Let's solve the problem step by step.

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 $n = 146$, the degrees of freedom are: $df = 146 - 1 = 145$

Part 2: Critical Value $\chi^2_L$

• The critical value $\chi^2_L$ corresponds to the lower tail of the chi-square distribution. Given an 80% confidence level, the area in the tails of the distribution is $\frac{1 - 0.80}{2} = 0.10$.

For $df = 145$, and using a chi-square table or calculator, $\chi^2_L$ corresponds to the 10th percentile of the chi-square distribution.

Given: $\chi^2_L = 123.65$

Part 3: Critical Value $\chi^2_R$

• The critical value $\chi^2_R$ corresponds to the upper tail of the chi-square distribution, with $\chi^2_R$ being the 90th percentile (since 80% confidence level leaves 10% in the upper tail).

For $df = 145$, using a chi-square table or calculator: $\chi^2_R = 167.21$

Part 4: 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:

• $n = 146$
• $s = 1.99$ (sample standard deviation)
• $\chi^2_L = 123.65$
• $\chi^2_R = 167.21$

Let's calculate the confidence interval:

$\text{Lower bound} = \sqrt{\frac{145 \times (1.99)^2}{167.21}}$ $\text{Upper bound} = \sqrt{\frac{145 \times (1.99)^2}{123.65}}$

Let's compute these values.

$\text{Lower bound} = \sqrt{\frac{145 \times 3.9601}{167.21}} = \sqrt{\frac{574.2145}{167.21}} \approx \sqrt{3.434} \approx 1.85$

$\text{Upper bound} = \sqrt{\frac{145 \times 3.9601}{123.65}} = \sqrt{\frac{574.2145}{123.65}} \approx \sqrt{4.644} \approx 2.15$

Final Answer:

The confidence interval estimate of $\sigma$ is: $1.85 \, (1000 \, \text{cells}/\mu L) < \sigma < 2.15 \, (1000 \, \text{cells}/\mu L)$

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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. How would the confidence interval change with a 95% confidence level instead of 80%?
2. What effect does increasing the sample size $n$ have on the width of the confidence interval?
3. How do you interpret the confidence interval in the context of this problem?
4. What are the assumptions underlying the use of the chi-square distribution for this interval estimate?
5. How does the shape of the chi-square distribution change as the degrees of freedom increase?
6. Can the chi-square distribution be used for non-normal populations? Why or why not?
7. How would you calculate the confidence interval for the variance instead of the standard deviation?
8. How would outliers in the data affect the confidence interval for $\sigma$?

Tip: The width of a confidence interval is directly related to the confidence level and sample size. Higher confidence levels or smaller sample sizes result in wider intervals.