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Caviar is an expensive delicacy, so companies that package it pay very 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 95% 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)}{9 n}}[/tex]

Answer :

To find a 95% confidence interval for the mean amount of caviar per tin, we need to follow a series of steps:

1. Identify the Sample Statistics:
- 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 the Standard Error (SE):
The standard error of the mean is calculated using the formula:

[tex]\[
SE = \frac{s}{\sqrt{n}}
\][/tex]

Substituting the values we have:

[tex]\[
SE = \frac{0.9}{\sqrt{20}} \approx 0.2014
\][/tex]

3. Choose the Appropriate Critical Value:
For a 95% confidence interval with a relatively small sample size (n = 20), we can use the t-distribution critical value. However, the problem provides options that use the z-distribution ([tex]\(1.96\)[/tex]) and an alternative t-distribution value ([tex]\(2.093\)[/tex]). For simplicity and based on the problem choices, we'll proceed with the z-value ([tex]\(1.96\)[/tex]).

4. Calculate the Margin of Error (ME):
Use the critical value and the standard error to find the margin of error:

[tex]\[
ME = \text{critical value} \times SE = 1.96 \times 0.2014 \approx 0.3944
\][/tex]

5. Determine the Confidence Interval:
Now, calculate the confidence interval by adding and subtracting the margin of error from the sample mean:

[tex]\[
\text{Lower limit} = \bar{x} - ME = 99.8 - 0.3944 \approx 99.4056
\][/tex]
[tex]\[
\text{Upper limit} = \bar{x} + ME = 99.8 + 0.3944 \approx 100.1944
\][/tex]

Now, compare the margin of error in each of the given options to the one we calculated.

- Option (A): [tex]\(99.8 \pm 1.96\left(\frac{0.9}{\sqrt{20}}\right)\)[/tex]

The margin of error computed here is [tex]\(0.3944\)[/tex], which matches our calculation.

Therefore, the correct choice is Option (A).

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