Logic of Hypothesis Testing I know that there is something wrong in the following reasoning but I cannot explain it, can you say what is wrong with the following reasoning:


*

*Null Hypothesis: The coin is fair (that is, the probability of a head is 0.5).

*The coin is thrown 100 times and 55 times it comes up heads.

*The probability of getting 55 heads with a fair coin is 0.048 

*Since 0.048 < 0.05 we can reject the null hypothesis.


I know that we should find P(X>=55) which is 0.18 but I really don't understand why are we finding P(X>=55).
Thanks
 A: The p-value is the probability of getting your observed result or a result more extreme. In your case, the more extreme cases are 56 heads, 57 heads, etc.
The probability of your particular observation can be quite low, but the p-value may be high because of the inclusion of many more extreme events. This is what happened in your case.
A: It seems you have modified your question. For earlier question, @Dave's answer was correct. Let me try to answer your new question: 

I know that we should find P(X>=55) which is 0.18 but I really don't understand why are we finding P(X>=55).

We are not calculating this. But this helps to get the p-value. You can ignore this calculation in fact. Formally, you have defined your random variable, $X$, to be number of heads. Clearly, X \sim Bin(100,p)$
$H_0: p=0.5$
$H_a: p \neq 0.5$
From sample, you calculate, $\hat{p}=0.55$. Although, exact distribution of $\hat{p}$ can also be obtained, given large sample size, we can use CLT.
$\hat{p}$ is the mean of 100 random variables each following $Bern(0.5)$ under null. Since the variance of $Bern(0.5)$ is $0.25$, we get from CLT:
$\sqrt{100}(\hat{p}-0.5)\xrightarrow{d} N(0,0.25)$
$ \implies 20(\hat{p}-0.5) \xrightarrow{d} N(0,1)$
So your critical values become: $0.5 \pm 1.96/20$, i.e. $0.598$ and $0.402$. Since you are getting $\hat{p} < 0.598$ you cannot reject the null that coin is fair. 
So as such we don't need to calculate P(X>=0.55), but we can see that if indeed, P(X>=55)>0.025 (assuming two-sided test), then we cannot reject the null. Because, then sample estimate falls in the 95% confidence zone of the null hypothesis. So this is just an alternate way.
Hope it is clear. 
