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 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? 0.60 0.80 0.11 0.92 1.36 0.54 0.83 what is the confidence interval estimate of the population mean μ? ppm < μ < ppm (round to three decimal places as needed.)

Explanation:

Step1: Calculate sample mean ($\bar{x}$)

$$ \bar{x}=\frac{0.60 + 0.80+0.11 + 0.92+1.36+0.54+0.83}{7}=\frac{5.16}{7}\approx0.737 $$

Step2: Calculate sample standard deviation ($s$)

First, find the deviations from the mean:
\((0.60 - 0.737)=- 0.137\), \((0.80 - 0.737)=0.063\), \((0.11 - 0.737)=-0.627\), \((0.92 - 0.737)=0.183\), \((1.36 - 0.737)=0.623\), \((0.54 - 0.737)=-0.197\), \((0.83 - 0.737)=0.093\)

Then, square the deviations: \((-0.137)^2 = 0.018769\), \((0.063)^2=0.003969\), \((-0.627)^2 = 0.393129\), \((0.183)^2=0.033489\), \((0.623)^2 = 0.388129\), \((-0.197)^2=0.038809\), \((0.093)^2=0.008649\)

Sum of squared deviations: \(0.018769+0.003969 + 0.393129+0.033489+0.388129+0.038809+0.008649=0.884943\)

Sample variance \(s^{2}=\frac{0.884943}{7 - 1}\approx0.1475\)

Sample standard deviation \(s=\sqrt{0.1475}\approx0.384\)

Step3: Determine the critical value ($t_{\alpha/2}$)

For a \(99\%\) confidence interval and \(n-1=7 - 1 = 6\) degrees of freedom, \(\alpha=1 - 0.99=0.01\), \(\alpha/2=0.005\). From the \(t\)-distribution table, \(t_{0.005,6}=3.707\)

Step4: Calculate the margin of error ($E$)

$$ E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.707\times\frac{0.384}{\sqrt{7}}\approx3.707\times0.145\approx0.538 $$

Step5: Construct the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
\(0.737- 0.538<\mu<0.737 + 0.538\)
\(0.199<\mu<1.275\)

Answer:

\(0.199\) ppm \(<\mu<1.275\) ppm