Calculate the smallest value of n that would result in rejec

Calculate the smallest value of n that would result in rejecting H₀ using the Neyman-Pearson lemma.

Solution

The form of the critical region from the Neyman-Pearson is the usual critical region for the one-sided Normal test:
X_-bar c.
For a sample of size n, X is Normal with mean µ and standard deviation 5/ sqrt(n) .
alpha=0.05 = Prob[ X c | µ = 20] = 1 - [ (c- 20) / (5/ sqrt(n ) ]
1.645 = (c- 20)/ (5/ sqrt( n ))
c = 20 + 8.225/ sqrt (n )
In order to H0 reject , when X_-bar = 21.49, we require that 21.49 c = 20 + 8.225/ sqrt(n) .
i.e, n 30.47
n = 31.
Smallest value of n = 31...