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 95​% ​confidence; nequals147​, sequals1.98 ​(1000 ​cells/mu​L).
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Part 1
dfequals
  
146 ​(Type a whole​ number.) 
Part 2
chi Subscript Upper L Superscript 2equals
  
enter your response here
​(Round to two decimal places as​ needed.)

### Part 1: Degrees of Freedom

The degrees of freedom (df) for the chi-squared distribution is calculated as:

\[
df = n - 1
\]

Where:
- \(n = 147\) (sample size)

Thus:

\[
df = 147 - 1 = 146
\]

So, the degrees of freedom \(df = 146\).

### Part 2: Critical Chi-Square Values \( \chi^2_L \) and \( \chi^2_R \)

Given that we have a 95% confidence level, the significance level \( \alpha = 1 - 0.95 = 0.05 \). For the two-tailed chi-square distribution, we split this significance level into two equal parts:
- \( \alpha/2 = 0.025 \) for the left tail
- \( 1 - \alpha/2 = 0.975 \) for the right tail

To find the critical values, we look up the chi-square distribution with \( df = 146 \).

Using chi-square tables or technology:

- \( \chi^2_L \) is the critical value for \( \alpha/2 = 0.025 \) with \( df = 146 \).
- \( \chi^2_R \) is the critical value for \( 1 - \alpha/2 = 0.975 \) with \( df = 146 \).

From chi-square distribution tables:
- \( \chi^2_L \approx 114.88 \) (left tail at 0.025 significance)
- \( \chi^2_R \approx 179.58 \) (right tail at 0.975 significance)

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

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

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

Where:
- \( n = 147 \)
- \( s = 1.98 \) (sample standard deviation)
- \( \chi^2_L = 114.88 \)
- \( \chi^2_R = 179.58 \)

First, compute \( (n - 1) s^2 \):

\[
(n - 1) s^2 = 146 \times (1.98)^2 = 146 \times 3.9204 = 572.38
\]

Now calculate the confidence interval bounds:

\[
\text{Lower bound} = \sqrt{\frac{572.38}{179.58}} = \sqrt{3.19} \approx 1.79
\]

\[
\text{Upper bound} = \sqrt{\frac{572.38}{114.88}} = \sqrt{4.98} \approx 2.23
\]

Thus, the 95% confidence interval for \( \sigma \) is approximately \( (1.79, 2.23) \).

### Final Answers:
1. Degrees of freedom: \( df = 146 \)
2. Critical value \( \chi^2_L \approx 114.88 \)
3. Critical value \( \chi^2_R \approx 179.58 \)
4. Confidence interval for \( \sigma \): \( (1.79, 2.23) \)

Would you like more details on any of these steps or calculations?

### Related Questions:
1. How do you interpret degrees of freedom in hypothesis testing?
2. What is the significance of the critical values in a chi-square distribution?
3. How is the confidence interval for variance derived from the chi-square distribution?
4. Can this method be used for populations not normally distributed?
5. How does increasing the sample size affect the width of the confidence interval?

### Tip:
When constructing confidence intervals using the chi-square distribution, always ensure the assumption of normality is reasonably met for accurate results.

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 95​% ​confidence; n = 147​, s = 1.98 ​(1000 ​cells/mu​L).

Learn how to calculate degrees of freedom, critical chi-square values, and a 95% confidence interval for the standard deviation of white blood counts in women. This problem involves chi-square distribution concepts and is suitable for college-level statistics.

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