Question

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 8. The hypotheses H0: μ = 75 and Ha: μ < 75 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.) 1.69 Correct: Your answer is correct. standard deviations

(b) If x = 72.3, what is the conclusion using α = 0.005? 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 = -1.69 Correct: Your answer is correct. P-value = .0458 Correct: Your answer is correct. State the conclusion in the problem context. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 75. Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 75. Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. Correct: Your answer is correct.

(c) For the test procedure with α = 0.005, what is β(70)? (Round your answer to four decimal places.) β(70) = .075 Incorrect: Your answer is incorrect.

(d) If the test procedure with α = 0.005 is used, what n is necessary to ensure that β(70) = 0.01? (Round your answer up to the next whole number.) n = 5 Incorrect: Your answer is incorrect. specimens

Answer 1

a) (75 - 72.3)/(8/) = 1.69 standard deviations.

b) The test statistic z = ()/()

                                 = (72.3 - 75)/(8/)

                                 = -1.69

P-value = P(Z < -1.69)

            = 0.0455

Since the P-value is greater than the significance level (0.0455 > 0.005), so we should not reject the null hypothesis.

Do not reject the null hypothesis . There is not sufficient evidence to conclude that the mean drying time is less than 75.

c) At alpha = 0.005, the critical value value is z_0.005 = -2.575

z_crit = -2.575

or, ()/() = -2.575

or, ( - 75)/(8/) = -2.575

or, = -2.575 * (8/) + 75

or, = 70.88

= P( > 70.88)

            = P(( - )/() > (70.88 - )/())

            = P(Z > (70.88 - 70)/(8/))

            = P(Z > 0.55)

            = 1 - P(Z < 0.55)

            = 1 - 0.7088

            = 0.2912

d) = 0.01

or, P(( - )/() > (70.88 - )/()) = 0.01

or, P(Z > (70.88 - 70)/(8/)) = 0.01

or, P(Z < (70.88 - 70)/(8/)) = 0.99

or, (70.88 - 70)/(8/) = 2.33

or, 0.88/(8/) = 2.33

or, n = (2.33 * 8/0.88)^2

or, n = 449