To construct a 95% confidence interval for the population standard deviation $\sigma$, based on the given sample data, we'll need to follow these steps:

Step 1: Calculate the sample variance and standard deviation

First, calculate the sample mean ($\bar{x}$) and then the sample variance ($s^2$) and sample standard deviation ($s$).

Step 2: Use the Chi-Square Distribution

The confidence interval for the population standard deviation is given by the formula:

$\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right)$

Where:

• $n$ is the sample size,
• $s^2$ is the sample variance,
• $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are critical values from the Chi-Square distribution with $n-1$ degrees of freedom.

Let me calculate these values.The sample standard deviation is approximately $4.2 \, \text{mi/h}$. The 95% confidence interval for the population standard deviation is:

$2.99 \, \text{mi/h} \, < \sigma < 7.17 \, \text{mi/h}$

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How does increasing the sample size affect the width of the confidence interval?
2. What are the assumptions behind using the Chi-Square distribution for confidence intervals?
3. How is the Chi-Square distribution related to the normal distribution?
4. How would the confidence interval change if we used a different confidence level, such as 90% or 99%?
5. What are some applications of confidence intervals in real-world statistics?

Tip: Confidence intervals provide a range of plausible values for a parameter, offering more insight than a single point estimate.