# Uniformly Most Powerful Test for Weibull Distribution

• MHB
• Ackbach
In summary, the Uniformly Most Powerful (UMP) test for Weibull distribution is a statistical hypothesis test that determines whether a random sample comes from a Weibull distribution. It is calculated using the likelihood ratio test and relies on the asymptotic properties of the chi-square distribution. The assumptions for this test include independent and identically distributed data, continuous and uncensored observations, and a known shape parameter. It is not suitable for small sample sizes and is interpreted by comparing the test statistic to a critical value to determine whether to reject the null hypothesis.
Ackbach
Gold Member
MHB
$\newcommand{\szdp}[1]{\!\left(#1\right)} \newcommand{\szdb}[1]{\!\left[#1\right]}$
Problem Statement: Let $Y_1,\dots,Y_n$ be a random sample from the probability density function given by
$$f(y|\theta)= \begin{cases} \dfrac1\theta\,m\,y^{m-1}\,e^{-y^m/\theta},&y>0\\ 0,&\text{elsewhere} \end{cases}$$
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 1: This is Problem 10.80 in Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Sheaffer.
Note 2: This is cross-posted here.

My Work So Far:

1. This is a Weibull distribution. We construct the likelihood function
$$L(\theta)=\szdp{\frac{m}{\theta}}^{\!\!n}\szdb{\prod_{i=1}^ny_i^{m-1}} \exp\szdb{-\frac1\theta\sum_{i=1}^ny_i^m}.$$
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}
\exp\szdb{-\frac{1}{\theta_0}\sum_{i=1}^ny_i^m}}
{\displaystyle \szdp{\frac{m}{\theta_a}}^{\!\!n}\prod_{i=1}^ny_i^{m-1}
\exp\szdb{-\frac{1}{\theta_a}\sum_{i=1}^ny_i^m}}&<k\\
\frac{\displaystyle \theta_a^n
\exp\szdb{-\frac{1}{\theta_0}\sum_{i=1}^ny_i^m}}
{\displaystyle \theta_0^n
\exp\szdb{-\frac{1}{\theta_a}\sum_{i=1}^ny_i^m}}&<k\\
\frac{\theta_a^n}{\theta_0^n}\,\exp\szdb{-\frac{\theta_a-\theta_0}
{\theta_0\theta_a}\sum_{i=1}^ny_i^m}&<k\\
n\ln(\theta_a/\theta_0)-\frac{\theta_a-\theta_0}
{\theta_0\theta_a}\sum_{i=1}^ny_i^m&<\ln(k)\\
n\ln(\theta_a/\theta_0)-\ln(k)&<\frac{\theta_a-\theta_0}
{\theta_0\theta_a}\sum_{i=1}^ny_i^m.
\end{align*}
The end result is
$$\sum_{i=1}^ny_i^m>\frac{\theta_0\theta_a}{\theta_a-\theta_0} \szdb{n\ln(\theta_a/\theta_0)-\ln(k)},$$
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)
&=f\szdp{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}\\
&=\frac1\theta\,w^{1-1/m}e^{-w/\theta}\,w^{(1/m)-1}\\
&=\frac1\theta\,e^{-w/\theta},
\end{align*}
which is the distribution of an exponential with parameter $\theta,$ as I claimed. It follows, then, that $\displaystyle\sum_{i=1}^ny_i^m$ is $\Gamma(n,\theta)$ distributed, and hence that $\displaystyle\frac{2}{\theta}\sum_{i=1}^ny_i^m$ 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&=P\szdp{U(\theta_0)>\chi_\alpha^2}\\
\beta&=P\szdp{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^2\theta_0}{2}&=\frac{\chi_\beta^2\theta_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?

Based on your work, it seems like you have a good understanding of the problem and have made some good progress. However, I would recommend double-checking your calculations and your assumptions to ensure that your solution is correct.

In part 1, you correctly identified the likelihood function and used the Neyman-Pearson Lemma to construct the inequality for the uniformly most powerful test. However, I would suggest checking your algebra and making sure that you have correctly simplified the inequality to get the final result. You may also want to consider looking at the solution for this problem in the textbook or consulting with a classmate or instructor to compare your work.

In part 2, you correctly identified the distribution of $\sum_{i=1}^ny_i^m$ as $\Gamma(n,\theta)$, but I believe your calculation for the critical region may be incorrect. The critical region should be based on the distribution of $U(\theta)$, not $\sum_{i=1}^ny_i^m$. You may also want to consider the fact that the critical region should be chosen to satisfy both $\alpha$ and $\beta$, not just one of them.

Overall, it seems like you are on the right track and have a good understanding of the problem, but I would recommend double-checking your calculations and assumptions to ensure that your solution is correct.

## 1. What is a Uniformly Most Powerful Test for Weibull Distribution?

A Uniformly Most Powerful (UMP) test for Weibull Distribution is a statistical hypothesis test that is designed to have the highest statistical power among all possible tests for a given significance level. It is used to determine whether a set of data follows a Weibull distribution or not.

## 2. How is a UMP test for Weibull Distribution different from other statistical tests?

A UMP test for Weibull Distribution is different from other statistical tests because it is specifically designed to have the highest power among all possible tests. This means that it is more likely to correctly detect a true difference or relationship between variables, making it a more powerful test.

## 3. What is the significance level in a UMP test for Weibull Distribution?

The significance level in a UMP test for Weibull Distribution is the probability of rejecting the null hypothesis when it is actually true. It is typically set at 5% or 0.05, meaning that there is a 5% chance of incorrectly rejecting the null hypothesis and concluding that there is a difference or relationship when there is not.

## 4. How is a UMP test for Weibull Distribution conducted?

A UMP test for Weibull Distribution is conducted by first setting the significance level and then calculating the test statistic using the data. The test statistic is then compared to a critical value from a statistical table to determine whether to reject or fail to reject the null hypothesis. If the test statistic is larger than the critical value, the null hypothesis is rejected and it can be concluded that the data follows a Weibull distribution.

## 5. What are the assumptions for conducting a UMP test for Weibull Distribution?

The assumptions for conducting a UMP test for Weibull Distribution include:

• The data is independent and identically distributed.
• The data follows a Weibull distribution.
• The sample size is large enough for the test to be valid.
If these assumptions are not met, the results of the test may not be accurate and other statistical tests may be more appropriate.

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