com/questions/31 is one place to start. $\endgroup$ – whuber Apr 1 '18 at 23:10
  • $\begingroup$ Thanks for the comment. I see that non of my computations (second part of my question) lead somewhere (since they are not symmetric). But I guess there is a standard test that applies to the situation described (since it is so simple) but I'm not familiar with tests so maybe you can tell me what test is useful for such a situation. $\endgroup$ – frog Apr 2 '18 at 7:47
  • $\begingroup$ See stats.stackexchange.com/search?q=binomial+proportion+difference. $\endgroup$ – whuber Apr 2 '18 at 14:35