# Uniformly Most Powerful Test for Weibull Distribution


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)}& 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?

• With respect to question 2 above - you can't fix both the probability of type I error AND the probability of type II error given a null and alternative hypothesis. The test is most powerful, so gives you the smallest type II error possible given $\alpha$. What is the actual question statement? Aug 31, 2021 at 19:30
• @jbowman The actual question statement is given word-for-word in the first bolded section entitled Problem Statement:, not including the notes. Aug 31, 2021 at 19:48
• Ah, I missed the sample size as a variable that can be controlled, that gives you the extra parameter you need to make it work. Aug 31, 2021 at 20:53

You've got $$L=\frac{\theta_a^n}{\theta_0^n}\,exp\left({-\frac{\theta_a-\theta_0} {\theta_0\theta_a}\sum_{i=1}^ny_i^m}\right), which is good. Now we take $$log$$ of both sides:

$$n log\left(\frac{\theta_a}{\theta_0}\right)+\left(\frac{\theta_0-\theta_a}{\theta_0\theta_a}\right)\sum_{i=1}^{n}{y_i^m} < log(k)$$

and so the test itself is in the form: $$\left\{ \sum_{i=1}^{n}{y_i^m} < c \right\}$$

(rejecting if $$\sum_{i=1}^{n}{y_i^m} > c$$).

Now, for part (b), there's something to note here: $$y^m$$ has an exponential distribution, and so the $$\sum{y^m_i}\sim \Gamma(n,\theta)$$. Under the null we get that $$\frac{2\sum_{i=1}^{n}{y_i^m}}{\theta_0} > \frac{2c}{\theta_0}$$ has a $$\chi^2$$ distribution with $$2n$$ degrees of freedom (look for the relation between gamma and chi-squared).

Now let's solve (b):

$$\theta_0=100,\theta_a=400,\alpha=0.05,\beta=0.05$$

When $$H_0$$ is true, we get $$\alpha$$ using:

$$\alpha=P\left(\frac{2\sum_{i=1}^{n}{y_i^m}}{100} > \chi^2_{0.05}\right)=0.05.$$

When $$H_a$$ is true, we get $$\beta$$ using:

$$\beta=P\left(\frac{2\sum_{i=1}^{n}{y_i^m}}{100} \le \chi^2_{0.05} \middle| \theta=400\right)=P\left(\frac{2\sum_{i=1}^{n}{y_i^m}}{400} \le \frac{1}{4}\chi^2_{0.05} \middle| \theta=400\right)=P\left(\chi^2\le\frac{1}{4}\chi^2_{0.05}\right)=0.05$$

So, we need to find the row in $$\chi^2$$ table where $$\frac{1}{4}\chi^2_{0.05}=\chi^2_{0.95}$$:

You can see that for $$12$$ degrees of freedom, $$\chi^2_{0.95}=5.226$$ and $$\chi^2_{0.05}=21.03$$, which is the closest we get for achieving $$\frac{1}{4}\chi^2_{0.05}=\chi^2_{0.95}$$. Recall that this has $$2n$$ degrees of freedom, so the appropriate sample size is $$6$$.

• Great, thanks! That is essentially my method of solution, so it's good to see it validated. I greatly appreciate your time in looking into these problems! Sep 8, 2021 at 17:34