Question

In a rural area in Ontario, excessive use of fertilizers has caused an increase in the Nitrate concentration (mg/L) of 2 public wells that are used as drinking water sources. The concentrations, however, are not beyond the regulatory limit yet. To ensure the safety of dirking water, the utility plans to close the well with the higher nitrate level and take it out of service until they can find a solution. Nine samples are collected from each well and the results are as follows: Well1 36 25 26 34 34 29 33 36 35 Well 2 34 24 31 34 32 25 33 27 30

a) Assuming that the concentrations in both wells are normally distributed and the actual variability of them is the same, find the 95% confidence interval for the difference between the mean concentrations of the wells.

b) Based on your result of part a, how do you compare the nitrate concentration of the two wells? Can you suggest which well has the higher concentration and should be out of service based on your results?

Answer 1

X :- Represent Well 1

Y :- Represent Well 2

Mean X̅ = Σ Xi / n
X̅ = 288 / 9 = 32
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 144 / 9 -1 ) = 4.2426

Mean Y̅ = ΣYi / n
Y̅ = 270 / 9 = 30
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1 )
SY = √ ( 116 / 9 -1) = 3.8079

Part )

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.05/2, 9 + 9 - 2) = 2.12

( 32 - 30 ) ± t(0.05/2 , 9 + 9 -2) 4.0311 √ ( (1/9) + (1/9))
Lower Limit = ( 32 - 30 ) - t(0.05/2 , 9 + 9 -2) 4.0311 √( (1/9) + (1/9))
Lower Limit = -2.0286
Upper Limit = ( 32 - 30 ) + t(0.05/2 , 9 + 9 -2) 4.0311 √( (1/9) + (1/9))
Upper Limit = 6.0286
95% Confidence Interval is ( -2.0286 , 6.0286 )

Part b)

Since the value 0 lies in the interval  ( -2.0286 , 6.0286 ) that indicate both well have same concentration and should not be out of service.