sol8 - Homework 8 due Mar 28 9-36(10-40 9-39(10-43...

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Homework 8 due Mar. 28 9-36 (10-40) 9-39 (10-43) 9-57 (10-64) 9-58 (10-65) 9-60 (10-67) 9-68 (10-75) Note: the numbers in brackets are for the new edition of the textbook 9-36(10.40). a) 1) The parameter of interest is the mean difference in natural vibration frequencies, μ d where d i = finite element - Equivalent Plate. 2) H 0 : μ d = 0 3) H 1 : μ d 0 4) α = 0.01 5) The test statistic is t d s n d 0 = / 6) Reject the null hypothesis if t 0 < - t 0 005 6 . , where - t 0 005 6 . , = - 3.707 or t 0 > t 0 005 6 . , where t 0 005 6 . , = 3.707 7) d = - 5.49 s d = 5.924 n = 7 t 0 549 5924 7 2 45 = - = - . . / . 8) Since - 3.707 < - 2.45 < 3.707, do not reject the null and conclude the data suggest that the two methods do not produce significantly different mean values for natural vibration frequency at the 0.01 level of significance. b) 99% confidence interval: d t s n d t s n n d d n d -  ≤ + - - α α μ / , / , 2 1 2 1 - -  ≤ ≤ - + 549 3707 5924 7 549 3707 5924 7 . . . . . . μ d - 13.790 μ d 2.810 With 99% confidence, we believe that the mean difference between the natural vibration frequency from the equivalent plate method and the natural vibration frequency from the finite element method is between - 13.790 and 2.810 cycle/s.
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9-39(10.43). 1) The parameter of interest is the difference in mean weight loss,
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sol8 - Homework 8 due Mar 28 9-36(10-40 9-39(10-43...

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