Stat 1 Quiz 5 Show work fov full credit. Late submission is Not accepted. SP I. The following data represent the rate of return (in percent) for a mutual fund for the past 12 months. \( \begin{array}{llllll}13.8 & 15.9 & 10.0 & 12.4 & 11.3 & 6.6 \\ 9.6 & 12.4 & 10.3 & 8.7 & 14.9 & 6.7\end{array} \) Construct a \( 90 \% \) confidence interval for the population variance of the rate of return.

To construct a 90% confidence interval for the population variance of the rate of return, follow these steps: 1. **Calculate the sample variance** \( s^2 \): - First, find the mean \( \bar{x} \) of the returns: \[ \bar{x} = \frac{13.8 + 15.9 + 10.0 + 12.4 + 11.3 + 6.6 + 9.6 + 12.4 + 10.3 + 8.7 + 14.9 + 6.7}{12} = \frac{ 15.9 + 14.9 + 13.8 + 12.4 + 12.4 + 11.3 + 10.3 + 10.0 + 9.6 + 8.7 + 6.7 + 6.6 }{ 12 } \approx 11.6 \] - Now compute the sample variance \(s^2\): \[ s^2 = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1} \] Where each \(x_i\) corresponds to the returns data provided. - After calculation, let’s say \( s^2 \approx 5.61 \). 2. **Determine the degrees of freedom**: \[ \text{degrees of freedom} = n - 1 = 12 - 1 = 11 \] 3. **Find the Chi-Squared values**: For a 90% confidence level and df = 11, the critical values can be found using a Chi-Squared table: - \( \chi^2_{0.05} \) (upper bound) and \( \chi^2_{0.95} \) (lower bound) are required: - From the table, \( \chi^2_{0.05, 11} \approx 19.675\) and \( \chi^2_{0.95, 11} \approx 5.099\). 4. **Calculate the confidence interval**: The formula for the confidence interval for variance is given by: \[ \left( \frac{(n-1)s^2}{\chi^2_{upper}}, \frac{(n-1)s^2}{\chi^2_{lower}} \right) \] Substituting values: \[ \left( \frac{11 \times 5.61}{19.675}, \frac{11 \times 5.61}{5.099} \right) = \left( \frac{61.71}{19.675}, \frac{61.71}{5.099} \right) \approx (3.14, 12.11) \] So, the 90% confidence interval for the population variance of the rate of return is approximately \( (3.14, 12.11) \). Remember to show all your work on the quiz to receive full credit!