Question

The following table contains observed frequencies for a sample of 200.

Row Variable	Column Variable
	A	B	C
P	16	40	46
Q	34	30	34

Test for independence of the row and column variables using α = 0.05.

State the null and alternative hypotheses.

H₀: Variable P is not independent of variable Q. H_a: Variable P is independent of variable Q.

H₀: Variable P is independent of variable Q.H_a: Variable P is not independent of variable Q.    

H₀: The column variable is independent of the row variable. H_a: The column variable is not independent of the row variable.

H₀: The column variable is not independent of the row variable. H_a: The column variable is independent of the row variable.

Find the value of the test statistic. (Round your answer to three decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Do not reject H₀. We cannot conclude that there is an association between the column variable and the row variable.

Reject H₀. We conclude that the column and row variables are independent.    

Do not reject H₀. We cannot conclude that the column and row variables are independent.

Reject H₀. We conclude that there is an association between the column variable and the row variable.

Answer 1

Row Variable	Column Variable
	A	B	C	Total
P	16	40	46	102
Q	34	30	34	98
Total	50	70	80	200

Row Variable	Column Variable    ( Expected Table )
	A	B	C	Total
P	(102 * 50) / 200 = 25.5	(102 * 70) / 200 = 35.7	(102 * 80) / 200 = 40.8	102
Q	(98 * 50) / 200 = 24.5	(98 * 70) / 200 = 34.3	(98 * 80) / 200 = 39.2	98
Total	50	70	80	200

	Oi	Ei	( Oi - Ei )	( Oi - Ei )2	( Oi - Ei )2 / Ei
	16	25.5	-9.5	90.25	3.5392
	40	35.7	4.3	18.49	0.5179
	46	40.8	5.2	27.04	0.6627
	34	24.5	9.5	90.25	3.6837
	30	34.3	-4.3	18.49	0.5391
	34	39.2	-5.2	27.04	0.6898
Total	200	200	0	271.56	9.6324

To Test :-

H₀: The column variable is independent of the row variable. H_a: The column variable is not independent of the row variable.

Test Statistic :-
X² = Σ (Oi - Ei )² / Ei
X² = 9.6324

Test Criteria
Reject null hypothesis if X² > X²_{(α, n-1)}
Critical value X²_(0.05,2-1) = X²_(0.05,1) = 3.841 ( From chi square table )
Since, 9.6324 > 3.841
Conclusion = Reject null hypothesis

Decision based on P value
P (X² > 9.6324) = 0.0019
Reject null hypothesis if P value < α = 0.05
P value = 0.0019 < 0.05, hence we reject null hypothesis
Conclusion = Reject null hypothesis

Reject H₀. We conclude that there is an association between the column variable and the row variable.