# Carl Steinhauer Consultant

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Carl Steinhauer Consultant
Statistical Analysis Carl Steinhauer Consultant

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

Theory Assumptions Normal Distribution
Tools: Average and Standard Deviation Percent Within Limits (PWL) Concept

Assumptions Limited # of test results Statistical Analysis Quality characteristics of large amount of material Test result variability Components: materials sampling-ERLPM testing-ERLPM Same Process Random sampling-Lot, Sublot Normal Distribution

Specific Procedures Sublots, Lots, Partial Lots Calculations Retesting
Outliers

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

96 tests n=100

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

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

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)

Given x1=2 x2=4 x3=6 x4=8 X = = 5 4

Sn = d1²+d2²+d3²=d4² n-1 Sn = 9+1+1+9 = 20 4-1 3 Sn =2.58
X=5 d1=2-5=-3 d1²=9 d2=4-5=-1 d2²=1 d3=6-5=1 d3²=1 d4=8-5=3 d4²=9 Sn = d1²+d2²+d3²=d4² n-1 Sn = = 20 Sn =2.58 (calculator n-1)

Roundout Rules ERLPM-page 47
Example-last digit to be kept-nearest 10th 4.61 4.62 4.64 4.6500 4.66 4.67 4.68 4.69 Even Digit-same Odd Digit-increase by 1 This case becomes 4.6 If it was it would become 4.8 becomes 4.7

MAT Density Manual Appendix E, page 4
Sublot 1. = 96.0 2. = 97.0 3. = 99.0 4. = 100.0 x=98.0 Sn=1.8 QL=x-L Sn QL= = .9444 1.8 Section 110-Table l, N=4 PL=82 Section f Spec Tables 5 page 50 ERLPM

Target Density 98.0 Achieved Sn=1.8 versus 1.3
MAT Density PL = 82% 98.0 18% 96.3 Target Density 98.0 Achieved Sn=1.8 versus 1.3 Acceptable QC Value

Effect of Quality Control
Sn = 1.3 Spec. pg 24 Sn = 1.8 82 PWL 90 PWL 96.3% 98% x

Air Voids App. D, page 3 Sublot 1= 2.1 2= 3.2 3=2.5 4=6.0 X= 3.4
Sn= 1.76 Spec. page

QL = x-L = = .7955 Sn PL(table 1) = 77% n=4 QU= U-x = = .909 Sn PU(table 1) = 81% PWL= PL + PU-100 PWL= = 58

Air Voids 58% 23% 19% 2 5 3.4 PWL= PL + PU-100 PWL= =58

Payment Spec-par 401-8.1a-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

Joint Density Appendix E, page 5
93.3 95.0 97.0 96.0 X= 95.3 Sn= 1.58 QL= ( ) = 1.58 PL= 93 Spec. page 21 par (b)(3) if < 71% there is a 5% penalty Table 5

Partial Lots spec page 20 Section P c

Partial lot situation-6 sublots Corrective Action! 401-5.2(b)(2)
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 QU= U-X = = 18.75; PL= 100 Sn PWL= = 88<90 Corrective Action! (b)(2)

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

Test for Outliers MAT Density 94.0 QL= 96.2-96.3 = -.0585 96.0 1.71
97.0 98.0 x= 96.2 Sn= 1.71 QL= = 1.71 PL= <50% ASTM E 178, par. 4 T1= (x-x1)/Sn T1= = 1.286 1.71

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

Sample Problem-Outliers
Air Voids 2.0, 4.8, 4.9, 5.0 X=4.2 Sn=1.45 QL= = ; PL= 100 1.45 QU= = ; PU= 69 PWL= (100+69)-100=69 ASTM E 178 par. 4 Tn= (x-x1)/Sn = = 1.517 Table 1, ASTM E 178, N=4, 5% significance T= 1.463<1.517 therefore 2.0 is the outlier and it is discarded

Resampling 401-5.3 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 QL= x-L = = .7337 Sn Table 1, N= 8 PWL= 77

Questions?

Thanks!