Is power always associated with hypothesis testing? Suppose I know that the true population proportion of a mutation is p = 0.3493119. I want to know that given power = 0.8, what's the proportion of of mutation in my sample of n = 30? Here's what I have so far: 
1 - cdf((x-p)/sqrt(p*(1-p)/30) = 0.8
Since p = 0.3493119, I can solve for x and I get x = 0.276055. So does this mean that the probability of having 27.6% of the sample be mutated = 0.8? Is it correct to say that?
From my experience I know that power is the probability of correctly rejecting the null. But since I already know the true population proportion of the mutation, is it still necessary to conduct a hypothesis test? I am leaning towards the negative, but without a hypothesis test, how can I calculate the power? 
If I had to perform a hypothesis test to get the power...would it be something like this:
H0: p = 0.5 vs. H1: p > 0.5
power = P(Z_statistic > critical_value | p > 0.5)
power = 1 - normal_CDF(critical_value | p > 0.5)
Now I'm confused by how to deal with the p > 0.5, if it were just p = 0.5, I could've standardized the critical value so that it follows a N(0, 1) distribution. 
Overall I think I just want to know the answer to the question, given power = 0.8, what is the proportion of mutation in my sample of n = 30? (true population proportion = 0.3493119). 
 A: I would say "Yes"power is alway associated with hypothesis testing it relates to  sample size, type I error and both H1 and H0. 
The followings are R code to show the power curve for OP's cases (one tailed at 0.05 level, by approximate a normal distribution. 
mu<-30*0.35
sd<-sqrt(30*0.35*(1-0.35))
c<-qnorm(0.95,mu,sd)
mus<-seq(10,30,1)
power = 1-pnorm(c, mus, sd)
plot(mus, power, type="l")


Two tailed case by approximating normal distribution
mu<-30*0.35
sd<-sqrt(30*0.35*(1-0.35))
mu
sd
c1<-qnorm(0.025,mu,sd)
c2<-qnorm(0.975,mu,sd)
mus<-seq(0,30,1)
power2 <- pnorm(c1,mus,sd)+1-pnorm(c2,mus,sd)
plot(mus, power2, type="l")


A: This is for binomail distribution not by approximation.
#under binomail distribution, one tail,
qbinom(0.05,30,0.35) #critical rigion
n<-seq(6,30,1) #numbers bigger than critical region
pow<-pbinom(n,30,0.35)#one tailed test
plot(n, pow, type="l")


A: Your question is not a question about power.  Assuming a given population mean, you are interested in predicting the sample mean. If you have a binomial distribution, you can find the density distribution centered over your population mean. 
Power would require specification of a criterion value for deviations of sample means from the population mean.  
