Math Problem Statement

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. LOADING... Click the icon to view the table of​ Chi-Square critical values. Question content area bottom Part 1 dfequals

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Solution

To find the degrees of freedom (df), the critical chi-square values χL2\chi^2_L and χR2\chi^2_R, and the confidence interval estimate for σ\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=n1df = n - 1 where nn is the sample size. Given n=25n = 25, df=251=24df = 25 - 1 = 24

Step 2: Find the critical values χL2\chi^2_L and χR2\chi^2_R

For a 90% confidence interval, the significance level α=10.90=0.10\alpha = 1 - 0.90 = 0.10. Since this is a two-tailed test, each tail will have an area of α/2=0.05\alpha/2 = 0.05.

From the chi-square table, look up the values for:

  • χL2\chi^2_L at α/2=0.05\alpha/2 = 0.05 with df=24df = 24
  • χR2\chi^2_R at 1α/2=0.951 - \alpha/2 = 0.95 with df=24df = 24

Using the table, you get: χL2=13.848andχR2=36.415\chi^2_L = 13.848 \quad \text{and} \quad \chi^2_R = 36.415

Step 3: Confidence Interval for σ\sigma

The confidence interval for the population standard deviation σ\sigma is given by: ((n1)s2χR2,(n1)s2χL2)\left( \sqrt{\frac{(n - 1) s^2}{\chi^2_R}}, \sqrt{\frac{(n - 1) s^2}{\chi^2_L}} \right) where n=25n = 25, s=0.28s = 0.28, χL2=13.848\chi^2_L = 13.848, and χR2=36.415\chi^2_R = 36.415.

First, calculate (n1)s2(n - 1)s^2: (n1)s2=24×(0.28)2=24×0.0784=1.8816(n - 1)s^2 = 24 \times (0.28)^2 = 24 \times 0.0784 = 1.8816

Now, find the confidence interval: Lower bound=1.881636.415=0.0517=0.2274\text{Lower bound} = \sqrt{\frac{1.8816}{36.415}} = \sqrt{0.0517} = 0.2274 Upper bound=1.881613.848=0.1359=0.3687\text{Upper bound} = \sqrt{\frac{1.8816}{13.848}} = \sqrt{0.1359} = 0.3687

Final Answer:

  • Degrees of freedom (df): 24
  • Critical values: χL2=13.848\chi^2_L = 13.848, χR2=36.415\chi^2_R = 36.415
  • Confidence interval for σ\sigma: (0.2274,0.3687)(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 σ\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 χL2\chi^2_L and χR2\chi^2_R will spread out, leading to a wider confidence interval.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Interval
Chi-Square Distribution

Formulas

df = n - 1
CI for σ: (√((n-1) * s² / χ²_R), √((n-1) * s² / χ²_L))

Theorems

Chi-Square Distribution
Confidence Interval Theorem

Suitable Grade Level

College Level Statistics