bAnd the 90 interval for the population mean The following d

The following data represent the concentration of dissolved organic carbon (me) 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. The sample mean is (Round to two decimal places as needed.) (a)Find the Standard deviation (b)And the 90% interval for the population mean

Solution

(a) mean=16.31

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(b) standard deviation =8.60

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(c) The degree of freedom =n-1=20-1=19

Given a=1-0.9=0.1, t(0.05, df=19) =1.73 (from student t table)

So the lower bound is

xbar -t*s/vn= 16.31 -1.73*8.60/sqrt(20) =12.9831

So the upper bound is

xbar +t*s/vn= 16.31 +1.73*8.60/sqrt(20) =19.63682