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

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. 7.40 8.81 30.91 19.80 29.80 11.90 4.86 14.86 27.10 20.46 17.50 8.09 16.51 14.90 15.35 20.46 5.20 33.67 9.72 18.30 (a) Find the sample mean The sample mean is (Round to two decimal places as needed.) (b) Find the sample standard deviation The sample standard deviation is (Round to two decimal places as needed.) (c) Construct a 90% confidence interval for the population mean The 90% confidence interval for the population mean is ( OD (Round to two decimal places as needed.)

Explanation / Answer

Solution:- give that

7.40,29.80,27.10,16.51,5.20,8.81,11.90,20.46,14.90,33.67,30.91,14.86,17.50,15.35,9.72,19.80,14.86,8.09,20.46,18.30

a) The sample mean is 17.28

=> mean = sum of term/no of terms = 345.6/20 = 17.28

b) The sample standard deviation is 8.09

c) The 90% confidence interval for the population mean is ( 14.30 , 20.26)

=> 17.28 +- 1.645 * (8.09/sqrt(20))

= ( 14.30 , 20.26)

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