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A food safety guideline is that the mercury in fish should be below one part per million (ppm). listed below are the amounts of mercury found in tuna sushi sampled at different stores in a major city. construct a 98% confidence interval estimate of the mean amount of mercury in the population. does it appear that there’s too much mercury in tuna sushi?0.58 0.82 0.10 0.88 1.32 0.50 0.92

User Aossey
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Answer:
Confidence\text{ Interval= }0.413\text{ < }\mu<1.047

Step-by-step explanation:

The amounts of mercury found in tuna sushi sampled at different stores are:

0.58, 0.82, 0.10, 0.88, 1.32, 0.50, 0.92

Number of samples, N = 7


\begin{gathered} \text{The mean, }\mu\text{ = }(0.58+0.82+0.10+0.88+1.32+0.50+0.92)/(7) \\ \mu\text{ = }(5.12)/(7) \\ \mu\text{ =}0.73 \end{gathered}

Standard deviation


\begin{gathered} \sigma\text{ = }\sqrt[]{\frac{\sum ^{}_{}{(x_1-\mu)^2}}{N}} \\ \sigma\text{ = }\sqrt[]{((0.58-0.73)^2+(0.82-0.73)^2+(0.10-0.73)^2+(0.88-0.73)^2+(1.32-0.73)^2+(0.50-0.73)^2+(0.92-0.73)^2)/(7)} \\ \sigma\text{ =}\sqrt[]{(0.9087)/(7)} \\ \sigma\text{ =}\sqrt[]{0.1298} \\ \sigma\text{ = }0.36 \end{gathered}

The confidence interval is given by the equation:


\begin{gathered} CI\text{ = }\mu\pm z\frac{\sigma}{\sqrt[]{N}} \\ CI=0.73\pm2.33(\frac{0.36}{\sqrt[]{7}}) \\ CI\text{ = }0.73\pm0.32 \\ CI\text{ = (0.73-0.317})\text{ to (0.73+0.317)} \\ CI\text{ = }0.413\text{ < }\mu<1.047 \end{gathered}

User Mike James
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