1. Carl Steinhauer Consultant
Statistical Analysis
Carl Steinhauer
Consultant

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

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

4. 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

5. Specific Procedures Sublots, Lots, Partial Lots Calculations Retesting
Outliers

6. 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

7. 96 tests
n=100

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

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

11. 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)

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

13. 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)

14. 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

15. 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

16. 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

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

18. 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

19. 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

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

21. 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

22. 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

23. Partial Lots spec page 20
Section P c

24. 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)

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

28. 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

29. 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!

30. 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

31. 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

32. Questions?

33. Thanks!