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Answer :
To determine the 95% confidence interval for the mean amount of caviar per tin, let's go through the process step-by-step:
1. Identify Key Information:
- 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. Choose the Correct Formula for Confidence Interval:
The confidence interval (CI) for the mean when the sample size is small (less than 30) typically uses the t-distribution. However, if we have a large sample or if we're using a standard normal distribution (z-distribution), the formula is:
[tex]\[
\bar{x} \pm z \times \left(\frac{s}{\sqrt{n}}\right)
\][/tex]
where [tex]\( z \)[/tex] is the z-score associated with the desired confidence level.
3. Determine the Appropriate Z-Score or T-Score:
- For a 95% confidence level using a z-distribution, the z-score is approximately 1.96.
- However, for a t-distribution with 19 degrees of freedom (since [tex]\( n-1 = 19 \)[/tex]), we often use a t-score. Here, one of the given options also uses a t-score of approximately 2.093.
4. Calculate the Standard Error (SE):
The standard error of the mean is calculated as:
[tex]\[
SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{20}} \approx 0.201
\][/tex]
5. Calculate the Margin of Error (ME) for Each Option:
- For Option A: Use a z-score of 1.96.
[tex]\[
ME = 1.96 \times 0.201 \approx 0.394
\][/tex]
- For Option B: Use a t-score of 2.093.
[tex]\[
ME = 2.093 \times 0.201 \approx 0.421
\][/tex]
6. Form the Confidence Intervals:
- For Option A:
[tex]\[
\text{CI} = 99.8 \pm 0.394 \rightarrow (99.4056, 100.1944)
\][/tex]
- For Option B:
[tex]\[
\text{CI} = 99.8 \pm 0.421 \rightarrow (99.3788, 100.2212)
\][/tex]
Upon reviewing the calculated margins of error and confidence intervals, it's clear that Option B, which uses the t-score of 2.093, provides a more appropriate interval considering the smaller sample size. Thus, the correct choice is:
(B) [tex]\(99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right)\)[/tex].
1. Identify Key Information:
- 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. Choose the Correct Formula for Confidence Interval:
The confidence interval (CI) for the mean when the sample size is small (less than 30) typically uses the t-distribution. However, if we have a large sample or if we're using a standard normal distribution (z-distribution), the formula is:
[tex]\[
\bar{x} \pm z \times \left(\frac{s}{\sqrt{n}}\right)
\][/tex]
where [tex]\( z \)[/tex] is the z-score associated with the desired confidence level.
3. Determine the Appropriate Z-Score or T-Score:
- For a 95% confidence level using a z-distribution, the z-score is approximately 1.96.
- However, for a t-distribution with 19 degrees of freedom (since [tex]\( n-1 = 19 \)[/tex]), we often use a t-score. Here, one of the given options also uses a t-score of approximately 2.093.
4. Calculate the Standard Error (SE):
The standard error of the mean is calculated as:
[tex]\[
SE = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{20}} \approx 0.201
\][/tex]
5. Calculate the Margin of Error (ME) for Each Option:
- For Option A: Use a z-score of 1.96.
[tex]\[
ME = 1.96 \times 0.201 \approx 0.394
\][/tex]
- For Option B: Use a t-score of 2.093.
[tex]\[
ME = 2.093 \times 0.201 \approx 0.421
\][/tex]
6. Form the Confidence Intervals:
- For Option A:
[tex]\[
\text{CI} = 99.8 \pm 0.394 \rightarrow (99.4056, 100.1944)
\][/tex]
- For Option B:
[tex]\[
\text{CI} = 99.8 \pm 0.421 \rightarrow (99.3788, 100.2212)
\][/tex]
Upon reviewing the calculated margins of error and confidence intervals, it's clear that Option B, which uses the t-score of 2.093, provides a more appropriate interval considering the smaller sample size. Thus, the correct choice is:
(B) [tex]\(99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right)\)[/tex].
Thank you for reading the article 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. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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