Chi Square Test for Independence

• Uses a cross classification table to examine the nature of the relationship between these variables
• Tables are sometimes referred to as contingency tables
• Determine variables are dependent on each other or not

Approach

• H0: chi square test for independence is conducted by assuming that there is no relationship between the two variables
• Ha: alternative hypothesis is that there is some relationship between the variables

The general formula for the degrees of freedom is the number of rows minus one, times the number of columns minus 1.

In terms of independence and dependence these hypotheses could be stated

• H0 : X and Y are independent
• H1 : X and Y are dependent

Expected Frequency = ((row total)*(column total))/Total Population

I liked the example provided in link  

Problem - Test for a Relationship between Sex and Class

X (Sex)

Y (Social Class) Male(M) Female(F) Total

Upper Middle (A) 33 29 62

Middle (B) 153 181             334

Working (C) 103 81 184

Lower (D) 16 14 30

Total 305 305            610

Table 10.12: Social Class Cross Classified by Sex of Respondents

Expected Frequency = ((row total)*(column total))/Total Population

1-pchisq(4.8748,df=3)

 0.1811978

Significance is greater than or equal to 0.05, you don't reject the null hypothesis

Results match with the problem although approach is different. The sum total sum is 610 (Total Sum)

Happy Learning!!!