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(a) Suppose you want to test Ho: μ = 7 versus H₁₂: μ = 7 using a = 0.05. What conclusion would be appropriate, and why? O Since the null value of 7 is in the 95% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 95% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is not in the 95% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. (b) If you wanted to use a significance level of 0.01 for the test in (a), what is your confidence interval? (Round your answers to three decimal places.) 6.848 x 6.912 x ) mm What conclusion would be appropriate? O Since the null value of 7 is in the 99% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 99% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is not in the 99% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. X X

(a) Suppose you want to test Ho: μ = 7 versus H₁₂: μ = 7 using a = 0.05. What conclusion would be appropriate, and why? O Since the null value of 7 is in the 95% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 95% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is not in the 95% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. (b) If you wanted to use a significance level of 0.01 for the test in (a), what is your confidence interval? (Round your answers to three decimal places.) 6.848 x 6.912 x ) mm What conclusion would be appropriate? O Since the null value of 7 is in the 99% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 99% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is not in the 99% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. X X

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A 95% CI for true average amount of warpage (mm) of laminate sheets under specified conditions was calculated as (6.83, 6.93), based on a sample size of n = 12 and the assumption that amount of warpage is normally distributed. USE SALT (a) Suppose you want to test Ho: μ = 7 versus H₂: μ # 7 using a = 0.05. What conclusion would be appropriate, and why? O Since the null value of 7 is in the 95% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 95% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. (b) If you wanted to use a significance level of 0.01 for the test in (a), what is your confidence interval? (Round your answers to three decimal places.) 6.848 X 6.912 1x ) m mm What conclusion would be appropriate? O Since the null value of 7 is in the 99% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. ● Since the null value of 7 is in the 99% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. X X
Transcribed Image Text:A 95% CI for true average amount of warpage (mm) of laminate sheets under specified conditions was calculated as (6.83, 6.93), based on a sample size of n = 12 and the assumption that amount of warpage is normally distributed. USE SALT (a) Suppose you want to test Ho: μ = 7 versus H₂: μ # 7 using a = 0.05. What conclusion would be appropriate, and why? O Since the null value of 7 is in the 95% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Since the null value of 7 is in the 95% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 95% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. (b) If you wanted to use a significance level of 0.01 for the test in (a), what is your confidence interval? (Round your answers to three decimal places.) 6.848 X 6.912 1x ) m mm What conclusion would be appropriate? O Since the null value of 7 is in the 99% confidence interval we reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. ● Since the null value of 7 is in the 99% confidence interval we fail to reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we reject Ho. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. O Since the null value of 7 is not in the 99% confidence interval we fail to reject Ho. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. X X
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