Solved – Help calculating the power function at a point \$mu=0.5\$ for the normal distribution

For some reason, I've been stuck on this question, even though other questions of this type weren't that much of a problem to me before (with for instance the binomial distribution). The situation is as follows:

Suppose $$X_1…X_{25}$$ is a sample from an $$N(mu,2^2)$$-distribution. We would like to test for $$H_0: muleq 0$$; $$H_1:mu>0$$, with significance $$a_o=0.05$$.

Calculate the power function of this test at the point $$mu=0.5$$.

My approach:

Earlier in this problem I had to calculate the critical values and found the following: $$K_T = {Tin mathbb{R}:Tgeq 1.64}$$} and $$K={muinmathbb{R}:mugeq0.656}$$} where $$T=sqrt{n}frac{mu-mu_0}{sigma}$$

For $$H_0$$ we use $$mu_0=0$$ and find (with a table or calculator) these values for $$c$$ in $$P(Tgeq c)leq0.05$$.

Now we have to use $$mu_0=0.5$$. Given this information, we have to calculate $$P_{0.5}(Xin K)$$ which is the power function. I thought of two things, but they both give me wrong answers.

First I tried to just calculate $$P(Tin K_T$$) given $$mu=0.5$$, but I stumbled upon a problem; I didn't seem to have used the fact that $$mu=0$$ in my calculation for the critical values for $$K_T$$. I just chose $$T=sqrt{n}frac{mu-mu_0}{sigma} =frac{5}{2}mu$$ and said that $$Tsim N(0,1)$$ and proceeded to use the table. Only for $$K$$ I used that $$T=frac{5}{2}mugeq1.64 iff mugeq0.656$$. (so here we used $$mu_0=0implies T=frac{5}{2}mu$$)

So, changing to $$mu_o=0.5$$ didn't really give me anything in this calculation, if I would just put it in there I would get that $$P(Tin K_T)=0.05=a_0$$ again.

Then I tried to calculate
$$P(Xin K)=P(N(0.5,2^2)geq1.64)=$$
$$P(4N(0,1)+0.5geq1.64)=P(N(0,1)geq 0.285)=0.39$$

This comes closer, but the correction model says it's $$0.34$$. I did find, that $$P(N(0,1)geq a)=0.34$$ for $$a=0.41$$ which is $$1.64/4$$, the original critical value divided by $$4$$.

I really feel like I'm missing a (relatively simple) key element of the way to approach this problem. Any corrections, tips, hints or explanations are highly appreciated!

Contents

Under $$H_o$$, find the critical value (C) for $$bar X$$, which is $$Pr(bar X > C) =Pr(frac {bar X}{2/5}>frac C{2/5}) = 0.05$$ We have $$frac C{2/5} = 1.64$$, so $$C= 1.64times frac 25=0.656$$. You did it correctly.

Then think about if $$H_a: mu = 0.5$$ is true, $$bar X sim Nleft(0.5, frac {2^2}{25}right)$$.

The Probability of $$bar X>0.656$$, which is power. is

$$Pr(bar X >0.656) = Pr(frac{bar X -0.5}{2/5}>frac {0.656-0.5}{2/5})$$ It equals $$Z = (0.656 – 0.5) times 5/2 = 0.39$$

From standard normal distribution, $$Pr(Z>0.39) = 0.3483$$

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