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 does it appear that there is too much mercury in tuna sushi? a. no, because it is possible that the mean is not greater than 1 ppm. also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. b. no, because it is not possible that the mean is greater than 1 ppm. also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. c. yes, because it is possible that the mean is greater than 1 ppm. also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury. d. yes, because it is possible that the mean is not greater than 1 ppm. also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury.

Explanation:

Step1: Calculate sample mean

First, find the sample mean $\bar{x}$.
$\bar{x}=\frac{0.60 + 0.80+0.11 + 0.92+1.36+0.54+0.83}{7}=\frac{5.16}{7}\approx0.74$

Step2: Calculate sample standard deviation

The formula for sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
$\sum_{i=1}^{7}(x_{i}-\bar{x})^{2}=(0.60 - 0.74)^{2}+(0.80 - 0.74)^{2}+(0.11 - 0.74)^{2}+(0.92 - 0.74)^{2}+(1.36 - 0.74)^{2}+(0.54 - 0.74)^{2}+(0.83 - 0.74)^{2}$
$=(- 0.14)^{2}+0.06^{2}+(-0.63)^{2}+0.18^{2}+0.62^{2}+(-0.2)^{2}+0.09^{2}$
$=0.0196 + 0.0036+0.3969+0.0324+0.3844+0.04+0.0081=0.885$
$s=\sqrt{\frac{0.885}{7 - 1}}=\sqrt{\frac{0.885}{6}}\approx0.38$

Step3: Determine the critical value

For a $99\%$ confidence interval and $n-1 = 6$ degrees of freedom, the critical value $t_{\alpha/2}$ from the t - distribution table is $t_{0.005,6}=3.707$

Step4: Calculate the margin of error

The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}=3.707\times\frac{0.38}{\sqrt{7}}\approx3.707\times0.144\approx0.54$

Step5: Calculate the confidence interval

The confidence interval is $\bar{x}-E<\mu<\bar{x}+E$
$0.74-0.54 <\mu<0.74 + 0.54$
$0.20<\mu<1.28$

Since the confidence interval includes values greater than $1$ ppm (it is possible that the population mean is greater than $1$ ppm) and at least one sample value ($1.36$ ppm) exceeds $1$ ppm.

Answer:

C. Yes, because it is possible that the mean is greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have too much mercury.