Caviar is an expensive delicacy, so companies that package it pay close attention to the amount of product in their tins. Suppose a company that produces over 1,000 tins of caviar per day took a simple random sample (SRS) of 20 tins from one day's production. The sample showed a mean of 99.8 g of caviar per tin with a standard deviation of 0.9 g. The amounts were roughly symmetric with no outliers.

Based on this sample, which of the following is a [tex]$95\%$[/tex] confidence interval for the mean amount of caviar (in grams) per tin from that day's production?

Choose one answer:

A. [tex]$99.8 \pm 1.96\left(\frac{0.9}{\sqrt{20}}\right)$[/tex]

B. [tex]$99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right)$[/tex]

C. [tex]$99.8 \pm 1.96 \sqrt{\frac{0.9(0.1)}{20}}$[/tex]

D. [tex]$99.8 \pm 1.96(0.9)$[/tex]

E. [tex]$99.8 \pm 2.093(0.9)$[/tex]

To find the 95% confidence interval for the mean amount of caviar per tin from the sample data provided, follow these steps:

1. Understand the given data:
- Sample mean ([tex]$\bar{x}$[/tex]): 99.8 grams
- Sample standard deviation ([tex]$s$[/tex]): 0.9 grams
- Sample size ([tex]$n$[/tex]): 20 tins

2. Determine which statistical distribution to use:
- Since the sample size is small ([tex]$n < 30$[/tex]), and we do not know the population standard deviation, we use the t-distribution.

3. Calculate the standard error of the mean (SEM):
- The formula for the standard error is:
[tex]\[
\text{SEM} = \frac{s}{\sqrt{n}}
\][/tex]
- Substituting the values:
[tex]\[
\text{SEM} = \frac{0.9}{\sqrt{20}} \approx 0.2012
\][/tex]

4. Find the critical t-value:
- For a 95% confidence interval with [tex]$n - 1 = 19$[/tex] degrees of freedom, the t-critical value is approximately 2.093.

5. Calculate the margin of error (ME):
- The formula for the margin of error is:
[tex]\[
\text{ME} = t_{\text{critical}} \times \text{SEM}
\][/tex]
- Substituting the values:
[tex]\[
\text{ME} = 2.093 \times 0.2012 \approx 0.4212
\][/tex]

6. Determine the confidence interval:
- The 95% confidence interval is calculated as:
[tex]\[
(\bar{x} - \text{ME}, \bar{x} + \text{ME})
\][/tex]
- Substituting the values:
[tex]\[
(99.8 - 0.4212, 99.8 + 0.4212) \approx (99.38, 100.22)
\][/tex]

Thus, the 95% confidence interval for the mean amount of caviar per tin is approximately [tex]$(99.38, 100.22)$[/tex] grams. Based on these steps, the correct choice is (B) [tex]$99.8 \pm 2.093 \left(\frac{0.9}{\sqrt{20}}\right)$[/tex].