Questions: Use the given confidence level and sample data to find a confidence interval for the population standard deviation σ. Assume that a simple random sample has been selected from a population that has a normal distribution. Salaries of college professors who took a geology course in college 90% confidence; n=71, x̄=73,900, s=16,124 (Round to the nearest dollar as needed.)

Solution Steps

The sample standard deviation \( s \) is given as \( 16124 \). The sample variance \( s^2 \) is calculated as follows:

\[ s^2 = 16124^2 = 259983376 \]

To find the confidence interval for the variance of a single population with an unknown population mean, we use the formula:

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

Substituting the values:

• Sample size \( n = 71 \)
• Sample variance \( s^2 = 259983376 \)
• Degrees of freedom \( n - 1 = 70 \)

The confidence interval for the variance is calculated as:

\[ CI = \left( \frac{(71 - 1) \times 259983376}{\chi^2_{\alpha/2}}, \frac{(71 - 1) \times 259983376}{\chi^2_{1 - \alpha/2}} \right) \]

Using the critical values from the Chi-Square distribution for \( \alpha = 0.10 \) (90% confidence level):

• \( \chi^2_{0.05} \) and \( \chi^2_{0.95} \) yield the interval:

\[ CI = (201022754.66, 351741211.05) \]

To find the confidence interval for the standard deviation, we take the square root of the variance confidence interval:

\[ \text{Lower bound: } \sqrt{201022754.66} \approx 14178 \] \[ \text{Upper bound: } \sqrt{351741211.05} \approx 18755 \]

Thus, the confidence interval for the standard deviation is:

\[ \text{Standard Deviation Confidence Interval: } (14178, 18755) \]

Final Answer