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

9595​%

​confidence;

nequals=148148​,

sequals=1.951.95

​(1000

​cells/muμ​L).

Question content area bottom

Part 1

dfequals=147147

​(Type a whole​ number.)

Part 2

chi Subscript Upper L Superscript 2χ2Lequals=enter your response here

​(Round to two decimal places as​ needed.)

To solve this problem and find the degrees of freedom, critical values \( \chi^2_L \) and \( \chi^2_R \), and the confidence interval estimate for the standard deviation \( \sigma \), we'll go step-by-step:

### Part 1: Degrees of Freedom

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

Given:
\[
n = 148
\]
So:
\[
df = 148 - 1 = 147
\]

This matches your given value, \( df = 147 \).

### 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 95% confidence interval, we need to use a chi-square distribution table or calculator. These critical values correspond to the lower and upper percentiles of the chi-square distribution with 147 degrees of freedom at the given confidence level.

For a 95% confidence interval, we need the chi-square values that capture the middle 95% of the distribution. This leaves 2.5% in each tail, so we need:
- The left-tail critical value \( \chi^2_L \) corresponding to a cumulative probability of 0.025.
- The right-tail critical value \( \chi^2_R \) corresponding to a cumulative probability of 0.975.

Using a chi-square table or calculator for \( df = 147 \):
\[
\chi^2_L \approx 109.97 \quad \text{(for 0.025)}
\]
\[
\chi^2_R \approx 186.57 \quad \text{(for 0.975)}
\]

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

To estimate the confidence interval for \( \sigma \) (the population standard deviation), we use the following formula:
\[
\left( \frac{(n-1)s^2}{\chi^2_R}, \frac{(n-1)s^2}{\chi^2_L} \right)
\]
where \( s \) is the sample standard deviation, \( n-1 \) is the degrees of freedom, and \( \chi^2_R \), \( \chi^2_L \) are the critical chi-square values.

Given:
- \( s = 1.95 \)
- \( df = 147 \)
- \( \chi^2_L = 109.97 \)
- \( \chi^2_R = 186.57 \)

We calculate the lower and upper bounds of the confidence interval for \( \sigma \).

1. Lower bound:
\[
\sqrt{\frac{(147)(1.95^2)}{186.57}} \approx \sqrt{\frac{559.6225}{186.57}} \approx \sqrt{2.999} \approx 1.73
\]

2. Upper bound:
\[
\sqrt{\frac{(147)(1.95^2)}{109.97}} \approx \sqrt{\frac{559.6225}{109.97}} \approx \sqrt{5.09} \approx 2.26
\]

So the 95% confidence interval estimate for \( \sigma \) is approximately:
\[
(1.73, 2.26)
\]

### Final Answers:
- **Degrees of freedom**: \( 147 \)
- **Critical values**: 
  - \( \chi^2_L \approx 109.97 \)
  - \( \chi^2_R \approx 186.57 \)
- **Confidence interval for \( \sigma \)**: \( (1.73, 2.26) \)

Would you like further explanation or details on any step?

### Related Questions:
1. What is the formula for finding the confidence interval for variance instead of standard deviation?
2. How do you interpret a chi-square distribution in hypothesis testing?
3. Why do we use the chi-square distribution for variance or standard deviation estimates?
4. How does the sample size affect the width of the confidence interval for \( \sigma \)?
5. How would the confidence interval change if the confidence level were 99% instead of 95%?

### Tip:
The larger the sample size, the narrower the confidence interval for the population standard deviation, meaning more precise estimates.

Use the given information to find the number of degrees of freedom, the critical values χ²_L and χ²_R, and the confidence interval estimate of σ. 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 = 148, s = 1.95 (1000 cells/μL).

Calculate the 95% confidence interval for the standard deviation of white blood cell counts for a sample of 148 women, with a sample standard deviation of 1.95 (1000 cells/μL). The solution involves using degrees of freedom, chi-square critical values, and a step-by-step method for finding the confidence interval.

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