Let $X_{1},..,X_{n}$ be a random sample from a normal distribution with mean ${\theta}$ and variance 1. We wish to test $H_{0}:{\theta}=0$ vs $H_{1}:{\theta}{\neq}0$. Write down likelihood ratio statistic and show it can be expressed $exp((n/2)m(x)^2){\geq}k$ for some non-negative $k$, where $m(x)$ is the mean of the sample values.
I get the test statistic to be $exp(1/2(x_{1}^2+...+x_{n}^2))$ but how do I get from the sum of squares to the square of the mean. Annoying thing is I'm sure I've done this question before.
First, let $$\bar{X}=m\left(\vec{X}\right)=m(X_1,X_2,...,X_n)=\frac{1}{n} \sum_{i=1}^{n} X_i \,\,.$$ Also, recall that the maximum likelihood estimator for the mean $\theta$, in this case, of a normal distribution is $\bar{X}$ (I can clarify this further if you commment) and we will call $\mathscr{L}$ the likelihood function. Now, we have the likelihood ratio, or likelihood ratio test statistic, $\Lambda \leq k$ for some $k \in \mathbb{R}$ determined by the significance level of the test, is given by $$\begin{align} \Lambda &= \frac{\mathscr{L}\left(\vec{X} \, \biggr| \, H_0\right)}{\mathscr{L}\left(\vec{X} \, \biggr| \, H_1\right)} \\&= \frac{\mathscr{L}(X_1,X_2,...,X_n \, | \, \theta = 0)}{\mathscr{L}(X_1,X_2,...,X_n \, | \, \theta = \bar{X} \neq 0)} \\&= \frac{\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}} \text{exp}\left( -\frac{1}{2}(X_i-0)^2\right)}{\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}} \text{exp}\left( -\frac{1}{2}(X_i-\bar{X})^2\right)} \\&= \frac{\text{exp}\left(-\frac{1}{2} \sum_{i=1}^{n} X_i^2\right)}{\text{exp}\left(-\frac{1}{2} \sum_{i=1}^{n} (X_i-\bar{X})^2\right)} \\&= \text{exp}\left(\frac{1}{2}\left(\sum_{i=1}^{n}(X_i -\bar{X})^2-\sum_{i=1}^{n}X_i^2\right)\right) \\&= \text{exp}\left(-\frac{n}{2}\bar{X}^2\right) \leq k \,\,, \end{align}$$ as desired (you are missing a negative sign), where the last equality comes from the fact that $$\begin{align} \sum_{i=1}^{n} (X_i-\bar{X})^2-\sum_{i=1}^{n}X_i^2 &= \sum_{i=1}^{n}(X_i^2-2\bar{X}X_i+\bar{X}^2)-\sum_{i=1}^{n}X_i^2 \\&= \sum_{i=1}^{n}X_i^2-2\bar{X}\sum_{i=1}^{n}X_i+\sum_{i=1}^{n}\bar{X}^2-\sum_{i=1}^{n}X_i^2 \\ &=\sum_{i=1}^{n}\bar{X}^2-2\bar{X}\sum_{i=1}^{n}X_i \\&=n\bar{X}^2 - 2\bar{X}(n\bar{X}) \\&= -n\bar{X}^2 \,. \end{align}$$