Question

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A consumer protection agent wants to test a paint manufacture's claim that the average drying time of his new fast-drying paint is 20 minutes with standard deviation 0 = 2.4 minutes. 36 cans of such paint were tested with the intention of rejecting the claim if the mean of the drying time exceeds 20.5 minutes. Otherwise, it will accept the claim. Find the probability of a type I error and the probability of a Type II error when u = 21 minutes. a. The Type I error is 0.1056 and the Type II error is 0.1056 b. None of these C. The Type I error is 0.1056 and the Type II error is 0.1251 d. The Type I error is 0.1251 and the Type II error is 0.1056

Answer 1

X ~ N ( µ = 20 , σ = 2.4 )
P ( X > 20.5 ) = 1 - P ( X < 20.5 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 20.5 - 20 ) / ( 2.4 / √ ( 36 ) )
Z = 1.25
P ( ( X - µ ) / ( σ / √ (n)) > ( 20.5 - 20 ) / ( 2.4 / √(36) )
P ( Z > 1.25 )
P ( X̅ > 20.5 ) = 1 - P ( Z < 1.25 )
P ( X̅ > 20.5 ) = 1 - 0.8944
P ( X̅ > 20.5 ) = 0.1056

P ( Type I error ) α = 0.1056

X ~ N ( µ = 21 , σ = 2.4 )
P ( X < 20.5 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 20.5 - 21 ) / ( 2.4 / √36 )
Z = -1.25
P ( ( X - µ ) / ( σ/√(n)) < ( 20.5 - 21 ) / ( 2.4 / √(36) )
P ( X < 20.5 ) = P ( Z < -1.25 )
P ( X̅ < 20.5 ) = 0.1056

P ( Type II error ) ß = 0.1056

Option a is correct.