Runs Captital Cement, San Antonio, TX Alamo, San Antonio, TX Holcim, Portland CO 1 <0.34 0.54 2 <0.34 0.66 3 <0.34 0.78 4 <0.34 0.25 5 <0.34 0.25 6 <0.34 0.27 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 E m

1 N i i 

## n

 

#  N 22 23 24 25 26 27 28 29 30 31 32 33 34 35 ni = number test runs = 660 0 0 Total Number of sources = 3
n =Total # test runs = 12 1 N i i 

## n

 

#  N11

This step is for QC only: means 0.34 0.46

0.40 Mean =

Pooled Variance = 0.0286 

#   2 2 1 

## 1 1 n i i 

## s x X n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m       

1 

### 1 N i i 

### df n

 

###        



# 

2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

m= number future runs = 30
Term1 0.1167

Term2 0.0033

Squared root Term2 0.0577707 NOTE: the pvalue for the t-statistic is calculated as: 2*alpha, where 1-alpha is desired confidence, so i f 

#   2 2 1 

## 1 1 n i i 

## s x X n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m       

1 

### 1 N i i 

### df n

 

###        



# 

2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# t‐ statistic = quantile t‐ distribution with df degrees of freedom at .99 confidence
UPL POOLED VARIANCE = 0.56 df 

## t

2

, 

## 1 1 df p 

## Xt s n m        m issions (ppmvd@7%O2))

0 00 0 0 0f 99% confidence is desired then alpha=0.01 and 2*alpha=2*(0.01 )level = 2.718079183UPL Pooled Variance= 

# 2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m        Runs Captital Cement, San Antonio, TX 1 <0.34 2 <0.34 Em

1 N i i 

## n

 

#  N 3 <0.34 4 <0.34 5 <0.34 6 <0.34 7 8 9

1 N i i 

## n

 

#  N 9 10 11 12 13 14 15

1 N i i 

## n

 

#  N 16 17 18 19 20 21 22

1 N i i 

## n

 

#  N 22 23 24 25 26 27 28

1 N i i 

## n

 

#  N 28 29 30 31 32 33 34 35

1 N i i 

## n

 

#  N 35 ni = number test runs = 600 0 0 Total Number of sources = 3

1 N i i 

## n

 

#  N
n =Total # test runs = 6 1 N i i 

## n

 

#  N5



#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

This step is for QC only: means 0.34



#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

0.34 Mean = 

#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

Pooled Variance = 0.0000 

#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

m= number future runs = 30
Term1 0.2000 

#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

Term2 0.0000 

#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

Squared root Term2 0 NOTE: the pvalue for the t-statistic is calculated as: 2*alpha, where 1-alpha is desired confidence, so 

#   2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# 

  2 2 1 

## 1 1 n i i 

## s xX n  

##  

 

#  1 1 

## 1 i n N ij i j 

## X x n   

##  

# 

1 1 n m



2 

### 1 1 s n m        1 

### 1 N i i 

### df n

 

###        



# 

df 

## t 2 

### 1 1 s n m        1 

## 1 i n i ik k i 

## x x n  

##  

#  1 N i i 

## n

 

# t ‐ statistic = quantile t ‐ distribution with df degrees of freedom at .99 confidenc e
UPL POOLED VARIANCE = 0.34 df 

## t

2

, 

## 1 1 df p 

## Xt s n m         df 

## t

2

, 

## 1 1 df p 

## Xt s n m        issions (ppmvd@7%O2))

0 00 0 0 0if 99% confidence is desired then alpha=0.01 and 2*alpha=2*(0.01)e level = 3.364929997UPL Pooled Variance= 

# 2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m         2

, 

# 1 1 df p 

# Xt s n m        