13-80.01 1.5 2 2.5 3 3.5 (for0.020.030.040.050.060.070.080.100.200.300.400.500.600.700.80φ α= 0.05)
23451.00Probability of accepting the hypothesisφ(forα= 0.01)ν 1 = 1α α9
10
12∞∞0.01 1 2 3 (for0.020.030.040.050.060.070.080.100.200.300.400.500.600.700.80φ α= 0.05)
13451.00Probability of accepting the hypothesisφ(forα= 0.01)ν 1 = 22ν 2 = 660
20
157ν 1 = numerator degrees of freedom. ν 2 = denominator degrees of freedom.= 0.05 = 0.01ν = 6
7
8
9
10
12
15
20
30
606 = ν 215
20
30
607
83012
10
9
8∞α = 0.05α = 0.012Source: These curves are adapted with permission from Biometrika Tables for Statisticians,Vol.
2, by E. S. Pearson and H. O. Hartley, Cambridge University Press, Cambridge, 1972.PQ220 6234F.Ch 13_CD 5/8/02 7:54 PM Page 8 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:PQ220 MONT 8/5/2002:Ch 13: