Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 8. The hypotheses H₀: μ = 73 and H_a: μ < 73 are to be tested using a random sample of n = 25 observations.

(a) How many standard deviations (of X) below the null value is x = 72.3? (Round your answer to two decimal places.)
standard deviations

(b) If x = 72.3, what is the conclusion using α = 0.006?
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	=

State the conclusion in the problem context.

Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73.    Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73.Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.

(c) For the test procedure with α = 0.006, what is β(70)? (Round your answer to four decimal places.)
β(70) =  

(d) If the test procedure with α = 0.006 is used, what n is necessary to ensure that β(70) = 0.01? (Round your answer up to the next whole number.)
n =  specimens

(e) If a level 0.01 test is used with n = 100, what is the probability of a type I error when μ = 76? (Round your answer to four decimal places.)

Answer 1

a) number of standard deviations =(73-72.3)*sqrt(25)/8 = 0.44

b)z= -0.44

p value =0.3300

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73.

c)

d)

e)P(type I error)=P(Xbar<71.1389)=P(Z<(71.1389-76)/0.8)=P(Z<-6.07)=0.0000