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)"