Hospital Noise Levels For a sample of 4 operating rooms taken in a hospital study, the mean nolse level was 39.2 decibels and the standard devlation was 10.6. Find the 90% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable Is normally distributed. Round your answers to at least two decimal places. くpく

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### Hospital Noise Levels 

For a sample of 4 operating rooms taken in a hospital study, the mean noise level was 39.2 decibels, and the standard deviation was 10.6. Find the 90% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable is normally distributed. Round your answers to at least two decimal places.

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The image shows a fill-in-the-blank 90% confidence interval calculation setup for user input, encapsulated with inequality symbols indicating the interval format.

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**Detailed explanation of calculations:**

1. **Identify the sample data:**
   - Sample mean (x̄): 39.2 decibels
   - Standard deviation (s): 10.6 decibels
   - Sample size (n): 4

2. **Determine the degrees of freedom (df):**
   - \( df = n - 1 = 4 - 1 = 3 \)

3. **Find the t-critical value (t*) for a 90% confidence interval with 3 degrees of freedom:**
   - From the t-table, t* for 90% confidence level and df = 3 is approximately 2.353

4. **Calculate the margin of error (ME):**
   - \( ME = t* \cdot \left( \frac{s}{\sqrt{n}} \right) \)
   - \( ME = 2.353 \cdot \left( \frac{10.6}{\sqrt{4}} \right) \)
   - \( ME = 2.353 \cdot 5.3 = 12.47 \)

5. **Determine the confidence interval:**
   - Lower bound = \( x̄ - ME = 39.2 - 12.47 = 26.73 \)
   - Upper bound = \( x̄ + ME = 39.2 + 12.47 = 51.67 \)

6. **90% confidence interval:**
   - \( 26.73 < μ < 51.67 \)

The result should be filled in the answer box provided on the webpage as follows: 

\[ 26.73 < μ < 51.67 \]

Save your work if you wish to continue later, or submit the assignment for evaluation.
Transcribed Image Text:### Hospital Noise Levels For a sample of 4 operating rooms taken in a hospital study, the mean noise level was 39.2 decibels, and the standard deviation was 10.6. Find the 90% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable is normally distributed. Round your answers to at least two decimal places. --- The image shows a fill-in-the-blank 90% confidence interval calculation setup for user input, encapsulated with inequality symbols indicating the interval format. --- **Detailed explanation of calculations:** 1. **Identify the sample data:** - Sample mean (x̄): 39.2 decibels - Standard deviation (s): 10.6 decibels - Sample size (n): 4 2. **Determine the degrees of freedom (df):** - \( df = n - 1 = 4 - 1 = 3 \) 3. **Find the t-critical value (t*) for a 90% confidence interval with 3 degrees of freedom:** - From the t-table, t* for 90% confidence level and df = 3 is approximately 2.353 4. **Calculate the margin of error (ME):** - \( ME = t* \cdot \left( \frac{s}{\sqrt{n}} \right) \) - \( ME = 2.353 \cdot \left( \frac{10.6}{\sqrt{4}} \right) \) - \( ME = 2.353 \cdot 5.3 = 12.47 \) 5. **Determine the confidence interval:** - Lower bound = \( x̄ - ME = 39.2 - 12.47 = 26.73 \) - Upper bound = \( x̄ + ME = 39.2 + 12.47 = 51.67 \) 6. **90% confidence interval:** - \( 26.73 < μ < 51.67 \) The result should be filled in the answer box provided on the webpage as follows: \[ 26.73 < μ < 51.67 \] Save your work if you wish to continue later, or submit the assignment for evaluation.
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