## Case Solution Mercury Data

Water specimens collected from a residential area that is served by the city water supply are indicated by subscript c; p indicates specimens taken from a residential area that is served by private wells. The averages, variances, standard deviations, and standard errors are:

City (nc = 13) yc = 0.157 ^g/L s2 = 0.0071 sc = 0.084 s^ = 0.023 Private (np = 10) yp = 0.151 ßg/L sp = 0.0076 sp = 0.087 s/ = 0.028

The difference in the averages of the measurements is yc - yp = 0.157 - 0.151 = 0.006 Mg/L. The

variances sc and sp of the city and private samples are nearly equal, so they can be pooled by weighting in proportion to their degrees of freedom:

The estimated variance of the difference between averages is:

V(yc - yp) = 4 + 4 = spoolj^-1 + ¡-1") = 0.00734^-3 + 1) = 0.0013 (Mg/L)2

and the standard error of yc - yp = 0.006 Mg/L is sy -y = 70.0013 = 0.036 Mg/ L.

The variance of the difference is estimated with v = 12 + 9 = 21 degrees of freedom. The 95% confidence interval is calculated using a/ 2 = 0.025 and t21>0025 = 2.080:

(yc - yp)± t21>0025s%-y = 0.006 ± 2.080(0.036) = 0.006 ± 0.075 Mg/L

It can be stated with 95% confidence that the true difference between the city and private water supplies falls in the interval of -0.069 Mg/L and 0.081 Mg/L. This confidence interval includes zero so there is no persuasive evidence in these data that the mercury contents are different in the two residential areas. Future sampling can be done in either area without worrying that the water supply will affect the outcome.

0 0