Question

1. Suppose we are about to sample 100 observations from a normally distributed population where it is known that σ = 20, but µ is unknown. We intend to test H0: µ = 30 against Ha: µ < 30 at α = 0.05.

(a) What values of the sample mean would lead to a rejection of the null hypothesis?

(b) What is the power of the test if µ = 28?

(c) What is the power of the test if µ = 26?

Answer 1

Part a)

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
Critical value Z(α/2) = Z( 0.05 /2 ) = ± 1.645
1.645 = ( X̅ - 30 ) / ( 20 / √( 100 ))
Rejection region X̅ <= 26.71

Part b)

X ~ N ( µ = 28 , σ = 20 )
P ( X > 26.71 ) = 1 - P ( X < 26.71 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 26.71 - 28 ) / ( 20 / √ ( 100 ) )
Z = -0.645
P ( ( X - µ ) / ( σ / √ (n)) > ( 26.71 - 28 ) / ( 20 / √(100) )
P ( Z > -0.64 )
P ( X̅ > 26.71 ) = 1 - P ( Z < -0.64 )
P ( X̅ > 26.71 ) = 1 - 0.2595
P ( X̅ > 26.71 ) = 0.7405

P ( Type II error ) ß = 0.7405
Power of test is 1 - ß = 0.2595

Part c)X ~ N ( µ = 26 , σ = 20 )
P ( X > 26.71 ) = 1 - P ( X < 26.71 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 26.71 - 26 ) / ( 20 / √ ( 100 ) )
Z = 0.355
P ( ( X - µ ) / ( σ / √ (n)) > ( 26.71 - 26 ) / ( 20 / √(100) )
P ( Z > 0.36 )
P ( X̅ > 26.71 ) = 1 - P ( Z < 0.36 )
P ( X̅ > 26.71 ) = 1 - 0.6387
P ( X̅ > 26.71 ) = 0.3613

P ( X̅ > 26.71 | µ = 26 ) = 0.3613
P ( Type II error ) ß  = 0.3613
Power of test is ( 1 - ß ) = 0.6387