In: Statistics and Probability
| 
 Culture\Custom  | 
 X  | 
 Y  | 
 Z  | 
 Total  | 
| 
 A  | 
 49  | 
 71  | 
 88  | 
 208  | 
| 
 B  | 
 102  | 
 24  | 
 24  | 
 150  | 
| 
 C  | 
 9  | 
 39  | 
 43  | 
 91  | 
| 
 Total  | 
 160  | 
 134  | 
 155  | 
 449  | 
Does this data provide enough evidence at the α = 0.05 confidence level that there is dependence based on someone’s culture and a particular custom they uphold? (X,Y,Z are customs and A,B,C are cultures)
Hypotheses are:
H0: There is no dependence based on someone’s culture and a particular custom they uphold.
Ha: There is dependence based on someone’s culture and a particular custom they uphold.
Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:
| Culture\Custom | X | Y | Z | Total | 
| A | 74.12 | 62.076 | 71.804 | 208 | 
| B | 53.452 | 44.766 | 51.782 | 150 | 
| C | 32.428 | 27.158 | 31.414 | 91 | 
| Total | 160 | 134 | 155 | 449 | 
Following table shows the calculations for chi square test statistics:
| O | E | (O-E)^2/E | 
| 49 | 74.12 | 8.513416082 | 
| 102 | 53.452 | 44.09392172 | 
| 9 | 32.428 | 16.92584137 | 
| 71 | 62.076 | 1.282907662 | 
| 24 | 44.766 | 9.632907921 | 
| 39 | 27.158 | 5.163596878 | 
| 88 | 71.804 | 3.653144894 | 
| 24 | 51.782 | 14.90555645 | 
| 43 | 31.414 | 4.273107404 | 
| Total | 108.4444004 | |
Following is the test statistics:

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (3-1)*(3-1) = 4
The p-value is: 0.0000
Conclusion: Since p-value is less than 0.05 so we reject the null hypothesis. That is we can conclude that there is dependence based on someone’s culture and a particular custom they uphold.
Excel function used for p-value: "=CHIDIST(108.44, 4)"