 3 x 3 Contingency Table for Chi-square Analysis

The equation below is the basic equation for the chi square test. Use the Chi square Webulator to enter your data and compute the Chi-square observed score.

 Variable 1 Condition 1 Condition 2 Condition 3 Row Sums Variable 2Condition 1 Variable 2Condition 2 Variable 2Condition 3 Sum ofColumn 1 Sum ofColumn 2 Sum ofColumn 3 GrandTotal

 Formula for column 1 Expected Responses Column 1 Formula for column 2 Expected Responses Column 2 Formula for column 3 Expected Responses Column 3 (row1 sum × col1 sum)grand total (row1 sum × col2 sum)grand total (row1 sum × col3 sum)grand total (row2 sum × col1 sum)grand total (row2 sum × col2 sum)grand total (row2 sum × col3 sum)grand total (row3 sum × col1 sum)grand total (row3 sum × col2 sum)grand total (row3 sum × col3 sum)grand total

 Chi Square Computations Column 1 Column 2 Column 3 (observed- expected)2expected (observed- expected)2expected (observed- expected)2expected Chi square formula Chi-Square=

The computed score is referred to as the chi-square observed. After computing the chi-square for the observed scores we next determine the chi-square critical score which represents the chi-square for the expected population.
The chi-square critical score for a three by three frequency table is determined by computing the “degrees of freedom” for our response set.

The computation of the degrees of freedom is:

degrees of freedom = (number of rows - 1) x (number of columns -1)

degrees of freedom = (3-1) x (3-1)

degrees of freedom = (2) x (2)

degrees of freedom = 4

and the "chi-square critical value” for degrees of freedom of "4” at p<0.05 = 9.49

Our null hypothesis in this scenario is that there is no association between the row and column variables. If the "chi-square observed value” is the "chi-square critical value of 9.49” then we would reject the null hypothesis and state that there is an association between the row and column variables. However, if the "chi-square observed value ” is the "chi-square critical value of 9.49”, we would ACCEPT the null hypothesis and state that the distributions ARE EQUAL. Click here to return to the Webulator Menu Page