SOLUTION: (1 point) The weights of unicorns are normally distributed. Suppose that a simple random sample of 12 unicorns have a standard deviation of 20.8. Estimate the standard deviation of

Question 1188789: (1 point) The weights of unicorns are normally distributed. Suppose that a simple random sample of 12 unicorns have a standard deviation of 20.8. Estimate the standard deviation of the entire population with 80% confidence.
LCL =
UCL =

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the confidence interval for the population standard deviation:
**1. Identify the given information:**
* Sample size (n) = 12
* Sample standard deviation (s) = 20.8
* Confidence level = 80%
**2. Find the degrees of freedom:**
Degrees of freedom (df) = n - 1 = 12 - 1 = 11
**3. Find the chi-square values:**
For an 80% confidence level, the alpha (α) is 1 - 0.80 = 0.20. We need to find the chi-square values for α/2 and 1-α/2.
* α/2 = 0.20 / 2 = 0.10
* 1 - α/2 = 1 - 0.10 = 0.90
Using a chi-square table or calculator, look up the values for df = 11:
* χ²(0.10, 11) ≈ 19.675
* χ²(0.90, 11) ≈ 5.578
**4. Calculate the lower confidence limit (LCL):**
LCL = sqrt[ (n-1) * s² / χ²(α/2, df) ]
LCL = sqrt[ (11 * 20.8²) / 19.675 ]
LCL = sqrt[ 4758.72 / 19.675 ]
LCL = sqrt[241.85]
LCL ≈ 15.55
**5. Calculate the upper confidence limit (UCL):**
UCL = sqrt[ (n-1) * s² / χ²(1-α/2, df) ]
UCL = sqrt[ (11 * 20.8²) / 5.578 ]
UCL = sqrt[ 4758.72 / 5.578 ]
UCL = sqrt[853.14]
UCL ≈ 29.21
**Answer:**
LCL ≈ 15.55
UCL ≈ 29.21
Therefore, we are 80% confident that the population standard deviation of unicorn weights is between approximately 15.55 and 29.21.

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