Binomial hypothesis test - Coin flip A coin was flipped 10 times and landed on head 9 times. We want to test $$H_0: p = \frac{1}{2} \ vs \ H_1: p > \frac{1}{2}, p = \mathbb{P}(head)$$ with a confidence level of $\alpha = 0.05$. The question specifically ask for:

*

*The test statistic

*Distribution under $H_0$

*The conclusion of the test and the $p$-value.

My attempt:
I don't know if the test statistic is given by $\hat{p} = 9/10 = 0.9$ or $$\mathbb{P}(X = 9) = \binom{10}{9} 0.5^9 \cdot 0.5^1 \approx 0.00977.$$
The distribution under the null hypothesis is $X \sim binomial(10, 1/2)$ and with a 0.05 signficance level, our conclusion would be to reject the null hypothesis. I think that the $p$-value would be $$p = \binom{10}{10} \frac{1}{2^{11}}$$ but i'm really confused about this question.

The test statistic and the sample mean are not the same thing. Recall the test statistic is what we compare to a known distribution to obtain inference - parts 1 and 2 of the question are essentially the same. The derivation is simple, or you can outline it if this a graduate class, or else (elementary non-calculus stats) you probably have covered tests of proportion in which case you just need to reference the correct formula.
Question 3 is about applying what you know of hypothesis tests. Having found the distribution under the null, you can find the area in the tail of that distribution (again using calculus or a lookup table) that's indexed by your test statistic.
