**– Make an estimate of the 95% confidence interval for the average mercury content of the population. Does tuna sushi seem to have too much mercury?**

**– What is the confidence interval estimate of the population mean?**

The question aims to find **confidence interval** estimates given sample mean and percentage confidence interval. The **confidence interval** estimate (CI) is a range of values for the **population parameters** based on the sample **mean** and **percentage.**

## Expert Answer

We need sample **mean** and **standard deviation** to find confidence intervals for the population.

**Step 1**: Calculate **sample mean** and **standard deviation:**

\[ \text{Total samples},\ n = 7 \]

\[ \sum x = 4.34\]

The **sample** **mean** is calculated as follows:

\[\bar x = \dfrac{\sum x}{n} = \dfrac{4.34}{7}=0.62\]

Now, we will find the **standard deviation** by using the formula:

\[S.D=\sqrt {\dfrac{\sum (x-\bar x)^2}{n-1}} \]

\[S.D=\sqrt{\dfrac{1.1716}{7-1}}=0.4419\]

The **standard deviation** is $0.4419$.

**Step 2: **The **confidence level** is given as $95\%$.

**Significance level** is calculated as:

\[\sigma=(100-95)\% =0.05\]

We can find the **degree** of **freedom** as follows:

\[d.f = n-1=7-1=6\]

The **critical value** is given as:

\[ t = 2.44469 \]

The **standard error** is calculated as:

\[S.E=\dfrac{S.D}{\sqrt n}=\dfrac{0.4419}{\sqrt 7}=0.167\]

The **margin** of **error** can be found as:

\[M.E=t\ast S.E = 0.40868\]

**Lower** and **Upper limit** are calculated as:

\[L.L=(\bar x-M.E)=0.62-0.40868\]

\[L.L=0.211\]

\[U.L=(\bar x+M.E)=0.62+0.40868\]

\[U.L=1.02868\]

## Numerical Result

The **sample mean** is given as:

\[\bar x=0.62\]

**Standard deviation** is given as:

\[S.D = 0.4419\]

**Lower limit** for the confidence interval is $L.L = 0.211$.

**Upper limit** for the confidence interval is $U.L = 1.02868$.

The $95\%$ **confidence interval** is $(0.211, 1.02868)$.

The **upper limit** of the confidence interval is greater than $1 ppm$ and the **mercury** must be less than $1 ppm$. That’s why there is too much mercury in **tuna sushi.**

## Example

**Food safety** guidelines stipulate that **fish mercury** must be less than **one part per million (ppm)**. Below is the **amount** of **mercury** (ppm) in tuna sushi tasted at various stores in major cities. Make an estimate of the $95\%$ **confidence interval** for the average mercury content of the population. Does it seem like there is too much mercury in tuna sushi?

The total **number** of **samples** is $7$.

The **sample mean** for **seven samples** is calculated as:

\[\bar x=0.714\]

**Standard deviation** is calculated as:

\[S.D=0.3737\]

The **confidence level** is given as $95\%$.

After calculating **standard error** and **margin** of **error, lower** and **upper limits** are calculated as:

\[L.L=(\bar x-margin\:of \:error)=0.3687\]

\[U.L=(\bar x+margin\: of \:error)=1.0599\]