*Chi-square TestDr. T. T. Kachwala

Using the Chi-Square Test*The following are the two Applications: Chi square as a test of IndependenceChi square as a test of Goodness of Fit

2 statistics measures difference between fo and feis always positiveif fe is small, it is over estimated (limitation of 2 statistics)

Chi-square as a Test of Independence

*ObjectiveIf we classify a population into two attributes with several classes, we can use Chi-square test to determine whether or not the two attributes are independent of each other.

Chi-square as a test of Independence

*Contingency Table: It is a table of classification of two attributes into a number of classes. One of the attributes is classified along the rows, while the second attribute is classified along the columns.

Chi-square as a test of Independence

*Observed and Expected Frequency Observed frequency is based on actual observation (Study or Survey)Expected frequency is based on theoretical calculations. The expected or theoretical frequency can be obtained for any cell as follows:fe =expected frequency of cellRT =Row total corresponding to that cellCT =Column total corresponding to that cellN =Total number of observation

Degree of Freedom (for Contingency table of size m * n)*Degree of Freedom is the number of independent observations that can be arbitrarily assigned without violating the restrictions of the problem. For example: Table of size 2*2; = 1Degree of Freedom for table of size m x n, is given by the following formula: = (m-1) (n-1){where m is the number of rows & n is the number of columns}. For example:Table of size 2*2 : m = 2, n = 2; = 1*1 = 1Table of size 3*2 : m = 3, n = 2; = 2*1 = 2

The Chi-Square Distribution*2CriticalThe sampling distribution of the statistic 2 can be closely approximated by a continuous curve known as Chi-square distribution. As in case of t distribution, there is different 2 distribution for each different number of degrees of freedom. However in practical research work, only a few values of are popular (0.05, 0.01, 0.1). Chi-square distribution is a skewed distribution & is defined by level of significance & degree of freedom as indicated below (for = 0.05) :0.05

Summarized procedure for Chi-Square Test of Independence

*Step (i)H0 : The two attributes are independentH1 : One attribute depends on the other attributeStep (ii)Assuming = 0.05, 2 distribution & = (m-1) * (n-1) 2Critical = {from table}

Step (iii) Calculate 2 statistic

Summarized procedure for Chi Square as a Test of Independence

*Step (iv & v) : Decision Rule & Conclusion2 CriticalIf 2 statistic is in acceptance area

Accept H0: the two attributes are independentIf 2 statistic is in rejection area

Reject H0 : the two attributes are independentAccept H1: One attribute depends on the other

Chi-square as a Test of Goodness of Fit *ObjectiveTo assess whether or not there is a significant difference between observed frequency fo & expected frequency feThe term Goodness of Fit signifies how well the theoretical distribution like Binomial, Poisson or Normal distribution fits or represents the observed frequency distribution

Chi-square as a Test of Goodness of Fit *The following is the summarized procedure: Step (i)H0 : fo = fe(Theoretical distribution is a good fit)H1: Not all fo are equal to fe (Theoretical distribution is not a good fit)Step (ii) Assuming =0.05, 2 critical = {from table depending on } ( depends on the Probability Distribution)Step (iii) Calculate 2 statistic

Chi-square as a Test of Goodness of Fit *Step (iv & v) : Decision Rule & Conclusion2 CriticalIf 2 statistic lies in acceptance area Accept H0: fo= fei..e Theoretical distribution is a Good Fit

If 2 statistic lies in rejection area Reject H0: fo= feAccept H1: Not all fo are equal to fe i.e. Theoretical distribution is not a Good Fit for the given fo

Precaution about using Chi-square Test*When the expected frequencies fe are too small, the value of 2 will be overestimated To avoid making incorrect inferences from 2 test follow the general rule that an expected frequency of less than five in one cell of the contingency table is too small to use.One of the suggested adjustments for small value of fe is combining frequency of the cells in the table. i.e. if the table contains more than one cell with an expected frequency of less than 5, we can combine them in order to get an expected frequency of 5 or more.

*Thanks and Good Luck Dr. T. T. Kachwala