mont4e_sm_ch16_supplemental

# mont4e_sm_ch16_supplemental - Supplementary Exercises 16-57...

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Supplementary Exercises 16-57 a) X-bar and Range - Initial Study -------------------------------------------------------------------------------- X-bar | Range ---- | ----- UCL: + 3.0 sigma = 64.0181 | UCL: + 3.0 sigma = 0.0453972 Centerline = 64 | Centerline = 0.01764 LCL: - 3.0 sigma = 63.982 | LCL: - 3.0 sigma = 0 | out of limits = 0 | out of limits = 0 -------------------------------------------------------------------------------- Chart: Both Normalize: No Estimated process mean = 64 process sigma = 0.0104194 mean Range = 0.01764 0 S u b g r o u p 5 1 01 52 02 5 63.98 63.99 64.00 64.01 64.02 Sample Mean Mean=64.00 UCL=64.02 LCL=63.98 0.00 0.01 0.02 0.03 0.04 0.05 Sample Range R=0.01764 UCL=0.04541 LCL=0 Xbar/R Chart for C1-C3 The process is in control. b) ± μ= = x6 4 0104 . 0 693 . 1 01764 . 0 ˆ 2 = = = d R σ c) 641 . 0 ) 0104 . 0 ( 6 98 . 63 02 . 64 ˆ 6 = = = LSL USL PCR 16-37

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The process does not meet the minimum capability level of PCR 1.33. d) [] 641 . 0 641 . 0 , 641 . 0 min ) 0104 . 0 ( 3 98 . 63 64 , ) 0104 . 0 ( 3 64 02 . 64 min ˆ 3 , ˆ 3 min = = = = σσ LSL x x USL PCR k e) In order to make this process a “six-sigma process”, the variance σ 2 would have to be decreased such that PCR k = 2.0. The value of the variance is found by solving PCR k = xL S L = 3 20 σ . for σ : 0033 . 0 6 98 . 63 0 . 64 98 . 63 0 . 64 6 0 . 2 3 98 . 63 64 = = = = σ Therefore, the process variance would have to be decreased to σ 2 = (0.0033) 2 = 0.000011. f) ± σ x = 0.0104 8295 . 0 0020 . 0 8315 . 0 ) 88 . 2 ( ) 96 . 0 ( ) 96 . 0 88 . 2 ( 0104 . 0 01 . 64 02 . 64 0104 . 0 01 . 64 98 . 63 ) 02 . 64 98 . 63 ( = = < < = < < = < < = < < Z P Z P Z P X P X P x μ The probability that this shift will be detected on the next sample is p = 1 0.8295 = 0.1705 87 . 5 1705 . 0 1 1 = = = p ARL 16-38
16-58 a) 0 Subgroup 5 10 15 20 25 63.98 63.99 64.00 64.01 64.02 Sample Mean Mean=64.00 UCL=64.02 LCL=63.98 0.00 0.01 0.02 Sample StDev S=0.009274 UCL=0.02382 LCL=0 Xbar/S Chart for C1-C3 b) ± μ= = x6 4 0104 . 0 8862 . 0 009274 . 0 ˆ 4 = = = c s σ c) Same as 16-57 641 . 0 ) 0104 . 0 ( 6 98 . 63 02 . 64 ˆ 6 = = = LSL USL PCR The process does not meet the minimum capability level of PCR 1.33. d) Same as 16-57 [] 641 . 0 641 . 0 , 641 . 0 min ) 0104 . 0 ( 3 98 . 63 64 , ) 0104 . 0 ( 3 64 02 . 64 min ˆ 3 , ˆ 3 min = = = = σσ LSL x x USL PCR k e) Same as 16-57 e). In order to make this process a “six-sigma process”, the variance σ 2 would have to be decreased such that PCR k = 2.0. The value of the variance is found by solving PCR k = xL S L = 3 20 σ . for σ : 0033 . 0 6 98 . 63 0 . 64 98 . 63 0 . 64 6 0 . 2 3 98 . 63 64 = = = = Therefore, the process variance would have to be decreased to σ 2 = (0.0033) 2 = 0.000011. 16-39

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f) Same as 16-57 ± σ x = 0.0104 8295 . 0 0020 . 0 8315 . 0 ) 88 . 2 ( ) 96 . 0 ( ) 96 . 0 88 . 2 ( 0104 . 0 01 . 64 02 . 64 0104 . 0 01 . 64 98 . 63 ) 02 . 64 98 . 63 ( = = < < = < < = < < = < < Z P Z P Z P X P X P x σ μ The probability that this shift will be detected on the next sample is p = 1 0.8295 = 0.1705 87 . 5 1705 . 0 1 1 = = = p ARL 16-59 a) 01 02 0 0.0 0.1 0.2 Sample Number Proportion P Chart for def P=0.11 UCL=0.2039 LCL=0.01613 There are no points beyond the control limits. The process is in control. b) 0 0.04 0.09 0.14 0.19 Sample Number P Chart for def2 1 P=0.11 UCL=0.1764 LCL=0.04363 There is one point beyond the upper control limit. The process is out of control. The revised limits are: 16-40
01 02 0 0.04 0.09 0.14 0.19 Sample Number Proportion P Chart for def2 P=0.1063 UCL=0.1717 LCL=0.04093 There are no further points beyond the control limits.

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## This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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mont4e_sm_ch16_supplemental - Supplementary Exercises 16-57...

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