Statistical Methods: Analysis of Normal Measurements Show that, if $\sigma$ is unknown, the likelihood ratio statistic for testing a value of $\alpha$ is given by $$D = n \log\left(1 + \frac{1}{n-1}T^2\right)\;,$$ where $$T = \frac{\hat{α} -\alpha}{\sqrt{s^2/n}}$$
So far, I have the following:
$\hat\alpha=\overline{y}$ and $\hat\sigma=\sqrt{\frac{\sum \left ( y_i-\bar{y} \right )^2}{n-1}}$.
Now, when I plug this in to my ratio, I have:
$$D=2\left [ l(\hat{\mu}, \hat{\sigma})-l(\mu_0,\hat{\sigma}) \right ]=\frac{n(\bar{y}-\mu_0)^2}{\hat{\sigma}}=\frac{(\bar{y}-\mu_0)^2}{c\hat{\sigma}}\text{ where }c=\frac{1}{n}$$
So, just to clarify the following things:


*

*I accidentally typed $n-1$. I meant $n$ in the denominator for $\sigma$.

*My log-likelihood function is: $l(\mu, \sigma)=-n\log(\sigma)-\frac{\sum{(y_i-\bar{y})^2}}{2\sigma^2}$

*When I expand my ratio statistic and simplify, the logs cancel because they are identical, and I am left with the equivalent expression of $T^2$. I don't understand how I am supposed to get the expression that I am asked for. Any ideas of what I am doing wrong?
 A: I'll use $\alpha_0$ for the value of $\alpha$ specified by the null hypothesis.
Under the null hypothesis $\alpha=\alpha_0$, you should get the MLE
$$\hat\sigma_\text{NH}^2 = \frac 1 n \sum_{i=1}^n (y_i - \alpha_0)^2. $$
Under the alternative hypothesis $\alpha\ne\alpha_0$, you should get two MLEs:
$$
\hat\alpha = \overline{y}=\frac{y_1+\cdots+y_n}{n}
$$
$$
\hat\sigma_\text{AH}^2 = \frac 1 n \sum_{i=1}^n (y_i-\overline{y})^2.
$$
(I don't know why you have $n-1$ instead of $n$.  We're looking for MLEs, not unbiased estimators.)
The likelihood function is
$$
L(\alpha,\sigma) = \text{constant}\cdot\frac{1}{\sigma^n} \exp\left( \frac{-1}{2} \sum_{i=1}^n \left(\frac{y_i-\alpha}{\sigma} \right)^2\right),
$$
so
$$
\ell = \log L = -n\log\sigma - \frac 1 2 \sum_{i=1}^n \left(\frac{y_i-\alpha}{\sigma} \right)^2.
$$
When you look at $\ell(\alpha_0,\hat\sigma)$, you might want to write
$$
\sum_{i=1}^n (y_i-\alpha_0)^2 = n(\alpha_0-\overline{y})^2 + \sum_{i=1}^n (y_i-\overline{y})^2.
$$
Could it be that you omitted the $-n\log\sigma$ term?
