A 95% CI for true average amount of warpage (mm) of laminate sheets under specified conditions was calculated as (6.82, 6.94), based on a sample size of n = 18 and the assumption that amount of warpage is normally distributed.USE SALT(a) Suppose you want to test H: 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.O 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 i 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 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.)mmWhat 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.O 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.

The 95% confidence interval for the true average amount of warpage (mm) of laminate sheets, based on…

A 95% CI for true average amount of warpage (mm) of laminate sheets under specified conditions was calculated as (6.82, 6.94), based on a sample size of n = 18 and the assumption that amount of warpage is normally distributed. LAUSE SALT (a) Suppose you want to test H: 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. warpage (mm) laminate sheets under specified conditions is not 7. O Since the null value of 7 is in the 95% confidence interval we fail to reject Ho. There is sufficient evidence conclude that the true average amount 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.) 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. O 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) 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.