In: Statistics and Probability
A poll classified respondents by gender and political party, as shown in the table. We wonder if there is evidence of an association between gender and party affiliation. (Assume a significance level of alpha=0.05)
D R I
Male 36 45 25
Female 51 38 13
calculate chi-square statistic x²=
Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: Two categorical variables are independent.
Alternative hypothesis: Ha: Two categorical variables are dependent.
We assume/given level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
We are given
Number of rows = r = 2
Number of columns = c = 3
Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2
α = 0.05
Critical value = 5.991465
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
|
Observed Frequencies |
||||
|
Column variable |
||||
|
Row variable |
Democratic |
Republic |
Independent |
Total |
|
Male |
36 |
45 |
25 |
106 |
|
Female |
51 |
38 |
13 |
102 |
|
Total |
87 |
83 |
38 |
208 |
|
Expected Frequencies |
||||
|
Column variable |
||||
|
Row variable |
Democratic |
Republic |
Independent |
Total |
|
Male |
44.33653846 |
42.29807692 |
19.36538462 |
106 |
|
Female |
42.66346154 |
40.70192308 |
18.63461538 |
102 |
|
Total |
87 |
83 |
38 |
208 |
|
Calculations |
||
|
(O - E) |
||
|
-8.33654 |
2.701923 |
5.634615 |
|
8.336538 |
-2.70192 |
-5.63462 |
|
(O - E)^2/E |
||
|
1.567508 |
0.172594 |
1.639466 |
|
1.628979 |
0.179362 |
1.703759 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 6.891667643
χ2 = 6.891667643
P-value = 0.03187817
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is sufficient evidence to conclude that there is an association between gender and party affiliation.