Symmetry test for a coin.

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We throw a coin 100 times. We have two hypotheses $H_0$: the coin is symmetrical and $H_1$: probability of head is equal to $\frac{6}{10}$. We used statistics that returned the number of head. Let $\alpha=0.1$. I need to construct a critical set and calculate the power of the test.

My attempt:

Critical set:

Our test: $T(x)=X-50$

$P(X-50>c)\approx 0.1 \Rightarrow \Phi(\frac{c}{5})\approx0.9 \Rightarrow c\approx 6.45$. So, $K=\{T(x)>6.45\}$ is a critical set. Now we will calculate $\beta=P(T(x)>6.45\mid p=\frac{6}{10})=P(X>56.45)=1-\Phi(\frac{56.45-60}{\sqrt(24)})\approx 0.6$. Power of test is equal to $1-\beta$.

I am not sure about my solution.

How to calculate p-value if we get 44 heads?

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If you assume binomial distribution, you could just use $T=\{57,\ldots 100\}$ since $\alpha=P(X \geq 57) \approx 9.7\% < 10\%$.

This yields $1-\beta=P(X \geq 57)\approx76.3\%$.

For the p-value you assume $T=\{0,1,2,\ldots,43\}$ and you would get a p-value of $P(X \geq 44) \approx 90.3\%$.