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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 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.

1.Properties of Exponential/Gamma distribution.

3.Order Statistics.

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.

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 > \(t\).

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 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.

1.Properties of Exponential/Gamma distribution.

3.Order Statistics.

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.

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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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?