Runs

Captital Cement, San Antonio, TX

Alamo, San Antonio, TX

Holcim, Portland CO

1 <0.34 0.54 1.18

2 <0.34 0.66 1.67

3 <0.34 0.78 1.48

4 <0.34 0.25 0.6

5 <0.34 0.25 0.79

6 <0.34 0.27 0.87

789

101112131415

16

Emissions (ppmvd@7%O2)

1718192021222324252627282930
31

32333435

ni = number test runs = 666 0 0 0 Total Number of sources = 3n =Total # test runs = 18

17

1 1

1 i N n

ij

i j

X x

n  

 

1

1

N

i

i

df n



 

      

1

1 i n

i ik

i k

x x

n 

 

1

N

i

i

n





N

This step is for QC only:means 0.34 0.46 1.10 #DIV/0! #DIV/0! #DIV/0! #DIV

0.63

Mean =

1 1

1 i N n

ij

i j

X x

n  

 

1

1

N

i

i

df n



 

      

1

1 i n

i ik

i k

x x

n 

 

1

N

i

i

n





N
Pooled Variance = 0.1839

m= number future runs = 30

Term1 0.0889Term2 0.0163

 2 2

1

1

1

n

i

i

s x X

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1 s

n m

      

2

,

1 1 Xt s df p n m        

Squared root Term2 0.1278639NOTE: the pvalue for the t-statistic is calculated as: 2*alpha, where 1-alpha is desired confidence, so if 99% confidence is desirt‐statistic = quantile t‐distribution with df degrees of freedom at .99 confidence level = 2.566UPL POOLED VARIANCE = 0.96

 2 2

1

1

1

n

i

i

s x X

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1 s

n m

      

2

,

1 1 Xt s df p n m        
UPL Pooled 

Variance=

2

,

1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 
00 0

#DIV/0! #DIV/0! #DIV/0!
nd 2*alpha=2*(0.01)
Runs

Captital Cement, San Antonio, TX

1 <0 34

Emissions (ppmvd@7%O2))

<0.34

2 <0.343 <0.344 <0.345 <0.346 <0 34 .789

10

1112131415161718192021222324252627282930
31

32333435

 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

i

i

df n



 

      

1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





N

ni = number test runs = 600 0 0 0 Total Number of sources = 3

 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

i

i

df n



 

      

1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





N

n =Total # test runs = 6

 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

i

i

df n



 

      

1

1 ni

i ik

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x x

n 

 

1

N

i

i

n





N

5

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

1 i N n

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1

1

N

i

i

df n



 

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1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





N

This step is for QC only:means 0.34 #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

1 i N n

ij

i j

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1

1

N

i

i

df n



 

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1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





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

i

i

df n



 

      

1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





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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

i

i

df n



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      

1

1 ni

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x x

n 

 

1

N

i

i

n

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

N

 2 2

1

1

1

n

i

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s xX

n 

 





1 1

1 i N n

ij

i j

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 

1

1

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i

i

df n



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      

1

1 ni

i ik

i k

x x

n 

 

1

N

i

i

n





N
Pooled Variance = 0.0000  2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

m= number future runs = 30

Term1 0.2000

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

Term2 0.0000

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

Squared root Term2 0

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

q

NOTE: the pvalue for the t-statistic is calculated as: 2*alpha, where 1-alpha is desired confidence, so if 99% confidence is desi

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

t‐statistic = quantile t‐distribution with df degrees of freedom at .99 confidence level = 3.36492UPL POOLED VARIANCE = 0.34

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        

 2 2

1

1

1

n

i

i

s xX

n 

 





1 1

n m



2 1 1

s

n m

      

df t

2 1 1

s

n m

      

2

,

1 1 Xt s df p n m        
UPL Pooled 

Variance=

2

,

1 

Xt s df p n 

 

   

 

Variance= 2 , 1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 

2 , 1 

Xt s df p n 

 

   

 
00 0

#DIV/0! #DIV/0! #DIV/0!
and 2*alpha=2*(0.01)