# 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? Apr 4, 2013 at 6:19
• Hint: what is the formula for $\hat{\sigma}^2$ here? Apr 4, 2013 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! Apr 4, 2013 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? Apr 4, 2013 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. Apr 4, 2013 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) Apr 5, 2013 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. Apr 5, 2013 at 0:29