Question

2. The following data represent the concentration of organic carbon (mg/L collected from organic soil. Construct a 99% confidence interval for the mean concentration of dissolved organic carbon collected from organic soil. Note that ? = 17.63mg/L and ? = 7.66mg/L

5.30 29.80 27.10 16.51 15.72

8.81 16.87 20.46 14.90 33.67

30.91 14.86 17.50 15.35 9.72

19.80 14.86 8.09 14.00 18.30

Step 1: Compute ?/2 for ? = (1 − confidence level) and determine the sample mean and sample standard deviation.

Step 2: Use Table IV to find ??/2 with ?? = ? − 1.

Step 3: Determine the CI.

Step 4: Interpret the CI.

Answer 1

Step 1

Mean X̅ = Σ Xi / n
X̅ = 352.53 / 20 = 17.6265

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 1114.6682 / 20 -1 ) = 7.6594

α = 0.01

Step 2

t(α/2, n-1) = t(0.01 /2, 20- 1 ) = 2.861

Step 3

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 20- 1 ) = 2.861
17.6265 ± t(0.01/2, 20 -1) * 7.6594/√(20)
Lower Limit = 17.6265 - t(0.01/2, 20 -1) 7.6594/√(20)
Lower Limit = 12.7265
Upper Limit = 17.6265 + t(0.01/2, 20 -1) 7.6594/√(20)
Upper Limit = 22.5265
99% Confidence interval is ( 12.7265 , 22.5265 )

Step 4

We are 99% confident that the truw population mean concentration of dissolved organic carbon collected from organic soil lies within the interval  ( 12.7265 , 22.5265 ).