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 = 0.73.

df =
P-value =

(b) A 4 by 4 table with χ² = 18.32.

df =
P-value =

(c) A 2 by 8 table with χ² = 23.66.

df =
P-value =

(d) A 5 by 3 table with χ² = 12.31.

df =
P-value =

I understand how to find df, just please explain in detail how to find the p-value. I am very confused, thank you.

Answer 1

(a) A 2 by 2 table with χ² = 0.73.

The df=1 chi-square value = 0.73 which is between 0.02 and 2.71 in the table, so we cannot calculate it from the table. the best way is to use excel.

=CHISQ.DIST.RT(0.73,1)=0.392883 which is > 0.05. Thus the result is not significant.

df =	1*1=1
P-value =	0.392883

(b) A 4 by 4 table with χ² = 18.32.

The df=9 chi-square value = 18.32 which is between 16.92 and 19.02 in the table, so p-value lies between 0.05 and 0.025.

We can calculate it with interpolation, but the best way is to use excel.

=CHISQ.DIST.RT(18.32,9)

df =	3*3=9
P-value =	0.031637

(c) A 2 by 8 table with χ² = 23.66.

The df=7 chi-square value = 23.66 which is greater than 18.48 in the table, so p-value is less than 0.01

We use excel.

=CHISQ.DIST.RT(23.66,7)=0.001307

df =	1*7=7
P-value =	0.001307

(d) A 5 by 3 table with χ² = 12.31.

The df=8 chi-square value = 12.31 which is between 3.49 and 13.36 in the table, so p-value lies between 0.9 and 0.1.

We can calculate it with interpolation, but the best way is to use excel.

=CHISQ.DIST.RT(12.31,8)=0.137899

df =	4*2=8
P-value =	0.137899