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# Life Testing Experiment | ISI MStat 2017 PSB Problem 5

This is a problem from the ISI MStat 2017 Entrance Examination and tests how good are your skills in modeling a life testing experiment using an exponential distribution.

## The Problem:

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.

## Prerequisites:

1.Properties of Exponential/Gamma distribution.

3.Order Statistics.

## Proof:

As it is clear from the arrangement of the bulbs, the first to fail(among 5 in a given location) has the smallest lifetime among the same.

That is, in more mathematical terms, for a location $$i$$, we can write $$Y_i = \text{min}(X_{i1},X_{i2},..,X_{i5})$$.

Here, $$X_{ij}$$ denotes the $$j$$ th unit in the $$i th$$ location where $$i=1,2,..,10$$ and $$j=1,2,..,5$$

It is given that $$X_{ij} \sim \text{Exp}(\lambda)$$.

Can you see that $$Y_i \sim \text{Exp}(5 \lambda)$$? You may try to prove this result for this:

If $$X_1,..,X_n$$ be a random sample from $$\text{Exp}(\lambda)$$ distribution,

then $$X_{(1)}=\text{min}(X_1,....,X_n) \sim \text{Exp}(n \lambda)$$.

So, now we have $$Y_1,Y_2,..Y_{10}$$ in hand each having $$\text{Exp}(5 \lambda)$$ distribution.

Let the joint pdf be $$f(\mathbf{y} )=\frac{1}{(5 \lambda)^{10}} e^{-\frac{\sum_{i=1}^{10} y_i}{5 \lambda}}$$.

For testing $$H_0: \lambda=1000$$ against $$H_1:\lambda=500$$, we use the Neyman Pearson Lemma.

We have the critical region of the most powerful test as $$\frac{f_{H_1}(\mathbf{y})}{f_{H_0}(\mathbf{y})} >c$$

which after simplification comes out to be $$\bar{Y} > K$$ where $$K$$ is an appropriate constant.

Also, see that $$\bar{Y} \sim \text{Gamma}(10,50 \lambda)$$.

Can you use this fact to find the value of $$K$$ using the size ($$\alpha$$) criterion ? (Exercise to the reader)

Also, find the power of the test.

## Challenge Problem:

The exponential distribution is used widely to model lifetime of appliances. The following scenario is based on such a model.

Suppose electric bulbs have a lifetime distribution with pdf $$f(t)=\lambda e^{-\lambda t}$$ where $$t \in [0, \infty)$$ .

These bulbs are used individually for street lighting in a large number of posts.A bulb is replaced immediately after it burns out.

Let's break down the problem in steps.

(i)Starting from time $$t=0$$ , the process is observed till $$t=T$$.Can you calculate the expected number of replacements in a post during the interval $$(0,T)$$ ?

(ii) Hence,deduce $$g(t) \text{dt}$$ ,the probability of a bulb being replaced in $$(t,t+ \text{dt})$$ for $$t < T$$,irrespective of when the bulb was put in.

(iii)Next,suppose that at the end of the first interval of time $$T$$,all bulbs which were put in the posts before time $$X < T$$ and have not burned out are replaced by new ones,but the bulbs replaced after ttime $$X$$ continue to be used,provided,of course,that they have not burned out.

Prove that with such a mixture of old and new bulbs, the probability of a bulb having an expected lifetime > $$\tau$$ in the second interval of length $$T$$ is given by

$$S_2(\tau)=\frac{1}{2}e^{-\lambda \tau}(1+ e^{-\lambda X})$$

Also, try proving the general case where the lifetimes of the bulbs follow the pdf $$f(t)$$ . Here, $$f(t)$$ need not be the pdf of an exponential distribution .

You should be getting: $$S_2(\tau)=(1-p)S_1(\tau) + \int_{0}^{x} g(T-x)S_1(x)S_1(\tau +x) \text{dx}$$ ; where $$\tau<T$$

where, $$p$$ is the proportion of bulbs not replaced at time $$t=T$$ and $$S_1(t)$$ is the probability that a bulb has lifetime > $$t$$.

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### One comment on “Life Testing Experiment | ISI MStat 2017 PSB Problem 5”

1. Rhritajit Sen says:

Here we have given that Xij follows exponential with mean lambda that is exponential distribution with parameter 1/ lambda. But it is written here that Xij follows exponential with parameter lambda. Is it not a mistake made in this solution?