Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.004 inch. Determine whether these randomly selected golf balls conform to this requirement at the a=0.10 level of significance. Assume that the population is normally distributed. Click the icon to view the chi-square distribution table. (Type integers or decimals. Do not round.) Find the sample standard deviation. s= 0.00277 (Round to five decimal places as needed.). Use s to calculate the value of the test statistic. x² = (Round to two decimal places as needed.). 1.681 1.678 1.682 1.681 1.677 1.676 1.681 1.682 1.683 1.676 1.684 1.682
Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.004 inch. Determine whether these randomly selected golf balls conform to this requirement at the a=0.10 level of significance. Assume that the population is normally distributed. Click the icon to view the chi-square distribution table. (Type integers or decimals. Do not round.) Find the sample standard deviation. s= 0.00277 (Round to five decimal places as needed.). Use s to calculate the value of the test statistic. x² = (Round to two decimal places as needed.). 1.681 1.678 1.682 1.681 1.677 1.676 1.681 1.682 1.683 1.676 1.684 1.682
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Example Problem: Determining Standard Deviation and Test Statistic
**Problem Context:**
Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.004 inch. Determine whether these randomly selected golf balls conform to this requirement at the α = 0.10 level of significance. Assume that the population is normally distributed.
**Data:**
| Diameter Measurements (in inches) |
|------------------------------------|
| 1.681 | 1.678 | 1.682 |
| 1.681 | 1.677 | 1.676 |
| 1.681 | 1.682 | 1.683 |
| 1.676 | 1.684 | 1.682 |
**Instructions:**
1. Find the sample standard deviation \( s \) (Round to five decimal places as needed).
2. Use \( s \) to calculate the value of the chi-square test statistic \( \chi_0^2 \) (Round to two decimal places as needed).
**Solution Steps:**
1. **Calculate Sample Standard Deviation \(s\):**
Using the provided data:
\( s = 0.00277 \)
2. **Calculate Test Statistic \( \chi_0^2 \):**
Using the sample standard deviation and the appropriate chi-square distribution:
\( \chi_0^2 = \) (value needs to be calculated based on provided information).
**Interactive Options:**
- Help me solve this
- View an example
- Get more help
**Tools:**
- Click the icon to view the chi-square distribution table.
**Additional Interface Elements:**
- Clear all
- Check answer
[Refer to the detailed steps and chi-square distribution table for exact calculations.]
#### Diagram Explanation:
In the context of this problem, diagrams or graphs are not provided directly within the data; however, referencing the chi-square distribution table might be necessary to obtain the critical value for comparison. This would typically be a chi-square distribution curve comparing the test statistic to the critical value at the 0.10 significance level.
---
**Note:** For more examples and practice problems, refer to the Statistics section on our Educational website.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfee5922-5e9b-4d5c-bb5d-6eee324be123%2F0655a646-ce33-448f-a4be-de0853e49855%2F5lzadhf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem: Determining Standard Deviation and Test Statistic
**Problem Context:**
Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.004 inch. Determine whether these randomly selected golf balls conform to this requirement at the α = 0.10 level of significance. Assume that the population is normally distributed.
**Data:**
| Diameter Measurements (in inches) |
|------------------------------------|
| 1.681 | 1.678 | 1.682 |
| 1.681 | 1.677 | 1.676 |
| 1.681 | 1.682 | 1.683 |
| 1.676 | 1.684 | 1.682 |
**Instructions:**
1. Find the sample standard deviation \( s \) (Round to five decimal places as needed).
2. Use \( s \) to calculate the value of the chi-square test statistic \( \chi_0^2 \) (Round to two decimal places as needed).
**Solution Steps:**
1. **Calculate Sample Standard Deviation \(s\):**
Using the provided data:
\( s = 0.00277 \)
2. **Calculate Test Statistic \( \chi_0^2 \):**
Using the sample standard deviation and the appropriate chi-square distribution:
\( \chi_0^2 = \) (value needs to be calculated based on provided information).
**Interactive Options:**
- Help me solve this
- View an example
- Get more help
**Tools:**
- Click the icon to view the chi-square distribution table.
**Additional Interface Elements:**
- Clear all
- Check answer
[Refer to the detailed steps and chi-square distribution table for exact calculations.]
#### Diagram Explanation:
In the context of this problem, diagrams or graphs are not provided directly within the data; however, referencing the chi-square distribution table might be necessary to obtain the critical value for comparison. This would typically be a chi-square distribution curve comparing the test statistic to the critical value at the 0.10 significance level.
---
**Note:** For more examples and practice problems, refer to the Statistics section on our Educational website.
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