Question

You are conducting a test of the claim that the row variable and the column variable are dependent in the following contingency table. х Y Z A[ 324 28 B27 21 35 Give all answers rounded to 2 places after the decimal point, if necessary. (a) Enter the expected frequencies below: х Y Z A B (b) What is the chi-square test-statistic for this data? Test Statistic: x2 (c) What is the p-value for this test of independence P-Value: = (d) What is the correct conclusion of this hypothesis test at the 0.05 significance level? There is sufficient evidence to support the claim that the row and column variables are dependent. There is not sufficient evidence to warrant rejection of the claim that the row and column variables are dependent. There is not sufficient evidence to support the claim that the row and column variables are dependent. There is sufficient evidence to warrant rejection of the claim that the row and column variables are dependent.

Answer 1

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: H_a: Two categorical variables are dependent.

We are 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	X	Y	Z	Total
A	32	4	28	64
B	27	21	35	83
Total	59	25	63	147

Part a

Answer:

Expected Frequencies
	Column variable
Row variable	X	Y	Z	Total
A	25.68707	10.88435	27.42857	64
B	33.31293	14.11565	35.57143	83
Total	59	25	63	147

Calculations
(O - E)
6.312925	-6.88435	0.571429
-6.31293	6.884354	-0.57143

(O - E)^2/E
1.551482	4.354354	0.011905
1.196323	3.357574	0.00918

Part b Answer:

Test Statistic = Chi square = ∑[(O – E)^2/E] = 10.48082

χ² = 10.48

P-value = 0.005298

(By using Chi square table or excel)

Part c Answer:

P-value = 0.01

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to support the claim that the row and column variables are dependent.