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}cx&$x-\bar{x}$&$(x-\bar{x})^2$\\[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}=(5.07)/(7)=0.7243\\ \\s= \sqrt{\frac{\Sigma(x-\bar{x})^2}{n-1}} = \sqrt{(0.9283)/(7-1) }=\sqrt{(0.9283)/(6) }=√(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)) (s)/(√(n)) \\ \\ 0.7243\pm t_((0.005,\ 6)) (0.3933)/(√(7))=0.7243\pm3.70743\left((0.3933)/(√(7))\right) \\ \\ =0.7243\pm3.70743(0.1487)=0.7243\pm0.5511=(0.1732,\ 1.2754)`