For a coin, there is no information whether it is fair or not. The following two hypothesis are supposed for getting tail : $H_0 : p = 0,5$ and $H_1 : p = 0,7$. This coin is tossed $10^4$ times and if the number of tails that appeared at the top is more than $4940$, $H_0$ is accepted. Find $\alpha $(type 1 error).
I do not understand why $H_0 $ is accepted,since we are in the non-critical region,then we can easily find $\beta$ but not $\alpha$
$\alpha=P(H_1:\text{accepted}|H_0 : true)=\frac{P(H_1\cap H_0)}{P(H_0:\text{true})}$
$P(H_0:\text{true})=\frac{4940}{10000}=0.494$ i dont know how to proceed from here...
The test statistic that you wrote is not consistent with the hypotheses you stated. If $H_0 : p = 0.5$ and $H_1 : p = 0.7$, and $p$ is the single trial probability of flipping a tail, then the test should reject $H_0$ if more than a certain number of tails are observed. What you have written is that $H_0$ is accepted if more than $4940$ tails are observed.
Okay, so say I take your question exactly as it is stated. Then your type I error is the conditional probability of rejecting $H_0$ when $H_0$ is true; i.e., $p = 0.5$ but you observed more than $4940$ tails out of $10^4$ trials. This is a binomial probability with $n = 10^4$, $p = 0.5$: $$\alpha = \sum_{x=4941}^{10000} \binom{10000}{x} (0.5)^x (1 - 0.5)^{10000-x}.$$ Of course, we can approximate this sum using a normal approximation: the binomial distribution can be approximated with a normal distribution with mean $np = 5000$ and variance $np(1-p) = 2500$, so $$\alpha \approx \Pr[4940.5 \le X] = \Pr[-1.19 \le Z] = 0.882977.$$ This is your probability of Type I error given the EXACT statement of your question. But as I said earlier, this makes no sense because your test statistic is not consistent with your hypotheses. You are basically saying that the more tails you observe, the more likely you will conclude the coin is fair. That is why the Type I error we calculated above is so large.