Why reject, not accept $H_0$ when $p$ value less than $\alpha$? Let's say ABC drug company claims their medicine cures 90% of patients. Now a doctor conducts a trial by taking a random sample of 15 patients. Result shows 11 of them are cured but 4 are not. 
In this case, $H_0$ should be: $p=0.9$. And $H_1$ is: $p<0.9$. If we use $X$ to represent the number of people cured in the sample, then $X \sim B(15,0.9)$. 
If significance level ($\alpha$) is set as 5%, the critical region in this case should be defined as:

$P(X<11)<\alpha$ where $\alpha=5\%$
  , $H_0$ will be rejected if the expression is true.

This really confuses me. The smaller of $P(X<11)$ is, the larger of $P(X>11)$ should be. Then the more patients should be cured and more likely H0 should be accepted. Anything wrong with this?  
To me, it seems to make more sense if $P(X>11) < \alpha$  proves to be true to reject $H_0$. 
 A: You are using your data to set your rejection region. This is incorrect.
If you want to test $H_0: p=0.9$ vs. $H_a: p<0.9$ using an iid sample of 15, then the cutoff is determined as follows:


*

*Let $X_i=1$ if patient $i$ is cured, $0$ o/w

*Assume $X_i \sim Ber(0.9) \implies T:=\sum X_i \sim Bin(15,0.9)$

*We want a lower cutoff, so how few people can be cured and still be acceptable with $p=0.9$? From a binomial probability table, we get:


$$P(T\leq 11)=0.0556, P(T\leq 10) = 0.0127$$
Notice that you can't get your desired $5\%$ significance exactly due to discreteness. 
Therefore, you will reject at the $5.56\%$ level if $T\leq 11$. In this case, it is, so you would reject. There is only approximately a $5\%$ chance that this few people would have been cured had the actual cure rate been $90\%$. 
Now you can see that having more people cured would make you less likely to reject.
A: We know that $$P(\text{Type I Error}) = P(\text{rejecting} \ H_0|H_{0} \ \text{is true})$$
$$ = 1-\sum_{x=x_c}^{15} \binom{15}{x} \left(\frac{9}{10} \right)^{x} \left(\frac{1}{10} \right)^{15-x}$$
For $\alpha = 0.05$, the rejection region is $\{x: x \leq 11 \}$. The p-value is the probability that we would get a result as extreme or more extreme than we did if the null hypothesis is true. This is:
$$\text{p-value} = 1-\sum_{x=11}^{15} \binom{15}{x} \left(\frac{9}{10} \right)^{x} \left(\frac{1}{10} \right)^{15-x} \approx 0.01$$
