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

Bounty: 50

$newcommand{szdp}[1]{!left(#1right)} newcommand{szdb}[1]{!left[#1right]}$
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)}&<k\
    frac{displaystyle szdp{frac{m}{theta_0}}^{!!n}prod_{i=1}^ny_i^{m-1}
    expszdb{-frac{1}{theta_0}sum_{i=1}^ny_i^m}}
    {displaystyle szdp{frac{m}{theta_a}}^{!!n}prod_{i=1}^ny_i^{m-1}
    expszdb{-frac{1}{theta_a}sum_{i=1}^ny_i^m}}&<k\
    frac{displaystyle theta_a^n
    expszdb{-frac{1}{theta_0}sum_{i=1}^ny_i^m}}
    {displaystyle theta_0^n
    expszdb{-frac{1}{theta_a}sum_{i=1}^ny_i^m}}&<k\
    frac{theta_a^n}{theta_0^n},expszdb{-frac{theta_a-theta_0}
    {theta_0theta_a}sum_{i=1}^ny_i^m}&<k\
    nln(theta_a/theta_0)-frac{theta_a-theta_0}
    {theta_0theta_a}sum_{i=1}^ny_i^m&<ln(k)\
    nln(theta_a/theta_0)-ln(k)&<frac{theta_a-theta_0}
    {theta_0theta_a}sum_{i=1}^ny_i^m.
    end{align*}

    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)<chi_beta^2}.
    end{align*}

    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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