A company that produces white bread is concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per slice. Construct a 96% confidence interval for the unknown mean sodium level assuming that the population standard deviation is 10 milligrams. mg << mg (Round values to the nearest tenth. There must be one digit after the decimal point. Do not write the units.)
A company that produces white bread is concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per slice. Construct a 96% confidence interval for the unknown mean sodium level assuming that the population standard deviation is 10 milligrams. mg << mg (Round values to the nearest tenth. There must be one digit after the decimal point. Do not write the units.)
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![### Confidence Interval Calculation for Sodium Levels in Bread
**Problem Statement:**
A company that produces white bread is concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per slice.
**Objective:**
Construct a 96% confidence interval for the unknown mean sodium level, assuming that the population standard deviation is 10 milligrams.
**Mathematical Formulation:**
We aim to find the confidence interval \(CI\) such that:
\[ \text{Lower Limit} < \mu < \text{Upper Limit} \]
Given:
- Sample Size (\(n\)) = 100
- Sample Mean (\(\bar{x}\)) = 103 mg
- Population Standard Deviation (\(\sigma\)) = 10 mg
- Confidence Level = 96%
The formula for the confidence interval for the mean when the population standard deviation is known is:
\[ \bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \]
1. Determine the Z-value:
- Since the confidence level is 96%, the significance level (\(\alpha\)) is 4%, or 0.04.
- Thus, \(\alpha/2 = 0.02\).
- For a 96% confidence level, the Z-value is approximately 2.05 (can be found from Z-tables or standard statistical tools).
2. Compute the margin of error (ME):
\[ ME = Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) = 2.05 \left( \frac{10}{\sqrt{100}} \right) = 2.05 \times 1 = 2.05 \]
3. Calculate the confidence interval limits:
- Lower Limit = \(\bar{x} - ME = 103 - 2.05 = 100.95\)
- Upper Limit = \(\bar{x} + ME = 103 + 2.05 = 105.05\)
**Result:**
\[ 100.9 \, \text{mg} < \mu < 105.1 \, \text{mg} \]
(Values are rounded to the nearest tenth with one digit after the decimal point,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdab6bfb8-cb1a-4a62-b8d6-065f4d11ace0%2Fb3a4bec3-5d2e-4e6f-9dd4-651638f6c084%2Fii9axbg_processed.png&w=3840&q=75)
Transcribed Image Text:### Confidence Interval Calculation for Sodium Levels in Bread
**Problem Statement:**
A company that produces white bread is concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per slice.
**Objective:**
Construct a 96% confidence interval for the unknown mean sodium level, assuming that the population standard deviation is 10 milligrams.
**Mathematical Formulation:**
We aim to find the confidence interval \(CI\) such that:
\[ \text{Lower Limit} < \mu < \text{Upper Limit} \]
Given:
- Sample Size (\(n\)) = 100
- Sample Mean (\(\bar{x}\)) = 103 mg
- Population Standard Deviation (\(\sigma\)) = 10 mg
- Confidence Level = 96%
The formula for the confidence interval for the mean when the population standard deviation is known is:
\[ \bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \]
1. Determine the Z-value:
- Since the confidence level is 96%, the significance level (\(\alpha\)) is 4%, or 0.04.
- Thus, \(\alpha/2 = 0.02\).
- For a 96% confidence level, the Z-value is approximately 2.05 (can be found from Z-tables or standard statistical tools).
2. Compute the margin of error (ME):
\[ ME = Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) = 2.05 \left( \frac{10}{\sqrt{100}} \right) = 2.05 \times 1 = 2.05 \]
3. Calculate the confidence interval limits:
- Lower Limit = \(\bar{x} - ME = 103 - 2.05 = 100.95\)
- Upper Limit = \(\bar{x} + ME = 103 + 2.05 = 105.05\)
**Result:**
\[ 100.9 \, \text{mg} < \mu < 105.1 \, \text{mg} \]
(Values are rounded to the nearest tenth with one digit after the decimal point,
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