Question

(1 point) Use software to test the null hypothesis of whether there is a relationship between the two classifications, A and B, of the 3×33×3 contingency table shown below. Test using α=0.05α=0.05. NOTE: You may do this by hand, but it will take a bit of time.

	B1B1	B2B2	B3B3	Total
A1A1	5454	4343	6868	165165
A2A2	5959	6565	7272	196196
A3A3	7575	6969	6565	209209
Total	188188	177177	205205	570570

(a)  χ2=χ2=

(b) Find the degrees of freedom.  

(c) Find the critical value.  χ2=χ2=

(d) The final conclusion is

A. We can reject the null hypothesis that A and B are not related and accept that there seems to be a relationship between A and B.  
B. There is not sufficient evidence to reject the null hypothesis that there is no relationship between A and B.

Answer 1

a)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	B1	B2	B3	Total
	A1	54.42	51.24	59.34	165
	A2	64.65	60.86	70.49	196
	A3	68.93	64.90	75.17	209
	total	188	177	205	570
chi square    χ2	=(Oi-Ei)2/Ei	B1	B2	B3	Total
	A1	0.003	1.324	1.263	2.5906
	A2	0.493	0.281	0.032	0.8065
	A3	0.534	0.259	1.375	2.1680
	total	1.0302	1.8643	2.6706	5.5651
test statistic X2 =		5.5651

b)

degree of freedom(df) =(rows-1)*(columns-1)=

4

c)

for 4 df and 0.05 level , critical value       χ2=

9.4877

d)

B. There is not sufficient evidence to reject the null hypothesis that there is no relationship between A and B.