Question

A) The drying time of a type of paint under specified test condition is known to be normally distributed with an assumed mean value of 75 and a standard deviation of 9. Test the hypothesis H0: µ=75 vs. Ha: µ<75 from a sample of 25 observations. For a test procedure with α= 0.002 what is the probability of making a type 2 error when the true mean is 70, ß(70) (to 4 decimals)?

B) The drying time of a type of paint under specified test condition is known to be normally distributed with an assumed mean value of 75 and a standard deviation of 9. Test the hypothesis H0: µ=75 vs. Ha: µ<75 from a sample of 25 observations. If the test procedure for α = 0.002 is used, what n is necessary to ensure that ß(70) = 0.01?

Answer 1

Part a)

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
Critical value Z(α/2) = Z( 0.002 /2 ) = ± 2.878
2.878 = ( X̅ - 75 ) / ( 9 / √( 25 ))
Rejection region X̅ <= 69.8196

X ~ N ( µ = 70 , σ = 9 )
P ( X > 69.8196 ) = 1 - P ( X < 69.8196 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 69.8196 - 70 ) / ( 9 / √ ( 25 ) )
Z = -0.1002
P ( ( X - µ ) / ( σ / √ (n)) > ( 69.8196 - 70 ) / ( 9 / √(25) )
P ( Z > -0.1 )
P ( X̅ > 69.8196 ) = 1 - P ( Z < -0.1 )
P ( X̅ > 69.8196 ) = 1 - 0.4601
P ( X̅ > 69.8196 ) = 0.5399
P ( Type II error ) ß = 0.5399

Part b)

Sample size can be calculated by below formula

Effect Size = | 75 - 70 | / 70 = 0.5556
Critical value Z(1 - α) = Z(1 - 0.002) = 2.878
Critical value Z(1 - β) = Z(0.99) = 2.33
n = 176
Required sample size is 176.