SLIDE 1
Power of the test that p ½
- Some anticipated p
A n – B n = – C (see solution in Normal approximation)
- For α =0.05 and power = 1 – β = 80%
– Zα = 1.96, Zβ = 0.84
β α
Z Z p np n np + ≥ − − ) 1 ( | 2 | 2 ) 1 ( ) ( n pn n p p Z Z − = − − +
β α