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 90% confidence interval estimate of the mean amount of mercury in the population. does it appear that there is too much mercury in tuna sushi?
0.55 0.81 0.10 0.97 1.30 0.49 0.82
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 $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Here, $x_1 = 0.55,x_2=0.81,x_3 = 0.10,x_4=0.97,x_5 = 1.30,x_6=0.49,x_7 = 0.82$, and $n = 7$.
$\sum_{i=1}^{7}x_{i}=0.55 + 0.81+0.10 + 0.97+1.30+0.49+0.82=5.04$
$\bar{x}=\frac{5.04}{7}=0.72$

Step2: Calculate the sample standard deviation $s$

The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
$(x_1-\bar{x})=(0.55 - 0.72)=- 0.17,(x_2-\bar{x})=(0.81 - 0.72)=0.09,(x_3-\bar{x})=(0.10 - 0.72)=-0.62,(x_4-\bar{x})=(0.97 - 0.72)=0.25,(x_5-\bar{x})=(1.30 - 0.72)=0.58,(x_6-\bar{x})=(0.49 - 0.72)=-0.23,(x_7-\bar{x})=(0.82 - 0.72)=0.10$
$\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}=(-0.17)^{2}+(0.09)^{2}+(-0.62)^{2}+(0.25)^{2}+(0.58)^{2}+(-0.23)^{2}+(0.10)^{2}$
$=0.0289+0.0081 + 0.3844+0.0625+0.3364+0.0529+0.01$
$=0.8832$
$s=\sqrt{\frac{0.8832}{7-1}}=\sqrt{\frac{0.8832}{6}}\approx0.383$

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

The confidence level is $C = 0.90$, so $\alpha=1 - C=0.10$ and $\frac{\alpha}{2}=0.05$.
The degrees of freedom is $df=n - 1=7-1 = 6$.
Using the $t$-distribution table or a calculator, $t_{\alpha/2,df = 6}=1.943$

Step4: Calculate the margin of error $E$

The formula for the margin of error for a $t$-distribution is $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$.
Substitute $t_{\alpha/2}=1.943,s = 0.383,n = 7$ into the formula.
$E=1.943\times\frac{0.383}{\sqrt{7}}\approx1.943\times\frac{0.383}{2.646}\approx0.282$

Step5: Construct the confidence interval

The confidence interval for the population mean $\mu$ when the population standard deviation $\sigma$ is unknown is $\bar{x}-E<\mu<\bar{x}+E$.
Substitute $\bar{x}=0.72$ and $E = 0.282$
$0.72-0.282<\mu<0.72 + 0.282$

Answer:

$0.438\ \text{ppm}<\mu<1.002\ \text{ppm}$

Since the upper limit of the confidence interval ($1.002$ ppm) is slightly above $1$ ppm, it appears that there might be too much mercury in tuna sushi.