Question

Consider the following hypothesis test. H0:  = 20 H1:  ≠ 20 The sample size is 200 and the standard deviation of the population is 10. Use a  = 0.05. What is the probability of making the Type II error if the real value of the population is:

a. μ = 18.0

b. μ = 22.5

c. μ = 21.0

Answer 1

Common Data:

Z = (X - ) / [ / Sqrt(n)]

= 20,

= 10, n = 200

= 0.05, Critical value (Left Tail) = -1.96 and Critical Value Right Tail = +1.96

Value of (Left) for which H0 gets rejected = 20 + (-1.96 * 10/sqrt(200)) = 18.61

Value of (Right) for which H0 gets rejected = 20 + (1.96 * 10/sqrt(200)) = 21.39

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(a) = 18,

For Left, The Z value for = 18; Z = (18.61 - 18) / 10/sqrt(200) = 0.87

The Probability, P(X < 18.61) = 0.8074

For Right, The Z value for = 18; Z = (21.39 - 18) / 10/sqrt(200) = 4.79

The Probability P(X < 21.39) = 1

Therefore the Probability P(X > 21.39) = 1 - 1 = 0

Power of the test = P(X < 18.61) + P(X > 21.39) = 0.8074 + 0 = 0.8074

P(Type II error) = 1 - 0.8074 = 0.1926

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(b) = 22.5,

For Left, The Z value for = 18; Z = (18.61 - 22.5) / 10/sqrt(200) = -5.50

The Probability, P(X < 18.61) = 0

For Right, The Z value for = 18; Z = (21.39 - 22.5) / 10/sqrt(200) = -1.58

The Probability P(X < 21.39) = 0.0576

Therefore the Probability P(X > 21.39) = 1 - 0.0575 = 0.9424

Power of the test = P(X < 18.61) + P(X > 21.39) = 0 + 0.9424 = 0.9424

P(Type II error) = 1 - 0.9424 = 0.0576

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(c) = 21,

For Left, The Z value for = 18; Z = (18.61 - 21) / 10/sqrt(200) = -3.37

The Probability, P(X < 18.61) = 0.0004

For Right, The Z value for = 18; Z = (21.39 - 21) / 10/sqrt(200) = 0.55

The Probability P(X < 21.39) = 0.7074

Therefore the Probability P(X > 21.39) = 1 - 0.7074 = 0.2926

Power of the test = P(X < 18.61) + P(X > 21.39) = 0.0004 + 0.2926 = 0.2930

P(Type II error) = 1 - 0.2930 = 0.7070

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