# Solved (Free): Listed below are the amounts of mercury (in parts per million, or ppm) found in the tuna sushi sampled at different stores

#### ByDr. Raju Chaudhari

Mar 14, 2021

Listed below are the amounts of mercury (in parts per million, or ppm) found in the tuna sushi sampled at different stores. The sample mean is 1.044 ppm and the sample standard deviation is 0.256. Use technology to construct a 90% confidence interval estimate of the mean amount of mercury in the population. 0.79, 0.99, 1.19, 0.68, 0.99, 1.33, 1.34. What is the confidence interval of the mean amount of mercury in the population?

#### Solution

Given that sample size $n = 7$, sample mean $\overline{X}= 1.044$, sample standard deviation $s = 0.256$.

The confidence level is $1-\alpha = 0.9$.

##### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

##### Step 2 Given information

Given that sample size $n =7$, sample mean $\overline{X}=1.044$, sample standard deviation $s=0.256$.

##### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}
where $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, and $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

##### Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t_{\alpha/2,n-1} = t_{0.05,7-1}= 1.943$.

##### Step 5 Compute the margin of error

The margin of error for mean is

 \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 1.943 \frac{0.256}{\sqrt{7}} \\ & = 0.188. \end{aligned}

##### Step 6 Determine the confidence interval

$90$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 1.044 - 0.188 & \leq \mu \leq 1.044 + 0.188\\ 0.856 &\leq \mu \leq 1.232. \end{aligned}
Thus, $90$% confidence interval estimate for population mean is $(0.856,1.232)$.