Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean. In Exercises 2-19,Mercury in SushiAn FDA guideline is that themercury 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 New York City. The study was sponsored by the New York Times, and the stores (in order) are D’Agostino, Eli’s Manhattan, Fairway, Food Emporium, Gourmet Garage, Grace’s Marketplace, and Whole Foods. Construct a 98% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi?

0.56 0.75 0.10 0.95 1.25 0.54 0.88

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:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{0.56+0.75+...+0.88}{7} \\ =0.719 \end{gathered}$$

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

The standard deviation is computed using the given formula:

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

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 $t_{\frac{\alpha }{2}}$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:

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

Therefore the margin of error is 5.198.

The confidence interval for the population mean is computed below:

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E \\ 0.719-0.432<\mu <0.719+0.432 \\ 0.287<\mu <1.151 \end{gathered}$$

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.