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,
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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