Question

You are conducting a test of independence for the claim that there is an association between the row variable and the column variable.

	X	Y	Z
A	26	52	43
B	54	23	46

What is the chi-square test-statistic for this data? χ 2 =

Report all answers accurate to three decimal places.

Answer 1

Solution Х 2 26 The pouled table is as follows: Y Total 43 lad 2 3 4 6 123 75 244 A 13 52 54. L. Total 89 Here we have to compute the value of chi-squeeze test statistic The expected values are computed in teams of www and column totals. In fact, the formula is Eij Rebel Rixli where Ri corresponds T to the total sum of elements in now i cj corresponds to the total sum of elements Column j, and I is the grand total. The table below shows the calculations to obtain the table with expected values : Expected blues X 4 Z Total gox12, -- 39.672 75X120 244 -37.193 89712T 244 =44.135 244 121 B 80X123 24.4 40.328 +5X123 29.4 = 37.8017 89x123 244 44-865 123 Jotal 75 89 244 Based on the obsewed, and expected values, the squared distances can be following formula 2 (-03%€ , The table coith ccording to the squared distances is shown below computed as

2 X y Squared Distances (26-39.673)= 4.712 652 – 37 193) 5.895 (43-44.135)?= 0.029 39.672 37.493 44.135 (54-40.323): 4.635 (23-37507) 5.799 (46-44.865){-0.029 4.0.328 37.807 44.865 The Chi-squared statistic is computed as follows in 72 I (aj-sip) Car-j- 4.712 + 4- 635+5.895+5.7994 Eig 0·029 to-029 21.ogg 21-099 Therefore the chi-square statistic is x?= 21.099