Why a condition of a null hypothesis always takes an exact value? $~ 10 ~$ samples were sampled from the glass partitions.
Each value of the following represents a refractive index of it.
$$ 1.77,~1.79,~1.78,~1.79,~1.79,~1.76,~1.8,~1.76,~1.79,~1.80 $$
As the standard deviation of refractive indices is less than or equal to $~ 0.008 ~$, the acceptance test can be passed, otherwise fails.
Judge the acceptability of the glasses with $~ \alpha=0.01 ~$
$$
\begin{cases}
\color{fuchsia}{H_0:\sigma^2=0.008^2} 
\\H_1:\sigma^2>0.008^2\end{cases}
$$
I think that the null-hypothesis should be $~ H_0:\sigma^2 \leq 0.008^2 ~$ since it is too diffuclt to infer the exact true variance of the population(infinite size?)
The table of test-statistics which attaches to the book of this problem statement only handles cases where  null-hypotheses take exact values.
I am really confusing.
Can anyone tell me why the pink eqn is adequate?
 A: The statement in words is the standard deviation of refractive indices is less than or equal to  $0.008$, the acceptance test can be passed, otherwise fails.
So the null hypothesis that you are testing and might reject is $H_0: \sigma \le 0.008$.  If the null hypothesis is in fact true, you want to reject it with probability no greater than $\alpha$.
Clearly in this example you are going to say the test fails if the sample standard deviation is too high, for example if the sample standard deviation $s_n$ exceeds some value $k$, where $k$ is determined by $\alpha$, the sample size and the value in the null hypothesis.  This will give you the most powerful test.
If the null hypothesis is in fact true, you want $\mathbb P(s_n>k)\le \alpha$.  Since $\mathbb P(s_n>k)$ is an increasing function of $\sigma$, the smallest value of $k$ which ensures $\mathbb P(s_n>k)\le \alpha$ for all $ \sigma \le 0.008$ is that which occurs when $\sigma=0.008$.  So you use that point value to decide $k$ and the critical region.

There is another issue to think about: you have data rounded to $0.01$ but are testing a standard deviation of $0.008$ so the result will be affected by the rounding.
A: $$
\begin{align}
\mathbb P(\chi^2>\chi_{\alpha}^2(n-1))&\leq\alpha\\ \iff
\mathbb P\left({n s^2\over \sigma^2}>\chi_{\alpha}^2(n-1)\right)&\leq\alpha
\\ \iff
\mathbb P\left({10 s^2\over \sigma^2}>\chi_{0.01}^2(9)\right)&\leq0.01
\end{align}
$$
$$
\begin{align}
\sigma^2&:0^2,\ldots, 0.001^2,\ldots,0.004^2,\dots,0.008^2\\
\text{P}_\text{r}~\text{of reject}&:1,\ldots,\text{large},\ldots,\text{intermediate},\ldots,\text{smallest}
\end{align}
$$
Hence it is sufficient to do only the hypothesis test with the easiest parameter(i.e. $~\sigma^2=0.008^2~$)
