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 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 0.74 0.10 0.93 1.25 0.52 0.90
what is the confidence interval estimate of the population mean μ?
□ ppm < μ < □ ppm
(round to three decimal places as needed.)

Explanation:

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

The formula for the sample mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Given data \(x = \{0.56,0.74,0.10,0.93,1.25,0.52,0.90\}\), \(n = 7\).
\(\sum_{i=1}^{7}x_{i}=0.56 + 0.74+0.10 + 0.93+1.25+0.52+0.90=5.0\)
\(\bar{x}=\frac{5.0}{7}\approx0.714\)

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(0.56 - 0.714)^{2}=(- 0.154)^{2}=0.023716\)
\((x_{2}-\bar{x})^{2}=(0.74 - 0.714)^{2}=(0.026)^{2}=0.000676\)
\((x_{3}-\bar{x})^{2}=(0.10 - 0.714)^{2}=(-0.614)^{2}=0.376996\)
\((x_{4}-\bar{x})^{2}=(0.93 - 0.714)^{2}=(0.216)^{2}=0.046656\)
\((x_{5}-\bar{x})^{2}=(1.25 - 0.714)^{2}=(0.536)^{2}=0.287296\)
\((x_{6}-\bar{x})^{2}=(0.52 - 0.714)^{2}=(-0.194)^{2}=0.037636\)
\((x_{7}-\bar{x})^{2}=(0.90 - 0.714)^{2}=(0.186)^{2}=0.034596\)
\(\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}=0.023716+0.000676 + 0.376996+0.046656+0.287296+0.037636+0.034596=0.807572\)
\(s=\sqrt{\frac{0.807572}{7 - 1}}=\sqrt{\frac{0.807572}{6}}\approx0.367\)

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

The confidence level is \(98\%\), so \(\alpha=1 - 0.98 = 0.02\), and \(\frac{\alpha}{2}=0.01\).
The degrees of freedom \(df=n - 1=7-1 = 6\).
Using the \(t\) - distribution table or a calculator, \(t_{0.01,6}=3.143\)

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

The formula for the margin of error for a \(t\) - distribution is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E=3.143\times\frac{0.367}{\sqrt{7}}\)
\(\sqrt{7}\approx2.646\), \(\frac{0.367}{2.646}\approx0.139\)
\(E=3.143\times0.139\approx0.437\)

Step5: Construct the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
\(0.714-0.437 <\mu<0.714 + 0.437\)
\(0.277<\mu<1.151\)

Answer:

\(0.277\) ppm \(<\mu<1.151\) ppm