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

1.06 < \sigma \sigma _ { 1 } ^ { 2 } / \sigma \frac { 2 } { 2 } < 7.75

1.03 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 9.07

1.23 < \sigma \frac { 2 } { 1 } / \sigma \frac { 2 } { 2 } < 8.02

Use the Traditional Method to Test the Given Hypothesis

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