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Chi-Square Test of Independence

1. Testing the association between two nominal variables

When measuring the association between two nominal variables, one can choose between using a Chi-square test or a Fisher’s Exact test. For instance, the Chi-square test is more appropriate when less than 20% of the cells to have expected count <5. If this assumption is violated, the Fisher’s Exact test could be more suitable.

Chi-Square Test of Independence

Null Hypothesis: The two variables are independent of each other.

Alternative Hypothesis: The two variables are related with each other.

2. Formula for Chi-Square Test of Independence

Chi-square statistics = Summation of (Oi – Ei)^2/Ei

3. Decision rules for Chi-Square Test of Independence

For such test, one can determine whether or not to reject the null hypothesis by comparing the test statistics they calcuated with the critical value. The null hypothesis will be rejected if the test statistics is more than the critical value.

 

Exercise

Conduct a Chi-Square test for the following set of data and determine whether the null hypothesis is rejected.

 

Question 1

 

v1 v2
2 2
1 2
2 1
1 1
1 2
2 2
2 1
2 2
1 2
2 2
1 2
2 2
1 1
1 1
2 2
1 1
2 1
2 1
1 1
1 1
2 1
1 1
1 2
1 1
2 2
1 2
2 1
1 2
2 2
2 2
1 2
2 2
2 1
1 2
1 2
1 2
1 2
2 2
1 1
1 2
2 2
2 2
2 2
1 1
2 2
2 1
2 1
1 1
2 1

 

Question 2

 

v3 v4
1 2
1 1
2 2
1 1
2 1
2 1
2 2
2 1
2 2
1 2
1 2
2 1
1 2
2 1
1 1
1 1
1 1
1 1
2 2
2 2
2 1
2 1
2 2
1 1
1 1
2 2
1 1
1 2
2 2
1 1
1 2
2 1
2 1
2 2
1 2
2 2
2 1
2 1
1 2
2 1
2 1
1 2
1 2
2 2
2 1
1 2
1 1
1 2
2 1

 

Question 3

 

v5 v6
2 2
2 1
1 1
1 2
1 2
1 1
1 1
1 1
2 1
2 1
2 2
2 2
2 1
2 2
2 1
1 2
1 2
2 1
1 1
2 1
1 2
1 2
2 1
2 2
1 2
1 2
2 2
2 2
1 2
1 2
2 2
1 1
2 2
1 2
2 2
1 1
1 2
2 1
2 2
1 1
1 1
2 2
1 2
1 2
2 2
2 1
1 2
2 2
2 1

 

Question 4

 

v7 v8
1 1
1 1
1 2
1 2
1 1
1 1
1 1
1 1
2 2
1 1
2 1
1 2
2 2
2 2
2 1
2 2
1 2
1 1
2 1
2 2
1 2
1 1
2 2
2 1
2 1
1 1
2 1
1 2
2 2
2 2
2 2
2 2
2 1
2 1
2 1
1 1
2 1
2 2
1 2
2 2
1 1
1 1
1 1
1 1
1 2
2 1
1 2
2 1
1 1

Question 5

 

v9 v10
2 2
1 2
1 2
1 2
2 2
1 2
1 1
1 2
2 1
1 2
1 2
2 2
2 2
1 1
2 1
2 1
1 2
2 2
1 1
1 2
1 1
1 2
1 1
1 2
2 2
2 1
1 1
2 2
1 1
2 1
1 2
1 2
1 2
2 2
1 2
2 2
2 1
1 2
1 2
2 1
1 2
2 2
2 2
2 1
2 1
1 2
2 2
2 1
1 1

Question 6

 

v11 v12
1 1
2 1
2 2
1 1
2 2
2 2
2 2
2 1
2 2
2 2
1 1
2 1
1 1
1 1
2 1
1 2
1 2
1 2
1 2
2 2
2 2
2 1
1 2
2 1
1 1
2 1
2 1
1 2
1 2
2 1
1 2
1 2
1 1
2 1
1 2
2 1
2 1
2 2
1 2
1 1
1 1
1 2
1 2
1 2
2 2
2 1
2 1
2 1
2 2

Answer

  1. p-value = 0.680, do not reject the null hypothesis
  2. p-value = 0.490, do not reject the null hypothesis
  3. p-value = 0.644, do not reject the null hypothesis
  4. p-value = 0.322, do not reject the null hypothesis
  5. p-value = 0.253, do not reject the null hypothesis
  6. p-value = 0.195, do not reject the null hypothesis

 

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