# Inference regarding the mean lifetime of a bulb using a new technique

The lifetime in hours of each bulb manufactured by a particular company follows an independent exponential distribution with mean $$\lambda$$. We need to test the null hypothesis $$H_0: \lambda=1000$$ against $$H_1:\lambda=500$$. A statistician sets up an experiment with $$50$$ bulbs, with $$5$$ bulbs in each of $$10$$ different locations, to examine their lifetimes.

To get quick preliminary results,the statistician decides to stop the experiment as soon as one bulb fails at each location.Let $$Y_i$$ denote the lifetime of the first bulb to fail at location $$i$$.Obtain the most powerful test of $$H_0$$ against $$H_1$$ based on $$Y_1,Y_2,...Y_{10}$$ and compute its power.

Now, I have problems regarding how to formulate the condition that a particular bulb has failed. Otherwise, how can I write the likelihood function in either of the hypotheses?

• Given the hypotheses $\sf{H_0:\lambda=\lambda_0}$ and $\sf{H_0:\lambda=\lambda_1}$ the likelihood ratio is $\sf{\frac{L(Y;\lambda_1)}{L(Y;\lambda_0)}}$. Note that the likelihood function for $\sf{\lambda_j}$ is just the product of each probability density function $\sf{f(y_i\mid\lambda_j)}$ for all $\sf{i}$ where $\sf{j=0,1}$. If you are fine with this, please add your attempts to your post to avoid its closure. – TheSimpliFire Apr 19 at 15:59
• Yes I know this but I can't write the condition for which a bulb fails – Legend Killer Apr 19 at 16:05

Let $$Y_i = \min(X_{i1}, \cdots, X_{i5})$$, where each $$X_{ij} \overset{\text{i.i.d}}{\sim} \text{Exp}(\lambda)$$, denote the failure time of the first bulb. By properties of the exponential distribution, you can show $$Y_i \sim \text{Exp}(5\lambda)$$.
From this, the LRT would show that you reject for large values of $$\overline{Y}$$ (i.e. $$\overline{Y} > c$$ for some critical value $$c$$), where $$\overline{Y} = \frac{1}{10}\sum_{i=1}^{10}Y_i \sim \text{Gamma}(10, 50\lambda)$$, where I'm using the shape/rate parametrization. Under $$H_0$$, $$\lambda = 1000$$, and use that to find the appropriate critical value $$c$$.
• @LegendKiller - how can the first to fail (among the $5$ in a given location) not have the smallest lifetime (among those same $5$ in the same given location)? – antkam Apr 22 at 12:27
• Agreeing with @antkam here, by definition the first to fail has the shortest lifetime. Before you observe this, it could be any of the $X_{i1}, \cdots, X_{i5}$, but we aren't designating which one and we are just treating this minimum lifetime as a random variable in itself. Does this make sense? – Tom Chen Apr 22 at 15:36