Question

s 1/3] 8/3] /3] 1.5/3) 3/3] 3/3] 3/3] 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 A 19 28 39 B 14 43 39 3/3] Give all answers rounded to 3 places after the decimal point, if necessary, (a) Enter the expected frequencies below: X Y Z 3/3) (2.7/3) (3/3) [3/3) [3/3) [3/3] [3/3) 1.2/45 A lon (b) What is the chi-square test-statistie for this data? Test Statistic: X (c) What is the critical value for this test of independence when using a significance level of a -0.025? Critical Value: x (d) What is the correct conclusion of this hypothesis test at the 0.025 significance level? 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 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 Remember to give all answers rounded to 3 places after the decimal point, if necessary

Answer 1

(a)

Observed Values: C1 R1 19 R2 14 Column Total 33 C2 28 43 71 СЗ 39 39 78 Row Total 86 96 182 Expected Values = [(Row total x Column total)/Grand total] C1 C2 C3 R1 15.593 33.549 36.857 R2 17.407 37.451 41.143

(b)

0 19 14 28 43 39 39 E 15.593 17.407 33.549 37.451 36.857 41.143 Total (0-E)?/E 0.744 0.667 0.918 0.822 0.125 0.112 3.387 x=2(0-E)? 3.387 E

(c)

Number of rows = r = 2 Number of columns =C = 3 D.F =(r-1)x(C-1) = 2 Significance level= a = 0.025 Critical Value = 7.378 (Using Excel command CHIINV(a,D.F))

(d) Since 3.387 < 7.378, we fail to reject H0.

There is not sufficient evidence to support the claim that the row and column variables are dependent