Answer :
The missing figures in the question can be seen below.
The average retirement age = 56.1 years ...
The number of a survey of retired citizen = 49
The standard deviation of the retirement age is 6 years.
Using alpha ∝ = 0.02
Answer:
Step-by-step explanation:
From the given options in the first question in the given information.
Type I error can take place when the researcher concludes the average retirement age increased, but the average retirement age did not increase.
A Type II error can take place when the researcher concludes that the average retirement age did not increase, but the average retirement age increased.
Recall that:
population mean = 56.1
sample size = 49
standard deviation = 6
At the level of significance of 0.02, using the Excel function (=Normsinv(0.02))
The critical value for z = 2.054
Standard error = [tex]\dfrac{\sigma}{\sqrt{n}}[/tex]
=[tex]\dfrac{6}{\sqrt{49}}[/tex]
= 6/7
= 0.857
The rejection region [tex]\overline X[/tex] = [tex]\mu +Z_{\alpha/0.02}*\sigma_x[/tex]
[tex]\overline X[/tex] = [tex]56.1+2.05374891*0.857[/tex]
[tex]\overline X[/tex] = 57.86
P(Type II error) is as follows:
[tex]P(\overline X < 57.86| \mu = 57.4) = P( Z< \dfrac{\overline X - \mu }{\sigma_x})[/tex]
[tex]= P( Z< \dfrac{57.86-57.4}{0.857})[/tex]
[tex]= P( Z< 0.537)[/tex]
From z tables;
P (Type II error) = 0.704
P(Type II error) is as follows:
[tex]P(\overline X < 57.86| \mu = 58.9) = P( Z< \dfrac{\overline X - \mu }{\sigma_x})[/tex]
[tex]= P( Z< \dfrac{57.86-58.9}{0.857})[/tex]
[tex]= P( Z<-1.214)[/tex]
From z tables;
P (Type II error) = 0.1124