Use the given information to find the number of degrees of freedom, the critical values x2 and x, and the confidence interval estimate of o. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Platelet Counts of Women 99% confidence; n = 26, s=65.8. Click the icon to view the table of Chi-Square critical values. df = (Type a whole number.) x²=0 (Round to three decimal places as needed.) x² = 0 (Round to three decimal places as needed.) The confidence interval estimate of o is <6< D

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**Understanding Degrees of Freedom and Confidence Intervals with Chi-Square Distribution**

This educational resource explains how to determine the degrees of freedom, find critical values from a Chi-Square distribution, and estimate confidence intervals for a population standard deviation. 

### Problem Context:
You are given data related to platelet counts of women with a 99% confidence level:
- Sample size (n) = 26
- Sample standard deviation (s) = 65.8

We assume this is a simple random sample from a normally distributed population.

### Steps:

1. **Degrees of Freedom (df):**
   - The degrees of freedom for Chi-Square tests are given by `n - 1`.
   - **df =** [Input field: Type a whole number]

2. **Chi-Square Critical Values:**
   - Use a Chi-Square distribution table to find `χ²_L` and `χ²_R`, the critical values for the left and right tails at a 99% confidence level.
   - **χ²_L =** [Input field: Round to three decimal places]
   - **χ²_R =** [Input field: Round to three decimal places]

3. **Confidence Interval for Standard Deviation (σ):**
   - The formula for the confidence interval for σ is based on the Chi-Square distribution:
     \[
     \sqrt{\frac{(n - 1) \cdot s^2}{χ²_R}} < σ < \sqrt{\frac{(n - 1) \cdot s^2}{χ²_L}}
     \]

   - **The confidence interval estimate of σ is:**
     - **Lower bound:** [Input field]
     - **Upper bound:** [Input field]

### Additional Resource:
To help find the Chi-Square critical values, click the provided icon link to access a Chi-Square distribution table. This tool will be essential for completing the calculations accurately.

This exercise is critical for understanding how statistical distributions can help estimate parameters from sample data and make inferences about a population with a specified level of confidence.
Transcribed Image Text:**Understanding Degrees of Freedom and Confidence Intervals with Chi-Square Distribution** This educational resource explains how to determine the degrees of freedom, find critical values from a Chi-Square distribution, and estimate confidence intervals for a population standard deviation. ### Problem Context: You are given data related to platelet counts of women with a 99% confidence level: - Sample size (n) = 26 - Sample standard deviation (s) = 65.8 We assume this is a simple random sample from a normally distributed population. ### Steps: 1. **Degrees of Freedom (df):** - The degrees of freedom for Chi-Square tests are given by `n - 1`. - **df =** [Input field: Type a whole number] 2. **Chi-Square Critical Values:** - Use a Chi-Square distribution table to find `χ²_L` and `χ²_R`, the critical values for the left and right tails at a 99% confidence level. - **χ²_L =** [Input field: Round to three decimal places] - **χ²_R =** [Input field: Round to three decimal places] 3. **Confidence Interval for Standard Deviation (σ):** - The formula for the confidence interval for σ is based on the Chi-Square distribution: \[ \sqrt{\frac{(n - 1) \cdot s^2}{χ²_R}} < σ < \sqrt{\frac{(n - 1) \cdot s^2}{χ²_L}} \] - **The confidence interval estimate of σ is:** - **Lower bound:** [Input field] - **Upper bound:** [Input field] ### Additional Resource: To help find the Chi-Square critical values, click the provided icon link to access a Chi-Square distribution table. This tool will be essential for completing the calculations accurately. This exercise is critical for understanding how statistical distributions can help estimate parameters from sample data and make inferences about a population with a specified level of confidence.
Expert Solution
Step 1

given data 

99% ci for standard deviation 

s = 65.8

n = 26

 

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