13. A county environmental agency suspects that the fish in a particular pollute

Question

13. A county environmental agency suspects that the fish in a particular polluted lake have elevated mercury level. To confirm that suspicion, five striped bass in that lake were caught and their tissues were tested for mercury. For the purpose of comparison, four striped bass in an unpolluted lake were also caught and tested. The fish tissue mercury levels in mg/kg are given below. Sampla 1 Bampla (from pollutad lako (Erom unpolluted lako) 0.580 0.382 0.376 0.570 0.366 0.571 0.666 0.50 a. Construct the 95% confidence interval for the difference in the population means based on these data. b. Test, at the 5% level of significance, whether the data provide sufficient evidence to conclude that fish in the polluted lake have elevated levels of mercury in their tissue.

Explanation / Answer

a)

x1(bar) = 0.6062

x2(bar) = 0.4235

s1 = 0.0824

s2 = 0.0979

n1 = 5

n2 = 4

SE = sqrt[ (s1^2/n1) + (s2^2/n2) ]

(s1^2/n1) = 0.0014

(s2^2/n2) = 0.0024

SE = 0.0613

x1bar - x2bar = 0.18

SE = 0.0613

CI = 95%

DF = 3

t-value = 3.1824

ME = t*SE

= 3.1824*0.0613

= 0.1949

Confidence Interval (x1bar - x2bar) +/- ME

Lower bound = 0.18 - 0.1949 = -0.01224

Upper bound = 0.18 + 0.1949 = 0.37764

Confidence Interval (-0.0122 , 0.3776 )

b)

As confidence interval does not lie to the right of zero, we fail to reject the null hypothesis.

There are not sufficient evidence to conclude that fish in the polluted lake have elevated levels of mercury in their tissue.

Given Data Sample 1 Sample 2 Name Polluted Unpolluted mean 0.6062 0.4235 Sample size 5 4 Std. dev. 0.0824 0.0979

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