Environmentalists claim that fracking is leading to an increase in metal content
ID: 3209572 • Letter: E
Question
Environmentalists claim that fracking is leading to an increase in metal content (of water) in nearby wells. Using total concentration (ppb) of lanthanide elements as the metric, two sample groups were identified, with the control group of wells being miles removed from any fracking site. Concentrations are as summarized below in Table. 1. State Ho and Ha. Apply the appropriate 2-sample t-test statistic. Specify a 95% confidence interval for the mean improvement of the entire population. What did you choose/calculate as the value for the degrees of freedom? Could another choice be valid or justified?Explanation / Answer
(1) Ho: 1 = 2 versus Ha: 1 2
(2)
Data:
n1 = 21
n2 = 23
x1-bar = 51.48
x2-bar = 41.52
s1 = 11.01
s2 = 17.15
Hypotheses:
Ho: 1 = 2
Ha: 1 2
Decision Rule:
= 0.05
Degrees of freedom = 21 + 23 - 2 = 42
Lower Critical t- score = -2.018081679
Upper Critical t- score = 2.018081679
Reject Ho if |t| > 2.018081679
Test Statistic:
Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = (((21 - 1) * 11.01^2 + (23 - 1) * 17.15^2)/(21 + 23 -2)) = 14.55
SE = s * {(1 /n1) + (1 /n2)} = 14.5529386657652 * ((1/21) + (1/23)) = 4.392416071
t = (x1-bar -x2-bar)/SE = 2.26754475
(3)
n1 = 21
n2 = 23
x1-bar = 51.48
x2-bar = 41.52
s1 = 11.01
s2 = 17.15
% = 95
Degrees of freedom = n1 + n2 - 2 = 21 + 23 -2 = 42
Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((21 - 1) * 11.01^2 + ( 23 - 1) * 17.15^2)/(21 + 23 -2)) = 14.55293867
SE = Pooled s * ((1/n1) + (1/n2)) = 14.5529386657652 * ((1/21) + (1/23)) = 4.392416071
t- score = 2.018081679
Width of the confidence interval = t * SE = 2.01808167886218 * 4.39241607081682 = 8.864254398
Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = 9.95999999999999 - 8.86425439845521 = 1.095745602
Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = 9.95999999999999 + 8.86425439845521 = 18.8242544
The 95% confidence interval is [1.096, 18.824]
(4) Degrees of freedom = 21 + 23 - 2 = 42
(5) Assuming unequal variances, degrees of freedom = 37.86
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