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. White Blood Counts of Women 90​% ​confidence; nequals150​, sequals1.95 ​(1000 ​cells/mu​L). Question content area bottom Part 1 dfequals    149 ​(Type a whole​ number.) Part 2 chi Subscript Upper L Superscript 2equals    121.79 ​(Round to two decimal places as​ needed.) Part 3 chi Subscript Upper R Superscript 2equals    enter your response here ​(Round to two decimal places as​ needed.)

Solution

Let's walk through the problem step-by-step to solve for the degrees of freedom (df), critical values χL2\chi_L^2 and χR2\chi_R^2, and the confidence interval estimate of σ\sigma (standard deviation).


Part 1: Degrees of Freedom (df)

The degrees of freedom dfdf are calculated as: df=n1df = n - 1 where nn is the sample size.

Given: n=150n = 150

df=1501=149df = 150 - 1 = 149

Answer: df=149df = 149


Part 2: Critical Values χL2\chi_L^2 and χR2\chi_R^2

These critical values correspond to a 90% confidence level for a chi-square distribution with 149 degrees of freedom. We need the left-tail and right-tail critical values such that:

P(χL2χ2χR2)=0.90P\left(\chi_L^2 \leq \chi^2 \leq \chi_R^2\right) = 0.90

This means:

  • Left-tail area = α/2=0.05\alpha / 2 = 0.05
  • Right-tail area = 1α/2=0.951 - \alpha / 2 = 0.95

Using a chi-square table or calculator for df=149df = 149: χL2=χ0.05,1492121.79\chi_L^2 = \chi_{0.05, 149}^2 \approx 121.79

Now let’s find the right critical value: χR2=χ0.95,1492\chi_R^2 = \chi_{0.95, 149}^2 Using a chi-square calculator for df=149df = 149: χR2178.16\chi_R^2 \approx 178.16

Answer: χL2=121.79andχR2=178.16\chi_L^2 = 121.79 \quad \text{and} \quad \chi_R^2 = 178.16


Part 3: Confidence Interval Estimate for σ\sigma

The confidence interval for the population standard deviation σ\sigma is given by:

(n1)s2χR2σ(n1)s2χL2\sqrt{\frac{(n - 1) \cdot s^2}{\chi_R^2}} \leq \sigma \leq \sqrt{\frac{(n - 1) \cdot s^2}{\chi_L^2}}

Given:

  • n=150n = 150
  • s=1.95s = 1.95
  • χL2=121.79\chi_L^2 = 121.79
  • χR2=178.16\chi_R^2 = 178.16

Step 1: Compute (n1)s2(n - 1) \cdot s^2: 149(1.95)2=1493.8025=566.5725149 \cdot (1.95)^2 = 149 \cdot 3.8025 = 566.5725

Step 2: Compute the lower and upper bounds:

Lower bound=566.5725178.16=3.181.78\text{Lower bound} = \sqrt{\frac{566.5725}{178.16}} = \sqrt{3.18} \approx 1.78

Upper bound=566.5725121.79=4.652.16\text{Upper bound} = \sqrt{\frac{566.5725}{121.79}} = \sqrt{4.65} \approx 2.16

Answer: 1.78σ2.161.78 \leq \sigma \leq 2.16


Final Answers:

  1. Degrees of freedom (df) = 149
  2. χL2\chi_L^2 = 121.79
  3. χR2\chi_R^2 = 178.16
  4. Confidence interval for σ\sigma = 1.78σ2.161.78 \leq \sigma \leq 2.16

Do you need further clarifications or additional steps?


Related Questions:

  1. How do degrees of freedom affect confidence intervals?
  2. What happens if the sample size increases—how does it impact the width of the interval?
  3. What is the difference between confidence intervals for σ\sigma and μ\mu?
  4. How can we compute chi-square critical values without a table?
  5. How does the confidence level (e.g., 90% vs. 95%) affect the critical values?

Tip: Always check that your data follows a normal distribution when using the chi-square method for variance or standard deviation estimates.

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Math Problem Analysis

Mathematical Concepts

Chi-square distribution
Confidence intervals
Degrees of freedom

Formulas

df = n - 1
P(chi_L^2 ≤ chi^2 ≤ chi_R^2) = 0.90
Confidence interval for sigma: sqrt((n - 1) * s^2 / chi_R^2) ≤ sigma ≤ sqrt((n - 1) * s^2 / chi_L^2)

Theorems

Chi-square distribution for confidence intervals
Degrees of freedom in statistics

Suitable Grade Level

Undergraduate Statistics