SOLUTION: Use the given confidence level and sample data to find a confidence interval for the population standard deviation o. Assume that a simple random sample has been selected from a po

Question 1182178: Use the given confidence level and sample data to find a confidence interval for the population standard deviation o. Assume that a simple random sample has been selected from a population that has a normal distribution.
Salaries of college graduates who took a geology course in college
95% confidence; n=51, x=$60100, s=$19008
$____< o < $____
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 (σ):
1. **Identify the given information:**
* Confidence level = 95%
* Sample size (n) = 51
* Sample mean (x̄) = $60,100 (This is not needed for the standard deviation confidence interval)
* Sample standard deviation (s) = $19,008
2. **Determine the degrees of freedom:**
Degrees of freedom (df) = n - 1 = 51 - 1 = 50
3. **Find the critical chi-square values:**
For a 95% confidence level, α = 1 - 0.95 = 0.05. We need to find the chi-square values for α/2 and 1-α/2.
* α/2 = 0.05 / 2 = 0.025
* 1 - α/2 = 1 - 0.025 = 0.975
Using a chi-square distribution table or calculator, look up the values for df = 50:
* χ²(0.025, 50) ≈ 71.42
* χ²(0.975, 50) ≈ 32.36
4. **Calculate the confidence interval:**
The formula for the confidence interval for σ is:
√[((n-1)s²) / χ²(α/2, df)] < σ < √[((n-1)s²) / χ²(1-α/2, df)]
Substitute the values:
√[((50) * (19008)²) / 71.42] < σ < √[((50) * (19008)²) / 32.36]
√[361304448.3 / 71.42] < σ < √[361304448.3 / 32.36]
√5059639.46 < σ < √11162002.11
$2249.36 < σ < $3340.96 (approximately)
Therefore, the 95% confidence interval for the population standard deviation (σ) is approximately **$16,942 < σ < $22,500**.

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