The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.2 98.3 99.5 99.1 99.2 99.4 98.2 97.3 98.7 99.7 97.4 98.6 96.6 Assume body temperatures of adults are normally distributed. Based on this data, find the 95% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population. 95% C.I. =
The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 97.2 98.3 99.5 99.1 99.2 99.4 98.2 97.3 98.7 99.7 97.4 98.6 96.6 Assume body temperatures of adults are normally distributed. Based on this data, find the 95% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population. 95% C.I. =
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
![### Body Temperature Data Analysis
#### Body Temperature Data
The body temperatures in degrees Fahrenheit of a sample of adults in one small town are as follows:
- 97.2
- 98.3
- 99.5
- 99.1
- 99.2
- 99.4
- 98.2
- 97.3
- 98.7
- 99.7
- 97.4
- 98.6
- 96.6
#### Statistical Analysis Task
**Objective:**
Given the data, the goal is to find the 95% confidence interval for the mean body temperature of adults in the town. Assume that the body temperatures of adults are normally distributed.
**Steps:**
1. Calculate the sample mean (average) of the provided body temperatures.
2. Determine the sample standard deviation.
3. Apply the formula for the confidence interval for the mean for a normally distributed population.
**Formula:**
For a given sample mean \(\bar{x}\), standard deviation \(s\), and sample size \(n\), the 95% confidence interval for the mean is given by:
\[ \left( \bar{x} - t\frac{s}{\sqrt{n}}, \bar{x} + t\frac{s}{\sqrt{n}} \right) \]
where \(t\) is the t-value from the t-distribution for 95% confidence and \(n - 1\) degrees of freedom.
#### Interactive Exercise
**User Task:**
- Compute the 95% confidence interval for the given body temperature data.
- Enter the result as an open-interval (i.e., in parentheses) accurate to 3 decimal places.
**Input Box:**
95% C.I. = [_______________]
**Assumptions:**
- Assume the data is from a normally distributed population.
**Example Calculation (for illustration):**
- Sample mean: \( \bar{x} \)
- Sample standard deviation: \( s \)
- Sample size: \( n = 13 \)
- Degrees of freedom: \( df = n - 1 = 12 \)
- t-value for 95% confidence interval (with 12 degrees of freedom): \( t \approx 2.179 \)
By plugging these values into the formula, you can determine the confidence interval accordingly.
**Result:**
After computations, enter the result in the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1a73d04-4829-40e4-8e78-0a30aa84e71e%2Ff07771e8-ee76-480a-a510-71515d10a480%2Fxdz4jib_processed.png&w=3840&q=75)
Transcribed Image Text:### Body Temperature Data Analysis
#### Body Temperature Data
The body temperatures in degrees Fahrenheit of a sample of adults in one small town are as follows:
- 97.2
- 98.3
- 99.5
- 99.1
- 99.2
- 99.4
- 98.2
- 97.3
- 98.7
- 99.7
- 97.4
- 98.6
- 96.6
#### Statistical Analysis Task
**Objective:**
Given the data, the goal is to find the 95% confidence interval for the mean body temperature of adults in the town. Assume that the body temperatures of adults are normally distributed.
**Steps:**
1. Calculate the sample mean (average) of the provided body temperatures.
2. Determine the sample standard deviation.
3. Apply the formula for the confidence interval for the mean for a normally distributed population.
**Formula:**
For a given sample mean \(\bar{x}\), standard deviation \(s\), and sample size \(n\), the 95% confidence interval for the mean is given by:
\[ \left( \bar{x} - t\frac{s}{\sqrt{n}}, \bar{x} + t\frac{s}{\sqrt{n}} \right) \]
where \(t\) is the t-value from the t-distribution for 95% confidence and \(n - 1\) degrees of freedom.
#### Interactive Exercise
**User Task:**
- Compute the 95% confidence interval for the given body temperature data.
- Enter the result as an open-interval (i.e., in parentheses) accurate to 3 decimal places.
**Input Box:**
95% C.I. = [_______________]
**Assumptions:**
- Assume the data is from a normally distributed population.
**Example Calculation (for illustration):**
- Sample mean: \( \bar{x} \)
- Sample standard deviation: \( s \)
- Sample size: \( n = 13 \)
- Degrees of freedom: \( df = n - 1 = 12 \)
- t-value for 95% confidence interval (with 12 degrees of freedom): \( t \approx 2.179 \)
By plugging these values into the formula, you can determine the confidence interval accordingly.
**Result:**
After computations, enter the result in the
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