## Confidence
Interval for the Ratio of Variances of Two Normally Distributed Populations

This file is part of a program based on the Bio 4835 Biostatistics class taught
at Kean University
in Union, New
Jersey. The course uses the following text:

Daniel, W. W. 1999. Biostatistics: a foundation for analysis in the
health sciences. New York:
John Wiley and Sons.

The file follows this text very closely and readers are encouraged to consult
the text for further information.

**F) ****Confidence interval for the ratio of variances of two
normally distributed populations**

A way to compare the variances of two normally distributed populations is to
use the variance ratio, / . The variance ratio is used, among other things, as the
test statistic for analysis of variance (ANOVA). If the two variances are
equal, then V. R. = 1.

**Sampling distribution**

The sampling distribution of ( /
)/( / ) is used.
Since the population variances are usually not known, the sample variances are
used. The assumptions are that and i are computed from
independent samples of size and , respectively, drawn
from two normally distributed populations. If the assumptions are
met, ( /
)/( / ) follows a
distribution known as the *F distribution* with *two* values used for
degrees of freedom.

**Degrees of freedom**

The F distribution uses two values for degrees of freedom. The
numerator degrees of freedom is the value of -1 which is
used in calculating .
The denominator degrees of freedom is the value of -1 which is
used in calculating .

**Reading F tables**

F tables come in denominations based on which
are , , , and
with one tail. For two tail intervals, the lower boundary, , must be calculated to give
values of , and .

**Confidence interval for / **

The distribution ( /
)/( / ) is used to
establish the 100(1- ) percent confidence interval for ** / **. The
staring point is

From this relation, it can be shown that the 100(1- ) percent
confidence interval for ** /** is

Example

Among 11 patients in a certain study, the standard deviation of the property of
interest was 5.8. In another group of 4 patients, the standard deviation
was 3.4. We wish to construct a 95 percent confidence interval for the
ratio of the variances of these two populations.

(1) Given

= 11
= = 33.64

= 4
= = 11.56

= .05

10, 3 = 14.42

= 1/ 3, 10 = 1/4.83 =
.20704

(2) Calculations

Calculation of the 95% confidence interval for **
/ **