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 = 0.97.

df =
P-value =

(b) A 4 by 4 table with χ² = 18.55.

df =
P-value =

(c) A 2 by 8 table with χ² = 23.16.

df =
P-value =

(d) A 5 by 3 table with χ² = 13.26.

df =
P-value =

Answer 1

The distribution of the statistic X² is chi-square with (r-1)(c-1)

a) A 2 by 2 table with χ² = 0.97

df=(2-1)(2-1)=1

P-value=P(X²>=0.97)

=CHIDIST(0.97,1)

= 0.32468

Since p-value is greater than 0.05, fail to reject the null hypothesis of no association between row and column variables.

b) A 4 by 4 table with χ2 = 18.55.

df=(4-1)(4-1)=3*3=9

P-value=P(X²>=18.55)

  =CHIDIST(18.55,9)

= 0.029303

Since p-value is lower than 0.05, we reject the null hypothesis of association between row and column variables.

c) A 2 by 8 table with χ2 = 23.16.

df=(2-1)(8-1)=1*7=7

P-value=P(X²>=23.16)

  =CHIDIST(23.16,7)

  = 0.001599

Since p-value is lower than 0.05, we reject the null hypothesis of association between row and column variables.

(d) A 5 by 3 table with χ2 = 13.26.

df=(5-1)(3-1)=4*2=8

P-value=P(X²>=13.26)

=CHIDIST(13.26,8)

=0.103211

Since p-value is greater than 0.05, fail to reject the null hypothesis of no association between row and column variables.