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To solve this problem, let's go through the details provided and calculate the Chi-square critical values.

### 1. Determining Degrees of Freedom
Given:
- Sample size \( n = 25 \)
- Degrees of freedom \( df = n - 1 = 25 - 1 = 24 \)

So, \( df = 24 \).

### 2. Finding Chi-square Critical Values (\( \chi^2_L \) and \( \chi^2_R \))
For a 95% confidence interval, we split the remaining 5% into two tails (2.5% in each tail), which means:
- Left tail probability = 0.025
- Right tail probability = 0.975

Using a Chi-square table or calculator for 24 degrees of freedom:
- \( \chi^2_L \) (for 0.975) ≈ 13.848
- \( \chi^2_R \) (for 0.025) ≈ 36.415

### Summary of Answers
- \( df = 24 \)
- \( \chi^2_L \approx 13.848 \) (rounded to three decimal places)
- \( \chi^2_R \approx 36.415 \) (rounded to three decimal places)

Would you like further details on how to use these values in calculating a confidence interval for \( \sigma \)? 

Here are some related questions to deepen understanding:
1. How are degrees of freedom calculated for different sample sizes?
2. What is the Chi-square distribution, and why is it used for variance estimates?
3. How do confidence intervals change with different confidence levels (e.g., 90%, 99%)?
4. What is the impact of sample size on Chi-square critical values?
5. How would you interpret the calculated interval in practical terms for this data?

**Tip:** Chi-square tables are essential in statistics for tests involving variances; familiarizing yourself with these values can make calculations quicker!

Use the given information to find the number of degrees of freedom, the critical values χ²_L and χ²_R, and the confidence interval estimate of σ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Given 95% confidence, n = 25, and s = 0.27 mg.

Calculate the degrees of freedom, chi-square critical values, and confidence interval estimate for the standard deviation of nicotine levels in menthol cigarettes at 95% confidence. Given n = 25 and s = 0.27 mg, find the values for a sample selected from a normal distribution.

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