Example
F0(x) – discrete uniform on the integers 1, 2, 3, 4, 5
n = 10, observe the values 1,1,1,2,2,2,3,3,3,3
Ho: population distribution is Fo(x)
H1: Not Ho
Draw the cumulative probability of Fo(x) and S(x)
D_obs = sup| Fo(x) – S(x) | = 0.4 at x =3
We want to find the associate p-value
Find P(D+ >= 0.4)
n(1-D_obs) = 6, so we need to find f0, f1,…f5
The horizontal line with ordinate 1 – D_obs = 1- 0.4 = 0.6 intersects Fo(x) right at the top of the
jump at x=3,so fo = 0.6
The horizontal line with ordinate 1-0.4-0.1 = 0.5 intersects Fo(x) at a jump, so f1 = equals the
height of Fo(x) at the bottom of the jump i.e., 0.4. Likewise f2=0.4, f3=f4=0.2, f5=0
Then
e_0 = 1,
e_1=1-0.6=0.4
e_2=0.32
e_3=0.208
e_4=0.295
P(D+ >=0.4) = long painful calculation = 0.021
P(D>= 0.4) =
P(D+ >=0.4)+ P(D- >=0.4) = 0.042
At level alpha = 0.05. He reject ho
Note: If we had use Table A13, n=10, alpha=0.05, two-sided test, the critical value is 0.409>0.4
=> Don’t reject Ho.