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# ISI MStat PSB 2008 Problem 10 | Hypothesis Testing This is a really beautiful sample problem from ISI MStat PSB 2008 Problem 10. It is based on testing simple hypothesis. This problem teaches me how observation, makes life simple. Go for it!

## Problem- ISI MStat PSB 2008 Problem 10

Consider a population with three kinds of individuals labelled 1,2 and 3. Suppose the proportion of individuals of the three types are given by , k=1,2,3 where 0< <1. Let be a random sample from this population. Find the most powerful test for testing versus . ( ).

### Prerequisites

Binomial Distribution.

Neyman-Pearson Lemma.

Test function and power function.

Hypothesis Testing.

## Solution :

This is a quite beautiful problem, only when you observe it closely. Here the distribution of X may seem non-trivial ( non-theoretical), but if one observes the distribution of Y=X-1 (say), instead of X , one will find that .

so, now let, p= 1- , so, 0<p<1, and let, and .

and since , , and our hypotheses, reduces to, versus .

so, under , our joint pmf ( of Y=X-1), is ; where and under , our joint pmf is, ; where So, now we can use, widely used Neyman-Pearson Lemma , and end up with, = = = .

now we define a test function, . for some positive constant k.

Where so, our test rule is, we reject if , and we choose k such that the for a give level , , for a given ,

with a power function , . Can you find the more subtle condition when, ? Try It!

## Food For Thought

Suppose, , can you verify, that there for any constant c, . Can you generalize the situation, what kind distribution must X follow ?? Think over it, until we meet again !

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This is a really beautiful sample problem from ISI MStat PSB 2008 Problem 10. It is based on testing simple hypothesis. This problem teaches me how observation, makes life simple. Go for it!

## Problem- ISI MStat PSB 2008 Problem 10

Consider a population with three kinds of individuals labelled 1,2 and 3. Suppose the proportion of individuals of the three types are given by , k=1,2,3 where 0< <1. Let be a random sample from this population. Find the most powerful test for testing versus . ( ).

### Prerequisites

Binomial Distribution.

Neyman-Pearson Lemma.

Test function and power function.

Hypothesis Testing.

## Solution :

This is a quite beautiful problem, only when you observe it closely. Here the distribution of X may seem non-trivial ( non-theoretical), but if one observes the distribution of Y=X-1 (say), instead of X , one will find that .

so, now let, p= 1- , so, 0<p<1, and let, and .

and since , , and our hypotheses, reduces to, versus .

so, under , our joint pmf ( of Y=X-1), is ; where and under , our joint pmf is, ; where So, now we can use, widely used Neyman-Pearson Lemma , and end up with, = = = .

now we define a test function, . for some positive constant k.

Where so, our test rule is, we reject if , and we choose k such that the for a give level , , for a given ,

with a power function , . Can you find the more subtle condition when, ? Try It!

## Food For Thought

Suppose, , can you verify, that there for any constant c, . Can you generalize the situation, what kind distribution must X follow ?? Think over it, until we meet again !

## Subscribe to Cheenta at Youtube

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