UkusakazaL

2021-08-01

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a $95\mathrm{%}$ confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?
$\begin{array}{|ccccccc|}\hline 0.60& 0.82& 0.09& 0.89& 1.29& 0.49& 0.82\\ \hline\end{array}$
What is the confidence interval estimate of the population mean $\mu ?$

### Answer & Explanation

ddaeeric

Step 1
$\begin{array}{|cccc|}\hline & x& \left(x-\overline{x}\right)& \left(x-\overline{x}{\right)}^{2}\\ & 0.6& -0.11429& 0.01306122\\ & 0.82& 0.10571& 0.01117551\\ & 0.09& -0.62429& 0.38973265\\ & 0.89& 0.17571& 0.03087551\\ & 1.29& 0.57571& 0.33144694\\ & 0.49& -0.22429& 0.05030408\\ & 0.82& 0.10571& 0.01117551\\ Total& 5& & 0.83777143\\ \hline\end{array}$
Here $n=7$
Sampke mean $=\stackrel{―}{x}=\frac{\sum x}{n}=0.714$
Sample $S.D.=s=\sqrt{\frac{1}{n-1}\sum {\left(x-\stackrel{―}{x}\right)}^{2}}=0.3737$
Step 2

Standard Error $=\frac{s}{\sqrt{n}}=0.1412$
Margin of Error (M.E) $={t}^{\cdot }\frac{s}{\sqrt{n}}=0.3456$
$\text{Lower Limit}=\stackrel{―}{x}-\left(M.E\right)=0.3687$
$\text{Upper Limit}=\stackrel{―}{x}+\left(M.E\right)=1.0599$

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