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 tun different stores in a major city. construct a 98% confidence interval estimate of the mean amount of mercury in the population.
0.55 0.82 0.09 0.91 1.35 0.59 0.88
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 $\bar{x}$

$$\bar{x}=\frac{0.55 + 0.82+0.09+0.91+1.35+0.59+0.88}{7}=\frac{5.19}{7}\approx0.741$$

Step2: Calculate sample standard deviation $s$

First, find the deviations from the mean:
\((0.55 - 0.741)^2=(- 0.191)^2 = 0.036481\)
\((0.82 - 0.741)^2=(0.079)^2=0.006241\)
\((0.09 - 0.741)^2=(-0.651)^2 = 0.423801\)
\((0.91 - 0.741)^2=(0.169)^2=0.028561\)
\((1.35 - 0.741)^2=(0.609)^2 = 0.370881\)
\((0.59 - 0.741)^2=(-0.151)^2=0.022801\)
\((0.88 - 0.741)^2=(0.139)^2=0.019321\)

Sum of squared deviations: \(0.036481+0.006241 + 0.423801+0.028561+0.370881+0.022801+0.019321=0.907087\)

Sample variance \(s^{2}=\frac{0.907087}{7 - 1}=\frac{0.907087}{6}\approx0.151181\)

Sample standard deviation \(s=\sqrt{0.151181}\approx0.389\)

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

Confidence level \(C = 0.98\), so \(\alpha=1 - C=0.02\), and \(\frac{\alpha}{2}=0.01\)

Degrees of freedom \(df=n - 1=7-1 = 6\)

From the \(t\) - distribution table, \(t_{0.01,6}=3.143\)

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

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.143\times\frac{0.389}{\sqrt{7}}\approx3.143\times\frac{0.389}{2.646}\approx3.143\times0.147\approx0.462\)

Step5: Construct the confidence interval

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

\(0.741-0.462 <\mu<0.741 + 0.462\)

\(0.279<\mu<1.203\)

Answer:

\(0.279\) ppm \(<\mu<1.203\) ppm