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Statistical Analysis Carl Steinhauer Consultant

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Overview P 401 Test Results Statistical Analysis; PWL Estimate Verify Production Process: Payment

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Theory 1. Assumptions 2. Normal Distribution 3. Tools: Average and Standard Deviation 4. Percent Within Limits (PWL) Concept

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Assumptions 1. Limited # of test results Statistical Analysis Quality characteristics of large amount of material 2. Test result variability Components:materialssampling-ERLPMtesting-ERLPM 3. Same Process 4. Random sampling-Lot, Sublot 5. Normal Distribution

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Specific Procedures 1. Sublots, Lots, Partial Lots 2. Calculations 3. Retesting 4. Outliers

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Analysis % Taller than 5-5 % between 5-5 and 6-5 Average Height Limit # of samples Statistical Analysis Estimate average and % within limits

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n= tests

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x+/- 1Sn=68% x+/- 1Sn=68% x+/-2Sn+95% x+/-3Sn+99.7% x

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L L=spec lower tolerance limit eg. Mat density 96.3 for P401 eg. Mat density 96.3 for P401 PWL-% of test result exceeding L

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PWL Calculation Procedures ERLPM-page 47 Section 110-AC A Method for Computing PWL and Examples Section Spec. P401 Table 5-L and U Spec. Limits (page 24)

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Given x 1 =2 x 2 =4 x 3 =6 x 4 =8 X = = 5 4

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X=5 d 1 =2-5=-3d 1 ²=9 d 2 =4-5=-1d 2 ²=1 d 3 =6-5=1d 3 ²=1 d 4 =8-5=3d 4 ²=9 Sn = d 1 ²+d 2 ²+d 3 ²=d 4 ² n-1 n-1 Sn = = Sn =2.58 (calculator n -1)

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Roundout Rules ERLPM-page 47 Example-last digit to be kept-nearest 10 th Even Digit-same Odd Digit-increase by 1 This case becomes 4.6 If it was it would become 4.8 becomes 4.7

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MAT Density Manual Appendix E, page 4 Sublot 1. = = = = x=98.0 Sn =1.8 Q L =x-L Sn Sn Q L = = Section 110-Table l, N=4 P L =82 Section f Spec Tables 5 page 50 ERLPM

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Target Density 98.0 Achieved Sn =1.8 versus 1.3 Acceptable QC Value 18% P L = 82% MAT Density

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Effect of Quality Control 90 PWL 82 PWL Sn = 1.8 Sn = 1.3 Spec. pg % 98% x

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Air Voids App. D, page 3 Sublot 1= 2.1 2= 3.2 3=2.54=6.0 X= 3.4 Sn = 1.76 Spec. page

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Q L = x-L = =.7955 Sn 1.76 P L (table 1) = 77% P L (table 1) = 77% n=4 n=4 Q U = U-x = =.909 Sn 1.76 Sn 1.76 P U (table 1) = 81% P U (table 1) = 81%n=4 PWL= P L + P U -100 PWL= = 58

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Air Voids 23% 58% 19% PWL= P L + P U -100 PWL= =58

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Payment Spec-par a-page 29 MAT Density PWL=82 Air Voids PWL= 58 Lot Pay Factor Air Voids- 1.4 x 58-12= 69.2% Mat Density- 0.5 x 82+55= 96% Use lower of 2 values- 69.2% Lower value

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Joint Density Appendix E, page X= 95.3 Sn = 1.58 Q L = ( ) = P L = 93 Spec. page 21 par (b)(3) if < 71% there is a 5% penalty Table 5

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Partial Lots spec page 20 Section P c

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Sample Problem Flow-Appendix D, page 5 Partial lot situation-6 sublots 8.0, 8.2, 8.5, 8.2, 8.9, 9.1 X= 8.5 Sn = 0.44 QL= x-L = = ; PL= 88 Sn 0.44 Sn 0.44 QU= U-X = = 18.75; PL= 100 Sn 0.44 Sn 0.44 PWL= = 88<90 Corrective Action! (b)(2)

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Outliers Spec -pg d -pg c -pg c MAT Density and Air Voids

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Test for Outliers MAT Density x= 96.2 Sn= 1.71 QL= = PL= <50% PL= <50% ASTM E 178, par. 4 T 1 = (x-x 1 )/ Sn T 1 = =

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Table 1-ASTM E 178 Table 1-ASTM E 178 N=4 N=4 Upper 5% significance level Upper 5% significance level Since 1.286<1.463 the 94.0 test value is not considered an outlier and is retained! Since 1.286<1.463 the 94.0 test value is not considered an outlier and is retained!

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Sample Problem-Outliers Air Voids 2.0, 4.8, 4.9, 5.0 X=4.2Sn=1.45 Q L = = ; PL= Q U = = ; PU= PWL= (100+69)-100=69 ASTM E 178 par. 4 T n = (x-x 1 )/ Sn = = Table 1, ASTM E 178, N=4, 5% significance T= 1.463<1.517 therefore 2.0 is the outlier and it is discarded

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Resampling page 25 MAT Density ONLY (Appendix E-pg 4) Prior MAT Density- 96, 97, 99, 100 PWL 82 4 new cores 96, 96, 97, 98 AVG-all 8, 97.4 Sn= 1.51 Q L = x-L = =.7337 Sn 1.51 Sn 1.51 Table 1, N= 8 PWL= 77

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Questions?

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Thanks!

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