Question

During the Flint, MI lead water crisis, you have developed a new method for determining lead in water that is faster and cheaper than the current conventional method. Use the data in the table to answer the following questions:

Method	Average Lead Concentration (ppb)	Standard deviation (ppb)	Number of replicates
Conventional	14.9	0.8	6
Your New Method	15.6	1.1	6

1. Identify if the standard deviations between the two methods are statistically the same or not. Show your work
2. Is the data from your method statistically different from that of the conventional method? Show your work
3. Are your new method measurements precise? Hint: What value can we relate to precision!

Are your new method measurements accurate if a standard reference water sample contains 15.0 ppb of lead (which is the legal limit in drinking water)? Hint: Which two numbers are you comparing! Give a brief explanation to receive

Answer 1

Soluion :

a)

Null and alternative hypothesis:  
Hₒ : σ₁ = σ₂  
H₁ : σ₁ ≠ σ₂  
Test statistic:
F = s₁² / s₂² = 0.8² / 1.1² =    0.5289
Degree of freedom:  
df₁ = n₁-1 =    5
df₂ = n₂-1 =    5
Critical value(s):  
Lower tailed critical value, FL = F.INV(0.05/2, 5, 5) = 0.1399
Upper tailed critical value, FU = F.INV(1-0.05/2, 5, 5) = 7.1464
P-value :  
P-value = 2*F.DIST.RT(0.5289, 5, 5) =    1.4986
Conclusion:  
As p-value > α, we fail to reject the null hypothesis.  

standard deviations between the two methods are statistically the same

b)

Ho :   µ1 - µ2 =   0              
Ha :   µ1-µ2 ╪   0              
                      
Level of Significance ,    α =    0.05              
                      
Sample #1   ----> Conventional
mean of sample 1,    x̅1=   14.90              
standard deviation of sample 1,   s1 =    0.80              
size of sample 1,    n1=   6              
                      
Sample #2   ----> New Method
mean of sample 2,    x̅2=   15.60              
standard deviation of sample 2,   s2 =    1.10              
size of sample 2,    n2=   6              
                      
difference in sample means =    x̅1-x̅2 =    14.9000   -   15.6   =   -0.70
                      
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.9618              
std error , SE = Sp*√(1/n1+1/n2) =    0.5553              
                      
t-statistic = ((x̅1-x̅2)-µd)/SE = (  -0.7000   -   0   ) / 0.56  =
                      
Degree of freedom, DF=   n1+n2-2 =    10              

p-value = 0.236061   (excel function: =T.DIST.2T(t stat,df) )          
Conclusion: p-value>α , Do not reject null hypothesis                  
                      
There is not enough evidence to Convemtional and New method are different.

c)

new method measurements is not precise as it has standard deviation of 1.1.

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