Question: Given specific sample data, such as the data given in Exercise 1, which confidence interval is wider: the $95\%$ confidence interval or the $80\%$ confidence interval? Why is it wider?
1. **Problem Statement:**
Given sample data, we want to compare the widths of the $95\%$ confidence interval and the $80\%$ confidence interval.
2. **Formula for Confidence Interval:**
A confidence interval for a population mean $\mu$ is given by:
$$\text{CI} = \bar{x} \pm z^* \times \frac{s}{\sqrt{n}}$$
where:
- $\bar{x}$ is the sample mean,
- $z^*$ is the critical value from the standard normal distribution corresponding to the confidence level,
- $s$ is the sample standard deviation,
- $n$ is the sample size.
3. **Important Rule:**
The width of the confidence interval depends on the critical value $z^*$, which increases as the confidence level increases.
4. **Critical Values:**
- For $95\%$ confidence, $z^* \approx 1.96$
- For $80\%$ confidence, $z^* \approx 1.28$
5. **Comparison of Widths:**
The width of the confidence interval is:
$$\text{Width} = 2 \times z^* \times \frac{s}{\sqrt{n}}$$
Since $1.96 > 1.28$, the $95\%$ confidence interval is wider than the $80\%$ confidence interval.
6. **Explanation:**
A higher confidence level means we want to be more certain that the interval contains the true population parameter. To achieve this higher certainty, the interval must be wider to cover more possible values.
**Final Answer:**
The $95\%$ confidence interval is wider than the $80\%$ confidence interval because it uses a larger critical value $z^*$, resulting in a wider range to ensure higher confidence.