Question

2. A researcher wants to study the average lifetime of a certain brand of batteries (in hours). In testing the hypotheses, H0: μ = 950 hours vs. H1: μ ≠ 950 hours, a random sample of 25 batteries is drawn from a normal population whose standard deviation is 200 hours.

Calculate β, the probability of a Type II error when μ = 1000 and α = 0.10.

Answer 1

2.

Given that,
Standard deviation, σ =200
Sample Mean, X =1000
Null, H0: μ=950
Alternate, H1: μ!=950
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-950)/200/√(n) < -1.6449 OR if (x-950)/200/√(n) > 1.6449
Reject Ho if x < 950-328.98/√(n) OR if x > 950-328.98/√(n)
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Suppose the size of the sample is n = 25 then the critical region
becomes,
Reject Ho if x < 950-328.98/√(25) OR if x > 950+328.98/√(25)
Reject Ho if x < 884.204 OR if x > 1015.796
Implies, don't reject Ho if 884.204≤ x ≤ 1015.796
Suppose the true mean is 1000
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(884.204 ≤ x ≤ 1015.796 | μ1 = 1000)
= P(884.204-1000/200/√(25) ≤ x - μ / σ/√n ≤ 1015.796-1000/200/√(25)
= P(-2.8949 ≤ Z ≤0.3949 )
= P( Z ≤0.3949) - P( Z ≤-2.8949)
= 0.6535 - 0.0019 [ Using Z Table ]
= 0.6516
For n =25 the probability of Type II error is 0.6516