Question

aminate sheets under specified conditions was calculated as 5.81 695 A 95% CI for true average mount of war age mm o warpage is normally distributed. ased on a sample size and the as surm on that amount r n (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 o 7 is in the 95% confidence interval we re ect 0 There is not su cent evidence to conclude that the true average amount of wa page mm or aminate 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 Since the null value of 7 is not in the 95% confidence interval we fail to reject HD There is not sufficient evidence to conclude that etre average amount of war age mm afar inate sheets under specified conditions is nat 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.) What conclusion would be appropriate? 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 H0. There is sufficient evidence to conclude that the true average amount of warpage (mm) of laminate sheets under specified conditions is nat 7. O Since the null value of 7 is not in the 99% 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. You may need to use the appropriate table in the Appendix of Tables to answer this question

Answer 1

Given aj Siro e tha null valued, 7 LL η0fin, -ne 95% CT we"reject tha There subbictet evidlence to coneludo hot h e omrage a mout wo pagc (mm) 2 2. 0.07 Χψια 2-101 19 = (6.79, C.98) rnm There ia sutbictent eviden to coneuda