The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be 2.6 grams per liter. Find the 95% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 gram per liter. (Z0.025=1.96,Z0.005=2.575)

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SkladanH · Accepted Answer

Step 1  The 100(1−α)% confidence interval for the population mean, μ when the population standard deviations, σ is known the z-confidence interval as follows: (x―−Zα2σn,x―+zα2σn)  In this case,x―,σ,μ, are the sample mean, population standard deviation, population mean and sample size of the population, respectively and zα2 is the critical value of the standard normal distribution, above which,100(α2)%orα2 proportion of the observations lie.  Step 2  The sample mean is, x―=2.6, the population standard deviation is σ=0.3 and the sample size is n=36. If 95% of the observations must lie within an interval, the remaining 5% must lie outside the interval. Further, due to symmetry, 2.5% will lie above the upper limit and the remaining 2.5% will lie below the lower limit of the interval. Thus, the upper limit of the interval is such that, 97.5% lie below it. As a result,zα2=z0.025≈1.96 [Using Excel formula: =NORM.S.INV(0.975)]. Thus, the confidence interval is, (x―−zα2σn,x―+zα2σn)=(2.6−(1.96)(0.336),2.6+(1.96)(0.336)) =(2.6−(1.96)(0.05),2.6+(1.96)(0.05)) (2.6−0.098,2.6+0.098) (2.502, 2.698) Thus, the 95% confidence interval for the mean zinc concentration in the river is (2.502 grams per liter, 2.698 grams per liter).

The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be

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