Which confidence interval to invert to get a hypothesis test?

I am trying to derive a hypothesis test from a confidence interval as the other direction seems to be more straightforward.

The case I am considering is when $$X_i$$ are iid $$N(\mu,\sigma^2)$$ and I want to test whether $$H_o:\sigma^2 = \sigma_0^2$$ or $$H_1: \sigma^2 > \sigma_0^2$$.

I know how to compute a confidence interval using the fact that $$\frac{(n-1)S^2}{\sigma^2}$$ has chi-square distribution. But do I invert an upper-confidence bound or a lower-confidence bound on $$\sigma^2$$?

It is not obvious to me how the fact that my $$H_1$$ is testing $$\sigma^2 > \sigma_0^2$$ plays a role here but I know it does.

By using the likelihood ratio test, I know the rejection region has to be something of the form $$S^2 > c$$.

Can anyone help me clarify this issue? Thanks.

EDIT: Let me construct the upper and lower confidence intervals to prevent confusion.

Let $$F$$ be the $$\xi^2_{n-1}$$ CDF and $$y_1$$ be s.t. $$F(y_1) = \alpha$$ and $$y_2$$ be s.t. $$1-F(y_2) =\alpha$$ for small $$\alpha$$.

Then one confidence interval could be obtained from: $$P(\frac{(n-1)S^2}{\sigma^2} \le t_1) = \alpha$$ which implies that $$P(\frac{(n-1)S^2}{t_1} \ge \sigma^2) = 1-\alpha$$

while the other confidence interval is $$P(\frac{(n-1)S^2}{\sigma^2} \le t_2) = 1-\alpha$$ which implies that $$P(\frac{(n-1)S^2}{t_2} \le \sigma^2) = 1-\alpha$$.

I guess the first one is an upper bound CI while the second is a lower bound CI?

• Do you actually want a p-value or just "accept/reject"? The latter is easier: compare the null variance to the lower bound of the CI to reject the null hypothesis that $\sigma^2 \le \sigma_0^2$. Commented Feb 7, 2020 at 17:35
• I want accept reject only. I want to find the rejection region for $H_0$ based on some confidence interval. Basically, I just want to show they're the same at least for this case. @AdamO: can you expand more why lower bound and not upper bound? Commented Feb 7, 2020 at 17:36

1 Answer

You already know that $$T=(n-1)S^2/\sigma^2$$ has a $$\chi^2_1$$ distribution under the null. Base your hypothesis on this. Find the critical value $$c$$ such that $$F_{T}^{-1}(1-\alpha/2) = c$$. Using this math fact about $$T$$ to generate a confidence interval, then inverting that confidence interval to find a test, is like cutting the roof off a car to make a convertible, then welding it back on to make a sedan.

• Hi @AdamO, see edit above where I wrote down the upper and lower CI. It's still not clear to me which of the 2 tests I should invert to construct a hypothesis test. Why choose lower bound and not upper bound? That's what's confusing me. Commented Feb 7, 2020 at 17:55