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 $\sigma$: $1.80 \, (1000 \, \text{cells}/\mu L) < \sigma < 2.20 \, (1000 \, \text{cells}/\mu L)$

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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. 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.