Question 1168610
Let's calculate the 95% confidence interval for the population standard deviation.

**1. Calculate the Sample Standard Deviation (s)**

First, we need to find the sample mean and then the sample standard deviation.

* Data: 62, 63, 63, 56, 63, 52, 61, 60, 61, 69, 60, 66
* n = 12

* Sample mean (x̄):
    * Sum = 62 + 63 + 63 + 56 + 63 + 52 + 61 + 60 + 61 + 69 + 60 + 66 = 736
    * x̄ = 736 / 12 ≈ 61.33

* Sample standard deviation (s):
    * s = √[Σ(xᵢ - x̄)² / (n - 1)]
    * Calculating the differences:
        * (62 - 61.33)² ≈ 0.4489
        * (63 - 61.33)² ≈ 2.7889
        * (63 - 61.33)² ≈ 2.7889
        * (56 - 61.33)² ≈ 28.4089
        * (63 - 61.33)² ≈ 2.7889
        * (52 - 61.33)² ≈ 86.9489
        * (61 - 61.33)² ≈ 0.1089
        * (60 - 61.33)² ≈ 1.7689
        * (61 - 61.33)² ≈ 0.1089
        * (69 - 61.33)² ≈ 58.8289
        * (60 - 61.33)² ≈ 1.7689
        * (66 - 61.33)² ≈ 21.8089
    * Sum of squares ≈ 208.5732
    * s = √(208.5732 / 11) ≈ √18.9612 ≈ 4.354

**2. Determine the Chi-Square Values**

* Degrees of freedom (df) = n - 1 = 12 - 1 = 11
* Confidence level = 95%, so α = 0.05
* α/2 = 0.025
* 1 - α/2 = 0.975

* Using a chi-square distribution table or calculator:
    * χ²(0.025, 11) ≈ 21.920
    * χ²(0.975, 11) ≈ 3.816

**3. Calculate the Confidence Interval**

* Confidence interval for σ:
    * √[(n - 1)s² / χ²(α/2)] < σ < √[(n - 1)s² / χ²(1 - α/2)]

* Lower bound:
    * √[(11 * 4.354²) / 21.920] ≈ √[(11 * 18.9573) / 21.920] ≈ √(208.5303 / 21.920) ≈ √9.5132 ≈ 3.084

* Upper bound:
    * √[(11 * 4.354²) / 3.816] ≈ √[(11 * 18.9573) / 3.816] ≈ √(208.5303 / 3.816) ≈ √54.6463 ≈ 7.392

**4. Round to One Decimal Place**

* Lower bound: 3.1
* Upper bound: 7.4

**Therefore, the 95% confidence interval estimate is 3.1 mi/h < σ < 7.4 mi/h.**