Question

An engineer is investigating engine life for a new oil compound. He has built 12 engines and tested them to end of life in a test. The sample mean was 63,000km. He has enough data to show that the populations standard deviation is equal to 3150km. He is stesting H0: mu = 62000km against H1: mu does not = 62000km, and using P(Type 1 error) = .05. If the true population mean tire life is 61000km what is the P(type II error)?

Please show work and if calculator is needed please use Ti methods.

Answer 1

Given that,
Standard deviation, σ =3150
Sample Mean, X =63000
Null, H0: μ=62000
Alternate, H1: μ!=62000
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-62000)/3150/√(n) < -1.96 OR if (x-62000)/3150/√(n) > 1.96
Reject Ho if x < 62000-6174/√(n) OR if x > 62000-6174/√(n)
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Suppose the size of the sample is n = 12 then the critical region
becomes,
Reject Ho if x < 62000-6174/√(12) OR if x > 62000+6174/√(12)
Reject Ho if x < 60217.72 OR if x > 63782.28
Implies, don't reject Ho if 60217.72≤ x ≤ 63782.28
Suppose the true mean is 61000
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(60217.72 ≤ x ≤ 63782.28 | μ1 = 61000)
= P(60217.72-61000/3150/√(12) ≤ x - μ / σ/√n ≤ 63782.28-61000/3150/√(12)
= P(-0.86 ≤ Z ≤3.06 )
= P( Z ≤3.06) - P( Z ≤-0.86)
= 0.9989 - 0.1949 [ Using Z Table ]
= 0.804
For n =12 the probability of Type II error is 0.804