Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 9. 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?

(b) If x = 72.3, what is the conclusion using α = 0.004?
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	=

(c) For the test procedure with α = 0.004, what is β(70)? (Round your answer to four decimal places.)
β(70) =

(d) If the test procedure with α = 0.004 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?

Answer 1

(a)

The z-score is

Answer: -0.39

(b)

The p-value is :

p-value =P(z < -0.39) = 0.3483

Since p-value is greater than 0.004 so we fail to reject the null hypothesis.

(c)

(d)

(e)

Test is left tailed so critical value of for which we will reject the null hypothesis is -2.33.

The test statistics is

So type I error is

P(z < -4.11) = 0.0000