Normal distribution hypothesis testing

Question

We observe single random variable $X$ from a distribution $\frac{1}{\sqrt{2\pi}\theta}\exp^{\frac{-x^2}{2\theta^2}}$ for $x\in\mathbb{R}$ and $\theta>0$ as a parameter. We verify hypothesis $H_0: \theta=4, H_1:\theta=1$. What i need to do is find a critical region with significance level $\alpha=0.1$.

What i am doing is; $$\left\{ \left. \frac{1}{\sqrt{2\pi}}\exp^{\frac{-x^2}{2}} \right/ \frac{1}{\sqrt{2\pi}\cdot4}\exp^{\frac{-x^2}{2\cdot4^2}}>c\right\}$$ and then i am trying to find $$P(x<k\mid\theta=4)=0.1$$ But this would require calculating an integral from a normal distribution what is difficult but doable however i am looking form the solution without evaluating this integral. How to do that?

Question still waits for an answer.

Answer 1

Your likelihood ratio simplifies as \begin{align} \frac{\frac{1}{\sqrt{2\pi}}\exp^{\frac{-x^2}{2}} } {\frac{1}{\sqrt{2\pi}\cdot4}\exp^{\frac{-x^2}{2\cdot4^2}}} &> c \\ \exp\left(-\frac{x^2}{2}\left(1-\frac{1}{16}\right)\right) &> c/4 \\ \exp\left(-\frac{x^2}{2}\left(1-\frac{1}{16}\right)\right) &> c/4 \\ \exp\left(-\frac{15\,x^2}{32}\right) &> c/4 \\ -\frac{15\,x^2}{32} &> \ln\left(c/4\right)\\ \frac{x^2}{16} &< \frac{-2 \, \ln(c/4)}{15}\\ \end{align}

Note that if $H_0: \theta=4$ is true, then $X^2/16$ is the square of a standard normal, i.e., $X^2/16 \sim \chi^2_1$.

Answer 2

• Your two hypotheses are both of a normal distribution with mean $0$
• Your null hypothesis has a standard deviation of $4$ while your alternative hypothesis has a smaller standard deviation of $1$
• Your best test using a single observation will be to reject the null hypothesis when the observation is close to $0$ and not reject the null hypothesis when the observation is further away from $0$. This gives more or less what you have written, but you need to consider the absolute value, i.e. $P(|x|<k \mid \theta=4)=0.1$
• So you need to find a critical region close to $0$ which has probability $0.1$ if the null hypothesis is true. For a normal distribution with mean $0$ and standard deviation $1$ this would be based on $\Phi^{-1}\left(0.5+\frac{0.1}{2}\right)$ - as Sean Robinson says, you will need tables or something equivalent - and you then need to scale this to the null hypothesis