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Question 797:



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  1. Q797_1_q797_chi_square.xls

For this question we'll use the Chi-Square test of independence. We have a contingency table which shows the relationship between NATFARE and race. We want to know if there is an association (dependence) between these two variables.  The general formula for a Chi-Square is to calculate the observed count in each cell minus the expect count, all squared, then this result divided by the expected count.  See the attached excel file for the calculations.


The null hypothesis for this test is that there is no difference between NATFARE and race. We will reject the NULL if the p-value is less than .05.


  1. There are 4 degrees of freedom for this table. That is found as the (number of rows-1) times (number of columns-1) = (3-1)*(3-1) = 4.
  2. The Chi-Square test statistics is 10.097 on 4 degrees of freedom. The p-value associated with this is .038824. Since the p-value is less than .05 we reject the Null Hypothesis and conclude that there is an association between NATFARE (attitude toward welfare spending) and race. That, is, differenct race's have different attitudes toward welfare spending

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