You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 118.1 93.8 87.9 31.5 100.6 97.7 94 Find the 99% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 99% C.I. = Answer should be obtained without any preliminary rounding.
You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 118.1 93.8 87.9 31.5 100.6 97.7 94 Find the 99% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 99% C.I. = Answer should be obtained without any preliminary rounding.
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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![### Estimating Mean Temperature with a 99% Confidence Interval
**Sample Data in Degrees Fahrenheit:**
- 118.1
- 93.8
- 87.9
- 31.5
- 100.6
- 97.7
- 94
**Task:**
Estimate the mean temperature using the provided sample data and find the 99% confidence interval.
**Instructions:**
1. Enter your answer as an **open-interval** (i.e., parentheses) accurate to two decimal places (since the sample data are reported accurate to one decimal place).
**Example Format:**
```
99% C.I. = (lower bound, upper bound)
```
**Notes:**
- Answer should be obtained without any preliminary rounding.
**Steps to Calculate:**
1. Compute the sample mean (average) \( \overline{X} \) of the data.
2. Calculate the sample standard deviation \( S \).
3. Determine the standard error \( SE \) of the mean.
4. Find the critical value (z-score or t-score) for the 99% confidence interval.
5. Calculate the margin of error.
6. Form the 99% confidence interval using the sample mean and the margin of error.
**Formulae:**
- Sample mean:
\[ \overline{X} = \frac{\sum X_i}{n} \]
- Sample standard deviation:
\[ S = \sqrt{\frac{\sum (X_i - \overline{X})^2}{n-1}} \]
- Standard error:
\[ SE = \frac{S}{\sqrt{n}} \]
- Confidence interval:
\[ 99\% \, CI = \overline{X} \pm (critical \, value \times SE) \]
**Data Breakdown:**
The list of temperatures provided will be used to compute the required statistics, acknowledging variations such as the one extremely low value (31.5°F) and the highest value (118.1°F). These individual differences underscore the importance of the standard deviation in accounting for the spread of data around the mean.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd120f3b3-582e-4ce3-bf67-95261f57ecc8%2F3358518f-847a-46db-88bc-14119c957c02%2F2h0gf5n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Estimating Mean Temperature with a 99% Confidence Interval
**Sample Data in Degrees Fahrenheit:**
- 118.1
- 93.8
- 87.9
- 31.5
- 100.6
- 97.7
- 94
**Task:**
Estimate the mean temperature using the provided sample data and find the 99% confidence interval.
**Instructions:**
1. Enter your answer as an **open-interval** (i.e., parentheses) accurate to two decimal places (since the sample data are reported accurate to one decimal place).
**Example Format:**
```
99% C.I. = (lower bound, upper bound)
```
**Notes:**
- Answer should be obtained without any preliminary rounding.
**Steps to Calculate:**
1. Compute the sample mean (average) \( \overline{X} \) of the data.
2. Calculate the sample standard deviation \( S \).
3. Determine the standard error \( SE \) of the mean.
4. Find the critical value (z-score or t-score) for the 99% confidence interval.
5. Calculate the margin of error.
6. Form the 99% confidence interval using the sample mean and the margin of error.
**Formulae:**
- Sample mean:
\[ \overline{X} = \frac{\sum X_i}{n} \]
- Sample standard deviation:
\[ S = \sqrt{\frac{\sum (X_i - \overline{X})^2}{n-1}} \]
- Standard error:
\[ SE = \frac{S}{\sqrt{n}} \]
- Confidence interval:
\[ 99\% \, CI = \overline{X} \pm (critical \, value \times SE) \]
**Data Breakdown:**
The list of temperatures provided will be used to compute the required statistics, acknowledging variations such as the one extremely low value (31.5°F) and the highest value (118.1°F). These individual differences underscore the importance of the standard deviation in accounting for the spread of data around the mean.
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