Question about p-value for one-sided hypothesis testing Is it true that the p-value is symmetric?
Let's say for a null hypothesis $H_{0}$: p = 0.75. Given that the z-score test statistics equals 1.44. Using the R command, I found out that
pnorm(1.44, lower=F) 
## [1] 0.0749337

I want to test $H_{\alpha}: p > 0.75$. I concluded that p-value is equal to 0.0749337 because by definition, assume H0 is true, then the p-value is the probability that the test statistic z takes a value (in support of $H_{\alpha}$) as or more extreme than the one we observed. For a two-sided hypothesis testing, $H_{0}: p \neq 0.75$, the p-value would be 2*0.0749337. 
Now, my quesiton is: can I claim that the p-value for $H_{\alpha}: p < 0.75$ is also 0.0749337?
Thank you!
 A: If you are doing an approximate normal test of $H_0: p = 3/4$ vs $H_a: p \ne 3/4,$ rejecting when $|Z_{\text{obs}}| > 1.96,$ where
$$Z_{\text{obs}} = \frac{\hat p = 3/4}{\sqrt{\frac{(3/4)[1-(3/4)]}{n}}},$$
then the two-sided P-value is $P(|Z| > Z_{\text{obs}}),$ where $Z \sim \mathsf{Norm}(0,1).$
So if $Z_{\text{obs}} = 1.44,$ then the P-value is $P(Z \le -1.44) + P(Z\ge 1.44) = 2P(Z \ge 1.44) \approx 0.15.$
However, if you do an exact 2-sided binomial test, then the null distribution of
the number $X$ of successes is $\mathsf{Binom(n, 3/4)},$ which is not a
symmetrical distribution. Opinions vary how to find the P-value in that case. Some authors and software packages interpret "more extreme" than $X_{\text{obs}}$ to mean farther in absolute value from the mean $np_0 = n(3/4)$
than $X_{\text{obs}}$ in either direction. Then finding the P-value may involve
adding two somewhat different tail probabilities, and I suppose this would not match your idea of 'symmetry'. Some simply double the probability from the 'observed tail`. 
Note: I can't illustrate the exact binomial test for your case, because you have
not given the values of $X_{\text{obs}}$ and $n.$ However, since you have R,
you can look at the help screen at ? binom.test where the example shows
the same null and alternative hypotheses as yours with $X = 682$ and $n = 925.$
The P-value for the exact binomial test is given as 0.3825. With the
approximate normal test, I obtain a P-value $\approx 2(.186) = 0.372.$
