Question

For each of the following situations give the degrees of freedom and an appropriate bound on the P-value (give the exact value if you have software available) for the χ² statistic for testing the null hypothesis of no association between the row and column variables.

(a) A 2 by 2 table with χ² = 0.94.

df =	1
P-value =	2

(b) A 4 by 4 table with χ² = 17.32.

df =	3
P-value =	4

(c) A 2 by 8 table with χ² = 22.47.

df =	5
P-value =	6

(d) A 5 by 3 table with χ² = 13.26.

df =	7
P-value =	8

Answer 1

(a). For a table with = 0.94,

(1). the value of  the degrees of freedom is = (k-1)(l-1) = $(2-1)×(2-1)=1×1=1$ .

(2). The P-value of test is , at 0.05 level of significance.

(b). For a table with = 17.32,

(3). The value of  the degrees of freedom is = (k-1)(l-1) =$(4-1)×(4-1)=3×3=9$

(4). The P-value of test is , at 0.05 level of significance

(c). For a table with = 22.47,

(5). ​The value of  the degrees of freedom is = (k-1)(l-1) =$(2-1)×(8-1)=1×7=7$

(6). The P-value of test is , at 0.05 level of significance

(d). For a table with = 13.26,

  (7). The value of  the degrees of freedom is = (k-1)(l-1) =$(5-1)×(3-1)=4×2=8$

(8). The P-value of test is
PlX >= X2) = 0.103211
, at 0.05 level of significance