Chi square test for independence of attribute , Operation Research

Chi Square Test for Independence  of Attribute

The chi square  test  can be  used to find out  whether two  or more attributes are associated or not. This  test helps  in finding the association or  independence of two  or more  attributes. In  order  to test  the independence  of two  attributes  you should take  the null  hypothesis that the two attributes  are  independent. The second  step under  null  hypothesis is to  compute the expected frequencies for  each  cell of  contingency table. In  case of  contingency table  the expected frequency  for ( A1, B1) will be  calculated as:

E ( A1, B1)  = (A1) (B1) / N

After computing expected frequencies you can  calculate the value  of chi square . the   calculated  value  of chi square  will be then compared with  table value at certain level of significance ( usually at 5% for  ( r- 1) ( c-1) degree of freedom. If  the calculated value  is lower than table  value then you will  accept the  null hypothesis  that two  attributes are independent. In case  calculated  value is greater than  the table value  then the  null hypothesis will be  rejected and you will  conclude that two  attributes are associated.

Posted Date: 2/26/2013 11:36:56 PM | Location : United States







Related Discussions:- Chi square test for independence of attribute , Assignment Help, Ask Question on Chi square test for independence of attribute , Get Answer, Expert's Help, Chi square test for independence of attribute Discussions

Write discussion on Chi square test for independence of attribute
Your posts are moderated
Related Questions

These models are applied to the management ( planning controlling and scheduling ) of large scale projects. PERT/ CPM techniques help in identifying potential trouble spots in

A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper i

b. A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y pape

Observation and Data Collection for Better Understanding of the Problem: Many times actual observations by trained observers at the scene of operation may be difficult and da

LIMITS OF TRANSPOTATION PROBLEM

A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper i

Introduction to Probability Distribution By  theoretical distribution we mean  a frequency distribution  which  is obtained  in relation to a random  variable by some  mathem

Use  of the Graphs a . It simplifies  the complexities of large numbers  or that of  large mass  of data. In general  graphical  or some  other method or representation  is ad

A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper i