Question

- Answer the following questions about Type II errors. When the marketing staff is σ=20 and n=100, we want to test the following hypothesis. Ho: μ=100 vs H!: μ≠ 100

1. Obtain the probability of type II error if you are actually μ=102 with a significant level σ=0.10.

2. Obtain the probability of type II error if you are actually μ=102 with a significant level σ=0.05.

3. Describe the effect of the reduction of σ on type II error β.

Answer 1

1.
Given that,
Standard deviation, σ =20
Sample Mean, X =102
Null, H0: μ=100
Alternate, H1: μ!=100
Level of significance, α = 0.1
From Standard normal table, Z α/2 =1.6449
Since our test is two-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-100)/20/√(n) < -1.6449 OR if (x-100)/20/√(n) > 1.6449
Reject Ho if x < 100-32.898/√(n) OR if x > 100-32.898/√(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 100-32.898/√(100) OR if x > 100+32.898/√(100)
Reject Ho if x < 96.7102 OR if x > 103.2898
Implies, don't reject Ho if 96.7102≤ x ≤ 103.2898
Suppose the true mean is 102
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(96.7102 ≤ x ≤ 103.2898 | μ1 = 102)
= P(96.7102-102/20/√(100) ≤ x - μ / σ/√n ≤ 103.2898-102/20/√(100)
= P(-2.6449 ≤ Z ≤0.6449 )
= P( Z ≤0.6449) - P( Z ≤-2.6449)
= 0.7405 - 0.0041 [ Using Z Table ]
= 0.7364
For n =100 the probability of Type II error is 0.7364
2.
Given that,
Standard deviation, σ =20
Sample Mean, X =102
Null, H0: μ=100
Alternate, H1: μ!=100
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-100)/20/√(n) < -1.96 OR if (x-100)/20/√(n) > 1.96
Reject Ho if x < 100-39.2/√(n) OR if x > 100-39.2/√(n)
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Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 100-39.2/√(100) OR if x > 100+39.2/√(100)
Reject Ho if x < 96.08 OR if x > 103.92
Implies, don't reject Ho if 96.08≤ x ≤ 103.92
Suppose the true mean is 102
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(96.08 ≤ x ≤ 103.92 | μ1 = 102)
= P(96.08-102/20/√(100) ≤ x - μ / σ/√n ≤ 103.92-102/20/√(100)
= P(-2.96 ≤ Z ≤0.96 )
= P( Z ≤0.96) - P( Z ≤-2.96)
= 0.8315 - 0.0015 [ Using Z Table ]
= 0.83
For n =100 the probability of Type II error is 0.83
3.
the effect of the reduction of σ on type II error β.
large population standard deviation reduces power and type 2 error is also decreases.
reduction of standard deviation then type 2 error is increases.