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The amounts of mercury present in tuna sushi are given for 7 different stores in New York.

Let x represent the amount of mercury present in tuna sushi.

The mean is computed below:

$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n} \\ =\frac{0.56+0.75+...+0.88}{7} \\ =0.719 \end{gathered}$$

Therefore, the sample mean amount of mercury in tuna sushi is 0.719 ppm.

The standard deviation is computed using the given formula:

$$\begin{gathered} s=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-\stackrel{炉}{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(0.56-0.719\right)}^2+{\left(0.75-0.719\right)}^2+...+{\left(0.88-0.719\right)}^2}{7-1}} \\ =0.366 \end{gathered}$$

Therefore, the standard deviation of the amount of mercury is 0366 ppm.

The degrees of freedom are computed as follows:

$$\begin{gathered} n-1=7-1 \\ =6 \end{gathered}$$

The confidence level is 98%. Thus, the level of significance is equal to 0.02.

Referring to the t distribution table, the critical value of $t_{\frac{伪}{2}}$with 6 degrees of freedom at a 0.02 level of significance is equal to 3.1427.

The margin of error is computed as follows:

$$\begin{gathered} E=t_{\frac{伪}{2}}\frac{s}{\sqrt{n}} \\ =3.1427脳\frac{0.366}{\sqrt{7}} \\ =5.198 \end{gathered}$$

Therefore the margin of error is 5.198.

The confidence interval for the population mean is computed below:

$$\begin{gathered} \stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ 0.719-0.432<渭<0.719+0.432 \\ 0.287<渭<1.151 \end{gathered}$$

The 98% confidence interval of the population mean amount of mercury in tuna sushi is equal to (0.287 ppm, 1.151 ppm).

As per FDA,the amount of mercury in fish should be below 1 ppm.

Upon observing the confidence interval, the mean amount of mercury can hold values greater than 1 ppm.

Therefore, it can be said that the amount of mercury in tuna sushi is high.