Question

8. If the population of interest has µ = 60 and σ = 40 and we draw a sample of 64 people with M = 72, what would be the Power for a two-tailed test with alpha = .05?

Report the standard error

Report the critical value for the null distribution

Report the sample mean value that corresponds to the critical value from part b) and its z-score for the alternative distribution

Report the Power

Report the probability of making a Type II Error

what would be the Power of a two-tailed test using alpha = .01?

Answer 1

8.
a.
Given that,
Standard deviation, σ =40
Sample Mean, X =72
Null, H0: μ=60
Alternate, H1: μ!=60
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-60)/40/√(n) < -1.96 OR if (x-60)/40/√(n) > 1.96
Reject Ho if x < 60-78.4/√(n) OR if x > 60-78.4/√(n)
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Suppose the size of the sample is n = 64 then the critical region
becomes,
Reject Ho if x < 60-78.4/√(64) OR if x > 60+78.4/√(64)
Reject Ho if x < 50.2 OR if x > 69.8
Implies, don't reject Ho if 50.2≤ x ≤ 69.8
Suppose the true mean is 72
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(50.2 ≤ x ≤ 69.8 | μ1 = 72)
= P(50.2-72/40/√(64) ≤ x - μ / σ/√n ≤ 69.8-72/40/√(64)
= P(-4.36 ≤ Z ≤-0.44 )
= P( Z ≤-0.44) - P( Z ≤-4.36)
= 0.33 - 0 [ Using Z Table ]
= 0.33
For n =64 the probability of Type II error is 0.33
power of the test =1-type 2 error
power= 1-0.33
power= 0.67
standard error = standard deviation /sqrt(n)
standard error = 60/sqrt(64)
standard error= 60/8 = 7.5
b.
Given that,
Standard deviation, σ =40
Sample Mean, X =72
Null, H0: μ=60
Alternate, H1: μ!=60
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-60)/40/√(n) < -2.5758 OR if (x-60)/40/√(n) > 2.5758
Reject Ho if x < 60-103.032/√(n) OR if x > 60-103.032/√(n)
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Suppose the size of the sample is n = 64 then the critical region
becomes,
Reject Ho if x < 60-103.032/√(64) OR if x > 60+103.032/√(64)
Reject Ho if x < 47.121 OR if x > 72.879
Implies, don't reject Ho if 47.121≤ x ≤ 72.879
Suppose the true mean is 72
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(47.121 ≤ x ≤ 72.879 | μ1 = 72)
= P(47.121-72/40/√(64) ≤ x - μ / σ/√n ≤ 72.879-72/40/√(64)
= P(-4.9758 ≤ Z ≤0.1758 )
= P( Z ≤0.1758) - P( Z ≤-4.9758)
= 0.5698 - 0 [ Using Z Table ]
= 0.5698
For n =64 the probability of Type II error is 0.5698
power = 1-type 2 error
power= 1-0.5698
power = 0.4302
type 2 error = 0.5698