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.
0.62 0.73 0.11 0.87 1.33 0.49 0.92
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 $\bar{x}$

$$\bar{x}=\frac{0.62 + 0.73+0.11 + 0.87+1.33+0.49+0.92}{7}=\frac{5.07}{7}\approx0.724$$

Step2: Calculate sample standard deviation $s$

First, calculate the deviations from the mean:
$(0.62 - 0.724)^2=(- 0.104)^2 = 0.010816$, $(0.73 - 0.724)^2=(0.006)^2=0.000036$, $(0.11 - 0.724)^2=(-0.614)^2 = 0.376996$, $(0.87 - 0.724)^2=(0.146)^2=0.021316$, $(1.33 - 0.724)^2=(0.606)^2 = 0.367236$, $(0.49 - 0.724)^2=(-0.234)^2=0.054756$, $(0.92 - 0.724)^2=(0.196)^2=0.038416$
Sum of squared deviations: $0.010816+0.000036 + 0.376996+0.021316+0.367236+0.054756+0.038416=0.869572$
Sample variance $s^{2}=\frac{0.869572}{7 - 1}\approx0.144929$
Sample standard deviation $s=\sqrt{0.144929}\approx0.381$

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

For a $99\%$ confidence interval and $n-1 = 6$ 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 $E$

$$E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.707\times\frac{0.381}{\sqrt{7}}\approx3.707\times0.144\approx0.534$$

Step5: Construct the confidence interval

The confidence interval is $\bar{x}-E<\mu<\bar{x} + E$
$$0.724-0.534<\mu<0.724 + 0.534$$
$$0.190<\mu<1.258$$

Answer:

$0.190$ ppm $<\mu<1.258$ ppm