Question

The average retirement age for a certain country was reported to be 56.3 years according to an international group dedicated to promoting trade and economic growth. With the pension system operating with a​ deficit, a bill was introduced by the government during the summer to raise the minimum retirement age from 60 to 62. Suppose a survey of 50 retiring citizens is taken to investigate whether the new bill has raised the average age at which people actually retire. Assume the standard deviation of the retirement age is 6 years. Using alpha α equals = 0.10​, answer parts a through c below.

a. Explain how Type I and Type II errors can occur in this hypothesis test.

A Type I error can occur when the researcher concludes the average retirement age increased, but the average retirement age did not increase. A Type II error can occur when the researcher concludes that the average retirement age did not increase, when, in​ fact, the average retirement age increased.

b. Calculate the probability of a Type II error occurring if the actual population age is 57.6 years old.

The probability of committing a Type II error is 0.401.

​(Round to three decimal places as​ needed.)

c. Calculate the probability of a Type II error occurring if the actual population age is 59.1 years old.

The probability of committing a Type II error is 0.022.

​(Round to three decimal places as​ needed

I have the answer, in italic. I just need to know how to solve part b and c. Thank you!

Answer 1

b)

for 0.1 level with right tail test , critical z=

1.282

(from excel:normsinv(0.1)

sample size    n=		50
std deviation σ=		6.00
std error ='σx=σ/√n=6/√50=		0.8485

rejection region: Xbar >=μ+Zα*σx or Xbar>= 56.3+1.28155157*0.849=57.3874

P(Type II error) =P(Xbar<57.387|μ=57.6)=P(Z<(57.3874-57.6)/0.849)=P(Z<-0.25)=0.401

(from excel: normsdist(-0.25))

c)

P(Type II error) =P(Xbar<57.387|μ=59.1)=P(Z<(57.3874-59.1)/0.849)=P(Z<-2.02)=0.022

(from excel: normsdist(-2.02))