Solve the Problem $\sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 }$

Solve the problem.
-A confidence interval estimate of the ratio $\sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 }$ can be found using the following expression:
$\left(\frac{\mathrm{s}_{1}^{2}}{\mathrm{~s}_{2}^{2}} \cdot \frac{1}{\mathrm{~F}_{\mathrm{R}}}\right) <\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}<\left(\frac{\mathrm{s}_{1}^{2}}{\mathrm{~s}_{2}^{2}} \cdot \frac{1}{\mathrm{~F}_{\mathrm{L}}}\right)$
where $\mathrm { F } _ { \mathrm { R } }$ is found in the standard way and $\mathrm { F } _ { \mathrm { L } }$ is found as follows: interchange the degrees of freedom, and then take the reciprocal of the resulting F value found in table A-5.
A manager at a bank is interested in the standard deviation of the waiting times when a single waiting line is us when individual lines are used. Obtain a $95 \%$ confidence interval for $\sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 }$ given the following sample data:
Sample 1: multiple waiting lines: $\mathrm { n } _ { 1 } = 13 , \mathrm {~s} _ { 1 } = 2.1$ minutes
Sample 2: single waiting line: $\mathrm { n } _ { 2 } = 16 , \mathrm {~s} _ { 2 } = 1.2$ minutes

A) $1.03 < \sigma \frac { 2 } { 1 } / \sigma 2 < 9.74$
B) $1.06 < \sigma \sigma _ { 1 } ^ { 2 } / \sigma \frac { 2 } { 2 } < 7.75$
C) $1.03 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 9.07$
D) $1.23 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 8.02$