# 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. (Z_{0.025} = 1.96, Z_{0.005} = 2.575)

Question
Confidence intervals
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. (Z_{0.025} = 1.96, Z_{0.005} = 2.575)

2021-01-11
Step 1 The $$\displaystyle{100}{\left({1}-\alpha\right)}\%$$ confidence interval for the population mean, $$\displaystyle\mu$$ when the population standard deviations, $$\displaystyle\sigma$$ is known the z-confidence interval as follows: $$\displaystyle{\left(\overline{{{x}}}-{Z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}$$ In this case,$$\displaystyle\overline{{{x}}},\sigma,\mu,$$ are the sample mean, population standard deviation, population mean and sample size of the population, respectively and $$\displaystyle{z}_{{\frac{\alpha}{{2}}}}$$ is the critical value of the standard normal distribution, above which,$$\displaystyle{100}{\left(\frac{\alpha}{{2}}\right)}\%{P}{S}{K}{\quad\text{or}\quad}{P}{S}{K}\frac{\alpha}{{2}}$$ proportion of the observations lie. Step 2 The sample mean is, $$\displaystyle\overline{{{x}}}={2.6}$$, the population standard deviation is $$\displaystyle\sigma={0.3}$$ and the sample size is $$\displaystyle{n}={36}$$. 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,$$\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.025}}}\approx{1.96}$$ [Using Excel formula: =NORM.S.INV(0.975)]. Thus, the confidence interval is, $$\displaystyle{\left(\overline{{{x}}}-{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}},\overline{{{x}}}+{z}_{{\frac{\alpha}{{2}}}}{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}={\left({2.6}-{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)},{2.6}+{\left({1.96}\right)}{\left({\frac{{{0.3}}}{{\sqrt{{{36}}}}}}\right)}\right)}$$
$$\displaystyle={\left({2.6}-{\left({1.96}\right)}{\left({0.05}\right)},{2.6}+{\left({1.96}\right)}{\left({0.05}\right)}\right)}$$
$$\displaystyle{\left({2.6}-{0.098},{2.6}+{0.098}\right)}$$
PSK(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).

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