Question

	Member	Not a Member	Total
Male	225	158	383
Female	306	112	418
Total	531	270	801

what is the Calculation.  of the test statistic for a Chi-Square test for independence using the chart above?

Answer 1

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: Two variables gender and whether member or not are independent.

Alternative hypothesis: Ha: Two variables gender and whether member or not 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 = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1

α = 0.05

Critical value = 3.841459149

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	Member	Not a Member	Total
Male	225	158	383
Female	306	112	418
Total	531	270	801

Expected Frequencies
	Column variable
Row variable	Member	Not a Member	Total
Male	253.8988764	129.1011236	383
Female	277.1011236	140.8988764	418
Total	531	270	801

Calculations
(O - E)
-28.8989	28.89888
28.89888	-28.8989

(O - E)^2/E
3.289282	6.468922
3.013864	5.927266

Test Statistic = Chi square = ∑[(O – E)^2/E] = 18.69933329

χ2 statistic = 18.69933329

P-value = 0.0000153

(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 two categorical variables gender and whether member or not are dependent.