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
[tex]\begin{gathered} \text{The mean, }\mu\text{ = }\frac{0.58+0.82+0.10+0.88+1.32+0.50+0.92}{7} \\ \mu\text{ = }\frac{5.12}{7} \\ \mu\text{ =}0.73 \end{gathered}[/tex]Standard deviation
[tex]\begin{gathered} \sigma\text{ = }\sqrt[]{\frac{\sum ^{}_{}{(x_1-\mu)^2}}{N}} \\ \sigma\text{ = }\sqrt[]{\frac{(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[]{\frac{0.9087}{7}} \\ \sigma\text{ =}\sqrt[]{0.1298} \\ \sigma\text{ = }0.36 \end{gathered}[/tex]The confidence interval is given by the equation:
[tex]\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}[/tex]