b construct a 99 confidence interval for the population mean

The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organ 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 sample standard deviation (b) construct a 99% confidence interval for the population mean m.

Solution

(a) mean=17.54

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(b) standard deviation=8.05

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

Given a=1-0.99=0.01, t(0.005, df=19) =2.86 (from student t table)

So the lower bound is

xbar - t*s/vn =17.54 -2.86*8.05/sqrt(20) =12.3919

So the upper bound is

xbar + t*s/vn =17.54 +2.86*8.05/sqrt(20) =22.6881