Questions: With 95% confidence, you can say that the population variance is between 6722229 and 38616699. What is the confidence interval for the population standard deviation σ ? (2593, 6214 ) (Round to the nearest integer as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to the nearest integer as needed.) A. With 95% confidence, you can say that the population standard deviation is between and .

Solution Steps

The mean \( \mu \) of the dataset is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{8807}{2} = 4403.5 \]

The variance \( \sigma^2 \) is computed using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = 3277910.25 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{3277910.25} = 1810.5 \]

The confidence interval for the variance of a single population with unknown population mean is given by:

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

Substituting the values, we have:

\[ CI = \left(\frac{(30 - 1) \times 3277910.25}{\chi^2_{\alpha/2}}, \frac{(30 - 1) \times 3277910.25}{\chi^2_{1 - \alpha/2}}\right) = (2079060.48, 5923784.67) \]

The confidence interval for the standard deviation is obtained by taking the square root of the variance interval:

\[ CI_{\sigma} = \left(\sqrt{2079060.48}, \sqrt{5923784.67}\right) = (1441.8948, 2433.8826) \]

Final Answer