# A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Construct a 95\​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?

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 $$\displaystyle{95}​\%$$ confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?
$$\begin{array}{|c|c|}\hline 0.60 & 0.82 & 0.09 & 0.89 & 1.29 & 0.49 & 0.82 \\ \hline \end{array}$$
What is the confidence interval estimate of the population mean $$\displaystyle\mu?$$
$$\displaystyle?\ {p}\pm{<}\mu{<}\ ?\ {p}\pm$$

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ddaeeric

Step 1
$$\begin{array}{|c|c|c|}\hline & x & (x-\bar{x}) & (x-\bar{x})^{2} \\ \hline & 0.6 & -0.11429 & 0.01306122 \\ \hline & 0.82 & 0.10571 & 0.01117551 \\ \hline & 0.09 & -0.62429 & 0.38973265 \\ \hline & 0.89 & 0.17571 & 0.03087551 \\ \hline & 1.29 & 0.57571 & 0.33144694 \\ \hline & 0.49 & -0.22429 & 0.05030408 \\ \hline & 0.82 & 0.10571 & 0.01117551 \\ \hline Total & 5 & & 0.83777143 \\ \hline \end{array}$$
Here $$\displaystyle{n}={7}$$
Sampke mean $$\displaystyle=\overline{{{x}}}={\frac{{\sum{x}}}{{{n}}}}={0.714}$$
Sample $$\displaystyle{S}.{D}.={s}=\sqrt{{{\frac{{{1}}}{{{n}-{1}}}}\sum{\left({x}-\overline{{{x}}}\right)}^{{{2}}}}}={0.3737}$$
Step 2
$$\begin{array}{|c|}\hline \text{Confidence Level}=95 \\ \hline \text{Significance Level}=\alpha=(100-95)\%=0.05 \\ \hline \text{Degrees of freedom}=n-1=7-1=6 \\ \hline \text{Critical value}=t^{*}=2.447 [\text{using Excel}=TINV(0.05,\ 6)]\\ \hline \end{array}$$
Standard Error $$\displaystyle={\frac{{{s}}}{{\sqrt{{{n}}}}}}={0.1412}$$
Margin of Error (M.E) $$\displaystyle={t}^{{\cdot}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}={0.3456}$$
$$\displaystyle\text{Lower Limit}=\overline{{{x}}}-{\left({M}.{E}\right)}={0.3687}$$
$$\displaystyle\text{Upper Limit}=\overline{{{x}}}+{\left({M}.{E}\right)}={1.0599}$$

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