Question

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 1

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.