We are going to test the null and the alternative hypotheses

We are going to test the null and the alternative hypotheses h0: mu - 80 and Ha: mu > 80 based on a simple random sample of size 50 and the population standard deviation = 20.

a)   Compute the probability of a type II error if actually, mu = 84.

b)   Compute the power of the test if actually, mu = 84.

c)   Compute the probability of a type II error if actually, mu = 90.

d)   Compute the power of the test if actually, mu = 90.

Solution

a) sigma=20/sqrt(50)=2.8284

this is a right tailed test. we fail to reject if z statisitc is greater than 1.64.

P(z>=1.64)=P(xbar>=xbar/mean=80,sigma=2.8284)=0.95

1.64=xbar-80/2.8284

xbar=84.6385

P(xbar>84.6385/mean=84,sigma=2.8284)

=P(z>84.6385-84/2.8284)=P(z>0.22577)=0.41068. this is P(type 2 error)

this is calculated using NORMDIST fucntion in excel and subtract with 1

b)power of the test is 1-0.41068=0.5893

c)

sigma=20/sqrt(50)=2.8284

this is a right tailed test. we fail to reject if z statisitc is greater than 1.64.

P(z>=1.64)=P(xbar>=xbar/mean=80,sigma=2.8284)=0.95

1.64=xbar-80/2.8284

xbar=84.6385

P(xbar>84.6385/mean=90,sigma=2.8284)

=P(z>84.6385-90/2.8284)=P(z>-1.89559)=0.9999. this is P(type 2 error)

this is calculated using NORMDIST fucntion in excel and subtract with 1

d)power of the test is 1-0.9999=0.000