A New Sequential Non-Parametric Test for the Two Sample Problem

 

Keerthi M. Mathad^*, I. D. Shetty­­

Department of Statistics, Karnatak University, Dharwad-580003, INDIA.

*Corresponding Address:

[email protected]

 

 

1. Introduction

Phatarfod and Sudbury (1988) considered a Wald-type test for the two-sample problem. Unlike in Bradley, Merchant and Wilcoxon (1966) they exploited the normal approximation to the Wilcoxon Mann-Whitney statistic. They first considered the Lehmann alternative and later on wider type of alternatives.

In this paper we consider a Wald-type test for the two-sample problem. We take   , . Here we wish to make a judgment between two new treatments and there is no prior preference involved. There would be a region of indifference, where, if the true situation was, it would not matter which of the two treatments was chosen. We propose the following Sequential test of strength. Observations are taken from the X and Y populations in pairs. At each stage, the U-statistic

Where             

=Median of X₁, X₂, X₃ and      =Median of Y₁,Y₂, Y₃
is calculated.

Sampling is continued as long as

          Where                      and                  

And  or  is accepted according as the R.H.S or the L.H.S inequality is the first not satisfied.

 

2. Mean and Variance of the U-statistic when

 MEAN of U is given by .

VARIANCE of U is given by

Var (U)   =

Where, = Covariance between the two kernels wherein cX observations and dY observations are in common between the two kernels.                                

Var (U)                                                      (2.1)

3. Mean and variance of the U-statistic when  and

MEAN is given by      

VARIANCE of U is given by

Var(U)=

Where, =Covariance between the two kernels wherein cX observations and dY observations are in common between the two kernels.        

Var(U)

                                               (3.1)

Where, ,,,,,,,,are as defined earlier. Details are omitted for the sake of brevity.     

4. The Sequential Non-Parametric Test

We have developed the SPRT based on the statistic U using the normal approximation to our statistic U when n is large for testing the simple hypothesis   Vs.

I.e. the continuation region at stage m is given by

Where  and

 

5. The Average Sample Number

The sample size needed to reach a decision in a sequential or a multiple sampling plan is a random variable N, because at any stage of the experiment the decision to terminate the process depends on the results of the observations made earlier. The distribution of this random variable depends on the true distribution of the observations during the sampling process.

The ASN is given by

E (n) = L () log B+ [1-L ()] log A/ E (z)      where  is the parameter

If E (z) =0 then, E (n) =-logAlogB/E (z­²)

Where    E (z­²) = ( ₀- ₁)²/ ²   i.e. E (n) = (-log A log B/( ₀- ₁)²)* ²

For this test the ASN function was obtained. To derive the ASN function, we note that the ASN function of our test is given by

E(N/K)=( L(K)log B+[1-L(K)]log A)/ E(z)          Where, E(z)=

 The following tables and graphs give the ASN for different values of k₀

 

The values of ASN function for  and the corresponding graph.

Table 1

                       

The values of ASN function for  and the corresponding graph

Table 2

 

The values of ASN function for  and the corresponding graph

 

 

Table 3

 

The values of ASN function for  and the corresponding graph

Table 4

 

The values of ASN function for  and the corresponding graph

Table 5

The test has greater ASN for the case K­₀=0.9

6. Average Sample Number for Uniform Distribution

We have compared our test with other tests for the same problem. We have considered two distributions, namely Uniform and Exponential distributions. We have found the mean and variance for the two distributions and later found the ASN function.

We have the uniform distribution

                        

                        

We have obtained the mean and variance of the statistic U under uniform distribution. The calculation is omitted for the sake of brevity.

 

 

 

 

The ASN for different values of

For

 

Table 6

 

 

Table 7

 

For

 

Table 8

 

For

Table 9

 

 

 

For

Table 10

 

7. Average Sample Number for Exponential Distribution

We now consider the exponential distribution we have found the mean and variance for the distribution and later found the ASN function.

   We have the exponential distribution

                                

                         

We have obtained the mean and variance of the statistic U under exponential distribution. The calculation is omitted for the sake of brevity.

 

The ASN for different values of δ

For

Table 11

 

For  

Table 12

 

  

 

 

For

Table 13

 

For

Table 14

 

For

 

Table 15

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