Test on a parameter in which $H_1$ is not complementary to $H_0$. Is this possible? I have the following problem and I don't know how to solve it.
We have a population of $N=20$ enterprises. Each enterprise has a probability $p$ to reduce his customers independently from the others.
The exercise asks to test the hypothesis  $H_0: p=0.4$ against the hypothesis $H_1: p=0.6$.
This is my argument:
If I call $X$ (that we suppose to be $12$ in our population) the number of enterprises that reduced their costumers I know that $X$ should be distributed as a binomial of parameters $N$ and $p$. Therefore if the hypothesis  $H_0$ is true
I would have that
$\frac{X-Np}{\sqrt{Np(1-p)}}$ (with $p=0.4$) maybe can be approximated by a standard Gaussian (I am not sure that $N=20$ is sufficiently large for this approximation).
My main problem anyway is that I don't know how to construct a test in which the alternative hypothesis  is not complementary to the hypothesis $H_0$.
Therefore my main problem is that the hypothesis $H_1$ is not $p\neq 0.4$ but is the one above.
 A: Here is a plot of PDFs of $\mathsf{Binom}(12, 0.4),$ corresponding to $H_0,$ and $\mathsf{Binom}(12, 0.6),$ corresponding to $H_a.$

R code for figure:
x = 0:20; PDF = dbinom(x,20,.4)
PDF.a = dbinom(x,20,.6)
hdr = "PDFs of BINOM(20, .4) [blue] and BINOM(20, .6)"
plot(x-.1, PDF, type="h", xlab="x", col="blue", lwd=2, main=hdr)
 lines(x+.1, PDF.a, type="h", col="maroon", lwd=2)
 abline(h=0, col="green2")

Suppose you have results $X$ from $n=20$ Bernoulli trials. If you reject $H_0$ when $X \ge c = 11.5,$ then the significance level of the test is
$$\alpha = P(X \ge 12\,|\, p=0.4) = 1 - P(X \le 11\,|\, p = 0.4) = 0.0565,$$ as computed in R, where pbinom is a binomial CDF. The value $c$ is sometimes called the critical value of the test.
1 - pbinom(11, 20, .4)
[1] 0.05652637

Sometimes one says that $\alpha = P(\mathrm{Type\,I\,Error}) = 0.0565.$ This is is the probability of rejecting $H_0$ when it is true.
1 - pbinom(11, 20, .4)
[1] 0.05652637

In this test
$$\beta = P(\mathrm{Type\,II\,Error}) = 
P(X \le c=11.5\,|\,p=0.6)
= 0.4044.$$
pbinom(11, 20, .6)
[1] 0.4044013

By changing the critical value $c$ of the test, one can decrease $\beta,$ while increasing $\alpha.$
One compromise to make error probabilities $\alpha$ and $\beta$ more nearly equal would be to let $c = 10.5.$
Then $\alpha = 0.1275, \beta = 0.2447.$
1 - pbinom(10, 20, .4)
[1] 0.1275212
pbinom(10, 20, .6)
[1] 0.2446628

Notes: (1) You could get normal approximations
for $\alpha$ and $\beta$ as you suggest in your question. Sample size $n = 20$ is just barely large
enough to get accurate normal approximations in this problem. I used R to get exact binomial probabilities, but I suppose most students
in your class will use normal approximations. If you need
practice doing normal approximations, maybe you should be one of them.
(2) The two distributions are too similarly located to
get a test with $\alpha$ and $\beta$ both small.
Using $n = 60$ observations would make it easier
to distinguish between the two success probabities $p = 0.4$ and $p = 0.6.$

