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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?

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  • $\begingroup$ 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. $\endgroup$ – TheSimpliFire Apr 19 at 15:59
  • $\begingroup$ Yes I know this but I can't write the condition for which a bulb fails $\endgroup$ – Legend Killer Apr 19 at 16:05
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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 $c$.

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  • $\begingroup$ How can you infer only based on the first order statistic? It may happen that the second order life time has also failed? $\endgroup$ – Legend Killer Apr 22 at 3:00
  • $\begingroup$ What do you mean by second-order lifetime? Do you mean the lifetime of the second light bulb to fail? That would be great if we could observe this information, but we don't, we're only given the first to fail. $\endgroup$ – Tom Chen Apr 22 at 3:32
  • $\begingroup$ So, according to your approach you claim that the first to fail has the least lifetime which may not be true necessarily $\endgroup$ – Legend Killer Apr 22 at 5:42
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    $\begingroup$ @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)? $\endgroup$ – antkam Apr 22 at 12:27
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    $\begingroup$ 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? $\endgroup$ – Tom Chen Apr 22 at 15:36

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