Question

You are conducting a test of the claim that the row variable and the column variable are dependent in the following contingency table. X Y Z A24 51 | 31 B 24 54 29 Give all answers rounded to 3 places after the decimal point, if necessary. (a) Enter the expected frequencies below: x Y Z 00 (b) What is the chi-square test-statistic for this data? Test Statistic: x (c) What is the critical value for this test of independence when using a significance level of a = 0.005? Critical Value: x² = (d) What is the correct conclusion of this hypothesis test at the 0.005 significance level? There is not sufficient evidence to support the claim that the row and column variables are dependent. There is 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. There is not sufficient evidence to warrant rejection of the claim that the row and column variables are dependent. Remember to give all answers rounded to 3 places after the decimal point, if necessary.

Answer 1

Solution-:

Here, we want to test

The row variabler and column variable  are independent of each other.

Но

Vs

$H_1:$ The row variabler and column variable  are dependent of each other.

The given frequencies table are observed frequencies $(O_{ij})$

	X	Y	Z	Total
A	24	51	31	106
B	24	54	29	107
Total	48	105	60	213

(a) The corresponding expected frequencies $(e_{ij})$ are obtained using formula

$e_{ij}=\frac{(A_{i})*(B_{j})}{N }$   for i=1,2,; j=1,2,3

Note: are not to be rounded-off to the integers.

S

The given frequencies table are expected frequencies $(e_{ij})$

	X	Y	Z
A	23.887	52.254	29.859
B	24.113	52.746	30.141

$N=\sum (O_{ij})=\sum (e_{ij})=213$

(b) For the chi-square test statistic for this data we proceed furtuer;

Table for  $O_{ij}^2/e_{ij}$

	X	Y	Z
A	24.113	49.777	32.184
B	23.888	55.283	27.902

$\sum_{i=1}^{2} \sum_{j=1}^{3}\frac{O_{ij}^2}{e_{ij}}=213.148$

The test statistic is

$\chi^2_{stat}=\chi^2_{(2-1)*(3-1)}=\chi^2_{2}=\sum_{i=1}^{2} \sum_{j=1}^{3}\frac{O_{ij}^2}{e_{ij}}-N$

$=213.148-213$

  $=0.148$

(c) Table value / Critical value :-

$\chi^2_{2,0.005}=10.597$

( From statistical table or By using MS-Excel Command "=CHIINV(0.005,2)" )

Decision Rule:-

Here $\chi^2_{cal}=0.148 <\chi^2_{2,0.005}=10.597$ hence we accept $H_o$ at l.o.s.

g= 0,05

(d) Conclusion:- There is not sufficient evidence to support the claim that the row and column variables are dependent.

Option (1) is correct.