Question

A company that receives shipments of batteries tests a random sample of nine of them before agreeing to take a shipment. The company is concerned that the true mean lifetime for all batteries in the shipment should be at least 50 hours. From past experience it is safe to conclude that the population distribution of lifetimes is normal with a standard deviation of 3 hours. For one particular shipment the mean lifetime for a sample of nine batteries was 48.2 hours. Find the power of a 10%- level test of the null hypothesis that the mean lifetime is at least 50 hours when the true mean lifetime of batteries is 49 hours. The power is

• A.

  0.3452

• B.

  0.6548

• C.

  0.3897

• D.

  0.7881

Answer 1

Given that,
Standard deviation, σ =3
Sample Mean, X =48.2
Null, H0: μ<=50
Alternate, H1: μ>=50
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-50)/3/√(n) < -1.2816 OR if (x-50)/3/√(n) > 1.2816
Reject Ho if x < 50-3.8448/√(n) OR if x > 50-3.8448/√(n)
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Suppose the size of the sample is n = 9 then the critical region
becomes,
Reject Ho if x < 50-3.8448/√(9) OR if x > 50+3.8448/√(9)
Reject Ho if x < 48.7184 OR if x > 51.2816
Implies, don't reject Ho if 48.7184≤ x ≤ 51.2816
Suppose the true mean is 49
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(48.7184 ≤ x ≤ 51.2816 | μ1 = 49)
= P(48.7184-49/3/√(9) ≤ x - μ / σ/√n ≤ 51.2816-49/3/√(9)
= P(-0.2816 ≤ Z ≤2.2816 )
= P( Z ≤2.2816) - P( Z ≤-0.2816)
= 0.9887 - 0.3891 [ Using Z Table ]
= 0.5996
For n =9 the probability of Type II error is 0.5996
power = 1-type 2 error
power = 1-0.5996
power = 0.4004
approximately 0.3897
option:C