The following data represent the concentration of organic carbon (mg/L) collecte

Question

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: x = 18.38 mg/L and s= 7.24 mg/L) 11.90 8.81 30.91 19.80 29.80 16.87 14.86 14.86 27.10 20.46 22.49 8.09 16.51 14.90 15.35 17.50 15.72 33.67 9.72 18.30 page 1 of the table of critical values for the t distribution. the Construct a 99% confidence interval for the mean concentration of dissolved organic carbon collected from organic soil mg/L tomg/L (Use ascending order. Round to two decimal places as needed.)

Explanation / Answer

Solution:- x = 18.38mg/L, and s = 7.24 mg/L, n = 20 ,

CL = (1-)•100%
99% = (1-)•100%
= 0.01
/2 = 0.005

df = n-1
df = 20-1 = 19

Use table to find t(sub 0.005) with a df of 19 = 2.861

=>  x  ± t *(s/n)

= 18.38 ± 2.861*(7.24/20)

= 18.38  ± 2.861*(1.62)

= 18.38  ±  4.63

= (13.75 , 23.01)

=====>   13.75 mg/L to 23.01 mg/L.

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