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

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. $\left({Z}_{0.025}=1.96,{Z}_{0.005}=2.575\right)$

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Step 1

The $100\left(1-\alpha \right)\mathrm{%}$ confidence interval for the population mean, $\mu$ when the population standard deviations, $\sigma$ is known the z-confidence interval as follows: $\left(\stackrel{―}{x}-{Z}_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}},\stackrel{―}{x}+{z}_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\right)$

In this case,$\stackrel{―}{x},\sigma ,\mu ,$ are the sample mean, population standard deviation, population mean and sample size of the population, respectively and ${z}_{\frac{\alpha }{2}}$ is the critical value of the standard normal distribution, above which,$100\left(\frac{\alpha }{2}\right)\mathrm{%}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{\alpha }{2}$ proportion of the observations lie.

Step 2

The sample mean is, $\stackrel{―}{x}=2.6$, the population standard deviation is $\sigma =0.3$ 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,${z}_{\frac{\alpha }{2}}={z}_{0.025}\approx 1.96$ [Using Excel formula: =NORM.S.INV(0.975)]. Thus, the confidence interval is, $\left(\stackrel{―}{x}-{z}_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}},\stackrel{―}{x}+{z}_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\right)=\left(2.6-\left(1.96\right)\left(\frac{0.3}{\sqrt{36}}\right),2.6+\left(1.96\right)\left(\frac{0.3}{\sqrt{36}}\right)\right)$
$=\left(2.6-\left(1.96\right)\left(0.05\right),2.6+\left(1.96\right)\left(0.05\right)\right)$
$\left(2.6-0.098,2.6+0.098\right)$
(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).