To find the power of the test against hypothesis test

Question

This problem is all about probability density function and hypothesis which I am familiar with. But the problem is that I can't figure out how to approach the problem. A rigorous proof is needed. Thank you.

Answer 1

(a) $~~$At first, we need to find the distribution of $X = \max(0 , Z)$ . Clearly, if $x < 0$ , then $F_X(x) = 0$ , and if $x \geqslant 0$ , $F_X(x) = F_Z(x)$ .

Therefore, given that $\mu = 0$ , for $x \geqslant 0$ , $$F_X(x) = \int_{-\infty}^x \frac{1}{2}e^{-\vert z \vert} ~dz = \frac{1}{2} + \int_0^x \frac{1}{2}e^{-\vert z \vert}~dz = 1 - \frac{1}{2} e^{-x}$$ And, for $x < 0$ , we can see that $F_X(x) = 0 $ very clearly.

Since this is a one-tailed test, thus : $$F_X(c) = 0.95\quad\iff\quad e^{-c} = 0.1 \quad \iff\quad \boxed{~c = \log_e 10~}$$

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(b) $~~$ Once (a) is done, (b) is straightforward. Power of the test will be : $$P(X > c~\vert~ \mu = 2) = F_X(c - 2) = 1 - \frac{1}{2} e^{-(\log_e 10 - 2)} = \boxed{~0.6305}$$

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Hope this helps. Let me know if my answer is erroneous.

Answer 2

In a very rigorous way it is enough to apply the definitions.

(a) Find $c$ such that

$$P[X>c |\mu=0]=0.05$$

(b) Fixed $c$ found in (a), calculate

$$P[X>c|\mu=2]$$

These are the rigorous definitions of "size" and "power" . All you have to do is to do the calculations