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# sol6 - MIT OpenCourseWare http/ocw.mit.edu 2.830J 6.780J...

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MIT OpenCourseWare http://ocw.mit.edu 2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303) Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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MIT 2.830/6.780 Problem Set 6 (2008) — Solutions Problem 1 See the following pages for exemplary solutions (courtesy X. Su and K. Umeda) For the t-test part of the question, two approaches were accepted: one in which the significance of each effect was tested in turn by looking for evidence of a mean shift (interactions are not probed in this approach); the other is to define a standard error as in example 12-7 of Montgomery and perform a t-test on all effects, including interactions. This second approach is exactly equivalent to doing ANOVA.
XIANGYONG SU 2 2 15 b c 2 SS T = SS A + SS B + SS AB + SS E = 0.002386 , or SS T = X X y ijk @ y ffff , where: i = 1 X j = 1 k = 1 ` a` a 2 b c 2 SS A = 2 15 X y ffff i AA @ y ffff B ` i = 1 a 2 ` a 2 C = (2)(15) 2.0377 @ 2.03935 + 2.041 @ 2.03935 = 0.000163 ` a` a 2 b c 2 SS B = 2 15 X j = 1 y ffff A j A @ y ffff B ` a 2 ` a 2 C = (2)(15) 2.035567 @ 2.03935 + 2.043133 @ 2.03935 = 0.000859 ` a 2 2 b c 2 SS AB = 15 X y ffff ij A @ y ffff i AA @ y ffff A j A + y ffff = 0.0000837 i = 1 X j = 1 2 2 15 b c 2 SS E = X X y ijk @ y ffff ij A = 0.001279 i = 1 X j = 1 k = 1 SUMMARY LV(i=1) Count Sum Average Variance HV(i=2) Count Sum Average Variance Total Count Sum Average Variance LH (j=1) HH(j=2) Total 15 15 30 30.491 30.64 61.131 30.491 30.64 61.131 y ffff 11 A = fffffffffffffffffffff = 2.032733 y ffff 12 A = fffffffffffffffff = 2.042667 y ffff 1 AA = fffffffffffffffffffff = 2.0377 15 15 30 9.5E-06 2.24E-05 4.09E-05 15 15 30 30.576 30.654 61.23 30.576 30.654 61.23 = = 2.0384 = = 2.0436 = = 2.041 y ffff 21 A fffffffffffffffffffff 15 y ffff 22 A fffffffffffffffffffff 15 y ffff 2 AA fffffffffffffffff 30 3.74E-05 2.21E-05 3.57E-05 30 30 120 61.067 61.294 122.361 61.067 61.294 122 . 361 y ffff A 1 A = fffffffffffffffffffff 30 = 2.035567 y ffff A 2 A = fffffffffffffffffffff 30 = 2.043133 y ffff = fffffffffffffffffffffffff 120 = 2.03935 3.09E-05 2.17E-05 MANOVA table – Two-way with interactions: Source of Variation SS df MS(=SS/df) F P-value F crit Sample (Factor A) SS A = 0.000163 1 s 2 A = 0.000163 7.149541 0.009808 4.012973 Columns (Factor B) SS B = 0.000859 1 s B 2 = 0.000859 37.58889 9.3E-08 4.012973 Interaction SS AB = 8.4E-05 1 s 2 AB = 8.4E-05 3.677261 0.060265 4.012973 Within SS E = 0.001279 56 s E 2 = 2.28E-05 Total SS T = 0.002386 59 2

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XIANGYONG SU Comparing the computed F-ratios to a 5% upper critical value of the F distribution, where F 0.05,1,56 = 4.013, we conclude that since computed F-ratio 7.15 > 4.013 and 37.59 > 4.013, both the injection speed and hold time affects the output diameter. However, as the computed F-ratio for interaction is smaller than the critical F value, there is no indication of interaction between the two factors. b. Test on Factor A, injection speed input: Let H 0 be the hypothesis that μ HV = μ LV Let H 1 be the hypothesis that μ HV μ LV b c b c y ffff 1 @ y ffff 2 @ μ ffff 1 @ μ ffff 2 Test statistic: t 0 = ffffffffffffffff s ffffffffffffffffffffffffffffffffffffffffffffffffff s w n ffffffff 1 1 + n ffffffff 1 2 , where μ 1 = μ HV and μ 2 = μ LV . Test if the mean diameters under low and high velocity are equal 95% of the time ( α = 0.05) If they do not, then the injection speed affects the output diameter.
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