In: Statistics and Probability
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.
(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 |