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Part 1

Use the given information to find the number of degrees of​ freedom, the critical values 

chi Subscript Upper L Superscript 2χ2L 

and 

chi Subscript Upper R Superscript 2χ2R​, 

and the confidence interval estimate of 

sigmaσ. 

It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Nicotine in menthol cigarettes 

9090​% 

​confidence; 

nequals=2525​, 

sequals=0.280.28 

mg.

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Part 1

dfequals=2424 

​(Type a whole​ number.) 

Part 2

chi Subscript Upper L Superscript 2χ2Lequals=13.84813.848 

​(Round to three decimal places as​ needed.)

Part 3

chi Subscript Upper R Superscript 2χ2Requals=36.41536.415 

​(Round to three decimal places as​ needed.)

Part 4

The confidence interval estimate of 

sigmaσ 

is 

enter your response here 

mgless than<sigmaσless than<enter your response here 

mg.

​(Round to two decimal places as​ needed.)

Let's work through this step by step to find the confidence interval estimate of \(\sigma\) (the population standard deviation) for the given problem.

### Given information:
- Confidence level: \(90\%\)
- Sample size: \(n = 25\)
- Sample standard deviation: \(s = 0.28 \, \text{mg}\)
- Degrees of freedom: \(df = n - 1 = 24\)

The formula for the confidence interval of the population standard deviation \(\sigma\) is:

\[
\left( \sqrt{\frac{(n-1) s^2}{\chi^2_R}}, \sqrt{\frac{(n-1) s^2}{\chi^2_L}} \right)
\]

Where:
- \(n-1\) is the degrees of freedom (\(df = 24\)),
- \(\chi^2_L\) and \(\chi^2_R\) are the critical chi-square values for the left and right tails of the distribution at the given confidence level (90%).

From the question:
- \(\chi^2_L = 13.848\)
- \(\chi^2_R = 36.415\)

Now we can calculate the confidence interval for \(\sigma\).

### Step 1: Calculate the lower bound
\[
\sigma_{\text{lower}} = \sqrt{\frac{(25 - 1) \times (0.28)^2}{36.415}} = \sqrt{\frac{24 \times 0.0784}{36.415}}
\]

### Step 2: Calculate the upper bound
\[
\sigma_{\text{upper}} = \sqrt{\frac{(25 - 1) \times (0.28)^2}{13.848}} = \sqrt{\frac{24 \times 0.0784}{13.848}}
\]

Let's calculate these values.The confidence interval estimate for the population standard deviation \(\sigma\) is:

\[
0.23 \, \text{mg} < \sigma < 0.37 \, \text{mg}
\]

(Rounded to two decimal places as required).

Would you like further details on any part of the process, or have any additional questions?

Here are some related questions you might find helpful:
1. What is the interpretation of a confidence interval for a population standard deviation?
2. How does changing the confidence level affect the confidence interval?
3. What is the relationship between sample size and degrees of freedom in a chi-square distribution?
4. How are chi-square critical values determined from a chi-square table?
5. What assumptions must hold true for the confidence interval of \(\sigma\) to be valid?

**Tip:** Always ensure that the data comes from a normally distributed population when using chi-square intervals for the population variance or standard deviation.

Use the given information to find the number of degrees of freedom, the critical values chi^2_L and chi^2_R, and the confidence interval estimate of sigma. A simple random sample has been selected from a population with a normal distribution. The problem involves nicotine in menthol cigarettes, 90% confidence level, n = 25, s = 0.28 mg, df = 24, chi^2_L = 13.848, and chi^2_R = 36.415. Find the confidence interval for sigma.

Learn how to calculate the confidence interval for the standard deviation of nicotine content in menthol cigarettes, given a 90% confidence level, n = 25, and s = 0.28 mg. This problem involves the chi-square distribution and confidence interval estimation. Suitable for students in college-level statistics or advanced high school courses.

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