Question

Water quality measurements are taken daily on the Wyatt River. Shown below are the concentrations of phosphates in solution, in milligrams per liter (mg/L), for an 14-day period.

1.31      1.39      1.59      1.68       1.89      1.98      1.97     

1.74      1.59      1.67      1.59       1.78      1.97      1.84

1. a) Assuming an unknown variance, compute the 95% two-sided confidence interval of the mean concentration of phosphate.
2. b) Assuming the engineer cannot take any additional measurements, compute the confidence level α associated with a confidence interval of ±0.10 mg/L?

Answer 1

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 1.713571429

S = 0.209600614

n = 14

df = n – 1 = 13

Confidence level = 95%

Critical t value = 2.1604

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 1.713571429 ± 2.1604*0.209600614/sqrt(14)

Confidence interval = 1.713571429 ± 0.1210

Lower limit = 1.713571429 - 0.1210 = 1.5926

Upper limit = 1.713571429 + 0.1210 = 1.8346

Confidence interval = (1.5926, 1.8346)

Part b

We are given

Margin of error = E = 0.10

S = 0.209600614

n = 14

Margin of error = E = t*S/sqrt(n)

t = E*sqrt(n)/S

t = 0.10*sqrt(14)/ 0.209600614

t = 1.785137

α =0.005865

c = 1 - 0.005865 = 0.994135

Confidence level = 99.41%