Question

You are testing the null hypothesis that the average money that BC student has in his or her pocket is $20 against the alternative that it’s greater than that. You are testing at a 5 percent significance level. We know the population variance is 81 dollars squared. You take a random sample of 100 students.

i. Calculate the power of this test if the true mean is $23. Represent it graphically.

For parts, ii, iii, iv and v, please take the information above and change only the one item that is described. (i.e. compare part iii to our answer and values in part i, not to part ii.) Also, show me how your diagram would change if you were drawing the true distribution of sample means and the distribution under the null hypothesis.

ii. What does the power equal if we changed α to .10?

iii. What if we changed our sample size to 36?

iv. What if the population variance was really 25 dollars squared.

v. What if the true population mean is $22?

Answer 1

a.
Given that,
Standard deviation, σ =9
Sample Mean, X =23
Null, H0: μ=20
Alternate, H1: μ>20
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-20)/9/√(n) < -1.6449 OR if (x-20)/9/√(n) > 1.6449
Reject Ho if x < 20-14.8041/√(n) OR if x > 20-14.8041/√(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 20-14.8041/√(100) OR if x > 20+14.8041/√(100)
Reject Ho if x < 18.5196 OR if x > 21.4804
Implies, don't reject Ho if 18.5196≤ x ≤ 21.4804
Suppose the true mean is 23
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(18.5196 ≤ x ≤ 21.4804 | μ1 = 23)
= P(18.5196-23/9/√(100) ≤ x - μ / σ/√n ≤ 21.4804-23/9/√(100)
= P(-4.9782 ≤ Z ≤-1.6884 )
= P( Z ≤-1.6884) - P( Z ≤-4.9782)
= 0.0457 - 0 [ Using Z Table ]
= 0.0457
For n =100 the probability of Type II error is 0.0457
power of the test = 1- type 2 error
power of the test = 1-0.0457
i.
power of the test = 0.9543

ii.
level of significance =0.10
Given that,
Standard deviation, σ =9
Sample Mean, X =23
Null, H0: μ=20
Alternate, H1: μ>20
Level of significance, α = 0.1
From Standard normal table, Z α/2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-20)/9/√(n) < -1.2816 OR if (x-20)/9/√(n) > 1.2816
Reject Ho if x < 20-11.5344/√(n) OR if x > 20-11.5344/√(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 20-11.5344/√(100) OR if x > 20+11.5344/√(100)
Reject Ho if x < 18.8466 OR if x > 21.1534
Implies, don't reject Ho if 18.8466≤ x ≤ 21.1534
Suppose the true mean is 23
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(18.8466 ≤ x ≤ 21.1534 | μ1 = 23)
= P(18.8466-23/9/√(100) ≤ x - μ / σ/√n ≤ 21.1534-23/9/√(100)
= P(-4.6149 ≤ Z ≤-2.0518 )
= P( Z ≤-2.0518) - P( Z ≤-4.6149)
= 0.0201 - 0 [ Using Z Table ]
= 0.0201
For n =100 the probability of Type II error is 0.0201
power = 1-type 2 error
power = 1-0.0201
power = 0.9799

iii.
Given that,
Standard deviation, σ =9
Sample Mean, X =23
Null, H0: μ=20
Alternate, H1: μ>20
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-20)/9/√(n) < -1.6449 OR if (x-20)/9/√(n) > 1.6449
Reject Ho if x < 20-14.8041/√(n) OR if x > 20-14.8041/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 36 then the critical region
becomes,
Reject Ho if x < 20-14.8041/√(36) OR if x > 20+14.8041/√(36)
Reject Ho if x < 17.5327 OR if x > 22.4674
Implies, don't reject Ho if 17.5327≤ x ≤ 22.4674
Suppose the true mean is 23
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(17.5327 ≤ x ≤ 22.4674 | μ1 = 23)
= P(17.5327-23/9/√(36) ≤ x - μ / σ/√n ≤ 22.4674-23/9/√(36)
= P(-3.6449 ≤ Z ≤-0.3551 )
= P( Z ≤-0.3551) - P( Z ≤-3.6449)
= 0.3613 - 0.0001 [ Using Z Table ]
= 0.3612
For n =36 the probability of Type II error is 0.3612
power = 1-0.3612
power = 0.6388

iv.
Given that,
Standard deviation, σ =5
Sample Mean, X =23
Null, H0: μ=20
Alternate, H1: μ>20
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-20)/5/√(n) < -1.6449 OR if (x-20)/5/√(n) > 1.6449
Reject Ho if x < 20-8.2245/√(n) OR if x > 20-8.2245/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 20-8.2245/√(100) OR if x > 20+8.2245/√(100)
Reject Ho if x < 19.1776 OR if x > 20.8225
Implies, don't reject Ho if 19.1776≤ x ≤ 20.8225
Suppose the true mean is 23
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(19.1776 ≤ x ≤ 20.8225 | μ1 = 23)
= P(19.1776-23/5/√(100) ≤ x - μ / σ/√n ≤ 20.8225-23/5/√(100)
= P(-7.6448 ≤ Z ≤-4.355 )
= P( Z ≤-4.355) - P( Z ≤-7.6448)
= 0 - 0 [ Using Z Table ]
= 0
For n =100 the probability of Type II error is 0
power = 1-0 =1

v.
Given that,
Standard deviation, σ =9
Sample Mean, X =23
Null, H0: μ=20
Alternate, H1: μ>20
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-20)/9/√(n) < -1.6449 OR if (x-20)/9/√(n) > 1.6449
Reject Ho if x < 20-14.8041/√(n) OR if x > 20-14.8041/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 20-14.8041/√(100) OR if x > 20+14.8041/√(100)
Reject Ho if x < 18.5196 OR if x > 21.4804
Implies, don't reject Ho if 18.5196≤ x ≤ 21.4804
Suppose the true mean is 22
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(18.5196 ≤ x ≤ 21.4804 | μ1 = 22)
= P(18.5196-22/9/√(100) ≤ x - μ / σ/√n ≤ 21.4804-22/9/√(100)
= P(-3.8671 ≤ Z ≤-0.5773 )
= P( Z ≤-0.5773) - P( Z ≤-3.8671)
= 0.2819 - 0.0001 [ Using Z Table ]
= 0.2818
For n =100 the probability of Type II error is 0.2818
power = 1-0.2818
power = 0.7182