Hospital Noise Levels Noise levels at various area urban hospitals were measured in decibels. The mean noise level in 161 ward areas was 53.2 decibels, and the population standard deviation is 4.8. Find the 90% confidence interval of the true mean. Round your answers to at least one decimal place. ο <με X
Hospital Noise Levels Noise levels at various area urban hospitals were measured in decibels. The mean noise level in 161 ward areas was 53.2 decibels, and the population standard deviation is 4.8. Find the 90% confidence interval of the true mean. Round your answers to at least one decimal place. ο <με X
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
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![### Hospital Noise Levels
Noise levels at various area urban hospitals were measured in decibels. The mean noise level in 161 ward areas was 53.2 decibels, and the population standard deviation is 4.8. Find the 90% confidence interval of the true mean. Round your answers to at least one decimal place.
#### Confidence Interval Formula
The formula for calculating the confidence interval for a population mean when the population standard deviation is known is given by:
\[ \text{CI} = \bar{x} \pm Z \frac{\sigma}{\sqrt{n}} \]
Where:
- \( \bar{x} \) is the sample mean
- \( Z \) is the Z-score corresponding to the desired confidence level
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size
#### Given Data:
- Sample mean (\( \bar{x} \)) = 53.2 decibels
- Population standard deviation (\( \sigma \)) = 4.8
- Sample size (\( n \)) = 161
- Confidence level = 90%
To find the Z-score for a 90% confidence interval, we use the standard normal distribution table, which gives us a Z-score of approximately 1.645.
#### Calculation:
\[ \text{Margin of Error} = Z \frac{\sigma}{\sqrt{n}} = 1.645 \times \frac{4.8}{\sqrt{161}} \approx 0.621 \]
Therefore, the 90% confidence interval is:
\[ 53.2 \pm 0.621 \]
\[ \text{Lower limit} = 53.2 - 0.621 = 52.6 \]
\[ \text{Upper limit} = 53.2 + 0.621 = 53.8 \]
So, the 90% confidence interval for the true mean noise level is:
\[ 52.6 < \mu < 53.8 \]
#### Diagram Explanation:
At the bottom of the text, there is a visual representation that includes:
- Two open squares representing the lower and upper limits of the interval
- A placeholder box where users can input the calculated confidence interval values
- Buttons for submission and resetting the input
The boxes prompt the user to fill in the calculated confidence interval, \( 52.6 \) and \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c4dbd8d-055d-4a36-9216-35dfba9d5ff4%2Fe5baf162-9fc9-40f9-a12d-6e5e4cdc53a9%2Ftpq2az_processed.png&w=3840&q=75)
Transcribed Image Text:### Hospital Noise Levels
Noise levels at various area urban hospitals were measured in decibels. The mean noise level in 161 ward areas was 53.2 decibels, and the population standard deviation is 4.8. Find the 90% confidence interval of the true mean. Round your answers to at least one decimal place.
#### Confidence Interval Formula
The formula for calculating the confidence interval for a population mean when the population standard deviation is known is given by:
\[ \text{CI} = \bar{x} \pm Z \frac{\sigma}{\sqrt{n}} \]
Where:
- \( \bar{x} \) is the sample mean
- \( Z \) is the Z-score corresponding to the desired confidence level
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size
#### Given Data:
- Sample mean (\( \bar{x} \)) = 53.2 decibels
- Population standard deviation (\( \sigma \)) = 4.8
- Sample size (\( n \)) = 161
- Confidence level = 90%
To find the Z-score for a 90% confidence interval, we use the standard normal distribution table, which gives us a Z-score of approximately 1.645.
#### Calculation:
\[ \text{Margin of Error} = Z \frac{\sigma}{\sqrt{n}} = 1.645 \times \frac{4.8}{\sqrt{161}} \approx 0.621 \]
Therefore, the 90% confidence interval is:
\[ 53.2 \pm 0.621 \]
\[ \text{Lower limit} = 53.2 - 0.621 = 52.6 \]
\[ \text{Upper limit} = 53.2 + 0.621 = 53.8 \]
So, the 90% confidence interval for the true mean noise level is:
\[ 52.6 < \mu < 53.8 \]
#### Diagram Explanation:
At the bottom of the text, there is a visual representation that includes:
- Two open squares representing the lower and upper limits of the interval
- A placeholder box where users can input the calculated confidence interval values
- Buttons for submission and resetting the input
The boxes prompt the user to fill in the calculated confidence interval, \( 52.6 \) and \(
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