The average retirement age for a certain country was reported to be 57.2 years a

Question

The average retirement age for a certain country was reported to be 57.2 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 39 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 5 years. Using 0.10, answer parts a through c below. a. Explain how Type I and Type Il errors can occur in this hypothesis test A Type l error can occur when the researcher concludes the average retirement age | but the average retirement age | A Type II error can occur when the researcher | when, in concludes that the average retirement age fact, the average retirement age b. Calculate the probability of a Type ll error occurring if the actual population age is 58.5 years old. The probability of committing a Type Il error is (Round to three decimal places as needed.) c. Calculate the probability of a Type ll error occurring if the actual population age is 60.0 years old. The probability of committing a Type ll error is

Explanation / Answer

a.
i.
Given that,
Standard deviation, =5
Sample Mean, X =61
Null, H0: =57.2
Alternate, H1: >57.2
Level of significance, = 0.1
From Standard normal table, Z /2 =1.282
Since our test is right-tailed
Reject Ho, if Zo < -1.282 OR if Zo > 1.282
Reject Ho if (x-57.2)/5/(n) < -1.282 OR if (x-57.2)/5/(n) > 1.282
Reject Ho if x < 57.2-6.41/(n) OR if x > 57.2-6.41/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 39 then the critical region
becomes,
Reject Ho if x < 57.2-6.41/(39) OR if x > 57.2+6.41/(39)
Reject Ho if x < 56.174 OR if x > 58.226
Suppose the true mean is 57.2
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(56.174 < x OR x >58.226 | 1 = 57.2)
= P(56.174-57.2/5/(39) < x - / /n OR x - / /n >58.226-57.2/5/(39)
= P(-1.281 < Z OR Z >1.281 )
= P( Z <-1.281) + P( Z > 1.281)
= 0.1001 + 0.1001 [ Using Z Table ]
= 0.2

ii.
Given that,
Standard deviation, =5
Sample Mean, X =61
Null, H0: <57.2
Alternate, H1: >57.2
Level of significance, = 0.1
From Standard normal table, Z /2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-57.2)/5/(n) < -1.2816 OR if (x-57.2)/5/(n) > 1.2816
Reject Ho if x < 57.2-6.408/(n) OR if x > 57.2-6.408/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 39 then the critical region
becomes,
Reject Ho if x < 57.2-6.408/(39) OR if x > 57.2+6.408/(39)
Reject Ho if x < 56.174 OR if x > 58.226
Implies, don't reject Ho if 56.174 x 58.226
Suppose the true mean is 57.2
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(56.174 x 58.226 | 1 = 57.2)
= P(56.174-57.2/5/(39) x - / /n 58.226-57.2/5/(39)
= P(-1.281 Z 1.281 )
= P( Z 1.281) - P( Z -1.281)
= 0.8999 - 0.1001 [ Using Z Table ]
= 0.8
For n =39 the probability of Type II error is 0.8

b.
Given that,
Standard deviation, =5
Sample Mean, X =61
Null, H0: <57.2
Alternate, H1: >57.2
Level of significance, = 0.1
From Standard normal table, Z /2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-57.2)/5/(n) < -1.2816 OR if (x-57.2)/5/(n) > 1.2816
Reject Ho if x < 57.2-6.408/(n) OR if x > 57.2-6.408/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 39 then the critical region
becomes,
Reject Ho if x < 57.2-6.408/(39) OR if x > 57.2+6.408/(39)
Reject Ho if x < 56.174 OR if x > 58.226
Implies, don't reject Ho if 56.174 x 58.226
Suppose the true mean is 58.5
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(56.174 x 58.226 | 1 = 58.5)
= P(56.174-58.5/5/(39) x - / /n 58.226-58.5/5/(39)
= P(-2.905 Z -0.342 )
= P( Z -0.342) - P( Z -2.905)
= 0.3662 - 0.0018 [ Using Z Table ]
= 0.364
For n =39 the probability of Type II error is 0.364

c.
Given that,
Standard deviation, =5
Sample Mean, X =61
Null, H0: <57.2
Alternate, H1: >57.2
Level of significance, = 0.1
From Standard normal table, Z /2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-57.2)/5/(n) < -1.2816 OR if (x-57.2)/5/(n) > 1.2816
Reject Ho if x < 57.2-6.408/(n) OR if x > 57.2-6.408/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 39 then the critical region
becomes,
Reject Ho if x < 57.2-6.408/(39) OR if x > 57.2+6.408/(39)
Reject Ho if x < 56.174 OR if x > 58.226
Implies, don't reject Ho if 56.174 x 58.226
Suppose the true mean is 60
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(56.174 x 58.226 | 1 = 60)
= P(56.174-60/5/(39) x - / /n 58.226-60/5/(39)
= P(-4.779 Z -2.216 )
= P( Z -2.216) - P( Z -4.779)
= 0.0133 - 0 [ Using Z Table ]
= 0.013
For n =39 the probability of Type II error is 0.013

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