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.)

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

---

### Part 1: Degrees of Freedom (df)
The degrees of freedom \(df\) are calculated as:
\[
df = n - 1
\]
where \(n\) is the sample size.

Given:
\[
n = 150
\]

\[
df = 150 - 1 = 149
\]

Answer: \(df = 149\)

---

### Part 2: Critical Values \(\chi_L^2\) and \(\chi_R^2\)

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:

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

This means:
- **Left-tail area** = \( \alpha / 2 = 0.05 \)
- **Right-tail area** = \( 1 - \alpha / 2 = 0.95 \)

Using a **chi-square table** or calculator for \(df = 149\):
\[
\chi_L^2 = \chi_{0.05, 149}^2 \approx 121.79
\]

Now let’s find the **right critical value**:
\[
\chi_R^2 = \chi_{0.95, 149}^2
\]
Using a chi-square calculator for \(df = 149\):
\[
\chi_R^2 \approx 178.16
\]

Answer:
\[
\chi_L^2 = 121.79 \quad \text{and} \quad \chi_R^2 = 178.16
\]

---

### Part 3: Confidence Interval Estimate for \(\sigma\)

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

\[
\sqrt{\frac{(n - 1) \cdot s^2}{\chi_R^2}} \leq \sigma \leq \sqrt{\frac{(n - 1) \cdot s^2}{\chi_L^2}}
\]

Given:
- \(n = 150\)
- \(s = 1.95\)
- \(\chi_L^2 = 121.79\)
- \(\chi_R^2 = 178.16\)

Step 1: Compute \((n - 1) \cdot s^2\):
\[
149 \cdot (1.95)^2 = 149 \cdot 3.8025 = 566.5725
\]

Step 2: Compute the lower and upper bounds:

\[
\text{Lower bound} = \sqrt{\frac{566.5725}{178.16}} = \sqrt{3.18} \approx 1.78
\]

\[
\text{Upper bound} = \sqrt{\frac{566.5725}{121.79}} = \sqrt{4.65} \approx 2.16
\]

Answer:
\[
1.78 \leq \sigma \leq 2.16
\]

---

### Final Answers:
1. **Degrees of freedom (df)** = 149  
2. **\(\chi_L^2\)** = 121.79  
3. **\(\chi_R^2\)** = 178.16  
4. **Confidence interval for \(\sigma\)** = \(1.78 \leq \sigma \leq 2.16\)

---

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 \(\sigma\) and \(\mu\)?
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.

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; n equals 150​, s equals 1.95 ​(1000 ​cells/mu​L).

Solve for degrees of freedom, chi-square critical values, and confidence interval for sigma with a 90% confidence level, given a sample size of 150 and sample standard deviation of 1.95 (1000 cells/mu​L). Learn the chi-square method for variance and standard deviation estimates in a normal distribution.

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