The average retirement age for a certain country was reported to be 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 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 years. Using ​, 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 ? did not increase, increased, but the average retirement age ? increased. did not increase. A Type II error can occur when the researcher concludes that the average retirement age ? did not increase, increased, ?when, in? fact, the average retirement age ? did not increase. increased.b. Calculate the probability of a Type II error occurring if the actual population age is 57.4 years old. The probability of committing a Type II error is _______ . ?(Round to three decimal places as? needed.)c. Calculate the probability of a Type II error occurring if the actual population age is 58.9 years old. The probability of committing a Type II error is _______ . ?(Round to three decimal places as? needed.)

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

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