Question

Consider the following hypothesis test.

H_0: μ = 36.1

H_a: μ ≠ 36.1

A sample of 60 items will be taken and the population standard deviation is σ =12.2. Use α = .05. Compute the probability of making a Type II error if the population mean is:

a) μ = 32

b) μ = 38.5

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 ) = ± 1.96
1.96 = ( X̅ - 36.1 ) / ( 12.2 / √( 60 ))
Rejection region X̅1 <= 33.013
Rejection region X̅2 >= 39.187
P ( X̅1 <= 33.013 | µ = 32 ) = 0.74
P ( X̅2 >= 39.187 | µ = 32 ) = 0

X ~ N ( µ = 32 , σ = 12.2 )
P ( 33.013 < X < 39.187 )

Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 33.013 - 32 ) / ( 12.2 / √(60))
Z = 0.6432
Z = ( 39.187 - 32 ) / ( 12.2 / √(60))
Z = 4.5631
P ( 0.64 < Z < 4.56 )
P ( 33.013 < X̅ < 39.187 ) = P ( Z < 4.56 ) - P ( Z < 0.64 )
P ( 33.013 < X̅ < 39.187 ) = 1 - 0.74
P ( 33.013 < X̅ < 39.187 ) = 0.26

P ( Type II error ) ß = 0.2600

Part b)

X ~ N ( µ = 38.5 , σ = 12.2 )
P ( 33.013 < X < 39.187 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 33.013 - 38.5 ) / ( 12.2 / √(60))
Z = -3.4838
Z = ( 39.187 - 38.5 ) / ( 12.2 / √(60))
Z = 0.4362
P ( -3.48 < Z < 0.44 )
P ( 33.013 < X̅ < 39.187 ) = P ( Z < 0.44 ) - P ( Z < -3.48 )
P ( 33.013 < X̅ < 39.187 ) = 0.6687 - 0.0002
P ( 33.013 < X̅ < 39.187 ) = 0.6684

P ( Type II error ) ß = 0.6684