Question

The following data represent the concentration of dissolved organic carbon​ (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts​ (a) through​ (c) on the right. A( find the sample mean . b(Find the sample standard deviation. c(construct a 90% confidence interval for the population mean

5.20 29.80 27.10 16.51 16.87 8.81 5.30 20.46 14.90 33.67 30.91 14.86 11.90 15.35 9.72 19.80 14.86 8.09 11.90 18.30

Answer 1

The following data represent the concentration of dissolved organic carbon​ (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed.

Solution:

5.20 29.80 27.10 16.51 16.87 8.81 5.30 20.46 14.90 33.67 30.91 14.86 11.90 15.35 9.72 19.80 14.86 8.09 11.90 18.3

a) find the sample mean

Mean, x̄ = Σx/n

Mean x̄ = (5.20+....+18.3)/20

Mean, x̄ = 334.31/20

Mean, x̄ = 16.7155

b) the sample standard deviation.

s = √Σ(xi - x̄)^2/n-1

s = √Σ(xi - 16.7155)^2/20-1

s = √1301.27/19

s = 8.2757

c) 90% confidence interval for the population mean

At 90% confidence interval, α = 0.10

df = n-1 = 20 - 1 = 19

t critical = t(α/2,df) = t(0.01,12)

t critical = 1.7291

the 90% confidence interval for mean:

CI = x̄ ± t critical * s/√n

CI = 16.7155 ± 1.7291 * 8.2757 / √20

CI = 16.7155 ± 3.1997

CI = (13.5158, 19.9152)

Therefore, the 90% confidence interval for mean is between 13.5158 and 19.9152.