The concentration of nitrate (C) is measured in 8 groundwater monitoring wells a

Question

The concentration of nitrate (C) is measured in 8 groundwater monitoring wells at an abandoned landfill. The sample mean is 5 mg/1, while the sample variance is 20 (mg/1)^2. Assume the following: (i) C is normally distributed, (ii) the variance of C is equal to the sample variance, and (iii) the measured concentrations are independent between wells. a) Determine the 95 percent confidence limits for the average nitrate concentration at the site. b) If we want to estimate the average nitrate concentration to within plusminus 2 mg/1 with 95 percent confidence, how many additional wells should be installed? c) In order to determine it the landfill has contaminated the groundwater, the average nitrate concentration in the site wells will be compared to that in 4 off- site ("clean") wells. Assume that the concentration of nitrate in the "clean" wells is also normally distributed with a variance of 20 (mg/1)^2. We will conclude that the site is contaminated it the average nitrate concentration in the site wells exceeds that in the "clean" wells by more than 3 mg/1. How likely are we to conclude incorrectly that the site wells are contaminated (i.e., conclude mu_site > mu_clean when mu_site = mu_clean)?

Explanation / Answer

a)

std error = std devn/ sqrt(n) = sqrt (20/8) = 1.5811

95% lower CI = mean - z0.95* std error = 8 -1.96*1.5811 = 4.9

95% upper CI = mean + z0.95* std error = 8 +1.96*1.5811 = 11.1

b) margin of error, moe = 2

moe = critical value * std devn/ sqrt(n)

n = 1.96^2 * std devn^2/moe^2

=1.96^2* 20/4 = 19.21 or approx 20

already 8 are monitored. additional required are 12

c) mu(s) - mu(c) > 3

assume mu(s) - mu(c) = 3

then t statistic = mu(s)-mu(c) ]/ sqrt ( s(s)^2 /n1 + s(c)^2 /n2 )

= 2.323

t (2.323,8+4-2 df) = t(2.323,10) = 2.125%

which is less than 0.05 and hence mu(site) is equal to mu(clean).

However there is less than 2.125% chance we can conclude incorrectly that mu(site) > mu(clean) by 3 mg/l

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