The PCB concentration of a fish caught in Lake Michigan was measured by a techni

Question

The PCB concentration of a fish caught in Lake Michigan was measured by a technique that is known to result in an error of measurement that is normally distributed with a standard deviation of .08ppm (parts per million). Supose the results of 10 independent measurements of this fish are 11.2, 12.4, 10.8, 11.6, 12.5, 10.1, 11.0, 12.2, 12.4, 10.6

a) Give a 95 percent confidence interval for the PCB level of this fish.

b) Give a 95 percent lower confidence interval.

c) Give a 95 percent upper confidence interval.

Explanation / Answer

Given n=10, xbar=11.48, s= 0.86(based on the data)

a. a=0.05, |t(0.025,df=n-1=9)|=2.26 (check student t table)
So 95% CI is
xbar±t*s/n

--> 11.48±2.26*0.86/sqrt(10)

--> (10.86538, 12.09462)

b.a=0.05, |t(0.05,df=n-1=9)|=1.83 (check student t table)

S0 95 percent lower confidence interval is

xbar-t*s/vn

=11.48-1.83*0.86/sqrt(10)

=10.98232

c. So 95 percent upper confidence interval is

xbar-t*s/vn

=11.48+1.83*0.86/sqrt(10)

=11.97768

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