Basic hypothesis testing of two normal populations Consider this problem in Montgomery's book:

Two types of plastic are suitable for use by an electronics component
manufacturer. The breaking strength of this plastic is important. It
is known that $\sigma_1=\sigma_2=1$ psi. From a random sample of size
$n_1=10$ and $n_2=12$, we obtain $\bar x_1=162.5$ and $\bar
 x_2=155.0$. The company will not adopt plastic 1 unless its mean
breaking strength exceeds that of plastic 2 by at least 10 psi.
(a) Based on the sample information, should it use plastic 1?

In the Solution manual, the hypotheses of the test are:
$H_0: \mu_1-\mu_2=10$ against $H_1:\mu_1-\mu_2>10$. Rejecting the null hypothesis means evidence to support the use of plastic 1.
In my interpretation, the company has to decide whether $\mu_1-\mu_2\geq 10$ holds ("at least 10"). It seems to me that both hypotheses corresponds to $\mu_1-\mu_2\geq 10$.
Question Why the alternative hypothesis is not $H_1:\mu_1-\mu_2<10$ ?

If the null hypothesis is $\mu_1-\mu_2<10$, then rejecting the null indicates evidence that plastic 1 is not sufficiently strong to warrant use. This means that failing to reject the null puts you in a position where you lack evidence against the strength of plastic 1. However, this is not the same as having evidence against the strength of plastic 1; it means that you don’t know.
The hypothesis given as correct allows for appropriate skepticism about plastic 1. Unless you have considerable evidence that plastic 1 is stronger by at least the required 10 psi, the company will continue using plastic 2.
