To find the 95% confidence interval for the population standard deviation based on the given data, we need to use the formula:

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

Where:

• $n$ is the sample size
• $s$ is the sample standard deviation
• $\chi^2_{\text{left}}$ and $\chi^2_{\text{right}}$ are the chi-square critical values for a 95% confidence level
• The degrees of freedom $df = n-1$

Step-by-step Solution:

1. Sample data:
   $63, 65, 65, 56, 55, 65, 55, 61, 60, 61, 69, 67$

2. Sample size $n$:
   There are 12 data points, so $n = 12$.

3. Calculate the sample standard deviation $s$: $s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$ First, compute the sample mean $\bar{x}$, then find the deviations and square them, sum them up, and divide by $n-1$, then take the square root.

4. Find chi-square critical values:

   • Use a chi-square distribution table or calculator for a 95% confidence level with $df = 11$ degrees of freedom.

5. Construct the interval:

   • Apply the formula to calculate the confidence interval for the population standard deviation.

Let me calculate these values for you.The sample standard deviation is approximately $s = 4.7$ mi/h. Using this value, the 95% confidence interval for the population standard deviation is:

$3.3 \, \text{mi/h} < \sigma < 8.0 \, \text{mi/h}$

Let me know if you would like more details or have any questions.

Here are 5 related questions you might find helpful:

1. How is the chi-square distribution used in constructing confidence intervals?
2. What assumptions must hold for constructing this confidence interval for the standard deviation?
3. How does increasing the sample size affect the confidence interval?
4. What would the confidence interval look like for a 99% confidence level?
5. Can you explain the difference between the confidence interval for the mean and for the standard deviation?

Tip: Always check the assumptions behind the statistical method you're using—here, we assume the data is normally distributed!