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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?

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  • $\begingroup$ 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 $\sigma^2 \le \sigma_0^2$. $\endgroup$
    – AdamO
    Commented Feb 7, 2020 at 17:35
  • $\begingroup$ 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? $\endgroup$
    – secondrate
    Commented Feb 7, 2020 at 17:36

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – secondrate
    Commented Feb 7, 2020 at 17:55

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