Small Sample Tests : Tests based on:-1. Chi square (χ2) distribution 2. t- distribution 3. F- distribution

 

Small Sample Tests

            We have seen large sample tests for testing hypothesis about population mean, population proportion and population correlation coefficient. However in number of situations exact tests or small sample tests can be used for testing hypothesis about a population parameter under some assumptions.

           In small sample tests, the statistic follows exact sampling distribution viz. chi square (χ2), t or F for sample of any size. Hence small sample tests can be regarded as exact tests. The procedure of testing the null hypothesis H0 in small sample tests is almost identical to that of large sample test. We divide these tests into three classes according to the distribution of test statistic as follows.

Tests based on:-

                            1. Chi square (χ2) distribution

                            2.  t- distribution

                            3. F- distribution


 Chi-square test for goodness of fit:

       The Chi-square test developed by Prof. Karl Pearson in 1900, is a powerful test for testing the significance of the difference between observed frequencies and theoretical (expected) frequencies.

          Let Oi, i=1, 2, 3,.....k is a set of observed frequencies and Ei, i=1, 2, 3,.....k is the corresponding set of expected frequencies. Under H0, the sum of observed and expected frequencies is equal.

Here we have to test null hypothesis,

          H0: There is no significant difference between the observed and expected frequencies or the fitting is good.

Against, H1: There is significant difference between the observed and expected frequencies or fitting is not good.

Under the null hypothesis, Karl Pearson proved that the statistic,


Note:

1. We can apply this test, if expected frequencies are greater than or equal to 5 and total frequency is greater than 50.

2. When expected frequency of a class is less than 5, the class is merged into neighboring class along with its observed and expected frequencies until total of expected frequencies becomes greater than or equal to 5. This is called ‘pooling the classes’ and number of class frequencies after pooling is the value of k.

3. If any parameter is not estimated while fitting a probability distribution or obtaining expected frequencies the value of the p is zero.

ii) Test for Independence of Attributes:

       Let us suppose that the given population consisting of items is divided into m classes A1,A2,A3,.........., Am w. r. t. the attribute A, so that randomly selected items belongs to one and only one of the attributes A1,A2,A3,.........., Am. Similarly, Let us suppose that the given population consisting of items is divided into n classes B1,B2,B3,.........., Bn  w. r. t. the another attribute B, so that randomly selected items belongs to one and only one of the attributes B1,B2,B3,.........., Bn. The frequency distribution of the items belonging to the classes A1,A2,A3,..........,Am and B1,B2,B3,..........,Bn can be represented in m x n  manifold contingency table. 


A/B

B1

B2

................

Bj

................

Bn

Total

A1

(A1B1)

(A1B2)

................

(A1Bj)

................

(A1Bn)

(A1)

A2

(A2B1)

(A2B2)

................

(A2Bj)

................

(A2Bn)

(A2)

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

Ai

(AiB1)

(AiB2)

................

(AiBj)

................

(AiBn)

(Ai)

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

:

Am

(AmB1)

(AmB2)

................

(AmBj)

................

(AmBn)

(Am)

Total

(B1)

(B2)

................

(Bj)

................

(Bn)

N

 e Ai, i= 1, 2, 3,........., m, (Bj) is the number of persons possessing the attribute Bj, j= 1,2,3,.........,n and (AiBj) is the number of persons possessing both the attributes Ai and Bj, i= 1, 2, 3,......., m, j=1,2,3,.....,n. Also, is the total frequency. Now, we have to test whether the two attributes A and B are independent or not.

i.e. H0: Attributes A and B are independent  against the alternative

      H1: Attributes A and B are not independent.

Under the null hypothesis that the attributes A and B are independent, the expected frequencies are calculated as follows;

 2× 2 contingency table

For 2 ×2 contingency table under the hypothesis of independence of A and B,


          B

A

B1             

B2

Total

A1

  a             

      b

  (a+b)

A2

  c             

     d

  (c+d)

Total

(a+c)

(b+d)

a+b+c+d= N

 

***

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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