Formulation of hypothesis and alternative. Your desired test is not completely specified until you state
both the null and alternative hypotheses. It seems you have $n = 100$
independent tosses of a coin with $P(\text{Head}) = \theta.$
Perhaps your null hypothesis is $H_0: \theta = 1/2$ (fair coin),
to be tested against the alternative $H_a: \theta > 1/2$ (biased
in favor of Heads).
If the null hypothesis is true, then the expected number of heads in 100 tosses is 50, but you have observed 70. The question is whether
70 is so extremely greater than 50 that the null hypothesis should
be rejected.
Exact P-value. The P-value in this this case is the probability
of seeing a result as extreme or more extreme than the 70 Heads actually. This probability is computed under the the assumption
that $H_0$ is true (fair coin).
The exact probability that $P(X \ge 70),$ where $X \sim Binom(100, 1/2),$ can be found by software. In R, the computation can be
done as follows:
1 - pbinom(69, 100, .5)
## 3.92507e-05
Thus the P-value is very small: 0.000039. One ordinarily rejects
the null hypothesis $H_0$ against $H_a$ if the P-value is smaller
than 0.05. Essentially, you have a choice of what to believe:
either (a) the coin is fair and something very unlikely happened
when you tossed it 100 times, or (b) the coin is not fair and
that explains the extremely large number 70 of heads observed.
The usual statistical judgment is the believe (b).
Normal approximation of P-value. Because $n = 100$ is a 'large' number of tosses, one could use
a normal approximation to get the P-value, as follows:
For $X \sim Binom(100, 1/2),$ we have $m = E(X) = 50$ and
$\sigma^2 = V(X) = 25$ so that $\sigma = SD(X) = 5.$ Then
$$P(X \ge 70) \approx P\{Z = (X - 50)/5 \ge (70 - 50)/5 = 4\},$$
where $Z \sim Norm(0,1)$. From normal tables one can see
that $P(X \ge 70) \approx P(Z \ge 4) \approx 0.\;$ So the P-value is again found to
be extremely small, and $H_0$ is rejected in favor of the
alternative $H_a.$
Required for power. You also ask about the power of this test. That requires you to
select a particular alternative value $\theta_a.$ Perhaps you want to know
the probability of rejecting $H_0,$ given that $P(\text{Heads}) = \theta_a = 0.60.$ In turn, that requires you to specify
the significance level of your test, so that we know for which
values of $X$ we will reject $H_0.$ If you supply the significance
level $\alpha$ (perhaps 5%) and the alternative value $\theta_a$ you have
in mind, then I (or someone else) can answer your question about
the power of the test.
Addendum on power. You say you want the significance level
to be $\alpha = 0.05.$ However, because the binomial distribution
is discrete it is not possible to have exactly $\alpha = 0.05.$
$P(X \ge 59|\theta=1/2) = 0.0443$ and
$P(X \ge 58|\theta=1/2) = 0.0666,$ so the closest we can get to
$\alpha = 0.05$ is to use $\{X \ge 59\}$ as the rejection region
and have $\alpha = 0.443.$
1 - pbinom(58, 100, .5)
## 0.04431304
1 - pbinom(57, 100, .5)
## 0.06660531
sum(dbinom(59:100, 100, .5))
## 0.04431304
sum(dbinom(58:100, 100, .5))
## 0.06660531
Then the 'power function' for various values of $\theta_a$ is
found by evaluating $P(X \ge 59 | \theta_a)$ for various
values of $\theta_a$. For example, the power against alternative
$\theta_a = .60$ is
$$P(\text{Reject } H_0 |\theta_a = .6) = P(X \ge 59 | \theta_a = .6) = 0.623.$$
This is computed in R using the binomial CDF 'pbinom'.
1 - pbinom(58, 100, .60)
## 0.6225327
The left-hand panel of the figure below shows the null distribution (fair coin) with the critical value 59 cutting off $\alpha=4.43\%$ of the
probability from the upper tail, and the right-hand panel shows
power values against various values of $\theta_a,$ where the power
against the specific alternative $\theta_a = 0.6$ is emphasized by a dotted line. The power against alternative values increases
as the alternative values get farther from the null value $\theta_0 = 1/2.$
Best Answer
I think the Central Limit Theorem is a poor approximation here for 5 samples. The exact probability is $P(T \ge t) = P(5T \ge 4) = P(5T = 5) + P(5T = 4) = \frac{1}{2^5} (\binom{5}{0} + \binom{5}{1}) = 0.1875$.