Why a condition of a null hypothesis always takes an exact value?

Question

$~ 10 ~$ 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?

Answer 1

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.

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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.

Answer 2

$$ \begin{align} \mathbb P(\chi^2>\chi_{\alpha}^2(n-1))&\leq\alpha\\ \iff \mathbb P\left({n s^2\over \sigma^2}>\chi_{\alpha}^2(n-1)\right)&\leq\alpha \\ \iff \mathbb P\left({10 s^2\over \sigma^2}>\chi_{0.01}^2(9)\right)&\leq0.01 \end{align} $$

$$ \begin{align} \sigma^2&:0^2,\ldots, 0.001^2,\ldots,0.004^2,\dots,0.008^2\\ \text{P}_\text{r}~\text{of reject}&:1,\ldots,\text{large},\ldots,\text{intermediate},\ldots,\text{smallest} \end{align} $$

Hence it is sufficient to do only the hypothesis test with the easiest parameter(i.e. $~\sigma^2=0.008^2~$)