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.White Blood Counts of Women 90​% ​confidence; nequals=148​, sequals=1.98 ​(1000 ​cells/muμ​L).Question content area bottomPart 1dfequals=enter your response here ​(Type a whole​ number.) 

Let's go through the steps to solve the problem.

### Part 1: Degrees of Freedom (df)
The degrees of freedom \( df \) for the chi-square distribution is given by:
\[
df = n - 1
\]
where \( n \) is the sample size.

Given that \( n = 148 \), the degrees of freedom are:
\[
df = 148 - 1 = 147
\]
So, \( df = 147 \).

### Part 2: Critical Values \( \chi^2_L \) and \( \chi^2_R \)
To find the critical values \( \chi^2_L \) and \( \chi^2_R \) for a 90% confidence interval, we need to find the chi-square values that correspond to the lower and upper tails of the distribution.

- For a 90% confidence interval, the area in the tails is \( \frac{1 - 0.90}{2} = 0.05 \) in each tail.

Thus:
- \( \chi^2_L \) corresponds to the 5th percentile of the chi-square distribution with 147 degrees of freedom.
- \( \chi^2_R \) corresponds to the 95th percentile of the chi-square distribution with 147 degrees of freedom.

Using a chi-square table or calculator:

- \( \chi^2_L \approx 118.88 \) (rounded to two decimal places)
- \( \chi^2_R \approx 177.80 \) (rounded to two decimal places)

### Part 3: Confidence Interval Estimate of \( \sigma \)
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:
- \( n = 148 \)
- \( s = 1.98 \) (sample standard deviation)
- \( \chi^2_L = 118.88 \)
- \( \chi^2_R = 177.80 \)

Let's calculate the confidence interval:

\[
\text{Lower bound} = \sqrt{\frac{147 \times (1.98)^2}{177.80}}
\]
\[
\text{Upper bound} = \sqrt{\frac{147 \times (1.98)^2}{118.88}}
\]

Let's compute these values:

\[
\text{Lower bound} = \sqrt{\frac{147 \times 3.9204}{177.80}} = \sqrt{\frac{576.2988}{177.80}} \approx \sqrt{3.241} \approx 1.80
\]

\[
\text{Upper bound} = \sqrt{\frac{147 \times 3.9204}{118.88}} = \sqrt{\frac{576.2988}{118.88}} \approx \sqrt{4.848} \approx 2.20
\]

### Final Answer:
- **Degrees of Freedom**: \( df = 147 \)
- **Critical Value \( \chi^2_L \)**: \( 118.88 \)
- **Critical Value \( \chi^2_R \)**: \( 177.80 \)
- **Confidence Interval for \( \sigma \)**: \( 1.80 \, (1000 \, \text{cells}/\mu L) < \sigma < 2.20 \, (1000 \, \text{cells}/\mu L) \)

---

Do you have any questions, or would you like more details on any part of the solution?

Here are some related questions to expand your understanding:

1. What would the critical values \( \chi^2_L \) and \( \chi^2_R \) be for a 95% confidence interval?
2. How does changing the sample size \( n \) affect the confidence interval estimate of \( \sigma \)?
3. What assumptions are required for the chi-square distribution to be applicable in this context?
4. How would you interpret the 90% confidence interval in a practical context?
5. What happens to the confidence interval if the standard deviation \( s \) is larger?
6. How would the confidence interval for \( \sigma \) change if the confidence level were increased to 99%?
7. How can outliers impact the confidence interval for \( \sigma \)?
8. How do you calculate a confidence interval for the variance \( \sigma^2 \) instead of the standard deviation \( \sigma \)?

**Tip:** Always check whether the sample size is sufficiently large when applying the chi-square distribution, as the distribution becomes more accurate with larger degrees of freedom.

Learn how to calculate degrees of freedom, critical values chi-square L and R, and the confidence interval estimate of sigma based on a sample with normal distribution. Practice solving problems involving statistical inference and confidence intervals.

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