Question

A 95% CI for true average amount of warpage (mm) of laminate sheets under specified conditions was calculated as (6.81, 6.95), based on a sample size of n = 17 and the assumption that amount of warpage is normally distributed. (a) Suppose you want to test H0: μ = 7 versus Ha: μ ≠ 7 using α = 0.05. What conclusion would be appropriate, and why? Since the null value of 7 is in the 95% confidence interval we reject H0. 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 H0. 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 H0. 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 fail to reject H0. There is not sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is not 7. Correct: Your answer is correct. (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

Answer 1

At 0.01 significance level, the confidence interval is ( 6.788, 6.972)