Hypothesis testing of normal distribution, known mean unknown variance

I've been working on review problems, and this one has me completely stumped.

Let $X_1 ... X_{10}$ be a random sample from a $N(3,\sigma^2)$ distribution, where $\sigma^2$ is unknown. Using the likelihood ratio test, determine a 5%-level critical region test for $H_0 : \sigma^2 = 1$ vs. $H_1 : \sigma^2 \neq 1$ (and, trivially, $\sigma^2 >0$).

It appears that in the general case, when one is testing a hypothesis about the variance, a chi-square statistic is used, which gives me something of an end-goal, but I'm not sure how to get there.

The joint pdf for the 10 r.v.s should be $\large(\frac{1}{\sqrt{2\pi\sigma^2}})^{10}\cdot e^-\frac{\sum_{i=1}^{10} (X_i - 3)^2}{2\sigma^2}$

Under the null hypothesis, this yields $\large(\frac{1}{\sqrt{2\pi}})^{10}\cdot e^-\frac{\sum_{i=1}^{10} (X_i - 3)^2}{2}$, since $\sigma^2 = 1$

Under the alternative hypothesis, we have $\large(\frac{1}{\sqrt{2\pi\hat\sigma^2}})^{10}\cdot e^-\frac{\sum_{i=1}^{10} (X_i - 3)^2}{2\hat\sigma^2}$

Setting these as numerator and denominator, respectively, I get

$\LARGE\frac{\exp(^-\frac{\sum_{i=1}^{10} (X_i - 3)^2}{2})}{(\frac{1}{\hat{\sigma}})^{10}\cdot \exp(^-\frac{\sum_{i=1}^{10} (X_i - 3)^2}{2\hat\sigma^2})} = \Lambda$

I believe the numerator has 0 free parameters, and the denominator has 1.

In order to get the log-likelihood, I apply $ln(\Lambda)$, and we know that $\hat\sigma^2$ can be represented as $\frac{1}{10}\sum_{i=1}^{10} (X_i-3)^2$, so further simplification yields

$-2Ln(\Lambda) = \sum(X_i-3)^2-10+10ln(\frac{10}{\sum(Xi-3)^2})$

According to the problem, this should be a $\chi_{10}^2$ statistic, but I don't know how to justify this (probably graphically)?

Again, I greatly appreciate the help!

Edit (and my proposed answer): If I instead put everything in terms of $\hat\sigma^2$, I end up with the following:

$10(\hat\sigma^2-ln(\sigma^2)-1)$, and since I'm purely looking to see if this monotonic, I can simplify this to $\hat\sigma^2-ln(\hat\sigma^2)$, which a quick graph shows to be not-monotonic. This means we are going to do a two sided test under the null hypothesis. We know $\hat\sigma^2$ follows a $\chi^2_10$ distribution so we reject $H_0$ at when $n*\sigma^2<$$\chi_{.025,10}^2$ and at $n*\sigma^2>\chi_{.975,10}^2$

• How did you get from your ratio to that result at the end? – Glen_b Apr 4 '13 at 6:19
• Hint: what is the formula for $\hat{\sigma}^2$ here? – Glen_b Apr 4 '13 at 23:13

Let's say you get some statistic, $\Lambda$, and let's imagine you don't make any errors.

Then if you can work out its distribution under the null hypothesis, you're done, you have a test.

More generally, you have to employ an asymptotic approximation:

http://en.wikipedia.org/wiki/Likelihood-ratio_test#Use

• Thanks for the feedback! As pointed out in your comment, the result at the end is indeed incorrect, and I guess I should be taking $-2log(\Lambda)$? I'm a bit confused as to what the degrees of freedom for the $H_0$ and $H_A$ are though? Thanks again for your help! – LindaT Apr 4 '13 at 16:27
• I've updated the question (hopefully my algebra is better this time around). I think I may still be confused conceptually though? – LindaT Apr 4 '13 at 17:54
• It doesn't matter whether you look at the likelihood ratio, or its log or minus twice its log or any other monotonic transformation of it... as long as you know/can figure out what the distribution of that statistic is. If you can't, then you can always take minus twice its log and use the asymptotic result. – Glen_b Apr 4 '13 at 23:12
• I know the statistic should be $\chi_{10}^2$, however I don't know how to justify that. Any thoughts? (I also improved the original question) – LindaT Apr 5 '13 at 0:17
• There's still errors in your mathematics. Personally, I'd have converted all to $\hat{\sigma}^2$ until later. You have two options: (i) make an argument that the statistic you end up with is monotonic in $\hat{\sigma}^2$ (an option already raised) and just work with the distribution of that, or (ii) apply the asymptotic argument I already mentioned. I seem to be saying the same thing over and over. Perhaps you should re-read the thread again, and check your notes again. – Glen_b Apr 5 '13 at 0:29