The width decreases (c) Construct a 98% confidence interval for o² if the sample size, n, is 20. The lower bound is (Round to two decimal places as needed.)

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Part C please
**Statistics Practice Problem: Confidence Intervals**

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**Problem Statement**

A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, \( s^2 \), is determined to be 12.2. Complete parts (a) through (c).

**Question (a):**

How does increasing the sample size affect the width of the interval?

- \( \circ \) The width increases
- \( \circ \) The width does not change
- \( \bullet \) The width decreases

**Question (c):**

Construct a 98% confidence interval for \( \sigma^2 \) if the sample size, \( n \), is 20.

**The lower bound is ______ .** (Round to two decimal places as needed.)

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**Graphical Explanation:**

While there are no explicit graphs or diagrams in the image provided, the problem involves understanding the relationship between sample size and the width of a confidence interval. Here is an explanation:

When constructing a confidence interval for the population variance \( \sigma^2 \), the width of the interval is inversely related to the sample size. Mathematically, the confidence interval for the variance is:

\[ \left( \frac{(n-1)s^2}{\chi_{1-\alpha/2, n-1}^2}, \frac{(n-1)s^2}{\chi_{\alpha/2, n-1}^2} \right) \]

Where:
- \( n \) is the sample size.
- \( s^2 \) is the sample variance.
- \( \chi^2 \) are the critical values from the Chi-Square distribution.
- \( \alpha \) is the level of significance.

As the sample size \( n \) increases, the critical values from the Chi-Square distribution become smaller, which narrows the interval. Thus, increasing the sample size decreases the width of the confidence interval.

In the given problem:
- Sample size (\( n \)): 20
- Sample variance (\( s^2 \)): 12.2

The problem asks for the lower bound of the 98% confidence interval for \( \sigma^2 \). The exact computation would involve using Chi-Square distribution tables or software to find the critical values for the given confidence level and sample size. The interval calculation will then lead to the determination of the lower bound.
Transcribed Image Text:**Statistics Practice Problem: Confidence Intervals** --- **Problem Statement** A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, \( s^2 \), is determined to be 12.2. Complete parts (a) through (c). **Question (a):** How does increasing the sample size affect the width of the interval? - \( \circ \) The width increases - \( \circ \) The width does not change - \( \bullet \) The width decreases **Question (c):** Construct a 98% confidence interval for \( \sigma^2 \) if the sample size, \( n \), is 20. **The lower bound is ______ .** (Round to two decimal places as needed.) --- **Graphical Explanation:** While there are no explicit graphs or diagrams in the image provided, the problem involves understanding the relationship between sample size and the width of a confidence interval. Here is an explanation: When constructing a confidence interval for the population variance \( \sigma^2 \), the width of the interval is inversely related to the sample size. Mathematically, the confidence interval for the variance is: \[ \left( \frac{(n-1)s^2}{\chi_{1-\alpha/2, n-1}^2}, \frac{(n-1)s^2}{\chi_{\alpha/2, n-1}^2} \right) \] Where: - \( n \) is the sample size. - \( s^2 \) is the sample variance. - \( \chi^2 \) are the critical values from the Chi-Square distribution. - \( \alpha \) is the level of significance. As the sample size \( n \) increases, the critical values from the Chi-Square distribution become smaller, which narrows the interval. Thus, increasing the sample size decreases the width of the confidence interval. In the given problem: - Sample size (\( n \)): 20 - Sample variance (\( s^2 \)): 12.2 The problem asks for the lower bound of the 98% confidence interval for \( \sigma^2 \). The exact computation would involve using Chi-Square distribution tables or software to find the critical values for the given confidence level and sample size. The interval calculation will then lead to the determination of the lower bound.
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