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I want to construct an NP-Test for simple Null- and Alternative-hypothesis. In particular the likelihood for the Nullhypothesis and Alternativehypothesis is given by $$h_0(x)=\chi_{(-2,2)}(x)\\h_1(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x^2)$$ respectively. This means I want to test by considering a sample, if the underlying distribution is uniform on $(-2,2)$ or (standard)normally distributed.

First I construct the Likelihoodquotient $T(x)=\frac{h_1(x)}{h_0(x)}$ which yields $$T(x)= \begin{cases} \frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x^2),\ -2\le x\le 2\\ \infty, \quad\quad\quad\quad\quad \ \quad\quad\quad \text{else} \end{cases}$$

In the examples I have seen so far it is used that $T$ is monotonic, because then I can just refer the the distribution of the data under the Nullhypothesis. But in this case I have no idea on how to proceed.

I somehow have to compute a $\gamma\in[0,1]$ and $k\in[0,\infty)$ such that $$\alpha=P_0(T>k)+\gamma P_0(T=k)$$ where $P_0$ is the probability distribution corresponding to $h_0$ and $\alpha$ is the significance level.

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1) The density of your uniform in (-2,2) is 1/4, not 1. This has an impact on your $T$ and $k$.

2) You can forget about $\gamma$, because both distributions are continuous, so $P_0(T=k)=0$.

3) This is somewhat nonstandard because $T$ indicates that $h_1$ is better outside $(-2,2)$, $h_0$ is better in a set of the shape $(-2,-a)\cup(a,2)$, and $h_1$ is better in $(-a,a)$, because of symmetry (you can make a drawing to see this). You need to find $a$ so that $P_0[(-2,-a)\cup(a,2)]=1-\alpha$, which is easy. If you have that, you don't even need $k$, or you could read it off easily.

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  • $\begingroup$ I like this answer! Could you elaborate on what you mean by "better"? How would I reach your conclusion from merely considering $\alpha=P_0(T>k)$? $\endgroup$ – EpsilonDelta Jan 7 '20 at 20:06
  • $\begingroup$ "$h_1$ is better" means that $T=h_1/h_0$ is rather large (or, in other words, $h_1$ is relatively large compared to $h_0$, where the word "relatively" is connected to the fact that what values of $T=h_1/h_0$ are taken as "large" depends on $\alpha$ through $k$). If you start thinking from "$\alpha=P_0(T>k)$" you realise that you must choose $k$ in such a way that the set in which $T$ is large enough has probability $\alpha$, or equivalently probability $1-\alpha$ where $T$ is small. This is what I did, I chose a region in which $T$ is small so that its probability is $1-\alpha$. $\endgroup$ – Lewian Jan 8 '20 at 16:29
  • $\begingroup$ I then realised that this is easier and clearer in terms of $a$ than in terms of $k$. If you have $a$ it isn't difficult to find $k$, but in order to define and run the test, $k$ isn't even needed if you know $a$. $\endgroup$ – Lewian Jan 8 '20 at 16:31

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