Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2 and chi Subscript Upper R Superscript 2​, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.
Nicotine in menthol cigarettes 90​% ​confidence; nequals25​, sequals0.28 mg.
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Part 1
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To find the degrees of freedom (df), the critical chi-square values \(\chi^2_L\) and \(\chi^2_R\), and the confidence interval estimate for \(\sigma\) (the population standard deviation), follow these steps:

### Step 1: Find the degrees of freedom (df)
The degrees of freedom are calculated using the formula:
\[
df = n - 1
\]
where \(n\) is the sample size. Given \(n = 25\),
\[
df = 25 - 1 = 24
\]

### Step 2: Find the critical values \(\chi^2_L\) and \(\chi^2_R\)
For a 90% confidence interval, the significance level \(\alpha = 1 - 0.90 = 0.10\). Since this is a two-tailed test, each tail will have an area of \(\alpha/2 = 0.05\).

From the chi-square table, look up the values for:
- \(\chi^2_L\) at \(\alpha/2 = 0.05\) with \(df = 24\)
- \(\chi^2_R\) at \(1 - \alpha/2 = 0.95\) with \(df = 24\)

Using the table, you get:
\[
\chi^2_L = 13.848 \quad \text{and} \quad \chi^2_R = 36.415
\]

### Step 3: Confidence Interval for \(\sigma\)
The confidence interval for the population standard deviation \(\sigma\) is given by:
\[
\left( \sqrt{\frac{(n - 1) s^2}{\chi^2_R}}, \sqrt{\frac{(n - 1) s^2}{\chi^2_L}} \right)
\]
where \(n = 25\), \(s = 0.28\), \(\chi^2_L = 13.848\), and \(\chi^2_R = 36.415\).

First, calculate \((n - 1)s^2\):
\[
(n - 1)s^2 = 24 \times (0.28)^2 = 24 \times 0.0784 = 1.8816
\]

Now, find the confidence interval:
\[
\text{Lower bound} = \sqrt{\frac{1.8816}{36.415}} = \sqrt{0.0517} = 0.2274
\]
\[
\text{Upper bound} = \sqrt{\frac{1.8816}{13.848}} = \sqrt{0.1359} = 0.3687
\]

### Final Answer:
- Degrees of freedom (df): 24
- Critical values: \(\chi^2_L = 13.848\), \(\chi^2_R = 36.415\)
- Confidence interval for \(\sigma\): \((0.2274, 0.3687)\)

Would you like further details or have any questions?

### 5 Related Questions:
1. How do you interpret the confidence interval for \(\sigma\) in this context?
2. What happens to the confidence interval if we increase the sample size?
3. How would the calculation change if the confidence level was 95%?
4. What is the relationship between the sample standard deviation and the population standard deviation?
5. Why do we assume a normal distribution when calculating chi-square values?

### Tip:
For higher confidence levels, the critical values \(\chi^2_L\) and \(\chi^2_R\) will spread out, leading to a wider confidence interval.

Use the given information to find the number of degrees of freedom, the critical values chi²_L and chi²_R, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 90% confidence; n = 25, s = 0.28 mg.

Calculate the 90% confidence interval for the population standard deviation using a sample size of 25 and a sample standard deviation of 0.28 mg. Learn how to determine degrees of freedom and critical chi-square values.

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