You are conducting a test of independence for the claim that there is an association between the row variable and the column variable.
X | Y | Z | |
---|---|---|---|
A | 30 | 34 | 21 |
B | 40 | 35 | 21 |
What is the chi-square test-statistic for this data?
Answer:
Chi-square test-statistic = 0.777427
Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: There is no association between the row variable and the column variable.
Alternative hypothesis: Ha: There is an association between the row variable and the column variable.
We assume 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 |
X |
Y |
Z |
Total |
A |
30 |
34 |
21 |
85 |
B |
40 |
35 |
21 |
96 |
Total |
70 |
69 |
42 |
181 |
Expected Frequencies |
||||
Column variable |
||||
Row variable |
X |
Y |
Z |
Total |
A |
32.87293 |
32.40331 |
19.72376 |
85 |
B |
37.12707 |
36.59669 |
22.27624 |
96 |
Total |
70 |
69 |
42 |
181 |
(O - E) |
||
-2.87293 |
1.596685 |
1.276243 |
2.872928 |
-1.59669 |
-1.27624 |
(O - E)^2/E |
||
0.251079 |
0.078677 |
0.08258 |
0.22231 |
0.069662 |
0.073118 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 0.777427
χ2 statistic = 0.777427
P-value = 0.677928
(By using Chi square table or excel)
P-value > α = 0.05
So, we do not reject the null hypothesis
There is not sufficient evidence to conclude that there is an association between the row variable and the column variable.
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