Question

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 9999​% confidence interval estimate of the mean amount of mercury in the population. does it appear that there is too much mercury in tuna​ sushi?

Answer 1

Given that a food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm).

Given below a table of the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city.

$\begin{tabular}{|c|c|c|}x& x-\bar{x} & (x-\bar{x})^2 \\[1ex]0.55&-0.17&0.0289\\0.74&0.02&0.0004\\0.09&-0.63&0.3969\\0.96&0.24&0.0576\\1.32&0.6&0.36\\0.50&-0.22&0.0484\\0.91&0.19&0.0361\\[1ex] \Sigma x=5.07&& \Sigma(x-\bar{x})^2=0.9283 \end{tabular}\\ \\ \bar{x}=\frac{5.07}{7}=0.7243\\ \\s= \sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}} = \sqrt{\frac{0.9283}{7-1} }=\sqrt{\frac{0.9283}{6} }=\sqrt{0.1547}=0.3933$

The 99​% confidence interval estimate of the mean amount of mercury in the population is given by:

$99\%\ C.I.=\bar{x}\pm t_{(\alpha/2,\ k)} \frac{s}{\sqrt{n}} \\ \\ 0.7243\pm t_{(0.005,\ 6)} \frac{0.3933}{\sqrt{7}}=0.7243\pm3.70743\left(\frac{0.3933}{\sqrt{7}}\right) \\ \\ =0.7243\pm3.70743(0.1487)=0.7243\pm0.5511=(0.1732,\ 1.2754)$