Construct a Confidence Interval For $\mu _ { \mathrm { d } }$

Construct a confidence interval for $\mu _ { \mathrm { d } }$ , the mean of the differences d for the population of paired data. Assume that the
population of paired differences is normally distributed.
-A confidence interval estimate of the ratio $\sigma \frac { 2 } { 1 } / \sigma 2$ can be found using the following expression:
$\left(\frac{\mathrm{s}_{1}^{2}}{\mathrm{~s}_{2}^{2}} \cdot \frac{1}{\mathrm{~F}_{\mathrm{R}}}\right) <\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}<\left(\frac{\mathrm{s}_{1}^{2}}{\mathrm{~s}_{2}^{2}} \cdot \frac{1}{\mathrm{~F}_{\mathrm{L}}}\right) \text {, }$

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 used and when individual lines are used. Obtain a $95 \%$ confidence interval for $\sigma \frac { 2 } { 1 } / { } _ { 2 } ^ { 2 }$ 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: $\mathrm { n } _ { 2 } = 16 , \mathrm {~s} _ { 2 } = 0.8$ minutes

A) $2.38 < \sigma \frac { 2 } { 1 } / \sigma 2 < 17.43$
B) $2.78 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 18.05$
C) $2.33 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 20.4$
D) $2.33 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 21.91$