Question

A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the arrounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 99% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury In tuna sushi? 

0.50 0.74 0.10 0.96 1.26 0.50 0.92 

What is the confidence interval estimate of the population mean u? 

ppm <u< ppm (Round to three decimal places as needed.)

Answer 1

\(\underline{\text { Solution: }}\)

From the given data \(\mathrm{s}=0.35, \mathrm{n}=7\), Mean \(\overline{\mathrm{x}}=0.71\)

Confidence interval \(=\bar{x} \pm t_{a / 2, n-1} " s / v n\)

Level of significance \(a=0.02\) \(\mathrm{t}_{a / 2, \mathrm{n}-1}=\mathrm{t}_{0.01,6}=3.1427\)

Now \(\mathrm{Cl}=\overline{\mathrm{x}} \pm \mathrm{t}_{\mathrm{a} / 2, \mathrm{n}-1}^{*} \mathrm{~s} / \mathrm{Vn}\)

\(=0.71 \pm 3.1427^{*} 0.35 / \sqrt{7}\)

\(=0.71 \pm 0.4157\)

\(=(0.294,1.126)\)

The confidence interval :

\(0.294 \mathrm{ppm}<\mu<1.126 \mathrm{ppm}\)