Testing the null hypothesis $H_0$ : $p = .6$ vs. the alternative $H_1$ : $p < .6$ for a binomial model In testing the null hypothesis $H_0$ : $p = .6$ vs. the alternative $H_1$ : $p < .6$ for a binomial model $b(n,p)$, the rejection region of a test that has structure $X ⩽ c$, where $X$ is the number of successes in n trials. For each of the following tests, determine the level of significance and the probability of type II error at the alternative $p$ = $.3$. 
$n = 10, c = 2$
Attempted Solution:
I am trying to make use of this definition:
Test $H_0$ : p $\geq$ $p_0$ ($H_0$ : $p$ = $p_0$) against $H_1$ : $p$ $\lt$ $p_0$
At the significance level of $\alpha$, most powerful test rejects $H_0$ if X $\leq$ c where c is the largest integer 
c $\lt$ $n$$p_0$ − $Z_α$$\sqrt{np_0(1−p_0)}$
In this case, since $c = 2, n = 10$, and $p_0$ = .6, we get,
$2$ $\lt$ $10(.6)$- $Z_α$$\sqrt{10*.6(1−.6)}$
$\Rightarrow$ $Z_α$ $\lt$ $2.58$
$\Rightarrow$ α = $.0049$
And the probability of type II error would be the probability that $p = .6$ is accepted given that
$p$ $\lt .6$. I think this would be 
data binom1;
binomcdf = 1-cdf('binomial', 2, 0.6, 10);
run;
proc print data = binom1;
run;

which gives $0.9877$.
Something I'm skeptical about is $p = .3$ is mentioned in the question but I never used that probability. I don't understand what it is associated with. Is that supposed to be the alternative hypothesis?
 A: You have $n = 10$ and wish to test $H_0: p = .6$ against $H_a: p < .6,$
using critical value $c = 2.$  Let $X \sim \mathsf{Binom}(10, p).$
So $\alpha = P(X \le 2 | p = .6) = 0.0123.$
The exact binomial probability can be computed in R statistical software
as follows:
pbinom(2, 10, .6)
## 0.01229455

Using a normal approximation (with continuity correction) with $n$ as small as ten is a bit risky, but
the computation would be as follows:
$$P(X \le 2) = P(X \le 2.5) 
= P\left(\frac{X-np}{\sqrt{np(1-p)}} \le \frac{2.5 - 6}{\sqrt{2.4}} \right)
\approx P(Z < -2.2596) = 0.0119,$$
which agrees with the exact value to two places, generally about what you can
expect with a normal approximation.
The normal approximation isn't bad for $p=0.6,$ but I wouldn't want to use
it for $p$ as small as $0.2.$

The power against the specific alternative $p = .3$ is
$P(X \le 2|p = .3) = 0.3838.$ So the probability of the Type II error
(failing to reject $H_0: p = .6,$ when $H_a: p = .3$ holds) is $1 - 0.3838 = 0.6172.$
pbinom(2, 10, .3)
## 0.3827828

This is not the answer you got, but I don't understand the method you used.
The probability of Type II error is a function of $p$ and so has many values.
p = seq(.1,.6, by= .01
p.rej = pbinom(2, 10, p)
plot(p, p.rej, type="l", ylim=0:1, col="blue", lwd=2, main="Power Curve")
   abline(h=0:1, col="green2"); abline(v=.6, col="green2")
   abline(v=.3, col="red")

The curve below shows power against various values of $p < .6$ A red line
indicates power against the specific alternative $p = .3$ computed above.
The Type II error probability is found by subtracting the power from 1.
 
