# #StackBounty: #hypothesis-testing #self-study #mathematical-statistics #statistical-power #weibull-distribution Uniformly Most Powerful…

### Bounty: 50

$$newcommand{szdp}{!left(#1right)} newcommand{szdb}{!left[#1right]}$$
Problem Statement: Let $$Y_1,dots,Y_n$$ be a random sample from the probability
density function given by
$$f(y|theta)= begin{cases} dfrac1theta,m,y^{m-1},e^{-y^m/theta},&y>0\ 0,&text{elsewhere} end{cases}$$
with $$m$$ denoting a known constant.

1. Find the uniformly most powerful test for testing
$$H_0:theta=theta_0$$ against $$H_a:theta>theta_0.$$
2. If the test in 1. is to have $$theta_0=100, alpha=0.05,$$ and
$$beta=0.05$$ when $$theta_a=400,$$ find the appropriate sample size and
critical region.

Note: This is Problem 10.80 in Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Sheaffer.

My Work So Far:

1. This is a Weibull distribution. We construct the
likelihood function
$$L(theta)=szdp{frac{m}{theta}}^{!!n}szdb{prod_{i=1}^ny_i^{m-1}} expszdb{-frac1thetasum_{i=1}^ny_i^m}.$$
Now we form the inequality indicated in the Neyman-Pearson Lemma:
begin{align*} frac{L(theta_0)}{L(theta_a)}&
The end result is
$$sum_{i=1}^ny_i^m>frac{theta_0theta_a}{theta_a-theta_0} szdb{nln(theta_a/theta_0)-ln(k)},$$
or
$$sum_{i=1}^ny_i^m>k’.$$
2. We have to discover the distribution of $$displaystyle sum_{i=1}^ny_i^m.$$
I claim that the random variable $$W=Y^m$$ is exponentially distributed with
parameter $$theta.$$ Proof:
begin{align*} f_W(w) &=fszdp{w^{1/m}}frac{dw^{1/m}}{dw}\ &=frac{m}{theta},(w^{1/m})^{m-1},e^{-w/theta}szdp{frac1m},w^{(1/m)-1}\ &=frac1theta,w^{1-1/m}e^{-w/theta},w^{(1/m)-1}\ &=frac1theta,e^{-w/theta}, end{align*}
which is the distribution of an exponential with parameter $$theta,$$ as I
claimed. It follows, then, that $$displaystylesum_{i=1}^ny_i^m$$ is
$$Gamma(n,theta)$$ distributed, and hence that
$$displaystylefrac{2}{theta}sum_{i=1}^ny_i^m$$ is $$chi^2$$ distributed with
$$2n$$ d.o.f. So the RR we can write as that region where
$$frac{2}{theta}sum_{i=1}^ny_i^m>chi_alpha^2,$$
with the $$2n$$ d.o.f. Let
$$U(theta)=frac{2}{theta}sum_{i=1}^ny_i^m.$$
Then we have
begin{align*} alpha&=Pszdp{U(theta_0)>chi_alpha^2}\ beta&=Pszdp{U(theta_a)
So now we solve
begin{align*} frac{2}{theta_0}sum_{i=1}^ny_i^m&=chi_alpha^2\ frac{2}{theta_a}sum_{i=1}^ny_i^m&=chi_beta^2\ frac{chi_alpha^2theta_0}{2}&=frac{chi_beta^2theta_a}{2}\ frac{chi_alpha^2}{chi_beta^2}&=frac{theta_a}{theta_0}. end{align*}
So we choose $$n$$ so that the $$chi^2$$ values corresponding to the ratio given
work out. The ratio of $$theta_a/theta_0=4,$$ and we choose $$chi_alpha^2$$ on
the high end, and $$chi_beta^2$$ on the low end so that their ratio is $$4,$$
by varying $$n$$. This happens at d.o.f. $$13=2n,$$ which means we must choose
$$n=7.$$ For this choice of $$n,$$ we have the critical region as
$$frac{2}{theta_0}sum_{i=1}^ny_i^m>23.6848.$$

My Question: This is one of the most complicated stats problems I’ve encountered yet in this textbook, and I just want to know if my solution is correct. I feel like I’m "out on a limb" with complex reasoning depending on complex reasoning. I’m fairly confident that part 1 is correct, but what about part 2?

Get this bounty!!!

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