Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 7. The hypotheses H₀: μ = 74 and H_a: μ < 74 are to be tested using a random sample of n = 25 observations.

(a) If x = 72.3, what is the conclusion using α = 0.003?
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z

=

P-value

=

(b) For the test procedure with α = 0.003, what is β(70)? (Round your answer to four decimal places.)
β(70) =

(c) If the test procedure with α = 0.003 is used, what n is necessary to ensure that β(70) = 0.01? (Round your answer up to the next whole number.)

n = specimens

  

Answer 1

Part a)

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 72.3 - 74 ) / ( 7 / √( 25 ))
Z = -1.21

P value = P ( Z < 1.2143 ) = 0.1123

Part b)

The values of sample mean X̅ for which null hypothesis is rejected
Z = ( X̅ - µ ) / ( σ / √(n))
Critical value Z(α/2) = Z( 0.003 /2 ) = ± 2.748
2.748 = ( X̅ - 74 ) / ( 7 / √( 25 ))
Rejection region X̅ <= 70.1528
P ( X̅ > 70.1528 | µ = 70 ) = 0.4566
P ( Type II error ) ß = 0.4566

Part c)

Sample size can be calculated by below formula

Critical value Z(1 - α) = Z(1 - 0.003) = 2.748
Critical value Z(1 - β) = Z(0.99) = 2.33
n = 158
Required sample size is 158.